Advances in Matrix Inequalities 3030760464, 9783030760465

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Table of contents :
Preface
Contents
Acronyms
1 Elementary Linear Algebra Review
1.1 Operators and Matrices in Hilbert Space
2 Interpolating the Arithmetic-Geometric Mean Inequality and Its Operator Version
2.1 Refinements of the Scalar Young and Heinz Inequalities
2.2 Operator Inequalities Involving Improved Young Inequality
2.3 Advanced Refinements of the Scalar Reverse Young Inequalities
2.4 Improvements of the Operator Reverse Young Inequality
3 Operator Inequalities for Positive Linear Maps
3.1 On an Operator Kantorovich Inequality for Positive Linear Maps
3.2 A Schwarz Inequality for Positive Linear Maps
3.3 Squaring the Reverse Arithmetic-Geometric Mean Inequality
3.4 Reverses of Ando's Inequality for Positive Linear Maps
3.5 Squaring the Reverse Ando's Operator Inequality
4 Operator Inequalities Involving Operator Monotone Functions
4.1 Young Inequalities Involving Operator Monotone Functions
4.2 Eigenvalue Inequalities Involving Operator Concave Functions
4.3 Operator Aczél Inequality Involving Operator Monotone Functions
4.4 Norm Inequalities Involving Operator Monotone Functions
5 Inequalities for Sector Matrices
5.1 Haynsworth and Hartfiel Type Determinantal Inequality
5.2 Inequalities with Determinants of Perturbed Positive Matrices
5.3 Analogue of Fischer's Inequality for Sector Matrices
5.4 Analogues of Hadamard and Minkowski Inequality for Sector Matrices
5.5 Generalizations of the Brunn Minkowski Inequality
5.6 A Lewent Type Determinantal Inequality
5.7 Principal Powers of Matrices with Positive Definite Real Part
5.8 Geometric Mean of Accretive Operators
5.9 Weighted Geometric Mean of Accretive Operators and Its Applications
5.10 Ficher Type Determinantal Inequalities for Accretive-Dissipative Matrices
5.11 Extensions of Fischer's Inequality for Sector Matrices
5.12 Singular Value Inequalities of Sector Matrices
5.13 Extension of Rotfel'd Inequality for Sector Matrices
5.14 A Further Extension of Rotfel'd Inequality for Accretive-Dissipative Matrices
5.15 Hilbert-Schmidt Norm Inequalities for Accretive-Dissipative Operators
5.16 Schatten p-Norm Inequalities for Accretive-Dissipative Matrices
5.17 Schatten p-Norm Inequalities for Sector Matrices
5.18 Schatten p-Norms and Determinantal Inequalities Involving Partial Traces
5.19 Ando-Choi Type Inequalities for Sector Matrices
5.20 Geometric Mean Inequality for Sector Matrices
5.21 Weighted Geometric Mean Inequality for Sector Matrices
6 Positive Partial Transpose Matrix Inequalities
6.1 Singular Value Inequalities Related to PPT Matrices
6.2 Matrix Inequalities and Completely PPT Maps
6.3 Hiroshima's Type Inequalities for Positive Semidefinite Block Matrices
6.4 Geometric Mean and Norm Schwarz Inequality
6.5 Inequalities Involving the Off-Diagonal Block of a PPT Matrix
6.6 Unitarily Invariant Norm Inequalities of PPT Matrices
6.7 On Symmetric Norm Inequalities for Positive Block Matrices
6.8 Matrix Norm Inequalities and Majorization Relation for Singular Values
Appendix References
Index
Recommend Papers

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Springer Optimization and Its Applications 176

Mohammad Bagher Ghaemi Nahid Gharakhanlu Themistocles M. Rassias Reza Saadati

Advances in Matrix Inequalities

Springer Optimization and Its Applications Volume 176

Series Editors Panos M. Pardalos My T. Thai

, University of Florida, Gainesville, FL, USA

, CSE Building, University of Florida, Gainesville, FL, USA

Honorary Editor Ding-Zhu Du, University of Texas at Dallas, Richardson, TX, USA Advisory Editors Roman V. Belavkin, Faculty of Science and Technology, Middlesex University, London, UK John R. Birge, University of Chicago, Chicago, IL, USA Sergiy Butenko, Texas A&M University, College Station, TX, USA Vipin Kumar, Dept Comp Sci & Engg, University of Minnesota, Minneapolis, MN, USA Anna Nagurney, Isenberg School of Management, University of Massachusetts Amherst, Amherst, MA, USA Jun Pei, School of Management, Hefei University of Technology, Hefei, Anhui, China Oleg Prokopyev, Department of Industrial Engineering, University of Pittsburgh, Pittsburgh, PA, USA Steffen Rebennack, Karlsruhe Institute of Technology, Karlsruhe, Baden-Württemberg, Germany Mauricio Resende, Amazon (United States), Seattle, WA, USA Tamás Terlaky, Lehigh University, Bethlehem, PA, USA Van Vu, Department of Mathematics, Yale University, New Haven, CT, USA Guoliang Xue, Ira A. Fulton School of Engineering, Arizona State University, Tempe, AZ, USA Yinyu Ye, Stanford University, Stanford, CA, USA Associate Editor Michael N. Vrahatis, Mathematics Department, University of Patras, Patras, Greece

Aims and Scope Optimization has continued to expand in all directions at an astonishing rate. New algorithmic and theoretical techniques are continually developing and the diffusion into other disciplines is proceeding at a rapid pace, with a spot light on machine learning, artificial intelligence, and quantum computing. Our knowledge of all aspects of the field has grown even more profound. At the same time, one of the most striking trends in optimization is the constantly increasing emphasis on the interdisciplinary nature of the field. Optimization has been a basic tool in areas not limited to applied mathematics, engineering, medicine, economics, computer science, operations research, and other sciences. The series Springer Optimization and Its Applications (SOIA) aims to publish state-of-the-art expository works (monographs, contributed volumes, textbooks, handbooks) that focus on theory, methods, and applications of optimization. Topics covered include, but are not limited to, nonlinear optimization, combinatorial optimization, continuous optimization, stochastic optimization, Bayesian optimization, optimal control, discrete optimization, multi-objective optimization, and more. New to the series portfolio include Works at the intersection of optimization and machine learning, artificial intelligence, and quantum computing. Volumes from this series are indexed by Web of Science, zbMATH, Mathematical Reviews, and SCOPUS.

More information about this series at http://www.springer.com/series/7393

Mohammad Bagher Ghaemi Nahid Gharakhanlu Themistocles M. Rassias Reza Saadati •



Advances in Matrix Inequalities

123



Mohammad Bagher Ghaemi School of Mathematics Iran University of Science and Technology Tehran, Iran

Nahid Gharakhanlu School of Mathematics Iran University of Science and Technology Tehran, Iran

Themistocles M. Rassias Department of Mathematics National Technical University of Athens Athens, Greece

Reza Saadati School of Mathematics Iran University of Science and Technology Tehran, Iran

ISSN 1931-6828 ISSN 1931-6836 (electronic) Springer Optimization and Its Applications ISBN 978-3-030-76046-5 ISBN 978-3-030-76047-2 (eBook) https://doi.org/10.1007/978-3-030-76047-2 Mathematics Subject Classification: 15-XX, 37-XX, 39-XX, 41-XX, 42-XX, 46-XX, 47-XX, 49-XX, 52-XX, 58-XX, 90-XX, 93-XX © Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

To my beloved parents: Thank you, Mom and Dad, for everything. You are the reason for all successes in my life. Nahid Gharakhanlu

Preface

The branch of Matrix Analysis is a research area of fundamental interest that is applicable to a broad spectrum of areas such as scientific computing, control theory, operations research, mathematical physics, statistics, economics, and engineering disciplines in addition to several areas of pure mathematics. Theorems in this field are commonly expressed in terms of inequalities. Subjects such as operator Young inequalities, operator inequalities for positive linear maps, operator inequalities involving operator monotone functions, norm inequalities, and inequalities for sector matrices are investigated thoroughly throughout this book which provides an account of a broad collection of classic and recent developments. The book is self-contained in the sense that it presents detailed proofs for all the main theorems and relevant technical lemmas. Thus, interested graduate and advanced undergraduate students will find it particularly useful. Besides frequent use of methods from operator theory, the reader will also appreciate techniques of classical analysis and algebraic arguments, as well as combinatorial methods. This book is arranged into six chapters. Chapter 1 reviews the basic concepts, definitions, and theorems used in the subsequent chapters. Most of the material within this chapter will be known to the reader who has a background in Operator Theory, Linear Algebra, and Matrix Analysis. Chapter 2 presents the classical Young inequalities for positive real numbers, as well as Young and Heinz inequalities for operators. The book presents multiple-term refinements of Young’s inequality for real numbers and operators. Furthermore, based on these inequalities, some operator inequalities and matrix inequalities for the Hilbert-Schmidt norm are established. Chapter 3 introduces main results on squaring the reverse arithmetic-geometric mean operator inequality, as well as the reverse Ando’s operator inequality. Chapter 4 improves known operator inequalities involving positive linear maps, geometric means, operator monotone functions, and doubly concave functions. Chapter 5 constitutes a crucial part of the book. Increasing recent studies on inequalities are devoted to the investigation of matrices with the numerical range in a sector of the complex plane. This includes the study of accretive dissipative vii

viii

Preface

matrices and positive definite matrices as special cases in which accretive dissipative matrices have attracted the attention of several researchers due to their rich applicability in various areas of mathematics and related fields. Matrix decomposition plays a fundamental role in these studies. A matrix whose numerical range is contained in a sector region is called a sector matrix. This chapter extends several known inequalities to this new class of matrices. The final chapter presents several inequalities related to 2  2 block PPT matrices and positive matrices partitioned in blocks. If these positive block matrices are PPT, accurate results can be obtained, motivated by Quantum Information Science. Tehran, Iran Tehran, Iran Athens, Greece Tehran, Iran December 2020

Mohammad Bagher Ghaemi Nahid Gharakhanlu Themistocles M. Rassias Reza Saadati

Acknowledgements We would like to express our thanks to the referees for reading the manuscript and providing valuable suggestions and comments which have helped to improve the presentation of the book. The authors are grateful to Professor Dorin Andrica for mentioning two straightforward proofs for Lemmas 2.1.2 and 2.1.4. Last but not least, it is our pleasure to acknowledge the superb assistance provided by the staff of Springer for the publication of the book.

Contents

1 Elementary Linear Algebra Review . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Operators and Matrices in Hilbert Space . . . . . . . . . . . . . . . . . . 2 Interpolating the Arithmetic-Geometric Mean Inequality and Its Operator Version . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Refinements of the Scalar Young and Heinz Inequalities . . . 2.2 Operator Inequalities Involving Improved Young Inequality 2.3 Advanced Refinements of the Scalar Reverse Young Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Improvements of the Operator Reverse Young Inequality . . 3 Operator Inequalities for Positive Linear Maps . . . . . . . . . . . . 3.1 On an Operator Kantorovich Inequality for Positive Linear Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 A Schwarz Inequality for Positive Linear Maps . . . . . . . . . 3.3 Squaring the Reverse Arithmetic-Geometric Mean Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Reverses of Ando’s Inequality for Positive Linear Maps . . . 3.5 Squaring the Reverse Ando’s Operator Inequality . . . . . . . . 4 Operator Inequalities Involving Operator Monotone Functions 4.1 Young Inequalities Involving Operator Monotone Functions 4.2 Eigenvalue Inequalities Involving Operator Concave Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Operator Aczél Inequality Involving Operator Monotone Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Norm Inequalities Involving Operator Monotone Functions .

1 1

.... .... ....

19 19 32

.... ....

36 53

....

61

.... ....

61 68

.... .... ....

72 78 84

.... ....

89 89

....

92

.... 97 . . . . 103

5 Inequalities for Sector Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 5.1 Haynsworth and Hartfiel Type Determinantal Inequality . . . . . . . 109 5.2 Inequalities with Determinants of Perturbed Positive Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

ix

x

Contents

5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17 5.18 5.19 5.20 5.21

Analogue of Fischer’s Inequality for Sector Matrices . . . . . . Analogues of Hadamard and Minkowski Inequality for Sector Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Generalizations of the Brunn Minkowski Inequality . . . . . . . A Lewent Type Determinantal Inequality . . . . . . . . . . . . . . . Principal Powers of Matrices with Positive Definite Real Part . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Geometric Mean of Accretive Operators . . . . . . . . . . . . . . . Weighted Geometric Mean of Accretive Operators and Its Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Ficher Type Determinantal Inequalities for AccretiveDissipative Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Extensions of Fischer’s Inequality for Sector Matrices . . . . . Singular Value Inequalities of Sector Matrices . . . . . . . . . . . Extension of Rotfel’d Inequality for Sector Matrices . . . . . . . A Further Extension of Rotfel’d Inequality for AccretiveDissipative Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Hilbert-Schmidt Norm Inequalities for Accretive-Dissipative Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Schatten p-Norm Inequalities for Accretive-Dissipative Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Schatten p-Norm Inequalities for Sector Matrices . . . . . . . . . Schatten p-Norms and Determinantal Inequalities Involving Partial Traces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Ando-Choi Type Inequalities for Sector Matrices . . . . . . . . . Geometric Mean Inequality for Sector Matrices . . . . . . . . . . Weighted Geometric Mean Inequality for Sector Matrices . . .

6 Positive Partial Transpose Matrix Inequalities . . . . . . . . . . . . . . 6.1 Singular Value Inequalities Related to PPT Matrices . . . . . . . 6.2 Matrix Inequalities and Completely PPT Maps . . . . . . . . . . . 6.3 Hiroshima’s Type Inequalities for Positive Semidefinite Block Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Geometric Mean and Norm Schwarz Inequality . . . . . . . . . . 6.5 Inequalities Involving the Off-Diagonal Block of a PPT Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 Unitarily Invariant Norm Inequalities of PPT Matrices . . . . . 6.7 On Symmetric Norm Inequalities for Positive Block Matrices 6.8 Matrix Norm Inequalities and Majorization Relation for Singular Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . 120 . . . 125 . . . 129 . . . 134 . . . 139 . . . 144 . . . 148 . . . .

. . . .

. . . .

156 160 168 173

. . . 176 . . . 181 . . . 191 . . . 197 . . . .

. . . .

. . . .

202 207 212 217

. . . 221 . . . 221 . . . 228 . . . 235 . . . 241 . . . 248 . . . 252 . . . 254 . . . 259

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275

Acronyms

R C H A; B; ::: h:; :i BðHÞ Mn ]‚ ] r‚ r Sð:Þ Kð:Þ Kð:; :Þ U kxk WðAÞ wðAÞ qðAÞ Sa ‚ðAÞ rðAÞ spðAÞ  w kAk k:ku k:k2

The real numbers The complex numbers A complex Hilbert space Linear operators on H Inner product Algebra of all bounded linear operators on H Full matrix algebra of all n  n matrices ‚-weighted geometric mean Geometric mean ‚ Arithmetic mean Specht’s ratio Kantorovich constant Generalized Kantorovich constant Positive linear map A norm of vector x Numerical range of A Numerical radius of A Spectral radius of A The sector in the complex plane Eigenvalue vector of A Singular value of A Spectrum of A Majorization Weakly majorization A norm of operator A Unitarily invariant norm Hilbert-Schmidt norm

xi

xii

k:kF k:kp k:kk

Acronyms

Frobenius norm of A Schatten p-norm Ky Fan k-norm

Chapter 1

Elementary Linear Algebra Review

In this chapter, we review the basic concepts, definitions, and theorems which are used throughout the book. Most of the material will be familiar to a reader who has had a standard Operator Theory, Linear Algebra, and Matrix Analysis courses. So it is presented quickly with no proofs. Some topics might be less familiar. These are treated here in somewhat greater detail.

1.1 Operators and Matrices in Hilbert Space Throughout the book, R will denote the set of real numbers, C the complex numbers, Z integers, and N the positive integers. Let H be a complex vector space. A functional ., . : H × H → C of two variables is called an inner product if it satisfies (i) (ii) (iii) (iv)

x + y, z = x, z + y, z (x, y, z ∈ H). λx, y = λx, y (λ ∈ C, x, y ∈ H). x, y = y, x (x, y ∈ H). x, x ≥ 0 for every x ∈ H and x, x = 0 only for x = 0.

Note that the inner product is linear in the first variable and it is conjugate linear in the second variable. The Schwarz inequality is |x, y|2 ≤ x, xy, y. The inner product determines a norm for the vectors as follows: x :=



x, x.

© Springer Nature Switzerland AG 2021 M. B. Ghaemi et al., Advances in Matrix Inequalities, Springer Optimization and Its Applications 176, https://doi.org/10.1007/978-3-030-76047-2_1

1

2

1 Elementary Linear Algebra Review

We recall that x is interpreted as the length of the vector x with the following properties: x + y ≤ x + y, |x, y| ≤ x.y. A further requirement in the definition of a Hilbert space H is that every Cauchy sequence must be convergent. That is, the space is complete. In the finite-dimensional case, the completeness always holds. In this book, the vector space is typically a finitedimensional complex Hilbert space. As an example, the linear space Cn of all n-tuples of complex numbers becomes a Hilbert space with the inner product ⎡ ⎤ y1 n ⎢ y2⎥  ⎢ ⎥ xi y i = [x1 , x2 , ..., xn ] ⎢ . ⎥ . x, y = ⎣ .. ⎦ i=1

yn Another example is the space of square integrable complex-valued functions on the real Euclidean space Rn . If f and g are such functions then

 f, g =

Rn

f (x)g(x)d x

gives the inner product. The latter space is denoted by L 2 (Rn ) and it is infinitedimensional contrary to the n-dimensional space Cn . Dimension of a Hilbert space H. If x, y = 0 for vectors x and y of a Hilbert space H, then x and y are called orthogonal, in notation x⊥y. When H ⊂ H, H ⊥ := {x ∈ H : x⊥h} for every h ∈ H is called the orthogonal complement of H . For any subset H ⊂ H, H ⊥ is a closed subspace. A family ei of vectors is called orthonormal if ei , ei  = 1 and ei , e j  = 0 if i = j. A maximal orthonormal system is called a basis or an orthonormal basis. The cardinality of a basis is called the dimension of the Hilbert space. Notice that the cardinality of any two bases is the same. In the space Cn , the standard orthonormal basis consists of the vectors δ1 = (1, 0, ..., 0), δ2 = (0, 1, ..., 0), ..., δn = (0, 0, ..., 1), where each vector has 0 coordinate n − 1 times and one coordinate equals 1. Orthonormal basis. Assume that in an n-dimensional Hilbert space, linearly independent vectors v1 , v2 , ..., vn are given. By the Gram-Schmidt procedure, an orthonormal basis can be obtained by linear combinations as follows:

1.1 Operators and Matrices in Hilbert Space

3

1 ν1 , ν1  1 e2 := ω2 with ω2 := ν2 − e1 , ν2 e1 , ω2  1 e3 := ω3 with ω3 := ν3 − e1 , ν3 e1 − e2 , ν3 e2 , ω3  .. . 1 ωn with ωn := νn − e1 , νn e1 − ... − en−1 , νn en−1 . en := ωn  e1 :=

Theorem 1.1.1 ([97]) Let e1 , e2 , ... be a basis in a Hilbert space H. Then for any vector x ∈ H the expansion  en , xen x= n

holds. Moreover, x2 =



|en , x|2 .

n

Let H and K be Hilbert spaces. A mapping Φ : H → K is called linear if it preserves linear combination Φ(λx + μy) = λΦ(x) + μΦ(y), (x, y ∈ H, λ, μ ∈ C). Let e1 , e2 , ..., en be a basis of the Hilbert space H and f 1 , f 2 , ..., f m be a basis of K. The linear mapping Φ : H → K is determined by the vectors Φ(e j ) , j = 1, 2, ..., n. Furthermore, the vector Φ(e j ) is determined by its coordinates as follows: Φ(e j ) = c1, j f 1 + c2, j f 2 + ... + cm, j f m . The numbers ci, j , 1 ≤ i ≤ m, 1 ≤ j ≤ n , form an m × n matrix, which is called the matrix of the linear transformation Φ with respect to the bases (e1 , e2 , ..., en ) and ( f 1 , f 2 , ..., f m ). Theorem 1.1.2 (Riesz-Fischer theorem) ([97]) Let Φ : H → C be a linear mapping on a finite-dimensional Hilbert space H. Then there is a unique vector v ∈ H such that Φ(x) = x, v for every vector x ∈ H. Proof. Let e1 , e2 , ..., en be an orthonormal basis in H. Then we need a vector v ∈ H such that Φ(ei ) = ei , v. So  ν= Φ(ei )ei , i

will satisfy the condition.



4

1 Elementary Linear Algebra Review

The linear mappings Φ : H → C are called functionals. If the Hilbert space is not finite dimensional, then in the previous theorem the boundedness condition |Φ(x)| ≤ cx should be added, where c is a positive number. Majorization of vectors. Let a = (a1 , ..., an ) ∈ Rn . We rearrange the components of a in decreasing order and obtain a vector a ↓ = (a1 ↓ , ..., an ↓ ), where a1 ↓ ≥ a2 ↓ ≥ ... ≥ an ↓ . For example, a = (−1, −2, 3), a ↓ = (3, −1, −2). Now, let a = (a1 , ..., an ) and b = (b1 , ..., bn ) be vectors in Rn . The majorization a ≺ b means that k k   ↓ ↓ ai ≤ bi (1 ≤ k ≤ n), (1.1.1) i=1

i=1

and the equality is required for k = n. The weak majorization a ≺ω b is defined by the inequality (1.1.1), where the equality for k = n is not required. The majorization a ≺ b is equivalent to the statement that a is a convex combination of permutations of the components of the vector b. This can be written as  λU U b, a= U

where the summation is over the n × n permutation matrices U and λU ≥ 0, λU = 1. The n × n matrix D = λU U has the property that all entries are positive and U

the sums of rows and columns are 1. Such a matrix D is called doubly stochastic. So a = Db. We recall that a matrix S is called doubly substochastic n × n matrix if n n Si j ≤ 1 for 1 ≤ i ≤ n and Si j ≤ 1 for 1 ≤ j ≤ n. j=1

i=1

Theorem 1.1.3 ([216]) The following conditions for a, b ∈ Rn are equivalent: (i) a ≺ b. n n (ii) |ai − r | ≤ |bi − r | for all r ∈ R. (iii)

i=1 n

i=1

f (ai ) ≤

i=1 ai , bi .

n

f (bi ) for any convex function f on an interval containing all

i=1

(iv) a is a convex combination of coordinate permutations of b. (v) a = Db for some doubly stochastic n × n matrix D. Note that the implication (v) ⇒ (iv) is seen directly from the well-known theorem of Birkhoff saying that any doubly stochastic matrix is a convex combination of permutation matrices. The previous theorem was about majorization and the next one is about weak majorization. Theorem 1.1.4 ([216]) The following conditions for a, b ∈ Rn are equivalent: (i) a ≺ω b.

1.1 Operators and Matrices in Hilbert Space

5

(ii) There exists a c ∈ Rn such that a ≤ c ≺ b, where a ≤ c means that ai ≤ ci , 1 ≤ i ≤ n. n n (ai − r )+ ≤ (bi − r )+ for all r ∈ R. (iii) (iv)

i=1 n

f (ai ) ≤

i=1 ai , bi .

n

i=1

f (bi ) for any convex function f on an interval containing all

i=1

(v) If a, b ≥ 0, then a = Sb for some doubly stochastic n × n matrix S. Example 1.1.5 Let a, b ∈ Rn and f be a convex function on an interval containing all ai , bi . We use the notation f (a) := ( f (a1 ), ..., f (an )) and similarly f (b). Assume that a ≺ b. Since f is a convex function, so is ( f (x) − r )+ for any r ∈ R. Hence, f (a) ≺ω f (b) follows from Theorems 1.1.3. Next assume that a ≺ω b and f is an increasing convex function, then f (a) ≺ω f (b) can be proved similarly. Let a, b ∈ Rn and a, b ≥ 0. We define the weak log-majorization a ≺ω(log) b when k

↓ ai



i=1

k



bi

(1 ≤ k ≤ n),

i=1

and the log-majorization a ≺(log) b when a ≺ω(log) b and equality holds for k = n. It is obvious that if a and b are strictly positive, then a ≺(log) b (resp.,a ≺ω(log) b) if and only if log a ≺ log b (resp., log a ≺ω log b ), where log a := (log a1 , ..., log an ). Theorem 1.1.6 ([216]) Let a, b ∈ Rn with a, b ≥ 0 and assume that a ≺ω(log) b. If f is a continuous increasing function on [0, ∞) such that f (e x ) is convex, then f (a) ≺ω f (b). In particular, a ≺ω(log) b implies a ≺(log) b. The space of all operator from H1 to H2 is denoted by B(H1 , H2 ) and we write B(H) for B(H, H). Throughout the book, B(H) denote the algebra of all bounded linear operators acting on a complex Hilbert space (H, ·, ·) and I be the identity operator. In the case when dim H = n, we identify B(H) with the full matrix algebra Mn = Mn (C) of all n × n matrices with entries in the complex field C and denote its identity by In . This means that to each linear operator A ∈ B(H), an n × n matrix A is associated and represented by [A]. The correspondence A → [A] is an algebraic isomorphism from B(H) to the algebra Mn of n × n matrices. This isomorphism shows that the theory of linear operators on an n-dimensional Hilbert space is the same as the theory of n × n matrices. For n, m ∈ N, Mn×m = Mn×m (C) denotes the space of all n × m complex matrices. A matrix M ∈ Mn×m is a mapping {1, 2, ..., n} × {1, 2, ..., m} → C. It is represented as an array with n rows and m columns as follows: ⎡

m 11 m 12 ⎢m 21 m 22 ⎢ M =⎢ . .. ⎣ .. . m n1 m n2

⎤ ... m 1m ... m 2m ⎥ ⎥ , . . .. ⎥ . . ⎦ ... m nm

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1 Elementary Linear Algebra Review

where m i j is the intersection of the i’th row and the j’th column. If the matrix is denoted by M, then this entry is denoted by m i j . If n = m, then we write Mn instead of Mn×n . A simple example is the identity matrix In ∈ Mn defined as m i j = δi, j where δi, j is the well-known Kronecker symbol, or ⎡

1 ⎢0 ⎢ In = ⎢ . ⎣ ..

0 1 .. .

... ... .. .

⎤ 0 0⎥ ⎥ .. ⎥ . .⎦

0 0 ... 1

Principal submatrix. Let A ∈ Mm×n . For index set α ⊆ {1, 2, ..., m} and β ⊆ {1, 2, ..., m}, we denote by A[α, β] the submatrix of entries that lie in the rows of A indexed by α and columns indexed by β. For example, ⎡



 123 ⎣4 5 6⎦ [{1, 3}, {1, 2, 3}] = 1 2 3 . 789 789 If α = β, the submatrix A[α, α]  is the principal submatrix of A and it is denoted by A[α]. An n-by-n matrix has nk principal submatrices of size k. For A ∈ Mn and k ∈ {1, 2, ..., n}, A[{1, ..., k}] is a leading principal submatrix and A{k, ..., n} is trailing principal submatrix. Definition 1.1.7 Let A ∈ Mn . The trace of A is the sum of the diagonal entries T r A :=

n 

Aii .

i=1

It is easy to show by definition that T r (AB) = T r (B A). The determinant of A ∈ Mn is defined by  det A := (−1)σ(π) A1π(1) A2π(2) ...Anπ(n) , π

where the sum is over all permutations π of the set 1, 2, ..., n and σ(π) is the parity of the permutation π. Therefore

 ab det = ad − bc, cd and another example is the following:

1.1 Operators and Matrices in Hilbert Space

7



  A11 A12 A13 A22 A23 A21 A23 ⎦ ⎣ det A21 A22 A23 = A11 (det ) − A12 (det ) A32 A33 A31 A33 A31 A32 A33 

A A + A13 (det 21 22 ). A31 A32 ⎡

It can be proven that det(AB) = det(A) det(B). Block diagonal matrix. A matrix A ∈ Mn of the form ⎡

A11 ⎢ 0 ⎢ A=⎢ . ⎣ .. 0 in which Aii ∈ Mni , i = 1, ..., k,

k

0 A22 .. .

... ... .. .

0 0 .. .

⎤ ⎥ ⎥ ⎥, ⎦

0 ... Akk

n i = n and all blocks above and below the block

i=1

diagonal are zero blocks, is called block diagonal and is denoted by A = A11 ⊕ A22 ⊕ k Aii . This is the direct sum of the matrices A11 , A22 , ..., Akk . Many ... ⊕ Akk = ⊕i=1 of properties of block diagonal matrices generalize those of diagonal matrices. The following theorem is immediate from definition. Theorem 1.1.8 ([104]) We have k K (i) det(⊕i=1 Aii ) = i=1 det Aii . K (ii) det(⊕i=1 Aii ) is nonsingular if and only if each Aii is nonsingular for i = 1, 2, ..., k. k k Aii and B = ⊕i=1 Bii in which each Aii and Bii is (iii) The direct sum A = ⊕i=1 the same size commute if and only if each pair Aii , Bii commutes. k K Aii = i=1 rank Aii . (iv) rank ⊕i=1 (v) If A ∈ Mn and B ∈ Mn are nonsingular then (A ⊕ B)−1 = A−1 ⊕ B −1 .

Schur complement. We consider A ∈ Mn to be partitioned as

A=

 A11 A12 , A21 A22

where diagonal blocks are square matrices. If A is nonsingular, then we partition A−1 conformally as A. If A11 is nonsingular, we have

A= where

   A11 0 I 0 I −A11 −1 A12 A = , 0 ( AA11 ) −A21 A11 −1 I 0 I

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1 Elementary Linear Algebra Review

A = A22 − A21 A−1 11 A12 A11 is called the Schur complement of A11 in A. By taking determinants,  det A = det A11 . det

A A11

 .

Operator norm. Let H be a finite-dimensional Hilbert space. The operator norm of a linear operator A ∈ B(H) is defined as A = sup{Ax : x ∈ H, x = 1}. It can be shown that A is finite. In addition to the common properties A + B ≤ A + B and λ A ≤ |λ|A, the submultiplicativity AB ≤ AB also holds. If A ≤ 1, then the operator A is called a contraction. Lemma 1.1.9 Let A ∈ B(H) and A < 1. Then I − A is invertible and (I − A)−1 =

∞ 

An .

n=0

Proof. (I − A)

N 

An = I − A N +1 and A N +1  ≤ A N +1 .

n=0

We can see that the limit of the first equation is (I − A)

∞ 

An = I,

n=0

and this shows the statement which is called a Neumann series.



Theorem 1.1.10 (Projection theorem) ([169]) Let M be a closed subspace of a Hilbert space H. Any vector h ∈ H can be written in a unique way in the form h = x + y, where x ∈ M and y ∈ M ⊥ . Note that a subspace of a finite-dimensional Hilbert space is always closed. The mapping P : h → x defined in the context of the previous theorem is called the orthogonal projection onto the subspace M. This mapping is linear as follows: P(λx + μy) = λP x + μP.

1.1 Operators and Matrices in Hilbert Space

9

Moreover, P 2 = P = P ∗ . The converse is also true: If P 2 = P = P ∗ , then P is an orthogonal projection onto its range. Adjoint of the operator. Let H and K be Hilbert spaces. If A : H → K is a linear operator, then its adjoint A∗ : K → H is determined by the formula x, AyK = A∗ x, yH , (x ∈ K, y ∈ H). Theorem 1.1.11 ([169]) We have the following properties for linear operators: (i) (ii) (iii) (iv) (v) (vi) (vii)

(A + B)∗ = A∗ + B ∗ . (λ A)∗ = λ A∗ (λ ∈ C). (A∗ )∗ = A. (AB)∗ = B ∗ A∗ . (A−1 )∗ = (A∗ )−1 if A is invertible. A = A∗ . A∗ A = A2 .

Example 1.1.12 Let A ∈ B(H) be a linear operator and e1 , e2 , ..., en be a basis in the Hilbert space H. The (i, j) element of the matrix of A is ei , Ae j . Since ei , Ae j  = e j , A∗ ei , this is the complex conjugate of the ( j, i) element of the matrix of A∗ . Definition 1.1.13 If A ∈ B(H), then (i) A is self-adjoint or Hermitian if A∗ = A; (ii) A is normal if A A∗ = A∗ A. In the analogy between adjoint and the complex conjugate, self-adjoint operator becomes the analogues of real numbers and normal operators are analogues of complex numbers. The following theorems give us some idea about the basic properties of self-adjoint operators which is an easy consequence of the definition of self-adjoint operators. Theorem 1.1.14 ([169]) If A, B ∈ B(H), then (i) If A is self-adjoint, then Ah, h is real. (ii) Let A, B ∈ B(H) are self-adjoint. AB is self-adjoint if and only if AB = B A. (iii) If (Bn ) ⊂ B(H) be a sequence of self-adjoint operator and Bn → B, then B is self-adjoint. (iv) If A is self-adjoint, then A = sup{Ah, h : h = 1}. The spectrum of the operator. Let A ∈ B(H). The spectrum of A is the set of λ ∈ C such that A − λI is not invertible in B(H). It is denoted by sp(A). Theorem 1.1.15 Let A ∈ B(H) be normal, denote by z the inclusive mapping of σ(A) in C. Then there is unique homomorphism φ : C(σ(A)) → B(H) such that φ(z) = A, φ(¯z ) = A∗ , for each f ∈ C(σ(A)) φ( f (z)) = f (A),  f (z) =  f (A), and σ( f (A)) = f (σ(A)).

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Proof. Let A be the commutative C ∗ -algebra generated by A and I and Ω(A) the set of all characteristic on A. Then the Gelfand representation Γ : A → C(σ(A)) ˆ where Aˆ : is a ∗-isomorphism. Define Λ : C(σ(A)) → C(Ω(A)) by Λ f = f ( A), Ω(A) → σ(A) taken each h ∈ Ω(A) to h(A) ∈ σ(A). Define φ : C(σ(A)) → A by φ( f ) = Γ −1 Λ( f ). Now φ is the desired homomorphism.



Functional Calculus. The homomorphism φ : C(σ(A)) → B(H) with property of Theorem 1.1.15 is called functional calculus. Definition 1.1.16 An operator A ∈ B(H) is called positive or positive semidefinite if Ax, x ≥ 0 for every x ∈ H and then we write A ≥ 0. If Ax, x > 0 for all nonzero x ∈ H, we will say A is strictly positive or positive definite and then we write A > 0. For self-adjoint operators A, B ∈ B(H), we say A ≤ B if B − A ≥ 0, i.e., Bx, x ≥ Ax, x. This is called the operator order. It is well known that a positive self-adjoint operator A is invertible if and only if Sp(A) ⊂ (0, ∞). Also, for a self-adjoint operator A, we have Sp(A) ⊂ [m, M] where m = inf Ax, x and x=1

M = sup Ax, x. Consequently, a positive self-adjoint operator A is invertible if x=1

and only if the smallest element of its spectrum, denoted by m, be strictly positive and in this case we can write A ≥ m I . Let A ∈ B(H) be self-adjoint. Then A2 = A A∗ is positive and also the operator and A− = (|A|−A) are both |A| = (A2 )1/2 is positive. The operators A+ = |A|+A 2 2 positive and A = A+ − A− , A+ A− = 0. Theorem 1.1.17 Let A ∈ B(H) be self-adjoint and t ∈ R. Then (i) if A − t I  ≤ t, then A ≥ 0. (ii) if A ≤ t and A ≥ 0, then A − t I  ≤ t. Proof. The above statement is true for C(σ(T )). Now the results follow from Theorem 1.1.15.  Definition 1.1.18 U ∈ B(H1 , H2 ) is partial isometry if U is isometric on (kerU )⊥ , that is, U x = x for all x ∈ (kerU )⊥ . Definition 1.1.19 (Polar decomposition) Let A ∈ B(H). Then there is a unique partial isometry U ∈ B(H) such that A = U |A| and U ∗ A = |A|. Definition 1.1.20 Let T : B(H1 , H2 ) be a linear operator. Then T is compact if and only if T (B(0, 1)) is compact, where B(0, 1) = {h ∈ H1 : h ≤ 1}. The space of all compact operator from H1 to H2 is denoted by K(H1 , H2 ) and we write K(H) for K(H, H).

1.1 Operators and Matrices in Hilbert Space

11

Theorem 1.1.21 ([169]) Let H be a Hilbert space. Then K(H) = B(H) if and only if H is finite dimensional. Theorem 1.1.21 shows that every A ∈ Mn×m is a compact operator. Example 1.1.22 Let H be a separable Hilbert space and {en } and { f n } be orthonormal sets in H, and (λn ) be a sequence of complex numbers so that λn → 0. The operator T defined by Tx =

∞ 

λn x, en  f n ,

x ∈ H.

n=1

is compact. Theorem 1.1.23 ([169]) Let H be a separable Hilbert space and T ∈ K(H) be selfadjoint. Then there is a sequence of real numbers (λn ) and there is an orthonormal set {en } ⊂ H such that λn → 0 and Tx =

∞ 

λn x, en en ,

x ∈ H.

n=1

Canonical decomposition. Let T be a compact operator on separable Hilbert space H. Then by polar decomposition in definition 1.1.19 we have T = U |T |, 1 where U is a partial isometry and |T | = (T T ∗ ) 2 is a positive operator. Since |T | is a self-adjoint, there are λn and an orthonormal set {en} such that λn → 0 and λ x, en en . By setting f n = U en we obtain another orthonormal set |T |x = ∞ n n=1 and conclude that ∞  Tx = λn x, en  f n , x ∈ H, n=1

which is called the canonical decomposition of T. The nonincreasing sequence (λn ) is called the singular value of T and is denoted by σn (T ). Definition 1.1.24 (Schatten p-class) Let H be a Hilbert space and 0 < p < ∞. Then the Schatten p-class denoted by S p , is the space of all compact operators T with (σn (T )) ∈ l p . For 0 ≤ p ≤ ∞, S p is a Banach space with the norm T  p = 1 ( |σn (T )| p ) p . The operators in S1 are called trace class operators and the operators n in S2 are referred to as Hilbert-Schmidt operators. Note that if T x = λn x, en  f n , n

then

T ∗x =



λn x, f n en .

n

In particular, for 1 ≤ p ≤ ∞, T ∈ S p if and only if T ∗ ∈ S p and also T  p = T ∗  p .

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For each T ∈ S1 and an orthonormal basis {en } ⊂ H, the trace of T is defined by T r (T ) =

∞  T en , en . n=1

If {e1 , ..., en } is an orthonormal basis of a finite-dimensional Hilbert space H, then the trace T r A of A ∈ B(H) is defined as T r A :=

n  ei , Aei .

(1.1.2)

i=1

The space Mn of matrices with the inner product A, B = T r (AB ∗ ) becomes Hilbert space which is called Hilbert-Schmidt space. It follows the HilbertSchmidt norm A2 :=

n   √ 1 A, A = T r A A∗ = ( |Ai j |2 ) 2 . i, j=1

Theorem 1.1.25 ([220]) Let 1 ≤ p ≤ ∞ and then (i) (ii) (iii) (iv)

1 p

+

1 q

= 1. If T ∈ S p and S ∈ Sq ,

T S ∈ S1 and ST ∈ S1 . T r (ST ) = T r (T S). |T r (T S)| ≤ T  p Sq . T  p = sup{|T r (K T ) : k p = 1}.

Theorem 1.1.26 Definition (1.1.2) is independent of the choice of an orthonormal basis {e1 , ..., en } and T r (AB) = T r (B A) for all A, B ∈ B(H). Proof. We have T r AB =

n n   ei , ABei  = A∗ ei , Bei  i=1

i=1

=

n  n  i=1 j=1

e j , A∗ ei e j , Bei 

1.1 Operators and Matrices in Hilbert Space

=

n  n 

13

ei , B ∗ e j ei , Ae j 

j=1 i=1 n  = e j , B Ae j  = T r B A. j=1

Now, let { f 1 , ..., f n } be another orthonormal basis of H. Then a unitary U is defined by U ei = f i , 1 ≤ i ≤ n, and we have n n    fi , A fi  = U ei , AU ei  = T r (U ∗ AU ) = T r (AUU ∗ ) = T r A, i=1

i=1

which says that the definition of T r A is actually independent of the choice of an orthonormal basis.  Unitary invariant norms on operators on Cn . We say a matrix norm .u on Mn is unitary invariant if for any A ∈ Mn and all unitary matrices U, V ∈ Mn U AV u = Au . The spectral norm defined on Mn by σmax (A) = ρ(A∗ A) is a matrix norm and it is unitary invariant. Two classes of unitary invariant norms are specially important. The first is the class of Schatten p-norms defined as A p =

n   p 1 σi (A) p ,

A ∈ Mn ,

i=1

for p ≥ 1. The second is the class of Ky Fan k-norm defined as A(k) =

k 

σi (A),

A ∈ Mn ,

i=1

for 1 ≤ k ≤ n. The Frobenius norm or the Hilbert-Schmidt norm or Schatten 2-norms is the matrix norm defined by  n  21  σi2 (A) , A F = i=1

which is a unitary invariant norm. The spectral norm and Frobenius norm are members of two larger family of invariant Ky Fan k-norm and the Schatten p-norms. We recall that a matrix vector norm on the vector space Mm×n is a function . from Mm×n to [0, ∞) satisfying

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(i) A ≥ 0 and A = 0 if and only if A = 0. (ii) c A = |c|A, c ∈ C. (iii) A + B ≤ A + B. Theorem 1.1.27 (Ky Fan Dominance Theorem) ([216]) Let A, B ∈ Mm×n . Then Au ≤ Bu for all unitary invariant norm on Mm×n if and only if A(k) ≤ B(k) ,

k = 1, 2, ..., q,

where q = min{m, n}. Theorem 1.1.28 Assume that A ∈ B(H) is normal and H is a finite-dimensional Hilbert spaces. Then there exist λ1 , λ2 , ..., λn ∈ Cn , and u 1 , u 2 , ..., u n ∈ H, such that {u 1 , u 2 , ..., u n } is an orthonormal basis of H and Au i = λi u i for all 1 ≤ i ≤ n. Proof. Let us prove by induction on n = dim(H). The case n = 1 trivially holds. Suppose the assertion holds for dimension n − 1 . Assume that dim(H) = n and A ∈ B(H) is normal. Choose a root λ1 of det(λI − A) = 0. As we know, λ1 is an eigenvalue of A so that there is an eigenvector u 1 with Au 1 = λ1 u 1 . One may assume that u 1 is a unit vector, i.e., u 1  = 1. Since A is normal, we have (A − λ1 I )∗ (A − λ1 I ) = (A∗ − λ1 I )(A − λ1 I ) = A ∗ A − λ1 A − λ1 A ∗ + λ1 λ1 I = A A ∗ − λ1 A − λ1 A ∗ + λ1 λ1 I = (A − λ1 I )(A − λ1 I )∗ .

That is, A − λ1 I is also normal. Therefore, (A∗ − λ1 I )u 1  = (A − λ1 I )∗ u 1  = (A − λ1 I )u 1  = 0, so that A∗ u 1 = λ1 u 1 . Let H1 := {u 1 }⊥ be the orthogonal complement of {u 1 }. If x ∈ H1 then Ax, u 1  = x, A∗ u 1  = x, λ1 u 1  = λ1 x, u 1  = 0, A∗ x, u 1  = x, Au 1  = x, λ1 u 1  = λ1 x, u 1  = 0,

1.1 Operators and Matrices in Hilbert Space

15

so that Ax, A∗ x ∈ H1 . Hence, we have AH1 ⊂ H1 and A∗ H1 ⊂ H1 . So one can define A1 |H1 ∈ B(H1 ). Then A∗1 = A∗ |H1 which implies that A1 is also normal. Since dim(H1 ) = n − 1, the induction hypothesis can be applied to obtain λ2 , ..., λn ∈ C and u 2 , ..., u n ∈ H1 such that {u 2 , ..., u n } is an orthonormal basis of H1 and A1 u i = λi u i for all i = 2, ..., n. Then {u 1 , u 2 , ..., u n } is an orthonormal basis of H and A1 u i = λi u i for all i = 1, 2, ..., n. Thus, the assertion holds for dimension n as well.  Remark 1.1.29 It is an important consequence that the matrix of a normal operator is diagonal in an appropriate orthonormal basis and the trace is the sum of the eigenvalues. Theorem 1.1.30 Assume that H is finite-dimensional Hilbert spaces and A ∈ B(H) is self-adjoint. If Aν = λν and Aω = μω with nonzero eigenvectors ν, ω and the eigenvalues λ and μ are different, then ν⊥ω and λ, μ ∈ R. Proof. First, we show that the eigenvalues are real as follows: λν, ν = ν, λν = ν, Aν = Aν, ν = λν, ν = λν, ν. The orthogonality ν, ω = 0 comes similarly in the following way: μν, ω = ν, μω = ν, Aω = Aν, ω = λν, ω = λν, ω.  Next, we give Poincare’s inequality. Theorem 1.1.31 Let A ∈ B(H) be a self-adjoint operator in a finite-dimensional Hilbert spaces H with eigenvalues λ1 ≥ λ2 ≥ ... ≥ λn (counted with multiplicity) and let K be a k-dimensional subspace of H. Then there are unit vectors x, y ∈ K such that x, Ax ≤ λk and y, Ay ≥ λk . Proof. Let νk , ..., νn be orthonormal eigenvectors corresponding to the eigenvalues λk , ..., λn . They span a subspace M of dimension n − k + 1 which must have an intersection with K. Take a unit vector x ∈ K ∩ M which has the expansion x=

n 

ci νi ,

i=k

and it has the required property x, Ax =

n  i=k

|ci |2 λi ≤ λk

n 

|ci |2 = λk .

i=k

To find the other vector y, the same argument can be used with the matrix −A.



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Operator monotone and convex functions. A continuous function f : R → R is said to be operator monotone if A and B be self-adjoint operator and A ≤ B implies f (A) ≤ f (B). f is operator convex if for every self-adjoint operators A and B and for all real number 0 ≤ λ ≤ 1, f ((1 − λ)A + λB) ≤ (1 − λ) f (A) + λ f (B).   ≤ Since f is continuous, therefore, the above condition can be replaced by f A+B 2 f (A) f (B) + 2 . A function f is called operator concave if the function − f is operator 2 convex. The set of operator monotone functions and the set of operator convex functions are both closed under positive linear combination and also under pointwise limits. Example 1.1.32 The function f (t) = α + βt is operator monotone on every interval for every α ∈ R and β ≥ 0. Example 1.1.33 The function f (t) = t 2 is not operator monotone. Let

A=

11 11

 and B =

 21 . 11

Then B − A is positive, but B 2 − A2 is not. Example 1.1.34 The function f (t) = t 2 is operator convex on every interval. Let A and B be two Hermitian operators. Then A2 + B 2 − 2



A+B 2

2 =

1 (A − B)2 ≥ 0. 4

This shows that the function f (t) = α + βt + γt 2 is operator convex for all α, β ∈ R and γ ≥ 0. Operator means. For positive operators on a complex Hilbert space H, the theory of operator means is established axiomatically by Kubo and Ando. A binary operation δ on the class of positive operators (A, B) → Aδ B is called a connection if the following monotonicity property holds: A ≤ C and B ≤ D imply Aδ B ≤ Cδ D. It also has the following two properties: C(Aδ B)C ≤ (C AC)δ(C BC), An ↓ A and Bn ↓ B imply (An δ Bn ) ↓ (Aδ B). A connection δ is a mean if it is normalized, i.e., I δ I = I where I denotes the identity operator. There exists an affine order isomorphism between the class of

1.1 Operators and Matrices in Hilbert Space

17

operator means δ and the class of positive operator monotone functions f defined on (0, ∞) with f (1) = 1 via f (t)I = I δ(t I ) for t > 0. This isomorphism δ ↔ f is characterized by  1  1 1 1 Aδ B = A 2 f A− 2 B A− 2 A 2 , for all strictly positive operators A, B. The operator monotone function f is called the representing function of δ. Using a standard limit argument, this notion can be extended for positive operators A, B. The operator mean corresponding to the operator monotone functions f (t) = t λ with λ ∈ [0, 1] is called the weighted operator geometric mean, which is indeed  1  1 1 1 λ Aλ B = A 2 A− 2 B A− 2 A 2 . The case where λ = 21 gives to the usual operator geometric mean . The operator mean corresponding to the operator monotone functions f (t) = (1 − λ) + λt with λ ∈ [0, 1] is called the weighted operator arithmetic mean, which is indeed A∇λ B = (1 − λ)A + λB. The case where λ = 21 gives to the usual operator arithmetic mean ∇. Positive linear Map. A linear map Φ : B(H) → B(K) is called positive if A ≥ 0 implies Φ(A) ≥ 0. It is said to be unital if Φ preserves the identity operator. If A is invertible, so does Φ(A). Now, we explain why? If A is invertible, then there is a real constant c > 0 such that Ah, h ≥ ch, h for all h ∈ H. Also, any positive linear map is monotone and preserves order relation. So Φ(A)h, h = Φ(Ah, h) ≥ Φ(ch, h) = Φ(c)h, h, and hence Φ(A) is invertible. In this book, what we mean by writing Φ −1 was not the inverse of the positive unital linear map φ. In fact, we meant (Φ(A))−1 as the inverse of the operator Φ(A) ∈ B(K). Numerical range and radius. Let A be an n × n complex matrix. For x = n 1 (x1 , x2 , ..., xn )T ∈ Cn , as usual, x = ( |xi |2 ) 2 is the norm of x. The numerii=1

cal range, also known as the field of values, of A is defined by W (A) = {x ∗ Ax : x = 1, x ∈ Cn }. For example, if

A=

 10 , 00

then W (A) is the closed interval [0, 1], and if

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1 Elementary Linear Algebra Review

A=

 00 , 11

then W (A) is the √ closed elliptical disk with foci at (0, 0) and (1, 0), minor axis 1, and major axis 2. One of the celebrated and fundamental results on the numerical range is the Toeplitz-Hausdorff convexity theorem as follows. Theorem 1.1.35 ([216]) The numerical range of a square matrix is a convex compact subset of the complex plane. When considering the smallest disk centered at the origin that covers the numerical range, we associate with W (A) a number ω(A) = sup{|z| : z ∈ W (A)} = sup |x ∗ Ax|, x=1

and call it the numerical radius of A ∈ Mn . Note that the sup can be attained by some z ∈ W (A). It is immediate that for any x ∈ Cn |x ∗ Ax| ≤ ω(A)x2 . We now make comparisons of the numerical radius w(A) to the largest eigenvalue ρ(A) in absolute value, or the spectral radius, i.e., ρ(A) = max{|λ| : λ is an eigenvalue of A}, and to the largest singular value σmax (A), also called the spectral norm. It is easy to see that Ax , σmax (A) = sup Ax = sup x=1 x =0 x and that for every x ∈ Cn

Ax ≤ σmax (A)x.

Theorem 1.1.36 ([216]) Let A be a square complex matrix. Then ρ(A) ≤ ω(A) ≤ σmax (A) ≤ 2ω(A). Notes and references. We refer the readers to [97], [104], [169], [216], and [220] for further details.

Chapter 2

Interpolating the Arithmetic-Geometric Mean Inequality and Its Operator Version

In this chapter, we gather refinements of the classical Young inequality for positive real numbers, and we use these refinements to establish improved Young and Heinz inequalities for operators. According to the definitions of operator entropies, relative operator entropies, and Tsallis operator entropy, the readers can employ the techniques in this chapter to get new upper and lower bounds of the operator entropies.

2.1 Refinements of the Scalar Young and Heinz Inequalities The weighted arithmetic-geometric mean inequality, which is also called Young’s inequality, states (2.1.1) (1 − λ)a + λb ≥ a 1−λ bλ , for a, b ≥ 0 and λ ∈ [0, 1]. This inequality, though very simple, has attracted researchers working in operator theory due to its applications in this field. If λ = 21 , we obtain the arithmetic-geometric mean inequality a+b √ ≥ ab. 2

(2.1.2)

The Heinz means, introduced in [19], are defined by Hλ (a, b) =

a 1−λ bλ + a λ b1−λ , 2

for a, b ≥ 0 and λ ∈ [0, 1]. It is easy to see that as a function of λ, Hλ (a, b) is convex and attains its minimum at λ = 21 . Thus © Springer Nature Switzerland AG 2021 M. B. Ghaemi et al., Advances in Matrix Inequalities, Springer Optimization and Its Applications 176, https://doi.org/10.1007/978-3-030-76047-2_2

19

20

2 Interpolating the Arithmetic-Geometric Mean Inequality …

√ a+b , ab ≤ Hλ (a, b) ≤ 2

λ ∈ [0, 1],

(2.1.3)

which is called Heinz inequality. Refining Young inequality has taken the attention of many researchers, where adding a positive term to the right side of the inequality becomes possible. We refer the reader to [7, 16, 52, 74–76, 84, 100, 101, 117–120, 123, 165, 188–190, 218] as a sample of the extensive use of Young and Heinz inequalities. One of the first refinements is a multiplicative-type refinement for the Young inequality with Specht’s ratio. We use the following lemmas to show refined Young inequality with Specht’s ratio. At first, we review the properties of Specht’s ratio. See [68, 196, 202], for example, as for the proof and details. Lemma 2.1.1 ([74]) Specht’s ratio 1

S(h) =

h h−1 1

e log h h−1

, (h = 1, h > 0)

has the following properties: (i) S(1) = 1 and S(h) = S( h1 ) > 1 for h > 0. (ii) S(h) is a monotone increasing function on (1, ∞). (iii) S(h) is a monotone decreasing function on (0, 1). Lemma 2.1.2 For x ≥ 1, we have 2(x − 1) x −1 ≤ log x ≤ √ . x +1 x Proof. We firstly prove the second inequality. We put f (t) =



x = t and

t2 − 1 − 2 log t, (t ≥ 1). t

 2 Then we have ddtf = t−1 and f (1) = 0. Thus, we have f (t) ≥ f (1) = 0 and then t t 2 −1 2 we have log t ≤ t which implies x −1 log x ≤ √ . x We also put g(x) = (x + 1) log x − 2(x − 1), (x ≥ 1). − 2, ddgx (1) = 0 and Thus g(1) = 0, ddgx = log x + x+1 x we have g(x) ≥ g(1) = 0 which implies

d2 g dx2

=

x−1 x2

≥ 0. Therefore,

2.1 Refinements of the Scalar Young and Heinz Inequalities

21

2(x − 1) ≤ log x, x +1 and this completes the proof.  Remark 2.1.3 The following proof for the second inequality in Lemma 2.1.2 is a more straightforward idea mentioned by Dorin Andrica. Put x = e2t for t ≥ 0. Then √ is equivalent to log x ≤ x−1 x e2t − 1 2t ≤ , et that is, t ≤ sinh t. If we want to provide an argument for the last inequality, just consider the function f (t) = sinh t − t (t ≥ 0), with

df = cosh t − 1 ≥ 0, hence f (t) ≥ f (0) = 0 and so sinh t ≥ t. dt

Lemma 2.1.4 For t ∈ (0, 1) ∪ (1, ∞), we have t

e(t 2 + 1) ≥ (t + 1)t t−1 . Proof. We firstly prove the inequality for t > 1. We put t

f (t) = e(t 2 + 1) − (t + 1)t t−1 . We have t

t

2t (t − 1)2 e + 2t (1 − t)t t−1 + t t−1 (t + 1) log t f (t) = (t − 1)2

t



t

2t (t − 1)2 e + 2t (1 − t)t t−1 + 2t t−1 (t − 1) ( by Lemma 2.1.2) (t − 1)2 1

2t (t − 1)2 e + 2t (t − 1)2 t t−1 = (t − 1)2 1



1

2t (t − 1)2 t t−1 − 2t (t − 1)2 t t−1 = 0. (t − 1)2 1

1

In the last inequality, we have used the fact that lim t t−1 = e and the function t t−1 is t→1

monotone decreasing for t ∈ (1, ∞). We also have limt→1 f (t) = 0 so that we have f (t) > 0 which proves the inequality t

e(t 2 + 1) ≥ (t + 1)t t−1 , t ≥ 1. Putting t =

1 s

in the above inequality with simple calculations, we have

22

2 Interpolating the Arithmetic-Geometric Mean Inequality … s

e(s 2 + 1) ≥ (s + 1)s s−1 , 0 < s < 1.  Remark 2.1.5 Thanks to Dorin Andrica we give a new different proof for Lemma 2.1.4 in the following way. The desired inequality in Lemma 2.1.4 is equivalent to (t − 1) + (t − 1) log(t 2 + 1) ≥ (t − 1) log(t + 1) + t log t, which can be extended for t ≥ 1. Let f : [1, ∞) → R, defined by f (t) = (t − 1) + (t − 1) log(t 2 + 1) − (t − 1) log(t + 1) − t log t. We have t −1 2t (t − 1) − log(t + 1) − − log t, 2 t +1 t +1 (t 5 − 1) + 7t (t 3 − 1) + 2t 2 (t − 1) f (t) = . t (t + 1)2 (t 2 + 1)2

f (t) = log(t 2 + 1) +









f (t) ≥ 0 for t ≥ 1. It follows that f is increasing, hence f (t) ≥ f (1) = 0. We get that f is increasing, i.e., f (t) ≥ f (1) = 0 and this completes the proof for t ≥ 1. Putting t = 1s in the above inequality we deduce the desired inequality for 0 < t ≤ 1. We have the following inequality which improves the classical Young inequality between λ-weighted geometric mean and λ-weighted arithmetic mean. Theorem 2.1.6 For a, b > 0 and λ ∈ [0, 1], (1 − λ)a + λb ≥ S

 r  b a 1−λ bλ , a

where r = min{λ, 1 − λ} and S(.) is Specht’s ratio. Proof. We prove the following inequality: (b − 1)λ + 1 e{(b − 1)λ + 1} log bλ =   bλ ≥ 1, λ λ b S(b ) bλ bλ −1 (bλ − 1) in the case of 0 ≤ λ ≤ 21 . From Lemma 2.1.2, we have 2 log bλ ≥ λ , b > 0. bλ − 1 b +1 Therefore, we have the following inequality:

2.1 Refinements of the Scalar Young and Heinz Inequalities

23

e{(b − 1)λ + 1} log bλ 2e{(b − 1)λ + 1} ≥   bλ ,   λbλ bλ b −1 (bλ − 1) bλ bλ −1 (bλ + 1) thus we have only to prove 2e{(b − 1)λ + 1} ≥ 1.   λbλ bλ b −1 (bλ + 1) For this purpose, we put the following function f b on λ ∈ [0, 21 ]:   bλ f b (λ) = 2e{(b − 1)λ + 1} − bλ bλ −1 (bλ + 1) for b > 0. Then, we have −1 (log b)2  λ  2bbλλ−1 d 2 f b (λ) =− λ b {(bλ − 1)2 (4b2λ − 5bλ − 1) 4 dλ (b − 1) − (bλ − 1)2 (3bλ + 1) log bλ + bλ (bλ + 1)(log bλ )2 }.

For the case of b ≥ 1, using the inequalities in Lemma 2.1.2, we have (bλ − 1)2 (4b2λ − 5bλ − 1) − (bλ − 1)2 (3bλ + 1) log bλ + bλ (bλ + 1)(log bλ )2 ≥ (bλ − 1)2 (4b2λ − 5bλ − 1) − (bλ − 1)2 (3bλ + 1) + bλ (bλ + 1) λ

=



2(bλ − 1) bλ + 1

λ

2

(b 2 − 1)4 (b 2 + 1)3 (4b2λ + b 2 + 4bλ + 1)

bλ − 1 λ

b2



λ

b 2 (bλ + 1)

≥ 0.

For the case of 0 < b ≤ 1, using the inequalities of Lemma 2.1.2, we also have (bλ − 1)2 (4b2λ − 5bλ − 1) − (bλ − 1)2 (3bλ + 1) log bλ + bλ (bλ + 1)(log bλ )2 1 1 2 = (bλ − 1)2 (4b2λ − 5bλ − 1) + (bλ − 1)2 (3bλ + 1) log + bλ (bλ + 1)(log ) bλ bλ ⎛ ⎞ 1 2( λ − 1) ⎠ ≥ (bλ − 1)2 (4b2λ − 5bλ − 1) + (bλ − 1)2 (3bλ + 1) ⎝ b1 +1 bλ ⎞2 ⎛ 2( 1λ − 1) b λ λ ⎠ + b (b + 1) ⎝ 1 λ +1 b

(bλ − 1)4 (4bλ + 1) ≥ 0. = bλ + 1

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2 Interpolating the Arithmetic-Geometric Mean Inequality …

f b (λ) Thus, we have d dλ ≤ 0 for b > 0. In addition, we have f b (λ) = 0 and f b ( 21 ) = √ √ √√b √ e(b + 1) − ( b + 1)( b) b−1 ≥ 0, applying Lemma 2.1.4 with t = b > 0. Therefore, we have f b (λ) ≥ 0 for λ ∈ [0, 21 ]. Thus, we have the following inequality: 2

1 (b − 1)λ + 1 ≥ 1, 0 ≤ λ ≤ , b > 0, λ λ b S(b ) 2 which implies

Replacing b by we have

(2.1.4)

λb + (1 − λ) ≥ S(bλ )bλ . b a

in the above inequality and then multiplying a to the both sides,

  b λ 1−λ λ 1 (1 − λ)a + λb ≥ S a b , 0 ≤ λ ≤ , a, b > 0. a 2 Finally, from the inequality (2.1.4), we have 1 (a − 1)μ + 1 ≥ 1, 0 ≤ μ ≤ , a > 0. μ μ a S(a ) 2 Putting λ = 1 − μ in the above inequality, we have λ + (1 − λ)a ≥ a 1−λ S(a 1−λ ), Replacing a by we have

a b

1 ≤ λ ≤ 1, a > 0. 2

in the above inequality and then multiplying b to the both sides,

(1 − λ)a + λb ≥ S

  a 1−λ b

a 1−λ bλ ,

1 ≤ λ ≤ 1, a, b > 0, 2

since S( h1 ) = S(h) for h > 0. Thus, the proof of the present theorem is completed.  Remark 2.1.7 Theorem 2.1.6 gives a tighter lower bound of the classical Young inequality since Specht’s ratio is greater than 1. That is  r  b a 1−λ bλ ≥ a 1−λ bλ . (1 − λ)a + λb ≥ S a What is common among the different appearing in the literature is the fact of having one refining term. That is, one positive term is added to the right side of Young’s inequality. In the following, we give known results with one and two refining

2.1 Refinements of the Scalar Young and Heinz Inequalities

25

terms. However, our main target is to present a refinement that has many terms as we wish! Theorem 2.1.8 If a, b ≥ 0 and λ ∈ [0, 1], then √ √ (1 − λ)a + λb ≥ a 1−λ bλ + r0 ( a − b)2 ,

(2.1.5)

where r0 = min{λ, 1 − λ}. Proof. If λ = 21 , the inequality (2.1.5) becomes an equality. Assume that λ ∈ [0, 21 ). Then r0 = λ and we have √ √ √ (1 − λ)a + λb − λ( a − b)2 = 2λ ab + (1 − 2λ)a ≥ (ab)λ a (1−2λ) ( by (2.1.1)) = a 1−λ bλ . So, we get

√ √ (1 − λ)a + λb ≥ a 1−λ bλ + λ( a − b)2 ,

for λ ∈ [0, 21 ). If λ ∈ ( 21 , 1], then 1 − λ
0 and λ ∈ [0, 1]. Given N ∈ N, consider the integers k j (λ) = [2 j−1 λ] and r j (λ) = [2 j λ], j = 1, 2, ..., N . Then (1 − λ)a + λb −

N  

(−1)r

j

(λ) j−1

2

λ + (−1)r

j

(λ)+1



j=1

r j (λ) + 1 2

 ×

2

2  j 2j j−1 j−1 × a 2 −k j (λ) bk j (λ) − a 2 −k j (λ)−1 bk j (λ)+1 N   2 = [2 N λ] + 1 − 2 N λ a 2 N −[2 N λ] b[2 N λ] + N   2 + 2 N λ − [2 N λ] a 2 N −[2 N λ]−1 b[2 N λ]+1 . Proof. We put  j  N   r j (λ) j−1 r j (λ)+1 r (λ) + 1 (−1) × S N (λ, a, b) = 2 λ + (−1) 2 j=1 2

2  j 2j j−1 j−1 × a 2 −k j (λ) bk j (λ) − a 2 −k j (λ)−1 bk j (λ)+1 , and N   2 a 2 N −[2 N λ] b[2 N λ] R N (λ, a, b) = [2 N λ] + 1 − 2 N λ    2N N N + 2 N λ − [2 N λ] a 2 −[2 λ]−1 b[2 N λ]+1 .

Thus, we prove (1 − λ)a + λb − S N (λ, a, b) = R N (λ, a, b). We proceed the proof by induction on N . When N = 1, then    2λ + 1 √ 1−[λ] [λ] √ −[λ] [λ]+1 2 a b − a b . S1 (λ, a, b) = (−1)[2λ] λ + (−1)[2λ]+1 2 For S1 (λ, a, b), we have three cases (i) if λ = 1, then S1 (1, a, b) = 0. Hence (1 − λ)a + λb − S1 (1, a, b) = b = R1 (1, a, b).

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2 Interpolating the Arithmetic-Geometric Mean Inequality …

(ii) If λ ∈ [0, 21 ), then [2λ] = 0 and S1 (λ, a, b) = λ

√ √ 2 a − b . Hence

√ (1 − λ)a + λb − S1 (λ, a, b) = (1 − 2λ)a + 2λ ab = R1 (λ, a, b). (iii) If λ ∈ [ 21 , 1), then [2λ] = 1 and S1 (λ, a, b) = (1 − λ)

√ √ 2 a − b . Hence

√ (1 − λ)a + λb − S1 (λ, a, b) = (2λ − 1)b + 2(1 − λ) ab = R1 (λ, a, b). Thus, we have (1 − λ)a + λb − S1 (λ, a, b) = R1 (λ, a, b) for λ ∈ [0, 1] and this completes the proof for N = 1. Now assume that (1 − λ)a + λb − S N (λ, a, b) = R N (λ, a, b) for some N ∈ N. We assert that (1 − λ)a + λb − S N +1 (λ, a, b) = R N +1 (λ, a, b). For this purpose, we consider the cases [2 N +1 λ] are odd and even. We have (1 − λ)a + λb − S N +1 (λ, a, b) = (1 − λ)a + λb − S N (λ, a, b)−  N +1   λ] + 1 [2 N +1 λ] N [2 N +1 λ]+1 [2 × 2 λ + (−1) − (−1) 2

2 2 N +1   2 N +1 × a 2 N −[2 N λ] b[2 N λ] − a 2 N −−[2 N λ]−1 b[2 N λ]+1 .

(2.1.8)

If [2 N +1 λ] is odd, then (2.1.8) becomes (1 − λ)a + λb − S N +1 (λ, a, b)    N +1 λ] + 1 [2 × = R N (λ, a, b) − −2 N λ + 2

2 2 N +1   2 N +1 × a 2 N −[2 N λ] b[2 N λ] − a 2 N −[2 N λ]−1 b−[2 N λ]+1 N   2 = [2 N λ] + 1 − 2 N λ a 2 N −[2 N λ] b[2 N λ]    N +1 N   2 λ] + 1 [2 × + 2 N λ − [2 N λ] a 2 N −[2 N λ]−1 b[2 N λ]+1 − −2 N λ + 2   2 N +1 a 2 N +1 −2[2 N λ] b2[2 N λ] − 2 a 2 N +1 −2[2 N λ]−1 b2[2 N λ]+1 +  2 N +1 + a 2 N +1 −2[2 N λ]−2 b2[2 N λ]+2 ). ×(

2 N +1

Now one can easily see that when [2 N +1 λ] is odd, then

(2.1.9)

2.1 Refinements of the Scalar Young and Heinz Inequalities

29

2[2 N λ] + 1 = [2 N +1 λ], [2 N +1 λ] + 1 −2 N λ + = −2 N λ + [2 N λ] + 1, 2   2 N +1 2 N +1 a 2 N +1 −2[2 N λ] b2[2 N λ] = a 2 N −[2 N λ] b[2 N λ] . Hence (2.1.9) becomes (1 − λ)a + λb − S N +1 (λ, a, b)    [2 N +1 λ] + 1 2 N +1 N N N a 2 N +1 −2[2 N λ]−2 b2[2 N λ]+2 = 2 λ − [2 λ] + 2 λ − 2    2 N +1 + [2 N +1 λ] + 1 − 2 N +1 λ a 2 N +1 −[2 N +1 λ] b[2 N +1 λ]    2 N +1 N +1 N +1 = −[2 N +1 λ] + 2 N +1 λ a 2 −[2 λ]−1 b[2 N +1 λ]+1    2 N +1 + [2 N +1 λ] + 1 − 2 N +1 λ a 2 N +1 −[2 N +1 λ] b[2 N +1 λ] = R N +1 (λ, a, b). This completes the proof for the case [2 N +1 λ] is odd. Now we consider the second case. If [2 N +1 λ] is even, then 

 [2 N +1 λ] + 1 [2 N +1 λ] = = [2 N λ], 2 2

and so (2.1.8) becomes (1 − λ)a + λb − S N +1 (λ, a, b)    N +1 λ] + 1 [2 N × = R N (λ, a, b) − 2 λ − 2

2 2 N +1   2 N +1 × a 2 N −[2 N λ] b[2 N λ] − a 2 N −[2 N λ]−1 b−[2 N λ]+1 N   2 = [2 N λ] + 1 − 2 N λ a 2 N −[2 N λ] b[2 N λ]    2N N N + 2 N λ − [2 N λ] a 2 −[2 λ]−1 b[2 N λ]+1

 N   2 2N − 2 N λ − [2 N λ] a 2 N −[2 N λ] b[2 N λ] + a 2 N −[2 N λ]−1 b[2 N λ]+1    2 N +1 + 2 N +1 λ − [2 N +1 λ] a 2 N +1 −[2 N +1 λ]−1 b[2 N +1 λ]+1

30

2 Interpolating the Arithmetic-Geometric Mean Inequality … N   2 = 2[2 N λ] + 1 − 2 N +1 λ a 2 N −[2 N λ] b[2 N λ] +    2 N +1 + 2 N +1 λ − [2 N +1 λ] a 2 N +1 −[2 N +1 λ]−1 b[2 N +1 λ]+1 = R N +1 (λ, a, b).

This completes the proof.  Theorem 2.1.12 Let a, b > 0 and λ ∈ [0, 1]. Then the following generalized refinement of Young’s inequality holds: (1 − λ)a + λb ≥ a 1−λ bλ + S N (λ, a, b). Proof. According to Lemma 2.1.11, we have (1 − λ)a + λb − S N (λ, a, b)  j  N   r j (λ) j−1 r j (λ)+1 r (λ) + 1 (−1) × = (1 − λ)a + λb − 2 λ + (−1) 2 j=1 2

2  j 2j j−1 j−1 × a 2 −k j (λ) bk j (λ) − a 2 −k j (λ)−1 bk j (λ)+1 N   2 = [2 N λ] + 1 − 2 N λ a 2 N −[2 N λ] b[2 N λ] +    2N N N + 2 N λ − [2 N λ] a 2 −[2 λ]−1 b[2 N λ]+1

[2 N λ]+1−2 N λ 2

2 N λ−[2 N λ] 2 N N ≥ a 2 N −[2 N λ] b[2 N λ] a 2 N −[2 N λ]−1 b[2 N λ]+1 = a 1−λ bλ , which is the last inequality obtained by (2.1.1).



Corollary 2.1.13 Let a, b > 0, λ ∈ [0, 1] and N ∈ N. Then (1 − λ)a + λb ≥ R N (λ, a, b) ≥ a 1−λ bλ , where R N (λ) is defined in Lemma 2.1.11. Proof. According to Theorem 2.1.12, we have (1 − λ)a + λb ≥ a 1−λ bλ + S N (λ, a, b). But S N (λa, b) = (1 − λ)a + λb − R N (λ, a, b), which implies

R N (λ, a, b) ≥ a 1−λ bλ .

Moreover, since S N (λ, a, b) ≥ 0 and R N (λ, a, b) = (1 − λ)a + λb − S N (λ, a, b), it follows that R N (λ, a, b) ≤ (1 − λ)a + λb.

2.1 Refinements of the Scalar Young and Heinz Inequalities

Hence (1 − λ)a + λb ≥ R N (λ, a, b) ≥ a 1−λ bλ .

31



Theorem 2.1.14 Let a, b > 0, λ ∈ [0, 1]. Then lim S N (λ, a, b) = (1 − λ)a + λb − a 1−λ bλ .

N →∞

Proof. According to Lemma 2.1.11, we have (1 − λ)a + λb − S N (λ, a, b) = R N (λ, a, b), where N   2 a 2 N −[2 N λ] b[2 N λ] R N (λ, a, b) = [2 N λ] + 1 − 2 N λ    2N N N + 2 N λ − [2 N λ] a 2 −[2 λ]−1 b[2 N λ]+1  2N = a 2 N −[2 N λ] b[2 N λ] +

 N   2 2N + 2 N λ − [2 N λ] a 2 N −[2 N λ]−1 b[2 N λ]+1 − a 2 N −[2 N λ] b[2 N λ] .

(2.1.10) If λ = 0 then lim R N (λ, a, b) = a = a 1−λ bλ . If λ = 0, then N →∞

lim

N →∞

   [2 N λ] [2 N λ] a 2 N −[2 N λ] b[2 N λ] = lim a 1− 2 N λ λ b 2 N λ λ

2N

N →∞

= a 1−λ bλ , [2 N λ] N N →∞ 2 λ

because lim

(2.1.11)

= 1. On the other hand

2

 N 2N a 2 N −[2 N λ]−1 b[2 N λ]+1 − a 2 N −[2 N λ] b[2 N λ] N →∞   [2 N λ]λ [2 N λ]λ [2 N λ]λ [2 N λ]λ 1 1 = lim a 1− 2 N − 2 N b 2 N + 2 N − a 1− 2 N b 2 N lim

N →∞

= a 1−λ bλ − a 1−λ bλ = 0. Now it follows from 0 ≤ (2 N λ − [2 N λ]) ≤ 1 that

 N   2 2N lim 2 N λ − [2 N λ] a 2 N −[2 N λ]−1 b[2 N λ]+1 − a 2 N −[2 N λ] b[2 N λ] = 0.

N →∞

(2.1.12) Then (2.1.11) and (2.1.12) invoked (2.1.10) imply lim R N (λ, a, b) = a 1−λ bλ and this completes the proof.



N →∞

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2 Interpolating the Arithmetic-Geometric Mean Inequality …

Notes and references. Lemma 2.1.2, Lemma 2.1.4, and Theorem 2.1.6 are due to [74]. Theorem 2.1.8 is proved by [119]. Theorem 2.1.10 is due to Zhao and Wu [218]. Sababheh and Choi obtained Lemma 2.1.11, Theorem 2.1.12, Corollary 2.1.13, and Theorem 2.1.14 in [188].

2.2 Operator Inequalities Involving Improved Young Inequality Let A, B > 0 and λ ∈ [0, 1]. We recall that the λ-weighted operator geometric mean of A and B, denoted by Aλ B, is defined as 1

1 1 1 λ Aλ B = A 2 A− 2 B A− 2 A 2 , and the λ-weighted operator arithmetic mean of A and B, denoted by A∇λ B, is A∇λ B = (1 − λ)A + λB. When λ = 21 , A 21 B, and A∇ 21 B are called operator geometric mean and operator arithmetic mean, and denoted by AB and A∇ B, respectively [124]. For positive operators A, B and λ ∈ [0, 1], we have [77, Definition 5.2] Aλ B = B1−λ A.

(2.2.1)

It is well known that if A, B > 0 and λ ∈ [0, 1], then [79, 80] A∇λ B ≥ Aλ B, which is the operator version of the scalar Young’s inequality. An operator version of Heinz means was introduced in [19] by Hλ (A, B) =

Aλ B + A1−λ B , 2

where λ ∈ [0, 1]. In particular, H1 (A, B) = H0 (A, B) = A∇ B. It is easy to see that the Heinz operator means interpolate between the arithmetic and geometric operator means AB ≤ Hλ (A, B) ≤ A∇ B, which is called the Heinz operator inequality [117, 118]. Now, we obtain the operator version of the scalar Young inequalities introduced in previous section. The techniques are based on the functional calculus described in the following lemma [77].

2.2 Operator Inequalities Involving Improved Young Inequality

33

Lemma 2.2.1 Let X ∈ B(H) be self-adjoint and let f and g be continuous functions such that f (t) ≤ g(t) for all t in the spectrum of X . Then f (X ) ≤ g(X ). Theorem 2.2.2 For two positive operators A, B and positive real numbers m, m , M, M satisfying the following condition (i) or (ii):



(i) 0 < m I ≤ A ≤ m I ≤ M I ≤ B ≤ M I , (ii) 0 < m I ≤ B ≤ m I ≤ M I ≤ A ≤ M I , with h =

M m



and h =



M m

, we have (1 − λ)A + λB ≥ S(h r )Aλ B,

where λ ∈ [0, 1], r = min{λ, 1 − λ} and Specht’s ratio S(.). Proof. From Theorem 2.1.6, we have λt + (1 − λ) ≥ S(t r )t λ , for any t > 0. Therefore, by functional calculus, we have λX + (1 − λ)I ≥ min S(t r )X λ , h≤t≤h



for the positive operator X such that h I ≤ X ≤ h I . We here put X = A− 2 B A− 2 . In the case of (i), we have M I ≤ B ≤ M I and m I ≤ A ≤ m I . So we easily find − 21 − 21 that h I ≤ A B A ≤ h I . Then we have 1

λ A− 2 B A− 2 + (1 − λ)I ≥ min S(x r )(A− 2 B A− 2 )λ . 1

1

1

1

h≤t≤h

Since S(t) is an monotone increasing function for t > 1, we have λA− 2 B A− 2 + (1 − λ)I ≥ S(h r )(A− 2 B A− 2 )λ . 1

1

1

1

Multiplying both sides of the inequality gives the desired inequality in (i). 1 1 In the case of (ii), we also have h1 I ≤ A− 2 B A− 2 ≤ h1 I . Then λA− 2 B A− 2 + (1 − λ)I ≥ min S(t r )(A− 2 B A− 2 )λ . 1

1

1

1 h

1

≤t≤ h1

Since S(t) is a monotone decreasing function for 0 < t < 1, we have  1 1 1 (A− 2 B A− 2 )λ hr   1 1 = S h r (A− 2 B A− 2 )λ .

λA− 2 B A− 2 + (1 − λ)I ≥ S 1

1



1

34

2 Interpolating the Arithmetic-Geometric Mean Inequality …

Multiplying both sides of the inequality gives the desired inequality in (ii).  Theorem 2.2.3 Let A, B > 0 and λ ∈ [0, 1], then A∇λ B ≥ Aλ B + 2r0 (A∇ B − AB), where r0 = min{λ, 1 − λ}. Proof. According to Theorem 2.1.8, we have the following inequality for t =

b a

> 0:

(1 − λ) + λt ≥ t λ + r0 (1 − 2t 2 + t). 1

By functional calculus, if we replace t with A− 2 B A− 2 and then multiplying both 1 sides of the inequality by A 2 , then 1

1

(1 − λ)A + λB ≥ Aλ B + r0 (A − 2 AB + B) = Aλ B + r0 (2 A∇ B − 2 AB) = Aλ B + 2r0 (A∇ B − AB), which is equivalent to A∇λ B ≥ Aλ B + 2r0 (A∇ B − AB).  As a direct consequence of Theorem 2.2.3, we have Corollary 2.2.4 Let A, B > 0 and λ ∈ [0, 1], then A∇ B ≥ Hλ (A, B) + 2r0 (A∇ B − AB), where r0 = min{λ, 1 − λ}. Proof. We have A∇λ B + B∇λ A ≥ Aλ B + 2r0 (A∇ B − AB) + Bλ A + 2r0 (B∇ A − BA), which is equivalent to A + B ≥ Aλ B + 2r0 (A∇ B − AB) + A1−λ B + 2r0 (A∇ B − AB), and so we have A+B Aλ B + A1−λ B ≥ + 2r0 (A∇ B − AB), 2 2 which is equivalent to the desired inequality. 

2.2 Operator Inequalities Involving Improved Young Inequality

35

Theorem 2.2.5 Let A, B > 0 and λ ∈ [0, 1], then (i) If λ ∈ [0, 21 ], then A∇λ B ≥ Aλ B + 2λ(A∇ B − AB) + r0 (AB − 2 A 14 B + A). (ii) If λ ∈ [ 21 , 1], then A∇λ B ≥ Aλ B + 2(1 − λ)(A∇ B − AB) + r0 (AB − 2 A 43 B + B), where r = min{λ, 1 − λ} and r0 = min{2r, 1 − 2r }. Proof.

(i) If λ ∈ [0, 21 ], then by Theorem 2.1.10 we have (1 − λ) + λt ≥ t λ + λ(1 −



t)2 + r0 (1 −

√ 4

t)2 ,

where t = ab > 0. By functional calculus, if we replace t with A− 2 B A− 2 and 1 then multiplying both sides of the inequality by A 2 , then 1

1

(1 − λ)A + λB ≥ Aλ B + λ(A − 2 AB + B) + r0 (A − 2 A 1 B + AB), 4

and so A∇λ B ≥ Aλ B + 2λ(A∇ B − AB) + r0 (AB − 2 A 14 B + A). (ii) If λ ∈ [ 21 , 1], then by Theorem 2.1.10 we have (1 − λ) + λt ≥ t λ + (1 − λ)(1 −

√ 2 √ √ t) + r0 ( t − 4 t)2 ,

where t = ab > 0. By functional calculus, if we replace t with A− 2 B A− 2 and 1 then multiplying both sides of the inequality by A 2 , then 1

1

(1 − λ)A + λB ≥ Aλ B + (1 − λ)(A − 2 AB + B) + r0 (B − 2 A 3 B + AB), 4

and so A∇λ B ≥ Aλ B + 2(1 − λ)(A∇ B − AB) + r0 (AB − 2 A 34 B + B). This completes the proof.  Theorem 2.2.6 Let A, B > 0 and λ ∈ [0, 1], then

36

2 Interpolating the Arithmetic-Geometric Mean Inequality …

 j  N   r (λ) + 1 j j (−1)r (λ) 2 j−1 λ + (−1)r (λ)+1 × 2 j=1   k (λ) 2k (λ)+1 k (λ)+1 B + A j B , × A j B − 2 A j

A∇λ B ≥ Aλ B +

2 j−1

2 j−1

2 j−1

where k j (λ) = [2 j−1 λ] and r j (λ) = [2 j λ] for j = 1, 2, ..., N . Proof. According to Theorem 2.1.12, we have the following inequality for a = 1 and b = t > 0: (1 − λ) + λt ≥ t λ + S N (λ, 1, t)  j  N   λ r j (λ) j−1 r j (λ)+1 r (λ) + 1 =t + 2 λ + (−1) (−1) × 2 j=1

2 2  j 2j t k j (λ) − t k j (λ)+1 ×  j  N   r (λ) + 1 j j (−1)r (λ) 2 j−1 λ + (−1)r (λ)+1 × = tλ + 2 j=1  k (λ)  2k j (λ)+1 k j (λ)+1 j × t 2 j−1 − 2t 2 j−1 + t 2 j−1 . By functional calculus, if we replace t with A− 2 B A− 2 and then multiplying both 1 sides of the inequality by A 2 , then 1

1

 j  N   r j (λ) j−1 r j (λ)+1 r (λ) + 1 (−1) A∇λ B ≥ Aλ B + × 2 λ + (−1) 2 j=1   × A k j (λ) B − 2 A 2k j (λ)+1 B + A k j (λ)+1 B .  2 j−1

2 j−1

2 j−1

Notes and references. Theorem 2.2.2 is proved by [74]. Note that the sketch of the proof of Theorem 2.2.3 and Corollary 2.2.4 are due to the authors. Theorem 2.2.5 proved by Zhao and Wu [218]. Theorem 2.2.6 is due to [188].

2.3 Advanced Refinements of the Scalar Reverse Young Inequalities Theorem 2.3.1 If a, b ≥ 0 and λ ∈ [0, 1], then √ √ (1 − λ)a + λb ≤ a 1−λ bλ + R0 ( a − b)2 ,

(2.3.1)

2.3 Advanced Refinements of the Scalar Reverse Young Inequalities

37

where R0 = max{λ, 1 − λ}. Proof. If λ = 21 , the inequality (2.3.1) becomes an equality. If λ ∈ ( 21 , 1], then R0 = λ and we have √ √ λ( a − b)2 + a 1−λ bλ − (1 − λ)a − λb √ = λ(a + b − 2 ab) + a 1−λ bλ − (1 − λ)a − λb √ = (2λ − 1)a + a 1−λ bλ − 2λ ab √ √ √ = (2λ − 1)a + a 1−λ bλ − 2λ ab + 2 ab − 2 ab √ √ = (2λ − 1)a + (2 − 2λ) ab + a 1−λ bλ − 2 ab √ ≥ a 2λ−1 a 1−λ b1−λ + a 1−λ bλ − 2 ab ( by (2.1.1)) √ = a λ b1−λ + a 1−λ bλ − 2 ab √ = 2Hλ (a, b) − 2 ab ≥ 0, and so

( by (2.1.3))

√ √ (1 − λ)a + λb ≤ a 1−λ bλ + λ( a − b)2 .

If λ ∈ [0, 21 ), then R0 = 1 − λ and we have √ √ (1 − λ)( a − b)2 + a 1−λ bλ − (1 − λ)a − λb √ = (1 − λ)(a + b − 2 ab) + a 1−λ bλ − (1 − λ)a − λb √ = (1 − 2λ)b + a 1−λ bλ − 2(1 − λ) ab √ √ = (1 − 2λ)b + 2λ ab + a 1−λ bλ − 2 ab √ ≥ b1−2λ a λ bλ + a 1−λ bλ − 2 ab ( by (2.1.1)) √ = a λ b1−λ + a 1−λ bλ − 2 ab √ = 2Hλ (a, b) − 2 ab ≥ 0, and so

( by (2.1.3))

√ √ (1 − λ)a + λb ≤ a 1−λ bλ + (1 − λ)( a − b)2 .

Hence, we deduce that √ √ (1 − λ)a + λb ≤ a 1−λ bλ + R0 ( a − b)2 , where R0 = max{λ, 1 − λ}.



As a direct consequence of Theorem 2.3.1, we get the following inequality:

38

2 Interpolating the Arithmetic-Geometric Mean Inequality …

√ √ a + b ≤ a 1−λ bλ + a λ b1−λ + 2R0 ( a − b)2 , and so we get a reverse Heinz inequality as follows. Corollary 2.3.2 If a, b ≥ 0 and λ ∈ [0, 1], then √ √ a+b ≤ Hλ (a, b) + R0 ( a − b)2 , 2 where R0 = max{λ, 1 − λ}. Theorem 2.3.3 Let a, b ≥ 0 and λ ∈ [0, 1]. (i) If λ ∈ [0, 21 ], then √ √ √ √ 4 (1 − λ)a + λb ≤ a 1−λ bλ + (1 − λ)( a − b)2 − r0 ( b − ab)2 , (ii) If λ ∈ [ 21 , 1], then √ √ √ √ 4 (1 − λ)a + λb ≤ a 1−λ bλ + λ( a − b)2 − r0 ( a − ab)2 , where r = min{λ, 1 − λ} and r0 = min{2r, 1 − 2r }. Proof. (i) If λ ∈ [0, 21 ], then we get √ √ a 1−λ bλ + (1 − λ)( a − b)2 − ((1 − λ)a + λb) √ √ = a 1−λ bλ + (1 − 2λ)b + 2λ ab − 2 ab ≥ a 1−λ bλ + (ab)λ b1−2λ

√ √ √ 4 + min{2λ, 1 − 2λ}( b − ab)2 − 2 ab ( by Theorem 2.1.8) √ √ √ 4 = a 1−λ bλ + a λ b1−λ + r0 ( b − ab)2 − 2 ab √ √ √ √ 4 ≥ 2 ab + r0 ( b − ab)2 − 2 ab ( by (2.1.3)) √ √ 4 = r0 ( b − ab)2 ,

and so √ √ √ √ 4 (1 − λ)a + λb ≤ a 1−λ bλ + (1 − λ)( a − b)2 − r0 ( b − ab)2 . (ii) Let λ ∈ [ 21 , 1]. Then (1 − λ) ∈ [0, 21 ] and so by changing the elements a, b and λ, (1 − λ) in (i), we get the desired inequality. This completes the proof.  Sababheh and Moslehian [189] gave a full description of all other refinements of the reverse Young’s inequality in the literature as follows. Theorem 2.3.4 Let a, b ≥ 0 and λ ∈ [0, 1].

2.3 Advanced Refinements of the Scalar Reverse Young Inequalities

39

(i) If λ ∈ [0, 21 ], then √ √ √ (1 − λ)a + λb ≤ a 1−λ bλ + (1 − λ)( a − b)2 − S N (2λ, ab, b). (ii) If λ ∈ [ 21 , 1], then √ √ √ (1 − λ)a + λb ≤ a 1−λ bλ + λ( a − b)2 − S N (2(1 − λ), ab, a). Here k j (λ) = [2 j−1 λ], r j (λ) = [2 j λ] and S N (λ, a, b) =

 j  N   r (λ) + 1 j j (−1)r (λ) 2 j−1 λ + (−1)r (λ)+1 × 2 j=1 2

2  j 2j j−1 j−1 × a 2 −k j (λ) bk j (λ) − a 2 −k j (λ)−1 bk j (λ)+1 .

Proof. (i) If λ ∈ [0, 21 ], then √ √ a 1−λ bλ + (1 − λ)( a − b)2 − ((1 − λ)a + λb) √ √ a 1−λ bλ + (1 − 2λ)b + 2λ ab − 2 ab

√ √ √ ≥ a 1−λ bλ + b1−2λ ( ab)2λ + S N (2λ, ab, b) − 2 ab (by Theorem 2.1.12) √ √ = a 1−λ bλ + a λ b1−λ − 2 ab + S N (2λ, ab, b) √ ≥ S N (2λ, ab, b) (by Heinz inequality), thus we have proved √ √ √ a 1−λ bλ + (1 − λ)( a − b)2 − ((1 − λ)a + λb) ≥ S N (2λ, ab, b). (ii) Similar computations give the desired inequality where λ ∈ [ 21 , 1].



In the study of Young’s inequalities, supplemental Young inequality is often discussed as follows. Theorem 2.3.5 Let a, b > 0 and λ ∈ / [0, 1]. Then (1 − λ)a + λb ≤ a 1−λ bλ . Proof. Assume that f (t) = t λ − (1 − λ) − λt for t > 0. f (t) has a minimum at t = 1 where t ∈ (0, ∞). Hence f (t) ≥ f (1) = 0 and so (1 − λ) + λt ≤ t λ .

40

2 Interpolating the Arithmetic-Geometric Mean Inequality …

By substituting t = ab , we get the desired inequality.  The main idea in the remaining of this chapter is to extend the range of λ and give the tighter bounds of the reverse Young’s inequalities proved above. Theorem 2.3.6 Let a, b ≥ 0 and λ ∈ [0, 1]. (i) If λ ∈ / [ 21 , 34 ], then √ √ √ √ 4 (1 − λ)a + λb ≤ a 1−λ bλ + (1 − λ)( a − b)2 + (2λ − 1)( ab − b)2 . (2.3.2) (ii) If λ ∈ / [ 41 , 21 ], then √ √ √ √ 4 (1 − λ)a + λb ≤ a 1−λ bλ + λ( a − b)2 − (2λ − 1)( ab − a)2 . (2.3.3) Proof. (i) Notice that (3 − 4λ) ∈ / [0, 1] where λ ∈ / [ 21 , 43 ]. Compute √ √ √ √ 4 (1 − λ)a + λb + (λ − 1)( a − b)2 − (2λ − 1)( ab − b)2 √ √ √ 4 = (1 − λ)a + λb + (λ − 1)(a − 2 ab + b) − (2λ − 1)( ab − 2 ab3 + b) √ √ 4 = (3 − 4λ) ab + (4λ − 2) ab3 √ √ 4 ≤ ( ab)3−4λ ( ab3 )4λ−2 = a 1−λ bλ , (by Theorem 2.3.5) and so √ √ √ √ 4 (1 − λ)a + λb + (λ − 1)( a − b)2 − (2λ − 1)( ab − b)2 ≤ a 1−λ bλ , which gives the inequality (2.3.2). / [0, 1]. We have (ii) If λ ∈ / [ 41 , 21 ], then (4λ − 1) ∈ √ √ √ √ 4 (1 − λ)a + λb − λ( a − b)2 + (2λ − 1)( ab − a)2 √ √ √ 4 = (1 − λ)a + λb − λ(a − 2 ab + b) + (2λ − 1)( ab − 2 a 3 b + a) √ √ 4 = (4λ − 1) ab + (2 − 4λ) a 3 b √ √ 4 ≤ ( ab)4λ−1 ( a 3 b)2−4λ = a 1−λ bλ , (by Theorem 2.3.5) so we get the desired inequality (2.3.3) as follows: √ √ √ √ 4 (1 − λ)a + λb − λ( a − b)2 + (2λ − 1)( ab − a)2 ≤ a 1−λ bλ . This completes the proof of Theorem.  We can rewrite Theorem 2.3.3 as follows.

2.3 Advanced Refinements of the Scalar Reverse Young Inequalities

41

Corollary 2.3.7 Let a, b ≥ 0 and λ ∈ [0, 1]. (i) If 0 ≤ λ ≤ 41 , then √ √ √ √ 4 (1 − λ)a + λb ≤ a 1−λ bλ + (1 − λ)( a − b)2 − 2λ( ab − b)2 . (ii) If

1 4

≤ λ ≤ 21 , then

√ √ √ √ 4 (1 − λ)a + λb ≤ a 1−λ bλ + (1 − λ)( a − b)2 + (2λ − 1)( ab − b)2 . (iii) If

1 2

< λ ≤ 43 , then √ √ √ √ 4 (1 − λ)a + λb ≤ a 1−λ bλ + λ( a − b)2 − (2λ − 1)( ab − a)2 .

(iv) If

3 4

≤ λ ≤ 1, then √ √ √ √ 4 (1 − λ)a + λb ≤ a 1−λ bλ + λ( a − b)2 + (2λ − 2)( ab − a)2 .

Remark 2.3.8 We here give the advantages of Theorem 2.3.6 in comparison with Theorem 2.3.3. For this purpose, we compare Theorem 2.3.6 and Corollary 2.3.7. (a)

Firstly, inequality (2.3.2) corresponds to (ii) of Corollary 2.3.7 when λ ∈ [ 41 , 21 ]. Notice that the range of (i) of Theorem 2.3.6 is wider than (ii) of Corollary 2.3.7. Namely, (i) of Theorem 2.3.6 also holds in cases such as λ ∈ [0, 41 ] and λ ∈ [ 43 , 1]. For the case of λ ∈ [0, 41 ], we easily find that the right-hand side of (i) of Theorem 2.3.6 is less than or equal to the right-hand side of (i) of Corollary 2.3.7. This means that (i) of Theorem 2.3.6 is tighter bound of (1 − λ)a + λb than (i) of Corollary 2.3.7 where λ ∈ [0, 41 ]. (a2) For the case of λ ∈ [ 43 , 1], we can claim that the right-hand side of (i) of Theorem 2.3.6 is less than or equal to the right-hand side of (iv) of Corollary 2.3.7. Since the inequality

(a1)

√ √ √ √ √ √ √ √ 4 4 λ( a − b)2 + 2(λ − 1)( ab − a)2 ≥ (1 − λ)( a − b)2 + (2λ − 1)( ab − b)2

is equivalent to the inequality   (t 1/4 − 1)2 (4λ − 2)t 1/4 + 4λ − 3 ≥ 0 for t > 0 and λ ∈ [ 43 , 1], this is obviously true. (b)

Secondly, inequality (2.3.3) corresponds to (iii) of Corollary 2.3.7 when λ ∈ [ 21 , 43 ]. Notice that (ii) of Theorem 2.3.6 also holds in the cases such as λ ∈ [0, 14 ] and λ ∈ [ 43 , 1].

42

2 Interpolating the Arithmetic-Geometric Mean Inequality …

For the case of λ ∈ [ 43 , 1], we easily find that the right-hand side of (ii) of Theorem 2.3.6 is less than or equal to the right-hand side of (iv) of Corollary 2.3.7. (b2) For the case of λ ∈ [0, 14 ], we can claim that the right-hand side of (ii) of Theorem 2.3.6 is less than or equal to the right-hand side of (i) of Corollary 2.3.7. Since the inequality

(b1)

√ √ √ √ √ √ √ √ 4 4 (1 − λ)( a − b)2 − 2λ( ab − b)2 ≥ λ( a − b)2 − (2λ − 1)( ab − a)2

is equivalent to the inequality   t 1/4 (t 1/4 − 1)2 (1 − 4λ)t 1/4 + 2 − 4λ ≥ 0 for t > 0 and λ ∈ [0, 41 ], this is obviously true. Thus, for all cases, the right-hand sides of both inequalities (2.3.2) and (2.3.3) in Theorem 2.3.6 give tighter upper bounds of λ-weighted arithmetic mean than those in Corollary 2.3.7. As a direct consequence of Theorem 2.3.6, we have the following improvement of the reverse Heinz inequalities. Corollary 2.3.9 Let a, b ≥ 0 and λ ∈ [0, 1]. (i) If λ ∈ / [ 21 , 34 ], then √ √ a+b ≤ Hλ (a, b) + (1 − λ)( a − b)2 2 √ √ √  1 √ 4 4 + (λ − ) ( ab − b)2 + ( ab − a)2 . 2 (ii) If λ ∈ / [ 41 , 21 ], then √ √ a+b ≤ Hλ (a, b) + λ( a − b)2 2 √ √ √  1 √ 4 4 − (λ − ) ( ab − b)2 + ( ab − a)2 . 2 Corollary 2.3.10 Let a, b ≥ 0 and λ ∈ [0, 1]. (i) If λ ∈ / [ 21 , 34 ], then 2  (1 − λ)a 2 + λb2 ≤ a 1−λ bλ

√ + (1 − λ)(a − b)2 + (2λ − 1)( ab − b)2 .

(ii) If λ ∈ / [ 41 , 21 ], then

(2.3.4)

2.3 Advanced Refinements of the Scalar Reverse Young Inequalities

2  (1 − λ)a 2 + λb2 ≤ a 1−λ bλ + λ(a − b)2 √ − (2λ − 1)( ab − a)2 .

43

(2.3.5)

Proof. Replacing a and b by their squares in Theorem 2.3.6 gives the desired inequalities.  Theorem 2.3.11 Let a, b ≥ 0 and λ ∈ [0, 1]. (i) If λ ∈ / [ 21 , 34 ], then √ 2  ((1 − λ)a + λb)2 ≤ a 1−λ bλ + (1 − λ)2 (a − b)2 + (2λ − 1)( ab − b)2 . (2.3.6) (ii) If λ ∈ / [ 41 , 21 ], then √ 2  ((1 − λ)a + λb)2 ≤ a 1−λ bλ + λ2 (a − b)2 − (2λ − 1)( ab − a)2 . (2.3.7) Proof. (i) By easy calculation, we have ((1 − λ)a + λb)2 − (1 − λ)2 (a − b)2 = (1 − λ)2 a 2 + λ2 b2 + 2λ(1 − λ)ab − (1 − λ)2 a 2 − (1 − λ)2 b2 + 2(1 − λ)2 ab = (1 − λ)a 2 + λb2 − (1 − λ)(a − b)2 √ 2  ≤ a 1−λ bλ + (2λ − 1)( ab − b)2 ,

(by (2.3.4))

which gives the inequality (2.3.6). (ii) According to the inequality (2.3.5), the proof can be completed by an argument similar to that used in (i).  Next, we represent the refinements of the reverse Young’s inequality for λ ∈ R. To get the main result, we show the following result. Corollary 2.3.12 Let a, b ≥ 0 and

1 2

= λ ∈ R.

(i) If λ ∈ / [0, 21 ], then √ √ (1 − λ)a + λb ≤ a 1−λ bλ + λ( a − b)2 . (ii) If λ ∈ / [ 21 , 1], then √ √ (1 − λ)a + λb ≤ a 1−λ bλ + (1 − λ)( a − b)2 . Proof. (i) If λ ∈ / [0, 21 ], then

44

2 Interpolating the Arithmetic-Geometric Mean Inequality …

√ √ (1 − λ)a + λb − λ( a − b)2 √ = (1 − 2λ)a + (2λ) ab √ ≤ a (1−2λ) ( ab)2λ ( by Theorem 2.3.5) = a 1−λ bλ . (ii) If λ ∈ / [ 21 , 1], then (1 − λ) ∈ / [0, 21 ]. So by changing two elements a, b and two weights λ, 1 − λ in (i), the desired inequality is obtained.  Theorem 2.3.13 Let a, b ≥ 0, n ∈ N such that n ≥ 2 and (i) If λ ∈ / [ 21 , 2

+1 ], 2n

n−1

1 2

= λ ∈ R. Then

then

√ √ (1 − λ)a + λb ≤ a 1−λ bλ + (1 − λ)( a − b)2 2

 n √  2k b k−2 −1 . + (2λ − 1) ab 2 a k=2 (ii) If λ ∈ / [2

−1 1 , 2 ], 2n

n−1

then

√ √ (1 − λ)a + λb ≤ a 1−λ bλ + λ( a − b)2 n √  + (1 − 2λ) ab 2k−2 k=2

Proof. (i) If λ ∈ / [ 21 , 2 [0, 1]. Now compute

(2.3.8)

+1 ], we have (2n λ 2n

n−1

 2k

2 a −1 . b

(2.3.9)

− 2n−1 ) ∈ / [0, 1] and (2n−1 − 2n λ + 1) ∈ /

⎧ 2 ⎫

 n ⎬ √ 2 √ ⎨ √ 2k b −1 (1 − λ)a + λb − (1 − λ)( a − b) − (2λ − 1) ab 2k−2 ⎩ ⎭ a k=2 √ √ = (1 − λ)a + λb − (1 − λ)( a − b)2 ⎧  2 2 2 ⎫



 ⎬ √ ⎨ 4 b 8 b 16 b −1 +2 −1 +4 −1 − (2λ − 1) ab ⎩ ⎭ a a a − ··· − 2

n−4

√ (2λ − 1) ab

 2n−2

2 b −1 a 2

2 b −1 −2 −2 a

  √ √ b 4 b −2 +1 = (1 − λ)a + λb − (1 − λ)(a − 2 ab + b) − (2λ − 1) ab a a n−3

√ (2λ − 1) ab

 2n−1

b −1 a

n−2

√ (2λ − 1) ab

 2n

2.3 Advanced Refinements of the Scalar Reverse Young Inequalities

45



   √ b 8 b 8 b 16 b −2 + 1 − 4(2λ − 1) ab −2 +1 a a a a

  √ 2n−3 b 2n−2 b n−4 − · · · − 2 (2λ − 1) ab −2 +1 a a

  √ 2n−2 b 2n−1 b n−3 −2 +1 − 2 (2λ − 1) ab a a

  √ 2n−1 b 2n b n−2 − 2 (2λ − 1) ab −2 +1 a a  n−2 √  √ 2n b √ l n−1 2 + 2 (2λ − 1) ab = 2(1 − λ) ab + (1 − 2λ) ab a l=0   n−2  √ 2n b √ = 2(1 − λ) + (1 − 2λ) 2l ab + 2n−1 (2λ − 1) ab a l=0  √ 2n b  √ n−1 n−1 = 2(1 − λ) + (1 − 2λ)(2 − 1) ab + 2 (2λ − 1) ab a  √ √ 2n b = (2n−1 − 2n λ + 1) ab + (2n λ − 2n−1 ) ab a

(2n λ−2n−1 )  √ (2n−1 −2n λ+1) √ 2n b ≤ ab ab (by Theorem 2.3.5) a

√ − 2(2λ − 1) ab

 4

= a 1−λ bλ . So, we get the following inequality: ⎧ 2 ⎫

 n ⎬ √ 2 √ ⎨ √ k b 2 −1 (1 − λ)a + λb − (1 − λ)( a − b) − (2λ − 1) ab 2k−2 ⎩ ⎭ a k=2

≤a

1−λ λ

b ,

which is equivalent to (2.3.8). n−1 n−1 / [ 21 , 2 2n+1 ]. Now by changing two elements (ii) If λ ∈ / [ 2 2n−1 , 21 ], then (1 − λ) ∈ a, b and replacing the weight λ with (1 − λ) in (i), the desired inequality (2.3.9) is deduced.  Remark 2.3.14 We would remark that if we rewrite Theorem 2.3.13 for n = 1, then we get Corollary 2.3.12.

46

2 Interpolating the Arithmetic-Geometric Mean Inequality …

Remark 2.3.15 From the equality of the proof in Theorem 2.3.13, the inequality (2.3.8) is equivalent to a



   √ 1 √ 2n b n −1 , b ≥ ab + 2 λ − ab 2 a

1−λ λ

(2.3.10)

which gives the following inequality:  λ− 21    n b b 1 2 −1 . ≥ 1 + 2n λ − a 2 a t r −1 r →0 r

Since lim

= log t, by putting r =

 lim 2n

n→∞

2n

1 , 2n

we have

b b − 1 = log . a a

Thus, we have the following inequality in the limit of n → ∞ for the inequality (2.3.8) in Theorem 2.3.13:  λ− 21  λ− 21 b b log ≤ − 1, a a for a, b > 0 and

1 2

= λ ∈ R, which comes from the condition λ ∈ /

(2.3.11) 

1 2n−1 +1 , 2n 2

 in

the limit of n → ∞. The above inequality recovers the equality in the case λ = 21 . Therefore, we have the inequality (2.3.11) for all λ ∈ R. We notice that the inequality  λ−1/2 in the inequality (2.3.11) can be proven directly by putting x = ab log x ≤ x − 1, (x > 0).

(2.3.12)

Similarly, in the limit of n → ∞ for the inequality (2.3.9) in Theorem 2.3.13, we get the inequality a 21 −λ a 21 −λ ≤ − 1, log b b by changing two elements a, b and replacing the weight λ with (1 − λ) in the inequality (2.3.11). Next, in Remark 2.3.8 and Remark 2.3.17, we show that Theorem 2.3.13 recovers the inequalities in Theorem 2.3.3. To achieve this, we compare Theorem 2.3.3 with Theorem 2.3.13 in the cases such as n = 2 and n = 3 where λ ∈ [0, 1]. As we noticed that Theorem 2.3.3 is equivalent to Corollary 2.3.7.

2.3 Advanced Refinements of the Scalar Reverse Young Inequalities

47

Remark 2.3.16 Consider Theorem 2.3.13 in the case n = 2 with λ ∈ [0, 1]. For a, b ≥ 0, we get Theorem 2.3.6 which we have showed that it gives tighter upper bounds of the λ-weighted arithmetic mean than those in Corollary 2.3.7. Remark 2.3.17 As a direct consequence of Theorem 2.3.13 in the case n = 3 with restricted range λ ∈ [0, 1], we have (i) If λ ∈ / [ 21 , 58 ], then √ √ √ √ 4 (1 − λ)a + λb ≤ a 1−λ bλ + (1 − λ)( a − b)2 + (2λ − 1)( b − ab)2 √ √ 8 4 + (4λ − 2)( ab3 − ab)2 . (2.3.13) (ii) If λ ∈ / [ 38 , 21 ], then √ √ √ √ 4 (1 − λ)a + λb ≤ a 1−λ bλ + λ( a − b)2 − (2λ − 1)( a − ab)2 √ √ 8 4 − (4λ − 2)( a 3 b − ab)2 . (2.3.14) We here give the advantages of inequalities (2.3.13) and (2.3.14) in comparison with Corollary 2.3.7. (a)

Firstly, we compare Corollary 2.3.7 with the inequality (2.3.13) which holds in the cases λ ∈ [0, 41 ], λ ∈ [ 41 , 21 ], λ ∈ [ 85 , 43 ], and λ ∈ [ 43 , 1] . (a1) In the case λ ∈ [0, 14 ], we have (2λ − 1) < (−2λ) and (4λ − 2) < 0. Indeed, the right-hand side of inequality (2.3.13) is less than the right-hand side of (i) in Corollary 2.3.7. For the case of λ ∈ [ 41 , 21 ], we have (4λ − 2) < 0. So we easily find that the right-hand side of the inequality (2.3.13) is less than the right-hand side of (ii) in Corollary 2.3.7. (a2) For the case of λ ∈ [ 85 , 43 ], we claim that the right-hand side of inequality (2.3.13) is less than or equal to the right-hand side of (iii) in Corollary 2.3.7. To prove our claim, we show that the following inequality holds: √ √ √ √ √ √ 8 4 4 (1 − λ)( a − b)2 + (2λ − 1)( b − ab)2 + (4λ − 2)( ab3 − ab)2 √ √ √ √ 4 ≤ λ( a − b)2 − (2λ − 1)( a − ab)2 , (2.3.15) which is equivalent to the inequality   2(1 − 2λ)t 1/4 3t 1/4 − 1 − 2t 3/8 ≥ 0, for t > 0 and λ ∈ [ 85 , 43 ]. To obtain this, it is enough to prove (3t 1/4 − 1 − 2t 3/8 ) ≤ 0. If t 1/8 = x, then we easily find that f (x) = 3x 2 − 2x 3 − 1 is increasing for 0 < x < 1 and decreasing where x > 1. Indeed, f (x) = 3x 2 − 2x 3 − 1 ≤ 0 where x > 0 and so 3t 1/4 − 1 − 2t 3/8 ≤ 0.

48

2 Interpolating the Arithmetic-Geometric Mean Inequality …

(a3) For the case of λ ∈ [ 43 , 1], we claim that the right-hand side of the inequality (2.3.13) is less than or equal to the right-hand side of (iv) in Corollary 2.3.7. To prove our claim, we show that the following inequality holds: √ √ √ √ √ √ 8 4 4 (1 − λ)( a − b)2 + (2λ − 1)( b − ab)2 + (4λ − 2)( ab3 − ab)2 √ √ √ √ 4 ≤ λ( a − b)2 − 2(1 − λ)( a − ab)2 , (2.3.16) which is equivalent to the inequality (4λ − 3) + (3 − 8λ)t 1/2 + (4 − 4λ)t 1/4 + (8λ − 4)t 5/8 ≥ 0,

(2.3.17)

for t > 0 and λ ∈ [ 43 , 1]. To obtain the inequality (2.3.17), it is sufficient to prove f (x, λ) ≥ 0 where x = t 1/8 > 0 and f (x, λ) ≡ (8λ − 4)x 5 + (3 − 8λ)x 4 + (4 − 4λ)x 2 + (4λ − 3). (x,λ) = 4(x − 1)2 (2x 3 + 2x 2 + 2x + 1) ≥ 0, we have f (x, λ) ≥ Since d f dλ 3 2 f (x, 4 ) = x (x − 1)2 (2x + 1) ≥ 0 for x > 0.

(b)

Secondly, we compare Corollary 2.3.7 with the inequality (2.3.14) which holds in the cases λ ∈ [0, 41 ], λ ∈ [ 41 , 38 ], λ ∈ [ 21 , 43 ], and λ ∈ [ 43 , 1]. (b1) For the case of λ ∈ [0, 41 ], we claim that the right-hand side of inequality (2.3.14) is less than or equal to the right-hand side of (i) in Corollary 2.3.7. To prove our claim, we give the following inequality: √ √ √ √ √ √ 8 4 4 λ( a − b)2 − (2λ − 1)( a − ab)2 − (4λ − 2)( a 3 b − ab)2 √ √ √ √ 4 ≤ (1 − λ)( a − b)2 − 2λ( b − ab)2 , which we get by replacing λ with (1 − λ) and changing the elements a, b in the inequality (2.3.16). (b2) For the case of λ ∈ [ 41 , 38 ], we claim that the right-hand side of the inequality (2.3.14) is less than or equal to the right-hand side of (ii) in Corollary 2.3.7. To prove our claim, we give the following inequality: √ √ √ √ √ √ 8 4 4 λ( a − b)2 − (2λ − 1)( a − ab)2 − (4λ − 2)( a 3 b − ab)2 √ √ √ √ 4 ≤ (1 − λ)( a − b)2 − (1 − 2λ)( b − ab)2 , which is deduced by replacing λ with (1 − λ) and changing the elements a, b in the inequality (2.3.15). (b3) For the case of λ ∈ [ 21 , 43 ], we have −(4λ − 2) < 0. So we easily find that the right-hand side of the inequality (2.3.14) is less than the right-hand side of (iii) in Corollary 2.3.7. In the case λ ∈ [ 43 , 1], we have −(2λ − 1) < (2λ −

2.3 Advanced Refinements of the Scalar Reverse Young Inequalities

49

2) and −(4λ − 2) < 0. That is, the right-hand side of inequality (2.3.14) is less than the right-hand side of (iv) in Corollary 2.3.7. Thus, according to Remark 2.3.16 and Remark 2.3.17, Theorem 2.3.13 recovers Theorem 2.3.3. We notice that the range of the reverse Young’s inequalities in Theorem 2.3.13 are wider than Theorem 2.3.3, namely, Theorem 2.3.13 holds in the case λ ∈ R and Theorem 2.3.3 holds for λ ∈ [0, 1]. Next, we compare Theorem 2.3.13 with Theorem 2.3.4. In Theorem 2.3.13, we n−1 n−1 n−1 have (I) λ ≤ 0, (II) 0 ≤ λ < 2 2n−1 , (III) 2 2n−1 ≤ λ ≤ 21 , (IV) 21 ≤ λ ≤ 2 2n+1 , (V) 2n−1 +1 < λ ≤ 1, and (VI) λ ≥ 1. 2n For the cases of (II) and (V), we claim that Theorem 2.3.13 has tighter upper bounds than those in Theorem 2.3.4, while Theorem 2.3.4 recovers Theorem 2.3.13 in the cases (III) and (IV). Therefore we conclude that Theorem 2.3.13 and Theorem 2.3.4 are different refinements of the reverse Young’s inequality which both of them recover Theorem 2.3.3. However, we emphasize that Theorem 2.3.13 gives the reverse Young’s inequality in the wider range than Theorem 2.3.4, namely, in the case λ ∈ R. This justifies why the refinement in Theorem 2.3.13 is better than Theorem 2.3.4. To prove our claims, we compare Theorem 2.3.13 with Theorem 2.3.4 in the same steps such as n = 2. For this purpose, we list up the following corollaries which are deduced directly from Theorem 2.3.13 and Theorem 2.3.4, respectively. Corollary 2.3.18 Let a, b ≥ 0 and λ ∈ [0, 1]. (i) If λ ∈ / [ 21 , 34 ], then √ √ √ √ 4 (1 − λ)a + λb − a 1−λ bλ ≤ (1 − λ)( a − b)2 + (2λ − 1)( b − ab)2 . (2.3.18) (ii) If λ ∈ / [ 41 , 21 ], then √ √ √ √ 4 (1 − λ)a + λb − a 1−λ bλ ≤ λ( a − b)2 − (2λ − 1)( a − ab)2 . (2.3.19) Corollary 2.3.19 Let a, b ≥ 0 and λ ∈ [0, 1]. (i) If λ ∈ [0, 14 ], then √ √ √ √ 4 (1 − λ)a + λb − a 1−λ bλ ≤ (1 − λ)( a − b)2 − 2λ( b − ab)2 √ √ 8 − 4λ( b − ab3 )2 . (2.3.20) (ii) If λ ∈ [ 41 , 21 ], then

50

2 Interpolating the Arithmetic-Geometric Mean Inequality …

√ √ √ √ 4 (1 − λ)a + λb − a 1−λ bλ ≤ (1 − λ)( a − b)2 + (2λ − 1)( b − ab)2 √ √ 8 4 − (4λ − 1)( ab3 − ab)2 . (2.3.21) (iii) If λ ∈ [ 21 , 34 ], then √ √ √ √ 4 (1 − λ)a + λb − a 1−λ bλ ≤ λ( a − b)2 − (2λ − 1)( a − ab)2 √ √ 8 4 + (4λ − 3)( ab3 − ab)2 . (2.3.22) (iv) If λ ∈ [ 43 , 1], then √ √ √ √ 4 (1 − λ)a + λb − a 1−λ bλ ≤ λ( a − b)2 + (2λ − 2)( a − ab)2 √ √ 8 − 4(1 − λ)( ab3 − a)2 . (2.3.23) Remark 2.3.20 Here we compare the upper bounds in Corollary 2.3.18 with those in Corollary 2.3.19. Firstly, we compare Corollary 2.3.19 with the inequality (2.3.18) which holds in cases λ ∈ [0, 41 ], λ ∈ [ 41 , 21 ], and λ ∈ [ 43 , 1]. For the case of λ ∈ [0, 14 ], we can find examples such that the right-hand side of (2.3.18) is tighter than that of (2.3.20) in Corollary 2.3.19. Actually, take a = 1, b = 16, and λ = 1/8, then the right-hand side of (2.3.18) is equal to 4.875, while the right-hand side of (2.3.20) is nearly equal to 6.2892. Indeed, the inequality (2.3.18) can recover Corollary 2.3.19 where λ ∈ [0, 41 ]. In the case λ ∈ [ 43 , 1], we claim that the right-hand side of the inequality (2.3.18) is less than or equal to the right-hand side of (2.3.23). According to (2.3.16), we have √ √ √ √ 4 (1 − λ)( a − b)2 + (2λ − 1)( b − ab)2 √ √ √ √ √ √ 8 4 4 ≤ λ( a − b)2 + (2λ − 2)( a − ab)2 − (4λ − 2)( ab3 − ab)2 . So to prove our claim, we show that the following inequality holds: √ √ √ √ 8 8 4 (2 − 4λ)( ab3 − ab)2 ≤ (4λ − 4)( ab3 − a)2 , which is equivalent to g(x, λ) = (4λ − 2)x 6 + (2 − 4λ)x 5 + (2λ − 1)x 4 + (2 − 4λ)x 3 + (2λ − 1) ≥ 0, 1

where x = t 8 > 0 and λ ∈ [ 43 , 1]. Since dg(x,λ) = 2(x − 1)2 (2x 4 + 2x 3 + 3x 2 + dλ 3 2x + 1) ≥ 0, we have g(x, λ) ≥ g(x, 4 ) = (x − 1)2 (x 4 + x 3 + 23 x 2 + x + 21 ) ≥ 0. This justifies that the inequality (2.3.18) also recovers Corollary 2.3.19 where λ ∈ [ 43 , 1]. However, in the case λ ∈ [ 41 , 21 ], we easily find that the right-hand side of (2.3.21) is less than or equal to the right-hand side of (2.3.18). That is, Corollary 2.3.19 is a refinement of (2.3.18) where λ ∈ [ 41 , 21 ].

2.3 Advanced Refinements of the Scalar Reverse Young Inequalities

51

Secondly, we compare Corollary 2.3.19 with the inequality (2.3.19) which holds in cases λ ∈ [0, 41 ], λ ∈ [ 21 , 34 ], and λ ∈ [ 43 , 1]. The comparison is done by the same way as in the first step and we omit it. Sababheh and Choi gave the following refinement of Theorem 2.3.5 which its complete proof can be found in [189]. Lemma 2.3.21 Let a, b > 0. Then we have (i) If λ ≤ 0, then (1 − λ)a + λb ≤ a 1−λ bλ + λ

n 

2k−1



2  2k a − a 2k−1 −1 b .

(2.3.24)

k=1

(ii) If λ ≥ 1, then (1 − λ)a + λb ≤ a 1−λ bλ + (1 − λ)

n 

2k−1



b−

2  2k ab2k−1 −1 .

k=1

(2.3.25) We can extend the ranges of λ in Lemma 2.3.21 to those in following theorem, by the similar way to the line of proof of Theorem 2.3.13. Theorem 2.3.22 Let a, b ≥ 0, n ∈ N and λ ∈ R. (i) If λ ∈ / [0, 21n ], then the inequality (2.3.24) holds. n (ii) If λ ∈ / [ 2 2−1 n , 1], then the inequality (2.3.25) holds. As we discussed the case of n → ∞ in Remark 2.3.15, we also give the following remark. Remark 2.3.23 The inequality (2.3.24) is equivalent to the inequality

  n b 1/2 a + λa2 − 1 ≤ a 1−λ bλ , a n

by the elementary computations. Using the formula lim

r →0

inequality log

t r −1 r

= log t, we have the

 λ  λ b b ≤ − 1, a a

for λ = 0 in the limit of n → ∞. The above inequality trivially holds for all λ ∈ R. It can be also obtained by the inequality (2.3.12). Similarly, the inequality (2.3.25) is equivalent to the inequality

52

2 Interpolating the Arithmetic-Geometric Mean Inequality …

b + (1 − λ)b2n

 n a 1/2 b

 − 1 ≤ a 1−λ bλ ,

so that we have the inequality log

a 1−λ b



a 1−λ b

− 1,

for λ = 1 in the limit of n → ∞. The above inequality trivially holds for all λ ∈ R. It can be also obtained by the inequality (2.3.12). As a direct consequence of Theorem 2.3.13 and Theorem 2.3.22, we have the following reverse inequalities with respect to the Heinz means. Corollary 2.3.24 Let a, b ≥ 0, n ∈ N such that n ≥ 2 and (i) If λ ∈ / [ 21 , 2

+1 ], 2n

n−1

1 2

= λ ∈ R.

then

√ √ a+b ≤ Hλ (a, b) + (1 − λ)( a − b)2 2 ⎧ 2 ⎫    2  n ⎬ k b 1 √  k−2 ⎨ 2k a 2 + λ− ab 2 . −1 + −1 ⎩ ⎭ 2 b a k=2 (ii) If λ ∈ / [2

−1 1 , 2 ], 2n

n−1

then

√ √ a+b ≤ Hλ (a, b) + λ( a − b)2 2 ⎧ 2 ⎫  2   n ⎨  a ⎬ √  k b 1 k 2 2 −λ −1 + −1 + ab 2k−2 . ⎩ ⎭ 2 b a k=2 Corollary 2.3.25 Let a, b ≥ 0, n ∈ N and λ ∈ R. (i) If λ ∈ / [0, 21n ], then a+b ≤ Hλ (a, b) 2  n

2 

2 √    √ 2k 2k k−1 −1 k−1 −1 k−2 2 2 . 2 a− a b + b − ab +λ k=1

(ii) If λ ∈ / [ 2 2−1 n , 1], then n

2.3 Advanced Refinements of the Scalar Reverse Young Inequalities

53

a+b ≤ Hλ (a, b) 2  n

2 

2 √    √ 2k 2k . 2k−2 a − a 2k−1 −1 b + b − ab2k−1 −1 + (1 − λ) k=1

Notes and references. Theorem 2.3.1 and Corollary 2.3.2 proved in [120]. Theorem 2.3.3 is due to [218]. Supplemental Young inequality in Theorem 2.3.5 is proved in [16]. We note that all the other main results in this section are proved in [84] and [73].

2.4 Improvements of the Operator Reverse Young Inequality Now, we obtain the operator version of the improved reverse Young inequalities by Lemma 2.2.1. Theorem 2.4.1 Let A, B > 0 and λ ∈ [0, 1], then A∇λ B ≤ Aλ B + 2R0 (A∇ B − AB), where R0 = max{λ, 1 − λ}. Proof. According to Theorem 2.3.1, we have the following inequality for t =

b a

> 0:

(1 − λ) + λt ≤ t λ + R0 (1 − 2t 2 + t). 1

By functional calculus, if we replace t with A− 2 B A− 2 and then on multiplying both 1 sides of the inequality by A 2 , we get 1

1

(1 − λ)A + λB ≤ Aλ B + R0 (A − 2 AB + B) = Aλ B + 2R0 (A∇ B − AB), which is equivalent to A∇λ B ≤ Aλ B + 2R0 (A∇ B − AB).  Corollary 2.4.2 Let A, B > 0 and λ ∈ [0, 1], then A∇ B ≤ Hλ (A, B) + 2R0 (A∇ B − AB), where R0 = max{λ, 1 − λ}. Proof. As a direct consequence of Theorem 2.4.1, we have

54

2 Interpolating the Arithmetic-Geometric Mean Inequality …

A∇λ B + B∇λ A ≤ Aλ B + 2R0 (A∇ B − AB) + Bλ A + 2R0 (B∇ A − BA), that is, A + B ≤ Aλ B + 2R0 (A∇ B − AB) + A1−λ B + 2R0 (A∇ B − AB), and so we have Aλ B + A1−λ B A+B ≤ + 2R0 (A∇ B − AB), 2 2 which is equivalent to the desired inequality.



Theorem 2.4.3 Let A, B > 0 and λ ∈ [0, 1], then (i) If λ ∈ [0, 21 ], then A∇λ B ≤ Aλ B + 2(1 − λ)(A∇ B − AB) − r0 (AB − 2 A 43 B + B). (ii) If λ ∈ [ 21 , 1], then A∇λ B ≤ Aλ B + 2λ(A∇ B − AB) − r0 (AB − 2 A 14 B + A), where r = min{λ, 1 − λ} and r0 = min{2r, 1 − 2r }. Proof.

(i) If λ ∈ [0, 21 ], then by Theorem 2.3.3 we have (1 − λ) + λt ≤ t λ + (1 − λ)(1 −

√ 2 √ √ t) − r0 ( t − 4 t)2 ,

where t = ab > 0. By functional calculus, if we replace t with A− 2 B A− 2 and 1 then on multiplying both sides of the inequality by A 2 , we get 1

1

(1 − λ)A + λB ≤ Aλ B + (1 − λ)(A − 2 AB + B) − r0 (B − 2 A 3 B + AB), 4

which is equivalent to A∇λ B ≤ Aλ B + 2(1 − λ)(A∇ B − AB) − r0 (AB − 2 A 34 B + B). (ii) If λ ∈ [ 21 , 1], then by Theorem 2.3.3 we have (1 − λ) + λt ≤ t λ + λ(1 −



t)2 + r0 (1 −

√ 4

t)2 ,

where t = ab > 0. By functional calculus, if we replace t with A− 2 B A− 2 and 1 then on multiplying both sides of the inequality by A 2 , we get 1

1

2.4 Improvements of the Operator Reverse Young Inequality

55

(1 − λ)A + λB ≤ Aλ B + λ(A − 2 AB + B) − r0 (A − 2 A 14 + AB), which is equivalent to A∇λ B ≤ Aλ B + 2λ(A∇ B − AB) − r0 (AB − 2 A 14 + A). This completes the proof.  Now, we obtain the operator version of Theorem 2.3.6 as follows. Theorem 2.4.4 Let A, B > 0 and λ ∈ [0, 1]. (i) If λ ∈ / [ 21 , 34 ], then A∇λ B ≤ Aλ B + 2(1 − λ)(A∇ B − AB) + (2λ − 1)(AB + B − 2 A 43 B).

(2.4.1)

A∇λ B ≤ Aλ B + 2λ(A∇ B − AB) − (2λ − 1)(AB + A − 2 A 14 B).

(2.4.2)

(ii) If λ ∈ / [ 41 , 21 ], then

Proof. (i) According to the inequality (2.3.2), we have the following inequality for t ≥ 0: √ √ √ (1 − λ) + λt ≤ t λ + (1 − λ)(1 − t)2 + (2λ − 1)( 4 t − t)2

1

1 3 = t λ + (1 − λ) 1 + t − 2t 2 + (2λ − 1) t 2 + t − 2t 4 , if we replace t with A− 2 B A− 2 and then multiplying both sides of the inequality by 1 A 2 , we get 1

1

(1 − λ)A + λB ≤ Aλ B + (1 − λ)(A + B − 2 AB) + (2λ − 1)(AB + B − 2 A 3 B), 4

since (1 − λ)A + λB = A∇λ B and

A+B 2

= A∇ B, so we have

A∇λ B ≤ Aλ B + 2(1 − λ)(A∇ B − AB) + (2λ − 1)(AB + B − 2 A 43 B), which is the desired inequality (2.4.1). (ii) The line of proof is similar to the one presented in (i) by applying the inequality (2.3.3), thus we omit it.  The operator version of Corollary 2.3.9 is deduced immediately by Theorem 2.4.4 as follows. Corollary 2.4.5 Let A, B > 0 and λ ∈ [0, 1].

56

2 Interpolating the Arithmetic-Geometric Mean Inequality …

(i) If λ ∈ / [ 21 , 34 ], then A∇ B ≤ Hλ (A, B) + 2(1 − λ)(A∇ B − AB) + (2λ − 1)(AB + A∇ B − 2H 34 (A, B)). (ii) If λ ∈ / [ 41 , 21 ], then A∇ B ≤ Hλ (A, B) + 2λ(A∇ B − AB) − (2λ − 1)(AB + A∇ B − 2H 14 (A, B)). We note that we use the following notations in what follows: 1

1 1 1 λ Aλ B ≡ A 2 A− 2 B A− 2 A 2 ,

Aλ B + A1−λ B , Hˆ λ (A, B) ≡ 2

for all λ ∈ R including the range λ ∈ / [0, 1]. Theorem 2.4.6 Let A, B > 0, n ∈ N such that n ≥ 2 and the following inequalities: (i) If λ ∈ / [ 21 , 2

+1 ], 2n

n−1

1 2

= λ ∈ R. Then we have

then

A∇λ B ≤ Aλ B + 2(1 − λ)(A∇λ B − AB) n

 2k−2 AB − 2 A 2k−1 +1 B + A 2k−2 +1 B . + (2λ − 1) 2k

k=2

(ii) If λ ∈ / [2

−1 1 , 2 ], 2n

n−1

2k−1

then

A∇λ B ≤ Aλ B + 2λ(A∇λ B − AB) n

 2k−2 AB − 2 A 2k−1 −1 B + A 2k−2 −1 B . + (1 − 2λ) 2k

k=2

2k−1

Proof. (i) According to the inequality (2.3.8), the following inequality holds for t ≥ 0: (1 − λ) + λt ≤ t λ + (1 − λ)(1 −



t)2

n √

2 √  k + (2λ − 1) t 2k−2 2 t − 1 . k=2

By functional calculus, if we replace t with A− 2 B A− 2 and then multiplying both 1 sides of the inequality by A 2 , the desired inequality is obtained. (ii) The line of proof is similar to (i) by applying the inequality (2.3.9).  1

1

2.4 Improvements of the Operator Reverse Young Inequality

57

Remark 2.4.7 We notice that Theorem 2.4.6 with λ ∈ [0, 1] recovers the inequalities obtained in Theorem 2.4.4, if we put n = 2. Theorem 2.4.8 Let A, B > 0, n ∈ N and λ ∈ R. Then we have the following inequalities: (i) If λ ∈ / [0, 21n ], then A∇λ B ≤ Aλ B + λ

n 

2k−1 A − 2 A 1k B + A 2

1 2k−1

B .

k=1

(ii) If λ ∈ / [ 2 2−1 n , 1], then n

A∇λ B ≤ Aλ B + (1 − λ)

n 



2k−1 B − 2 A 2k −1 B + A 2k−1 −1 B . 2k

k=1

2k−1

Proof. (i) According to Theorem 2.3.22 in the case λ ∈ / [0, 21n ], the following inequality holds for t ≥ 0: (1 − λ) + λt ≤ t λ + λ

n 



2 √ k 2k−1 1 − 2 t

k=1

= tλ + λ

n 



1 1 2k−1 1 − 2t 2k + t 2k−1 .

k=1

By functional calculus, if we replace t with A− 2 B A− 2 and then multiplying both 1 sides of the inequality by A 2 , the desired inequality is deduced. n (ii) By applying Theorem 2.3.22 for the case of λ ∈ / [ 2 2−1 n , 1], we get the desired inequality in the same way as in (i).  1

1

As a direct consequence of Theorem 2.4.6 and Theorem 2.4.8, we get the generalized reverse Heinz operator inequalities as follows. That is, Corollary 2.4.9 and Corollary 2.4.11 are operator versions of Corollary 2.3.24 and Corollary 2.3.25, respectively. Corollary 2.4.9 Let A, B > 0, n ∈ N such that n ≥ 2 and have the following inequalities: (i) If λ ∈ / [ 21 , 2

+1 ], 2n

n−1

1 2

= λ ∈ R. Then we

then

A∇ B ≤ Hˆ λ (A, B) + 2(1 − λ)(A∇ B − AB) n

 2k−2 AB − 2H 2k−1 +1 (A, B) + H 2k−2 +1 (A, B) . + (2λ − 1) k=2

2k

2k−1

58

2 Interpolating the Arithmetic-Geometric Mean Inequality …

(ii) If λ ∈ / [2

−1 1 , 2 ], 2n

n−1

then

A∇ B ≤ Hˆ λ (A, B) + 2λ(A∇ B − AB) n

 2k−2 AB − 2H 2k−1 −1 (A, B) + H 2k−2 −1 (A, B) . + (1 − 2λ) 2k

k=2

2k−1

Remark 2.4.10 Putting n = 2 in Corollary 2.4.9 with λ ∈ [0, 1] gives the inequalities obtained in Corollary 2.4.5. Corollary 2.4.11 Let A, B > 0, n ∈ N and λ ∈ R. Then we have the following inequalities: (i) If λ ∈ / [0, 21n ], then A∇ B ≤ Hˆ λ (A, B) +λ

n 

2k−1 A∇ B − 2H 1k (A, B) + H 2

1 2k−1

(A, B) .

k=1

(ii) If λ ∈ / [ 2 2−1 n , 1], then n

A∇ B ≤ Hˆ λ (A, B) + (1 − λ)

n 



2k−1 A∇ B − 2H 2k −1 (A, B) + H 2k−1 −1 (A, B) . 2k

k=1

2k−1

We recall that the Hilbert-Schmidt norm of A = [ai j ] ∈ Mn is defined by ⎛

A 2 = ⎝

n 

⎞ 21 |ai j |2 ⎠ .

i, j=1

This norm is unitarily invariant in the sense that U AV 2 = A 2 for all unitary matrices U, V ∈ Mn . Theorem 2.4.12 Let A, B, X ∈ Mn , A, B ≥ 0 and λ ∈ [0, 1]. (i) If λ ∈ / [ 21 , 34 ], then ! !2

(1 − λ)AX + λX B 22 ≤ ! A1−λ X B λ !2 + (1 − λ)2 AX − X B 22 ! 1 !2 1 ! ! + (2λ − 1) ! A 2 X B 2 − X B ! . (2.4.3) 2

(ii) If λ ∈ / [ 41 , 21 ], then

2.4 Improvements of the Operator Reverse Young Inequality

59

! !2

(1 − λ)AX + λX B 22 ≤ ! A1−λ X B λ !2 + λ2 AX − X B 22 ! 1 !2 1 ! ! − (2λ − 1) ! A 2 X B 2 − AX ! . 2

(2.4.4)

Proof. By the spectral theorem, there are unitary matrices U, V ∈ Mn such that A = U DU ∗ and B = V E V ∗ , where D = diag(λ1 , λ2 , ..., λn ) and E = diag(γ1 , γ2 , ..., γn ) with λi , γ j ≥ 0 for 1 ≤ i, j ≤ n. If Y = U ∗ X V = [yi j ], then we have # " (1 − λ)AX + λX B = U ((1 − λ)λi + λγ j )yi j V ∗ , # " A1−λ X B λ = U (λi1−λ γ λj )yi j V ∗ , " # AX − X B = U (λi − γ j )yi j V ∗ ,   1 1 1 A 2 X B 2 − X B = U ((λi γ j ) 2 − γ j )yi j V ∗ ,   1 1 1 A 2 X B 2 − AX = U ((λi γ j ) 2 − λi )yi j V ∗ . The proof of (ii) is similar to that of inequality in (i). Thus, we only prove (i). If λ ∈ [0, 21 ], then by applying Theorem 2.3.11 we have

(1 − λ)AX + λX B 22 ! " # !2 = !U ((1 − λ)λi + λγ j )yi j V ∗ !2 =

n   2 (1 − λ)λi + λγ j |yi j |2 i, j=1



n $

%   1−λ λ 2  2 1 λi γ j + (1 − λ)2 λi − γ j + (2λ − 1) ((λi γ j ) 2 − γ j |yi j |2 i, j=1

! 1 !2 ! !2 1 ! ! = ! A1−λ X B λ !2 + (1 − λ)2 AX − X B 22 + (2λ − 1) ! A 2 X B 2 − X B ! . 2

Thus, we get the following inequality: ! !2

(1 − λ)AX + λX B 22 ≤ ! A1−λ X B λ !2 + (1 − λ)2 AX − X B 22 ! 1 !2 1 ! ! + (2λ − 1) ! A 2 X B 2 − X B ! , 2

and this completes the proof.



Notes and references. We note that all the results in this section are proved in [84] and [73].

Chapter 3

Operator Inequalities for Positive Linear Maps

The main purpose of this chapter is to select the main results on squaring reverse arithmetic-geometric mean operator inequality and the reverse Ando’s operator inequality. We gathered the most important topics that showed the essential techniques to squaring operator inequalities and their p-power. The readers can obtain some new results in different ways of squaring the operator inequalities inspired by the ideas and techniques used in this chapter.

3.1 On an Operator Kantorovich Inequality for Positive Linear Maps The absolute value of operator A is denoted by |A| = (A∗ A) 2 , where A∗ stands for the adjoint of A. The Kantorovich inequality [115] (see also [103]) states that for every unit vector x ∈ H and 0 < m ≤ A ≤ M, 1

x, Axx, A−1 x ≤

(M + m)2 . 4Mm

The constant K (h) = (M+m) is called Kantorovich constant for h = M . An operator 4Mm m version of the Kantorovich inequality was first established by Marshall and Olkin [162] as follows. 2

Theorem 3.1.1 Let 0 < m ≤ A ≤ M. Then for every positive unital linear map Φ, Φ(A−1 ) ≤ K (h)Φ −1 (A), © Springer Nature Switzerland AG 2021 M. B. Ghaemi et al., Advances in Matrix Inequalities, Springer Optimization and Its Applications 176, https://doi.org/10.1007/978-3-030-76047-2_3

61

62

3 Operator Inequalities for Positive Linear Maps

where h =

M m

and K (h) =

(M+m)2 . 4Mm

Theorem 3.1.1 can be regarded as a counterpart to Choi’s inequality [44] which says that Theorem 3.1.2 Let 0 < m ≤ A ≤ M. Then for every positive unital linear map Φ, Φ −1 (A) ≤ Φ(A−1 ). We recall that the geometric mean is monotone, i.e., if 0 < A ≤ C and 0 < B ≤ D, then AB ≤ CD. Lemma 3.1.3 If 0 < A ≤ B, then AB −1 ≤ I . Proof. Operator reverse monotonicity of the inverse tells us that A−1 ≥ B −1 > 0. Then AB −1 ≤ AA−1 = I . To see why the converse fails, suppose AB −1 ≤ I , then A2 B 2 ≤ I .  Thus, we have Corollary 3.1.4 Let 0 < m ≤ A ≤ M. Then for every positive unital linear map Φ, M +m Φ(A−1 )Φ(A) ≤ √ . 2 Mm Proof. Theorem 3.1.1 states that Φ(A−1 ) ≤ K (h)Φ −1 (A). Now by Lemma 3.1.3 we have Φ(A−1 )K −1 (h)Φ(A) ≤ I, which is equivalent to the desired inequality since Φ(A−1 )K −1 (h)Φ(A)   21 1 1 1 1 = Φ(A−1 ) 2 Φ(A−1 )− 2 K −1 (h)Φ(A)Φ(A−1 )− 2 Φ(A−1 ) 2   21 1 1 1 1 1 = K −1 (h) 2 Φ(A−1 ) 2 Φ(A−1 )− 2 Φ(A)Φ(A−1 )− 2 Φ(A−1 ) 2  1  = K −1 (h) 2 Φ(A−1 )Φ(A) .  Remark 3.1.5 A more general case of Corollary 3.1.4 has been studied in the context of matrix reverse Cauchy-Schwarz inequality [125]. There is another way to get Corollary 3.1.4. We have (M − A)(m − A)A−1 ≤ 0, then

3.1 On an Operator Kantorovich Inequality for Positive Linear Maps

63

Mm A−1 + A ≤ M + m, and so

MmΦ(A−1 ) + Φ(A) ≤ M + m.

(3.1.1)

According to AM-GM operator inequality, we have 2MmΦ(A−1 )Φ(A) ≤ MmΦ(A−1 ) + Φ(A) ≤ M + m, ( by (3.1.1)), and hence

 M +m Φ(A−1 )Φ(A) ≤ √ = K (h).  2 Mm

It is well known that for two general positive operators A, B, A ≤ B  A2 ≤ B 2 . This simple observation already reveals the subtitle of operator inequalities. It is thus interesting to know for what kind of operator inequalities when they are squared, the inequality relation is preserved. In the following, we show that the operator Kantorovich inequality can be squared. For this purpose, we need the following Lemma. Lemma 3.1.6 For positive operators A and B we have AB ≤

1 A + B 2 . 4

Theorem 3.1.7 Let 0 < m ≤ A ≤ M. Then for every positive unital linear map Φ, Φ(A−1 )2 ≤ K 2 (h)Φ −2 (A). Proof. We need to prove the following inequality Φ(A−1 )Φ(A) ≤ K (h).

(3.1.2)

We have Mm Φ(A−1 )Φ(A) = MmΦ(A−1 )Φ(A) 1 ≤ MmΦ(A−1 ) + Φ(A) 2 ( by Lemma 3.1.6) 4 1 ≤ (M + m)2 , ( by (3.1.1)). 4

64

3 Operator Inequalities for Positive Linear Maps

So, we get Mm Φ(A−1 )Φ(A) ≤

1 (M + m)2 , 4

which is equivalent to (M + m)2 4Mm = K (h). 

Φ(A−1 )Φ(A) ≤

As f (t) = t p is operator monotone for 0 ≤ p ≤ 1, according to Theorem 3.1.7 we have Corollary 3.1.8 Let 0 < m ≤ A, B ≤ M. Then for every positive unital linear map Φ and 0 ≤ p ≤ 2, Φ p (A−1 ) ≤ K p (h)(Φ(A))− p . To prove the next theorem, the following lemma play an important role. To see the proof, we refer the reader to [10]. Lemma 3.1.9 Let A and B be positive operators. Then for r ≥ 1 Ar + B r ≤ A + B r . Theorem 3.1.10 Let 0 < m ≤ A, B ≤ M. Then for every positive unital linear map Φ and p ≥ 2,  p (M + m)2 p −1 Φ − p (A). (3.1.3) Φ (A ) ≤ 2 4 p Mm Proof. Inequality (3.1.3) is equivalent to (M + m)2 p 2 −1 2p . Φ (A )Φ (A) ≤ 2 4 p Mm Compute p p p p M 2 m 2 Φ 2 (A−1 )Φ 2 (A) 2 p 1 p p p ≤ M 2 m 2 Φ 2 (A−1 ) + Φ 2 (A) (by Lemma 3.1.6) 4 p 1 ≤ MmΦ(A−1 ) + Φ(A) (by Lemma 3.1.9) 4 1 ≤ (M + m) p , (by (3.1.1)). 4 Thus the desired inequality holds. 

3.1 On an Operator Kantorovich Inequality for Positive Linear Maps

65

Lemma 3.1.11 ([105]) For any bounded operator X ,  |X | ≤ t I ⇐⇒ X ≤ t ⇐⇒

tI X X∗ t I

 ≥ 0.

A remarkable property of the  geometricmean is the so called maximal characteriA AB zation [183], which says that is positive, and moreover, if the operator AB B   A X matrix , which X being self-adjoint, is positive, then X ≤ AB. X B Theorem 3.1.12 Let 0 < m ≤ A ≤ M. Then for every positive unital linear map Φ, (M + m)2 , (3.1.4) |Φ(A−1 )Φ(A) + Φ(A)Φ(A−1 )| ≤ 2Mm and Φ(A−1 )Φ(A) + Φ(A)Φ(A−1 ) ≤

(M + m)2 . 2Mm

(3.1.5)

Proof. By (3.1.2) and Lemma 3.1.11, we have 

K (h)I Φ(A−1 )Φ(A) −1 K (h)I Φ(A)Φ(A )



 ≥ 0 and

K (h)I Φ(A)Φ(A−1 ) −1 K (h)I Φ(A )Φ(A)

 ≥ 0.

Summing up these two operator matrices, we get

(M+m)2 2Mm

I Φ(A−1 )Φ(A) + Φ(A)Φ(A−1 ) (M+m)2 I, Φ(A−1 )Φ(A) + Φ(A)Φ(A−1 ) 2Mm

≥ 0,

by Lemma 3.1.11 again, inequality (3.1.4) follows. Moreover, As Φ(A−1 )Φ(A) + Φ(A)Φ(A−1 ) is self-adjoint, by the maximal characterization of geometric mean we get the following inequality Φ(A−1 )Φ(A) + Φ(A)Φ(A−1 ) ≤

(M + m)2 . 2Mm

Remark 3.1.13 As |X | ≥ X for any self-adjoint operator X , we find that (3.1.4) is stronger than (3.1.5). In the following, we present a refinement of Theorem 3.1.1 involving log convex function. For this purpose, we need some preliminaries. Definition 3.1.14 A positive function defined on the interval I (or, more generally, on a convex subset of some vector space) is called log convex if log f (x) is a convex function of x. We observe that such functions satisfy the elementary inequality

66

3 Operator Inequalities for Positive Linear Maps

f ((1 − v)a + vb) ≤ ( f (a))1−v ( f (b))v ,

0≤v≤1

for any a, b ∈ I . Because of the weighted arithmetic-geometric mean inequality, we also have f ((1 − v)a + vb) ≤ ( f (a))1−v ( f (b))v ≤ (1 − v) f (a) + v f (b), which says that any log convex function is a convex function. The following inequality is well known in the literature as the Choi-Davis-Jensen inequality [44, 50]. Theorem 3.1.15 Let A be a self-adjoint operator with spectrum in I and Φ be the unital positive linear map. If f is operator convex function on an interval I , then f (Φ(A)) ≤ Φ( f (A)). Though in the case of a convex function, Theorem 3.1.15 does not hold in general, we have the following estimate which is due to [166]. Theorem 3.1.16 Let A be a self-adjoint operator and 0 < m < M for some scalars m, M. Let Φ be a unital positive linear map. If f is nonnegative convex function, then 1 Φ( f (A)) ≤ f (Φ(A)) ≤ μ(m, M, f )Φ( f (A)), μ(m, M, f ) where μ(m, M, f ) is defined by μ(m, M, f ) = max

1 f (t)





M −t t −m f (m) + f (M) , m ≤ t ≤ M . M −m M −m

Next, we show an analogue of Theorem 3.1.16 which is a refinement of operator Kantorovich inequality proved in Theorem 3.1.1. Theorem 3.1.17 Let A > 0 with spectrum in (m, M) for some scalars m, M with m < M and Φ be the unital positive linear map. If f is nonnegative log convex function, then   M−A A−m Φ( f (A)) ≤ Φ ( f (m)) M−m ( f (M)) M−m ≤ μ(m, M, f ) f (Φ(A)), where μ(m, M, f ) is defined in Theorem 3.1.16. Proof. It can be easily find that

3.1 On an Operator Kantorovich Inequality for Positive Linear Maps

67

M −t ≤ 1, M −m t −m ≤ 1, 0≤ M −m t −m M −t + . 1= M −m M −m

0≤

Since f is log convex, by arithmetic-geometric mean inequality we have  f (t) = f

M −t t −m m+ M M −m M −m M−t



t−m

≤ ( f (m)) M−m ( f (M)) M−m t −m M −t f (m) + f (M). ≤ M −m M −m Applying functional calculus for the operator A, we get M−A

A−m

f (A) ≤ ( f (m)) M−m ( f (M)) M−m A−m M−A f (m) + f (M). ≤ M −m M −m Using the hypotheses made about Φ,   M−A A−m Φ( f (A)) ≤ Φ ( f (m)) M−m ( f (M)) M−m   A−m M−A f (m) + f (M) . ≤Φ M −m M −m Thanks to [166][Corollary 4.12] we have the following inequality  Φ

 A−m M−A f (m) + f (M) ≤ μ(m, M, f ) f (Φ(A)). M −m M −m

Thus, we obtain   M−A A−m Φ( f (A)) ≤ Φ ( f (m)) M−m ( f (M)) M−m ≤ μ(m, M, f ) f (Φ(A)), and the desired inequality is obtained.  Since f (t) = t p is log convex for p < 0, thus Corollary 3.1.18 Under the hypotheses of Theorem 3.1.17, let p < 0 and 0 < m < M. Then   M−A A−m Φ(A p ) ≤ Φ m p( M−m ) M p( M−m ) ≤ K (m, M, p)Φ p (A),

68

3 Operator Inequalities for Positive Linear Maps

where K (m, M, p) is the generalized Kantorovich constant defined by m M p − Mm p K (m, M, p) := ( p − 1)(M − m)



p − 1 Mp − mp . p m M p − Mm p

p .

Corollary 3.1.19 Let 0 < m ≤ A ≤ M for some scalars m, M with m < M and Φ be the unital positive linear map. Then  A−M m−A  (M + m)2 Φ −1 (A). Φ(A−1 ) ≤ Φ m M−m M M−m ≤ 4Mm Proof. This follows from Corollary 3.1.18 by putting p = −1. We should recall that 2 K (m, M, −1) = (M+m) .  4Mm Notes and references. Both (3.1.4) and (3.1.5) can be regarded as operator versions of Kantorovich inequality. We note that a similar version to (3.1.5) had been established earlier by Sun [197]. However, his line of proof is quite different from Theorem 3.1.12. Theorem 3.1.10 is proved by [65]. Theorem 3.1.17, Corollaries 3.1.18, and 3.1.19 are proved in [168]. All other results are due to M. Lin [133].

3.2 A Schwarz Inequality for Positive Linear Maps √ The geometric mean ab of two positive real numbers a, b is a concave operation. √ which is equivalent to the Cauchy-Schwarz inequality. The geoThat is ab ≤ a+b 2 metric mean of matrices share similar properties. Let A, B, ..., Z be n × n matrices, or operators on an n-dimensional space H. For A, B > 0, we know that √ 1. AB = B A implies AB = AB. 2. (X ∗ AX )(X ∗ B X ) = X ∗ (AB)X for any invertible X . As we mentioned in the previous section, a remarkable property of the geometric mean is a maximal characterization by Pusz and Woronowicz [183]. Here, we give the proof for the convenience of the readers. Theorem 3.2.1 Let A, B > 0. Then  

A X AB = max X > 0, ≥0 . X B 

 1 1 A X Proof. Notice that ≥ 0 means X = A 2 K B 2 , for some contraction K , X B 1 1 K ≤ 1. That is K = A− 2 X B − 2 ≤ 1 which deduce

3.2 A Schwarz Inequality for Positive Linear Maps

69



  1  1 1 ∗ 1 A− 2 X B − 2 A− 2 X B − 2 ≤ I  1  1  1 1 → B − 2 X A− 2 A− 2 X B − 2 ≤ I

→ X A− 2 A− 2 X ≤ B  1  1   1  1 1 1 → A− 2 X A− 2 A− 2 X A− 2 ≤ A− 2 B A− 2 . 1

Thus, we have

1



A− 2 X A− 2 1

1

2

  1 1 ≤ A− 2 B A− 2

1

by operator monotony of t 2 , we have 

  1  21 1 1 1 A− 2 X A− 2 ≤ A− 2 B A− 2 .

Hence X ≤ AB. 

 1 1 A AB ≥ 0. . Note that A− 2 (AB)B − 2 = 1 since AB B is unitary. This completes the proof. 

It remains to check that A− 2 (AB)B − 2 1

1

An immediate concavity corollary is deduced by Theorem 3.2.1 as follows. Corollary 3.2.2 The geometric mean AB is concave on pairs of positive matrices. m m Equivalently, for positive matrices {Ai }i=1 and {Bi }i=1 , 

Ai Bi ≤ (





Ai )(

Bi ).

This concavity property can be regarded as a matrix Cauchy-Schwarz inequality. In the following, we give the reverse inequalities for the geometric mean. The classical reverse Cauchy-Schwarz inequality states that Theorem 3.2.3 Let Ai , Bi > 0, i = 1, 2, ..., m with q Ai ≤ Bi ≤ p Ai for some p, q > 0. Then

(



Ai )(



  14 Bi ) ≤

p q

+ 2

  14 q p



Ai Bi .

Theorem 3.2.3 follows from a more general case whose proof need an elementary lemmas. Lemma 3.2.4 Let Z > 0 with external eigenvalues a, b. For all vectors h,

70

3 Operator Inequalities for Positive Linear Maps

 a  21 h Z h ≤

b

+ 2

 b  21 h, Z h.

a

This lemma is a reformulation of a Kantorovich inequality and can be essentially found in [33]. We also need two well-known facts about positive linear maps Φ : Mn −→ Mn . Lemma 3.2.5 Let Φ be a positive linear map on Mn . Then there exists X ≥ 0 such that Φ(A) = T r AX . Hence, if π(A) : Mn −→ Mn is the left multiplication by A, we can write Φ(A) = h, π(A)h, 1

where the inner product is the canonical inner product on Mn and h = X 2 . Lemma 3.2.6 For A, B > 0, all vectors h and all positive linear maps Φ, 1

1

h, Φ(AB)h ≤ h, Φ(A)h 2 h, Φ(B)h 2 . In particular, we have h, ABh  1  21 1 1 1 A 2 h = h, A 2 A− 2 B A− 2 1 2 1  1 1 1 = A 2 h, A− 2 B A− 2 A 2 h 1  1 1 1 2 −2 − 21 2 ≤ A h A B A A 2 h 1   21 1 1   1 1 1 1 1 1 1 2 = A 2 h 2  A− 2 B A− 2 A 2 h, A− 2 B A− 2 A 2 h 2  1  21  1  21 1 1 1 1 1 1 A− 2 B A− 2 = h, Ah 2 h, A 2 A− 2 B A− 2 A 2 h 2 1

1

= h, Ah 2 h, Bh 2 . That is

1

1

h, ABh ≤ h, Ah 2 h, Bh 2 . Lemma 3.2.6 is a special case of the operator Ando’s inequality which can be derived from Theorem 3.2.1, see [44, p. 107]. Theorem 3.2.7 Let A, B > 0 such that q A ≤ B ≤ p A for some p, q > 0 and let Φ be a positive linear map. Then   41 Φ(A)Φ(B) ≤

p q

+ 2

  41 q p

Φ(AB).

3.2 A Schwarz Inequality for Positive Linear Maps

Proof.

71

(i) Suppose that for a vector f , Φ(A) =  f, A f .

In Lemma 3.2.4, take  21  1 1 1 and h = A 2 f. Z = A− 2 B A− 2  21  1 1 1 1 ≤ p 2 , we get Since q A ≤ B ≤ p A implies q 2 ≤ A− 2 B A− 2   41 p q

1 2

{ f, A f  f, B f } ≤

+

  41 q p

 f, AB f ,

2

which is equivalent to   41 p q

Φ(A)Φ(B) ≤

+

  41 q p

Φ(AB).

2

(ii) We turn to the general case. Let h be any vector. Then, by Lemma 3.2.6, 1

1

h, Φ(A)Φ(B)h ≤ h, Φ(A)h 2 h, Φ(B)h 2 1

1

= ψ(A) 2 ψ(B) 2 , where ψ is defined by ψ(X ) = h, Φ(x)h. By Lemma 3.2.5, ψ(X ) = T r 1 1 Y X = Y 2 , π(X )Y 2  for some Y ≥ 0. Since q A ≤ B ≤ p A implies qπ(A) ≤ π(B) ≤ pπ(A), section (i) yields   41 1 2

p q

1 2

ψ(A) ψ(B) ≤

+

  41 q p

2

ψ(AB).

Combining with the previous inequality we get   14 h, Φ(A)Φ(B)h ≤

p q

+ 2

  14 q p

h, Φ(AB)h. 

Remark 3.2.8 Note that Theorem 3.2.7 implies Theorem 3.2.3 by setting

72

3 Operator Inequalities for Positive Linear Maps

A = diag(A1 , ..., Am ), B = diag(B1 , ..., Bm ), Φ(A) = Z ∗ AZ , where Z ∗ = (I, ..., I ). Notes and references. All results in this section are due to [125].

3.3 Squaring the Reverse Arithmetic-Geometric Mean Inequality We recall the arithmetic-geometric mean inequality as follows. Lemma 3.3.1 For positive invertible operators AB ≤ A∇ B. Can the arithmetic-geometric inequality be squared? The answer is no and this is the content of the next Proposition [138]. Proposition 3.3.2 There are examples that (AB)2 ≤ (A∇ B)2 fails for some A, B > 0. The next theorem is that a reverse version of the operator AM-Gm inequality can be squared. Theorem 3.3.3 Let 0 < m ≤ A, B ≤ M. Then for every positive unital linear map Φ, (i) Φ 2 (A∇ B) ≤ K 2 (h)Φ 2 (AB), (ii) Φ 2 (A∇ B) ≤ K 2 (h)(Φ(A)Φ(B))2 , where K (h) =

(1+h)2 4h

and h =

M . m

Proof. (i) Since m ≤ A, B ≤ M, then (m − A)(M − A)A−1 ≤ 0 and (m − B)(M − B)B −1 ≤ 0. By easy calculation we have m M A−1 + A ≤ m + M, m M B −1 + B ≤ m + M, which is equivalent to

3.3 Squaring the Reverse Arithmetic-Geometric Mean Inequality

73

Φ(A) + MmΦ(A−1 ) ≤ M + m, Φ(B) + MmΦ(B −1 ) ≤ M + m. Summing up, we get   (Φ(A) + Φ(B)) + Mm Φ(A−1 ) + Φ(B −1 ) ≤ 2(M + m) A+B A−1 + B −1 ) + MmΦ( ) ≤ (M + m) 2 2 ⇒ Φ(A∇ B) + MmΦ(A−1 ∇ B −1 ) ≤ M + m.

⇒ Φ(

(3.3.1)

Now we notice that (i) is equivalent to the following inequality Φ(A∇ B)Φ −1 (AB) ≤ K (h) =

(M + m)2 . 4Mm

We compute Mm Φ(A∇ B)Φ −1 (AB) 1 ≤ Φ(A∇ B) + MmΦ −1 (AB) 2 (by Lemma 3.1.6) 4 1 ≤ Φ(A∇ B) + MmΦ((AB)−1 ) 2 (by Theorem 3.1.2) 4 1 = Φ(A∇ B) + MmΦ(A−1 B −1 ) 2 4 1 ≤ Φ(A∇ B) + MmΦ(A−1 ∇ B −1 ) 2 (by Lemma 3.3.1) 4 1 ≤ (M + m)2 , (by (3.3.1)). 4 Thus, we get the desired inequality in (i). (ii) Since we have Φ −1 (A) + Φ −1 (B) 2 Φ(A−1 ) + Φ(B −1 ) (by Theorem 3.1.2) ≤ 2 A−1 + B −1 ) = Φ( 2 = Φ(A−1 ∇ B −1 ).

Φ −1 (A)∇Φ −1 (B) =

Thus

Φ −1 (A)∇Φ −1 (B) ≤ Φ(A−1 ∇ B −1 ).

(3.3.2)

74

3 Operator Inequalities for Positive Linear Maps

The inequality in (ii) is equivalent to Φ(A∇ B)(Φ(A)Φ(B))−1 ≤ K (h) =

(M + m)2 . 4Mm

We compute Mm Φ(A∇ B)(Φ(A)Φ(B))−1 1 ≤ Φ(A∇ B) + Mm(Φ(A)Φ(B))−1 2 (by Lemma 3.1.6) 4 1 = Φ(A∇ B) + Mm(Φ −1 (A)Φ −1 (B)) 2 4 1 ≤ Φ(A∇ B) + Mm(Φ −1 (A)∇Φ −1 (B)) 2 (by Lemma 3.3.1) 4 1 ≤ Φ(A∇ B) + MmΦ(A−1 ∇ B −1 ) 2 (by (3.3.2)) 4 1 ≤ (M + m)2 , (by (3.3.1)) 4 and this completes the proof.



Corollary 3.3.4 Let 0 < m ≤ A, B ≤ M and 0 ≤ p ≤ 2. Then for every positive unital linear map Φ, (i) Φ p (A∇ B) ≤ K (h) p Φ p (AB). (ii) Φ p (A∇ B) ≤ K (h) p (Φ(A)Φ(B)) p . Proof. The proof is easily followed by Theorem 3.3.3 by the fact that if A ≤ B, then A p ≤ B p for 0 ≤ p ≤ 1.  Next, we present a generalization of Theorem 3.3.3 for p ≥ 2. Theorem 3.3.5 Let 0 < m ≤ A, B ≤ M and p ≥ 2. Then for every positive unital linear map Φ,  (i) Φ (A∇ B) ≤ p

(ii) Φ p (A∇ B) ≤

(M + m)2 2

p

4 p Mm p  (M + m)2 2 p

4 Mm

Φ p (AB). (Φ(A)Φ(B)) p .

Proof. (i) To get the desired inequality, its sufficient to prove the following inequality (M + m) p p p 2 Φ (A∇ B)Φ − 2 (AB) ≤ p p . 4M 2 m 2 Compute

3.3 Squaring the Reverse Arithmetic-Geometric Mean Inequality

75

p p p p M 2 m 2 Φ 2 (A∇ B)Φ − 2 (AB) p p p p = Φ 2 (A∇ B)M 2 m 2 Φ − 2 (AB) 2 p p p 1 p ≤ Φ 2 (A∇ B) + M 2 m 2 Φ − 2 (AB) (by Lemma 3.1.6) 4 p 1 ≤ Φ(A∇ B) + MmΦ −1 (AB) (by Lemma 3.3.7) 4 1 ≤ (M + m) p , (by (3.1.1)). 4 (ii) The line of proof is similar to the one in Theorem 3.3.3 and we omit it.



Definition   3.3.6 A linear map Φ is 2-positive ifwhenever the 2 × 2 operator matrix A B Φ(A) Φ(B) is positive, then so is . B∗ C Φ(B ∗ ) Φ(C) The Wielandt inequality [103] states that if 0 < m ≤ A ≤ M, and x, y ∈ H with x⊥y, then   M −m 2 2 |x, Ay| ≤ x, Ayy, Ay. M +m Bhatia and Davis [26] proved the operator version of the Wielandt inequality as follows. Theorem 3.3.7 Let 0 < m ≤ A ≤ M and X, Y be two partial isometric operators whose final spaces are orthogonal to each other. Then for every 2-positive linear map Φ, ∗



−1



Φ(X AY )Φ(Y AY ) Φ(Y AX ) ≤



M −m M +m

2

Φ(X ∗ AX ).

In the next result, we obtain a squared version of Theorem 3.3.7. Theorem 3.3.8 Let 0 < m ≤ A ≤ M and X, Y be two partial isometric operators whose final spaces are orthogonal to each other. Then for every 2-positive linear map Φ, Φ(X ∗ AY )Φ(Y ∗ AY )−1 Φ(Y ∗ AX )Φ(X ∗ AX )−1

 2  1 M −m 2 1 ≤ M+ . 4 M +m m Proof. We have

76

3 Operator Inequalities for Positive Linear Maps

Φ(X ∗ AY )Φ(Y ∗ AY )−1 Φ(Y ∗ AX )Φ(X ∗ AX )−1

1 Φ(X ∗ AY )Φ(Y ∗ AY )−1 Φ(Y ∗ AX ) + Φ(X ∗ AX )−1 2 4 2  2 1 M −m ∗ ∗ −1 ≤ Φ(X AX ) + Φ(X AX ) , (by Theorem 3.3.7) 4 M +m ≤

where Φ(Y ∗ AY )−1 and Φ(X ∗ AX )−1 are positive. Moreover, since X is partial isometric and 0 < m ≤ A ≤ M, then m X ∗ X ≤ (X ∗ AX ) ≤ M X ∗ X and M1 X ∗ X ≤ (X ∗ AX )−1 ≤ m1 X ∗ X . So we have m ≤ Φ(X ∗ AX ) ≤ M, 1 1 ≤ Φ(X ∗ AX )−1 ≤ . M m Thus the inequality below holds Φ(X ∗ AY )Φ(Y ∗ AY )−1 Φ(Y ∗ AX )Φ(X ∗ AX )−1

2  2 M − m 1 1 ≤ M+ 4 M +m m

 2  1 M −m 2 1 = M+ .  4 M +m m Lemma 3.3.9 Let A and B be positive operators such that there exist positive numbers 1 < m < M with the property m A ≤ B ≤ M A. Then   (log m)2 AB ≤ A∇ B. 1+ 8 Proof. Firstly, we point out that for each a, b > 0 [221],   a+b (log b − log a)2 √ 1+ . ab ≤ 8 2 Note that if 0 < ma ≤ b ≤ Ma with 1 < m < M, then by monotonicity of logarithm function we have log b − log a ≥ log ma − log a = log m, and so

3.3 Squaring the Reverse Arithmetic-Geometric Mean Inequality

77

    (log m)2 √ (log b − log a)2 √ 1+ ab ≤ 1 + ab 8 8 a+b ≤ . 2 Taking a = 1 and b = t > 0 in the above inequality, we get   1+t (log m)2 √ . 1+ t≤ 8 2 Since m A ≤ B ≤ M A then m I ≤ A− 2 B A− 2 ≤ M I . By taking t = A− 2 B A− 2 and with functional calculus we have 1

1

1

1

  1 1  21 I + A− 2 B A− 2 1 (log m)2  − 1 A 2 B A− 2 1+ ≤ . 8 2 1

Multiplying both sides by A 2 , we deduce the desired inequality.  Next, we are devoted to obtaining a new bound of Theorem 3.3.3. Theorem 3.3.10 Let A and B be two positive operators such that 0 < m I ≤ A ≤   m I ≤ M I ≤ B ≤ M I . Then (i) Φ 2 (A∇ B) ≤

K 2 (h)

1+ (ii) Φ 2 (A∇ B) ≤

 M  m

2 2

Φ 2 (AB),

2 2

(Φ(A)Φ(B))2 ,

8

K 2 (h)

1+ where h =

 log

 log

 M  m

8

M . m

Proof. (i) It is not hard to see that the inequality in (i) is equivalent to Φ(A∇ B)Φ −1 (AB) ≤

(M + m)2

 4Mm 1 +

We have

log

 M  m

8

2 .

78

3 Operator Inequalities for Positive Linear Maps

⎛  2 ⎞  log M  m ⎜ ⎟ −1 Φ(A∇ B)Mm ⎝1 + ⎠ Φ (AB) 8 2 ⎛ ⎞   2  M log  1 m ⎜ ⎟ −1 ≤ Φ(A∇ B) + Mm ⎝1 + ⎠ Φ (AB) (by Lemma 3.1.6) 4 8 ⎛ ⎞   2  2 log M 1 m ⎜ ⎟ −1 ≤ Φ(A∇ B) + Mm ⎝1 + ⎠ Φ(AB) (by Theorem 3.1.2) 4 8 ⎛ ⎛ ⎞ ⎞ 2  2  M log  1 ⎜ m ⎜ ⎟ −1 ⎟ = Φ ⎝(A∇ B) + Mm ⎝1 + ⎠ (AB) ⎠ 4 8 ⎛ ⎛ ⎞ ⎞ 2  2  M log 1 ⎜ m ⎜ ⎟ −1 −1 ⎟ = Φ ⎝(A∇ B) + Mm ⎝1 + ⎠ (A B )⎠ 4 8  1  Φ A∇ B + Mm(A−1 ∇ B −1 ) (by Lemma 3.3.9) 4 1 = Φ(A∇ B) + MmΦ(A−1 ∇ B −1 ) 4 1 ≤ (M + m)2 , (by (3.3.1)). 4



Thus the desired inequality in (i) holds. (ii) It can be proved analogously as in (i).  Notes and references. Theorem 3.3.3 is due to M. Lin [138]. Theorem 3.3.5 and Theorem 3.3.8 are proved in [65]. Lemma 3.3.9 and Theorem 3.3.10 are due to [167].

3.4 Reverses of Ando’s Inequality for Positive Linear Maps The Cauchy-Schwarz inequality plays an essential role in mathematical analysis and its applications. In a semi inner product space H, the Cauchy-Schwarz inequality reads as follows: 1 1 |x, y| ≤ x, x 2 y, y 2 . There are interesting generalizations of the Cauchy-Schwarz inequality in various frameworks, e.g., finite sums, integrals, and inner product spaces [53–55, 77, 110, 111, 170, 175] and references therein. There are several reverses of the Cauchy-

3.4 Reverses of Ando’s Inequality for Positive Linear Maps

79

Schwarz inequality in the literature: Diaz-Metacalf, Pólya Szegó, Kantorovich, and Shisha-Mond’s inequalities. In the following, we show several reverse CauchySchwarz type inequalities for positive linear maps. The operator Ando’s inequality [11] states that Theorem 3.4.1 If A and B are positive operators and Φ is a positive linear map, then for each λ ∈ [0, 1] Φ(Aλ B) ≤ Φ(A)λ Φ(B). In the case of H be finite-dimensional space, since positive linear maps are regarded as a matrix version of integrals, Ando’s inequality with λ = 21 a genuine matrix Cauchy-Schwarz inequality. The reverse of this inequality [125] is known as matrix Casel or matrix reverse Cauchy-Schwarz inequality which asserts that if Φ is a positive linear map and A, B > 0 such that s A ≤ B ≤ t A for some scalars 0 < s ≤ t with w = st , then w 4 + w− 4 Φ(AB). 2 1

Φ(A)Φ(B) ≤

1

(3.4.1)

The study of this inequality has received considerable attention in recent years involving its generalizations, variations, and applications [31, 102, 116]. Theorem 3.4.2 Let A and B be positive invertible operators and Φ be a positive linear map. If m 2 A ≤ B ≤ M 2 A for some positive real numbers m < M, then the following inequalities hold. (i) Operator Diaz-Metacalf inequality of first type MmΦ(A) + Φ(B) ≤ (M + m)Φ(AB). (ii) Operator Cassels inequality M +m Φ(A)Φ(B) ≤ √ Φ(AB). 2 Mm (iii) Operator Klamkin-Mclenaghan inequality √ √ 1 1 1 1 Φ(AB)− 2 Φ(B)Φ(AB)− 2 −Φ(AB) 2 Φ(A)−1 Φ(AB) 2 ≤ ( M− m)2 . Proof. (i) If m 2 A ≤ B ≤ M 2 A for some positive numbers m < M, then m 2 ≤ 1 1 A− 2 B A− 2 ≤ M 2 . We have 

whence

  1  21   1  21 1 1 A− 2 B A− 2 − m ≥ 0, M − A− 2 B A− 2

80

3 Operator Inequalities for Positive Linear Maps

 21  1 1 1 1 Mm + A− 2 B A− 2 ≤ (M + m) A− 2 B A− 2 , and hence Mm A + B ≤ (M + m)AB. Since Φ is a positive linear map, we get the following desired inequality MmΦ(A) + Φ(B) ≤ (M + m)Φ(AB). (ii) We have √

Mm(Φ(A)Φ(B)) = MmΦ(A)Φ(B) 1 ≤ (MmΦ(A) + Φ(B)) (by AM-GM inequality) 2 1 ≤ (M + m)Φ(AB). (by Diaz-Metacalf inequality in (i)). 2

Thus, we get

M +m Φ(A)Φ(B) ≤ √ Φ(AB). 2 Mm

(iii) It follows from (i) that 1

1

1

1

Φ(AB)− 2 Φ(B)Φ(AB)− 2 − Φ(AB) 2 Φ(A)−1 Φ(AB) 2 1

1

1

1

≤ (M + m) − MmΦ(AB)− 2 Φ(A)Φ(AB)− 2 − Φ(AB) 2 Φ(A)−1 Φ(AB) 2

= (M + m)−

1  1 2  √ 1 1 2 1 1 2 − − −1 − Mm Φ(AB) 2 Φ(A)Φ(AB) 2 − Φ(AB) 2 Φ(A) Φ(AB) 2 − √ − 2 Mm

√ ≤ M + m − 2 Mm √ √ 2 M− m .  =

Corollary 3.4.3 Let A and B be positive invertible operators and Φ be a positive linear map. If m 21 ≤ A ≤ M12 and m 22 ≤ B ≤ M22 for some positive real numbers m 1 < M1 and m 2 < M2 , then the following inequalities hold. (i) Operator Diaz-Metacalf inequality of the second type M2 m 2 Φ(A) + Φ(B) ≤ M1 m 1 (ii) Operator Polya-Szeg ´ o¨ inequality



M2 m2 + m1 M1

 Φ(AB).

3.4 Reverses of Ando’s Inequality for Positive Linear Maps

1 Φ(A)Φ(B) ≤ 2



M1 M2 + m1m2



81

m1m2 M1 M2

.Φ(AB)

(iii) Operator Shisha-Mond inequality Φ(AB)− 2 Φ(B)Φ(AB)− 2 − Φ(AB) 2 Φ(A)−1 Φ(AB) 2

 2  M2 m2 ≤ + . m1 M1 1

1

1

1

Proof. If m 21 ≤ A ≤ M12 and m 22 ≤ B ≤ M22 , then m2 = Thus by putting m = ities. 

m2 M1

m 22 M22 − 21 − 21 ≤ A B A ≤ = M 2. M12 m 21

and M =

M2 m1

in Theorem 3.4.2, we get the desired inequal-

Corollary 3.4.4 Let A and B be positive invertible operators and Φ be a positive unital linear map. Then we have the operator Kantorovich inequality Φ(A)Φ(A−1 ) ≤

M 2 + m2 , 2Mm

where m 2 A ≤ B ≤ M 2 A. Proof. If A is invertible, Φ is unital and m 21 = m 2 ≤ A ≤ M 2 = M12 , then by putting m 22 = M12 ≤ B = A−1 ≤ m12 = M22 in (ii) of Corollary 3.4.3, we get the desired inequality. Now we present a different type of the reverse Ando’s operator inequality. Theorem 3.4.5 Let A and B be positive operators such that m A ≤ B ≤ M A for some scalars 0 < m ≤ M and let Φ be a positive linear map. Then for λ ∈ [0, 1] Φ(A)λ Φ(B) − Φ(Aλ B) ≤ −C(m, M, λ)Φ(A), where the Kantorovich constant for the difference C(m, M, λ) is defined by 

M λ − mλ C(m, M, λ) = (λ − 1) λ(M − m)

λ  λ−1

for any real number λ ∈ R. Proof. Put X = A− 2 B A− 2 and for each λ ∈ [0, 1], 1

1

+

Mm λ − m M λ M −m

82

3 Operator Inequalities for Positive Linear Maps

 α=

M λ − mλ λ(M − m)

1  λ−1

.

We have Φ(A)λ Φ(B) = αλ Φ(A)λ αλ−1 Φ(B) ≤ (1 − λ)αλ Φ(A) + λαλ−1 Φ(B).

(3.4.2)

Since m A ≤ B ≤ M A then m ≤ X ≤ M. It follows that (1 − λ)αλ I + λαλ−1 X M λ − mλ Mm λ − m M λ − C(m, M, λ)I + M −m M −m λ ≤ X − C(m, M, λ)I.

=

Thus

(1 − λ)αλ I + λαλ−1 X ≤ X λ − C(m, M, λ)I, 1

multiplying A 2 from both sides of the inequality gives (1 − λ)αλ A + λαλ−1 B ≤ Aλ B − C(m, M, α)A, and hence for positive linear map Φ (1 − λ)αλ Φ(A) + λαλ−1 Φ(B) ≤ Φ(Aλ B) − C(m, M, λ)Φ(A).

(3.4.3)

We get Φ(A)λ Φ(B) ≤ (1 − λ)αλ Φ(A) + λαλ−1 Φ(B) (by (3.4.2)) ≤ Φ(Aλ B) − C(m, M, λ)Φ(A), (by (3.4.3)) which completes the proof.  The following theorem is a ratio type reverse Ando’s operator inequality. Theorem 3.4.6 Let A and B be positive operators such that m A ≤ B ≤ M A for some scalars 0 < m ≤ M and let Φ be a positive linear map. Then for λ ∈ [0, 1] Φ(A)λ Φ(B) ≤ K (m, M, λ)−1 Φ(Aλ B), where the generalized Kantorovich constant K (m, M, α) is defined by

3.4 Reverses of Ando’s Inequality for Positive Linear Maps

K (m, M, λ) =

m M λ − Mm λ (λ − 1)(M − m)



83

λ−1 λ



M λ − mλ m M λ − Mm λ



for any real number λ ∈ R. Proof. Put X = A− 2 B A− 2 . If we put 1

λ0 =

1

λ 1−λ



M 1−λ − m 1−λ m −λ − M −λ

then

 and

K (m, M, λ)−1 =

μ0 =

μ0 λ01−λ

λ(M − m) , M λ − mλ

.

By easy calculation, we find that λt 1−λ + (1 − λ)λ0 t 1−λ ≤ μ0 ,

for all t ∈ [m, M].

Since m I ≤ X ≤ M I , we get λX + (1 − λ)λ0 I ≤ μ0 X λ , 1

and hence by multiplying A 2 from both sides we have λB + (1 − λ)λ0 A ≤ μ0 (Aλ B). This implies (1 − λ)λ0 Φ(A) + λΦ(B) ≤ μ0 Φ(Aλ B).

(3.4.4)

By the weighted arithmetic-geometric mean inequality, it follows that λ1−λ 0 Φ(A)λ Φ(B) ≤ (1 − λ)λ0 Φ(A) + λΦ(B).

(3.4.5)

Inequalities (3.4.4) and (3.4.5) deduced the desired inequalities.  Theorem 3.4.6 with λ = [125].

1 2

gives the following corollary which is due to E. Y. Lee

Corollary 3.4.7 Let A and B be positive operators such that m A ≤ B ≤ M A for some scalars 0 < m ≤ M and let Φ be a positive linear map. Then √ Φ(A)Φ(B) ≤

√ M+ m Φ(AB). √ 2 4 Mm

Notes and references. Note that for the difference and ratio type reverse Ando’s operator inequality see also [31] and [69].

84

3 Operator Inequalities for Positive Linear Maps

Theorem 3.4.2, Corollary 3.4.3 and Corollary 3.4.4 are proved in [171]. Theorem 3.4.5 and Theorem 3.4.6 are due to Y. Seo [191].

3.5 Squaring the Reverse Ando’s Operator Inequality Let A, B, C, D be positive operators and λ ∈ [0, 1]. Concavity property of geometric mean states that (3.5.1) (Aλ B) + (Cλ D) ≤ (A + C)λ (B + D). Theorem 3.5.1 Let Φ be the unital positive linear map and Let A and B be positive operators such that 0 < m 1 ≤ A ≤ M1 and 0 < m 2 ≤ B ≤ M2 for some positive real numbers m 1 < M1 and m 2 < M2 . Then for λ ∈ [0, 1]  (Φ(A)λ Φ(B))2 ≤

(M1 + m 1 )2(1−λ) (M2 + m 2 )2λ 4(m 2 M2 )λ (m 1 M1 )1−λ

2 Φ 2 (Aλ B).

(3.5.2)

Proof. Inequality (3.5.2) is equivalent to 2(1−λ) (M2 + m 2 )2λ (Φ(A)λ Φ(B))Φ −1 (Aλ B) ≤ (M1 + m 1 ) . 4(m 2 M2 )λ (m 1 M1 )1−λ

It is known that m 1 M1 Φ(A−1 ) + Φ(A) ≤ m 1 + M1 , m 2 M2 Φ(B −1 ) + Φ(B) ≤ m 2 + M2 . By the monotonicity of the λ-geometric mean we have (m 1 M1 Φ(A−1 ) + Φ(A))λ (m 2 M2 Φ(B −1 ) + Φ(B)) ≤ (m 1 + M1 )λ (m 2 + M2 ). Since the λ-geometric mean operation is subadditive, we can see

(3.5.3)

3.5 Squaring the Reverse Ando’s Operator Inequality

85

(Φ(A)λ Φ(B))((m 2 M2 )λ (m 1 M1 )1−λ Φ −1 (Aλ B)) 2 1 ≤ (Φ(A)λ Φ(B)) + ((m 2 M2 )λ (m 1 M1 )1−λ Φ −1 (Aλ B)) (by Lemma 3.1.6) 4 2 1 ≤ (Φ(A)λ Φ(B)) + ((m 2 M2 )λ (m 1 M1 )1−λ Φ(Aλ B)−1 ) (by Theorem 3.1.2) 4 2 1 = (Φ(A)λ Φ(B)) + ((m 2 M2 )λ (m 1 M1 )1−λ Φ(A−1 λ B −1 )) 4 2 1 ≤ (Φ(A)λ Φ(B)) + ((m 2 M2 )λ (m 1 M1 )1−λ Φ(A−1 )λ Φ(B −1 ))) (by Theorem 3.4.1) 4 2 1 = Φ(A)λ Φ(B) + (m 1 M1 Φ(A−1 )λ m 2 M2 Φ(B −1 )) 4    2  1 ≤ m 1 M1 Φ(A−1 ) + Φ(A) λ m 2 M2 Φ(B −1 ) + Φ(B) (by (3.5.1)) 4

1 (M1 + m 1 )λ (M2 + m 2 ) 2 (by (3.5.3)) 4 1 = (M1 + m 1 )2(1−λ) (M2 + m 2 )2λ , 4



which implies the desired inequality.  Next, we show a new reverse Ando’s operator inequality which is a squared Pólya -Szegó operator inequality. Theorem 3.5.2 Let Φ be the unital positive linear map. If 0 < m 21 ≤ A ≤ M12 and 0 < m 22 ≤ B ≤ M22 for some positive real numbers m 1 ≤ M1 and m 2 ≤ M2 . Then the following inequalities hold (Φ(A)Φ(B))2 ≤ βΦ(AB)2 , where β is defined as follows  1 M2 , then β = γ 4 . (i) If γ 2 ≤ M  m1m2     M1 M2 M1 M2 2 1 M2 (ii) If γ 2 ≥ M 2γ . , then β = − m1m2 m1m2 m1m2 and γ is the number given in (ii) of Corollary 3.4.3 by 1 γ := 2 Proof. Put



M1 M2 + m1m2



m1m2 M1 M2

 .

86

3 Operator Inequalities for Positive Linear Maps

M := M1 M2 , m = m1m2, C := Φ(AB), D := Φ(A)Φ(B),   1 M +m ≥ 1. γ := √ 2 Mm To get the desired inequality, we prove C −1 D 2 C −1 ≤ β. Since 0 < m 21 ≤ A ≤ M12 and 0 < m 22 ≤ B ≤ M22 , it follows from the monotone property of the geometric mean that 0 0 and λ ∈ [0, 1]. Then A∇λ B ≤ max{S( p), S(q)}(Aλ B), where S(t) is the so called Specht’s ratio. © Springer Nature Switzerland AG 2021 M. B. Ghaemi et al., Advances in Matrix Inequalities, Springer Optimization and Its Applications 176, https://doi.org/10.1007/978-3-030-76047-2_4

89

90

4 Operator Inequalities Involving Operator Monotone Functions

Proof. According to Lemma 4.1.1 for the invertible positive operator 0 < p I ≤ X ≤ q I , we have (1 − λ)X + λI ≤ max S(t)X 1−λ . p≤t≤q

Putting X = B − 2 AB − 2 we get 1

1

(1 − λ)A + λB ≤ max S(t)(B1−λ A). p≤t≤q

As we mentioned in (2.2.1), since Aλ B = B1−λ A, using the fact that max S(t) = max{S( p), S(q)},

p≤t≤q

gives the desired inequality.  Note that from Lemma 4.1.2, we deduce the following theorem which is due to M. m A≤B≤ M A. Since Tomimaga [202]. Because if 0 < m I ≤ A, B ≤ M I , then M m 1 m S(h) = S( h ), we obtain Theorem 4.1.3 by letting p = M and q = M in Lemma m 4.1.2. Theorem 4.1.3 If 0 < m I ≤ A, B ≤ M I with h =

M m

and λ ∈ [0, 1], then

A∇λ B ≤ S(h)(Aλ B), where the constant Specht’s ratio is defined by 1

S(t) =

t t−1 1

e log t t−1

,

for a positive real number t. Now, we introduce an improvement of Lemma 4.1.2 involving operator monotone functions. Theorem 4.1.4 Let f : [0, ∞) → [0, ∞) be operator monotone function and 0 < p A ≤ B ≤ q A, s, t > 0. Then fot all λ ∈ [0, 1] f (A)λ f (B) ≤ max{S( p), S(q)} f (Aλ B), where S(t) is the so called Specht’s ratio. Proof. First note that since f is analytic on (0, ∞), we may assume that f (x) > 0 for all x > 0. Otherwise f is identically zero. Also, since f is operator monotone function on [0, ∞), so it is operator concave. That is (1 − λ) f (A) + λ f (B) ≤ f ((1 − λ)A + λB).

(4.1.1)

4.1 Young Inequalities Involving Operator Monotone Functions

91

For the convenience we put M = max{S( p), S(q)}. So M ≥ 1. Now compute f (A)λ f (B) ≤ (1 − λ) f (A) + λ f (B) ≤ f ((1 − λ)A + λB) (by (4.1.1)) ≤ f (M(Aλ B)) (by Lemma 4.1.2) ≤ M f (Aλ B). Note that for every nonnegative concave function f and every z ≥ 1, f (zx) ≤ z f (x).  As an immediate result we have the following corollary. Corollary 4.1.5 Let f : [0, ∞) → [0, ∞) be operator monotone function and 0 < m I ≤ A, B ≤ M I . Then for all λ ∈ [0, 1] f (A)λ f (B) ≤ S(h) f (Aλ B), where S(t) is the so called Specht’s ratio and h =

M . m

Corollary 4.1.6 Let g be a nonnegative operator monotone decreasing function on (0, ∞) and 0 < m I ≤ A, B ≤ M I . Then for all λ ∈ [0, 1] g(Aλ B) ≤ S(h)(g(A)λ g(B)), where S(h) is the so called Specht’s ratio and h =

M . m

Proof. Since g is operator monotone decreasing on (0, ∞), so g1 = g −1 is operator monotone on (0, ∞). Now by applying Corollary 4.1.5 for f = g1 = g −1 , we have (g(A)λ g(B))−1 = g −1 (A)λ g −1 (B) ≤ S(h)g −1 (Aλ B). By reversing both sides, the desired inequality is deduced.  Furuichi and Minculete [76] gave another reverse inequality for Young’s inequality without using Specht’s ratio as follows. Lemma 4.1.7 Let 0 < m I ≤ A ≤ B ≤ M I with h =

M . m

Then for all λ ∈ [0, 1]



 1 2 (1 − λ)A + λB ≤ exp λ(1 − λ)(1 − ) (Aλ B). h Using this lemma deduce the next theorem. Theorem 4.1.8 Let f : [0, ∞) → [0, ∞) be operator monotone function and 0 < m I ≤ A ≤ B ≤ M I with h = M . Then for all λ ∈ [0, 1] m

92

4 Operator Inequalities Involving Operator Monotone Functions

  1 f (A)λ f (B) ≤ exp λ(1 − λ)(1 − )2 f (Aλ B). h Proof. For the convenience we put   1 M = exp λ(1 − λ)(1 − )2 ≥ 1. h Now compute f (A)λ f (B) ≤ (1 − λ) f (A) + λ f (B) ≤ f ((1 − λ)A + λB) (by (4.1.1)) ≤ f (M(Aλ B)) (by Lemma 4.1.7) ≤ M f (Aλ B).  Notes and references. Lemma 4.1.2, Theorem 4.1.4, Corollary 4.1.5, Corollary 4.1.6, and Theorem 4.1.8 are proved in [85].

4.2 Eigenvalue Inequalities Involving Operator Concave Functions In the following, we present an analogous of Theorem 4.1.4 with the generalized Kantorovich constant. For this purpose, we need the assumption of doubly concave functions. Definition 4.2.1 A nonnegative continuous function f (t) defined on a positive interval I ⊂ [0, ∞) is said doubly concave if (i) f (t) is concave in the usual sense. (ii) f (t) is geometrically concave, i.e., f (x λ y 1−λ ) ≥ f (x)λ f (y)1−λ for all x, y ∈ I and λ ∈ [0, 1]. If f (t) and g(t) are doubly concave on I , then so is their geometric mean f (t)λ g(t)1−λ for λ ∈ [0, 1] and their minimum min{ f (t), g(t)}. The most important example of doubly concave function on I = [0, ∞) is t −→ t p . We refer to [36] for details. In what follows, capital letters A, B means n × n matrices or bounded linear operators on an n-dimensional Hilbert space H. The following Lemmas will be used to prove the analogous of Theorem 4.1.4. Firstly, we recall the Minimax Principle [21] which can be proved by Poincare’s inequality.

4.2 Eigenvalue Inequalities Involving Operator Concave Functions

93

Lemma 4.2.2 Let A > 0. Then λk (A) = max min{ Ah, h , h ∈ F, h = 1}, dimF =k

where F is a subspace of H. Lemma 4.2.3 Let h be a vector such that h = 1, A > 0 and f (t) be any concave function defined on [0, ∞). Then

f (A)h, h ≤ f ( Ah, h ). J. C. Bourin et.al. [31] obtained the following lemma. Lemma 4.2.4 Let A, B > 0 with 0 < s A ≤ B ≤ t A fot some scalars 0 < s ≤ t with w = st . Then, for all vectors h and λ ∈ [0, 1]

(Aλ B)h, h ≤ Ah, h 1−λ Bh, h λ ≤ K (w, λ)−1 (Aλ B)h, h , where K (w, λ) is the generalized Kantorovich constant defined for w > 0 by K (w, λ) =

wλ − w (λ − 1)(w − 1)



λ − 1 wλ − 1 . λ wλ − w

λ .

(4.2.1)

It is known that K (w, λ) ∈ [0, 1) for λ ∈ [0, 1] [77]. Now we state our main theorem which provides a new reverse for Ando-Hiai inequality and thereupon a new reverse for Golden-Thompson type inequality. Theorem 4.2.5 Let f be a doubly concave function on [0, ∞) and 0 < s A ≤ B ≤ t A for some scalars 0 < s ≤ t with w = st . Then for all λ ∈ [0, 1] and k = 1, 2, ..., n, λk ( f (A)λ f (B)) ≤ K (w, λ)−1 λk ( f (Aλ B)), where K (w, λ) is the generalized Kantorovich constant defined as (4.2.1). Proof. By Lemma 4.2.3 for every vector h with h = 1, we can write

f (A)h, h ≤ f ( Ah, h ). Also, for any integer k less than or equal to the dimension of the space, we have a subspace F of dimension k such that

94

4 Operator Inequalities Involving Operator Monotone Functions

λk ( f (A)λ f (B)) = ≤ ≤ ≤ ≤

min

f (A)λ f (B)h, h (by Lemma 4.2.2)

min

f (A)h, h 1−λ f (B)h, h λ (by Lemma 4.2.4)

min

( f Ah, h )1−λ ( f Bh, h )λ (by Lemma 4.2.3)

h∈F , h =1 h∈F , h =1 h∈F , h =1

min

f ( Ah, h )1−λ · Bh, h λ ) (by Definition 4.2.1)

min

f (K (w, λ)−1 (Aλ B)h, h ) (by Lemma 4.2.4)

h∈F , h =1 h∈F , h =1

≤ K (w, λ)−1 = K (w, λ)−1

min

h∈F , h =1

min

h∈F , h =1

f ( (Aλ B)h, h )

f (Aλ B)h, h (by monotony of f )

≤ K (w, λ)−1 λk ( f (Aλ B)), (by Lemma 4.2.2). Note that for every nonnegative concave function f and every z ≥ 1, f (zx) ≤ z f (x). Notice that for λ ∈ [0, 1], K (w, λ)−1 ≥ 1. In the last inequality, we have used the minimax principle.  Remark 4.2.6 Note that the above statement is equal to the existence of a unitary operator U satisfying in the following inequality f (A)λ f (B) ≤ K (w, λ)−1 U f (Aλ B)U ∗ . Bourin et. al. [31] gave a new generalization of matrix reverse Cauchy-Schwarz inequality (3.4.1) as follows. Theorem 4.2.7 Let A, B > 0 such that s A ≤ B ≤ t A for some scalars 0 < s ≤ t with w = st and let Φ be a positive linear map. Then for all λ ∈ [0, 1] Φ(A)Φ(B) ≤ K (w, λ)−1 Φ(AB), where K (w, λ) is the generalized Kantorovich constant defined in (4.2.1). Theorem 4.2.8 Let Φ be a positive linear map, f be a doubly concave and operator concave function on [0, ∞) and 0 < s A ≤ B ≤ t A for some scalars 0 < s ≤ t with w = st . Then for all λ ∈ [0, 1] and k = 1, 2, ..., n λk ( f (Φ(A))λ f (Φ(B))) ≤ K (w, λ)−2 λk ( f (Φ(Aλ B))), where K (w, λ) defined in (4.2.1). Proof. By the minimax principle for any integer K less than or equal to the dimension of the space, we have a subspace F of dimension k such that

4.2 Eigenvalue Inequalities Involving Operator Concave Functions

95

λk ( f (Φ(A))λ f (Φ(B))) ≤ ≤ ≤ ≤ ≤

min

f (Φ(A))h, h 1−λ f (Φ(B))h, h λ (by Lemma 4.2.4)

min

( f ( Φ(A))h, h ))1−λ ( f ( Φ(B))h, h ))λ (by Lemma 4.2.3)

h∈F , h =1 h∈F , h =1

min

f ( Φ(A)h, h 1−λ Φ(B)h, h λ ) (by Definition 4.2.1)

min

f (K (w, λ)−1 Φ(A)λ Φ(B)h, h ) (by Lemma 4.2.4)

min

f (K (w, λ)−2 Φ(Aλ B)h, h ) (by Theorem 4.2.7)

h∈F , h =1 h∈F , h =1 h∈F , h =1

≤ K (w, λ)−2 min

f ( Φ(Aλ B)h, h )

= K (w, λ)−2

f (Φ(Aλ B)h, h (by monotony of f )

h∈, h =1

min

h∈F , h =1

≤ K (w, λ)−2 λk ( f (Φ(Aλ B))), (by Lemma 4.2.2).  We notice that Theorem 4.2.8 is equivalent to the existence of a unitary operator U satisfying in the following f (Φ(A))λ f (Φ(B)) ≤ K (w, λ)−2 U f (Φ(Aλ B))U ∗ . Corollary 4.2.9 Let Φ be the unital positive linear map, f be a doubly concave and operator concave function on [0, ∞) and m I ≤ A, B ≤ M I for some scalars 0 < m ≤ M with h = M . Then for all λ ∈ [0, 1] and k = 1, 2, ..., n m λk ( f (Φ(A))λ f (Φ(B))) − λk ( f (Φ(Aλ B))) ≤ (K (h 2 , λ)−2 − 1) f (M). Proof. Since m I ≤ A, B ≤ M I , by easy calculation we find that m I ≤ Φ(Aλ B) ≤ M I and so (4.2.2) f (m I ) ≤ f (Φ(Aλ B)) ≤ f (M I ). Also m I ≤ A, B ≤ M I implies

m M

A≤B≤

M m

A. We have

λk ( f (Φ(A))λ f (Φ(B))) − λk ( f (Φ(Aλ B))) ≤ (K (h 2 , λ)−2 − 1)λk ( f (Φ(Aλ B))) (by Theorem 4.2.8) ≤ (K (h 2 , λ)−2 − 1)λk ( f (M I )) (by (4.2.2)) ≤ (K (h 2 , λ)−2 − 1) f (M), (by the spectral mapping theorem).



Theorem 4.2.10 Let Φ be a positive linear map, f be a doubly concave and operator concave function on [0, ∞) and 0 ≤ s A ≤ B ≤ t A for some scalars 0 < s ≤ t with w = st . Then for all λ ∈ [0, 1] and k = 1, 2, ..., n λk ( f (Φ(A))λ f (Φ(Aλ B))) ≤ K (w λ , λ)−2 λk ( f (Φ(A2λ B))).

96

4 Operator Inequalities Involving Operator Monotone Functions

Proof. According to Theorem 4.2.5 we have λk ( f (A)λ f (B)) ≤ K (w, λ)−1 λk ( f (Aλ B)).

(4.2.3)

Also, 0 < s A ≤ B ≤ t A implies 0 < s λ A ≤ Aλ B ≤ t λ A and hence s λ Φ(A) ≤ Φ(Aλ B) ≤ t λ Φ(A). Note that the condition number of Φ(A) and Φ(Aλ B) is w λ . Now, replacing A and B by Φ(A) and Φ(Aλ B), respectively, in (4.2.3), we have λk ( f (Φ(A))λ f (Φ(Aλ B))) ≤ K (w λ , λ)−1 λk ( f (Φ(A))λ Φ(Aλ B))) ≤ K (w λ , λ)−2 λk ( f (Φ(Aλ (Aλ B))) (by Theorem 4.2.7) = K (w λ , λ)−2 λk ( f (Φ(A2λ B))).  In the following, we give the definition of Olson order ≺ols [177]. Definition 4.2.11 For positive operators A and B, we say A ≺ols B if and only if Ar ≤ B r for every r ≥ 1. Theorem 4.2.12 Let Φ be a positive linear map, f be a doubly concave and operator concave function on [0, ∞) and 0 < s A ≺ols B ≺ols t A for some scalars 0 < s ≤ t with w = st . Then for all λ ∈ [0, 1] and k = 1, 2, ..., n λk (Φ( f (A))λ Φ( f (B))) ≤ K (w, λ)−2 λk (Φ( f (Aλ B))). Proof. First, note that 0 < s A ≺ols B ≺ols t A gives s A ≤ B ≤ t A. This implies f (s A) ≤ f (B) ≤ f (t A) for every operator concave function f . So, for increasing concave function f on [0, ∞), we have the sandwich condition s f (A) ≤ f (B) ≤ t f (A), given that s ≤ 1 ≤ t. Now, applying Theorem 4.2.7 for f (A) and f (B) leads to Φ( f (A))λ Φ( f (B)) ≤ K (w, λ)−1 Φ( f (A)λ f (B)), and so

(4.2.4)

4.2 Eigenvalue Inequalities Involving Operator Concave Functions

97

λk (Φ( f (A))λ Φ( f (B))) ≤ K (w, λ)−1 λk (Φ( f (A)λ f (B))) ( by (4.2.4)) = K (w, λ)−1 Φ(λk ( f (A)λ f (B))) ≤ K (w, λ)−2 Φ(λk ( f (Aλ B))) ( by (4.2.3)) = K (w, λ)−2 λk (Φ( f (Aλ B))), where the spectral mapping theorem is used in the second and last equality.



Corollary 4.2.13 Let Φ be a positive linear map, f be a doubly concave and operator concave function on [0, ∞) and m I ≤ A, B ≤ M I for some scalars 0 < m ≤ M . Then for all λ ∈ [0, 1] and k = 1, 2, ..., n with h = M m λk (Φ( f (A))λ Φ( f (B))) − λk (Φ( f (Aλ B))) ≤ (K (h 2 , λ)−2 − 1)Φ( f (M)). Proof. Since m I ≤ A, B ≤ M I , we have f (m I ) ≤ f (Aλ B) ≤ f (M I ), and so Φ( f (m I )) ≤ Φ( f (Aλ B)) ≤ Φ( f (M I )). We notice that m I ≤ A, B ≤ M I deduce m 4.2.12 with M A ≺ols B ≺ols M A gives m

m M

A≤B≤

M m

A. Now applying Theorem

λk (Φ( f (A))λ Φ( f (B))) − λk (Φ( f (Aλ B))) ≤ (K (h 2 , λ)−2 − 1)λk (Φ( f (Aλ B))) ≤ (K (h 2 , λ)−2 − 1)λk (Φ( f (M I ))) =≤ (K (h 2 , λ)−2 − 1)Φ(λk ( f (M I ))) =≤ (K (h 2 , λ)−2 − 1)Φ( f (M)).  Notes and references. We notice that Theorem 4.2.5 is due to [86]. In [86], a new reverse for Ando-Hiai inequality and thereupon a new reverse for Golden-Thompson type inequality shown as an application of Theorem 4.2.5. Theorem 4.2.8, Corollary 4.2.9, Theorem 4.2.10, Theorem 4.2.12, Corollary 4.2.13 are proved in [113].

4.3 Operator Aczél Inequality Involving Operator Monotone Functions Aczél ’s inequality has important applications in the theory of functional equations in non-euclidean geometry. We refer the readers to [6, 51, 182, 201] and references

98

4 Operator Inequalities Involving Operator Monotone Functions

therein. The purpose of this section is to present some reverses of operator Aczél ’s inequality by using several reverse Young’s inequalities. We start this section with the operator version of the classical Aczél inequality deal with the weighted operator geometric mean. Lemma 4.3.1 Suppose that m t is a parametrized operator power mean not greater than the weighted arithmetic mean. If J is an interval of (0, ∞) and f : J −→ (0, ∞) is operator decreasing and operator concave on J and A, B are positive invertible operators with spectra contained in J , then f (Am t B) ≥ f (A)m t f (B). Proof. Since Am t B ≤ (1 − t)A + t B, it follows from the operator decreasing property of f that f (Am t B) ≥ f ((1 − t)A + t B) ≥ (1 − t) f (A) + t f (B) ( f is operator concave) ≥ f (A)m t f (B).



An operator Aczél inequality involving the weighted operator geometric mean is in the following form. Theorem 4.3.2 Let J be an interval of (0, ∞), let g : J −→ (0, ∞) be operator decreasing and operator concave on J , 1p + q1 = 1, p, q > 1 and let A, B > 0 with spectra contained in J . Then g(A p  q1 B q ) ≥ g(A p ) q1 g(B q ), 1 p

(4.3.1) 1 q

g(A p  q1 B q )x, x ≥ g(A p )x, x g(B q )x, x ,

(4.3.2)

for any vector x ∈ H. Proof. Lemma 4.3.1 yields inequality (4.3.1) by replacing A, B and m t with A p , B q ,  q1 respectively. Let x ∈ H be an arbitrary vector. We have 1 p 1 q )A + B )x, x (by monotone decreasing of f ) q q 1 p 1 q = g( A + B )x, x p q 1 1 ≥ ( g(A p ) + g(B q ))x, x (by concavity of f ) p q 1 1 = g(A p )x, x + g(B q )x, x p q

g(A p  q1 B q )x, x ≥ g((1 −

1

1

≥ g(A p )x, x p g(B q )x, x q ,

4.3 Operator Aczél Inequality Involving Operator Monotone Functions

99

where the last inequality follows from the weighted arithmetic-geometric mean inequality.  Corollary 4.3.3 Let 1p + q1 = 1, p, q > 1 and A, B be positive invertible operators. Let f be operator decreasing and operator concave. Then 1

1

1

g(A p  q1 B q ) 2 x ≥ g(A p ) 2 p x g(B q ) 2q x . Proof. The Holder-McCarthy inequality [77] asserts that if A is a positive operator, then

Ax, x r ≥ Ar x, x for all 0 < r < 1, and for all unit vector x ∈ H. It follows from (4.3.2) that 1

1

g(A p  q1 B q )x, x ≥ g(A p )x, x p g(B q )x, x q 1 p

1 q

≥ g(A p ) x, x g(B q ) x, x . Also for every positive operator A we have [21] 1

1

1

Ax, x = A 2 x, A 2 x = A 2 x 2 . So we obtain 1

1 2p

1 2q

g(A p  q1 B q ) 2 x ≥ g(A p ) x g(B q ) x , and this completes the proof.  Inequality (4.3.1) for commuting operators A, B gives the following corollary. Note that in the commuting case AB = B A, we have At B = A1−t B t . Corollary 4.3.4 Let J be an interval of (0, ∞), let g : J −→ (0, ∞) be operator decreasing and operator concave on J , 1p + q1 = 1, p, q > 1 and A, B be positive invertible commuting operators with spectra contained in J . Then 1  1  g(AB) ≥ g(A p ) p g(B q ) q . In the following, our purpose is to get the reverse of Aczél operator inequality involving operator monotone decreasing functions. To do this, we need some Lemmas and Theorems on the reverse Young inequalities. One of the reverse Young inequalities involving the Kantorovich constant is given by Liao et. al. [129] as follows. Lemma 4.3.5 Let A and B be positive operators satisfying the following conditions     0 < m I ≤ A ≤ m I ≤ M I ≤ B ≤ M I or 0 < m I ≤ B ≤ m I ≤ M I ≤ A ≤ M I   for some constants m, m , M, M . Then

100

4 Operator Inequalities Involving Operator Monotone Functions

A∇λ B ≤ K (h) R (Aλ B), where h =

M , m

λ ∈ [0, 1], R = max{1 − λ, λ} and Kantorovich constant K (h).

In the following, we generalize Lemma 4.3.5 with the more general sandwich condition 0 < s A ≤ B ≤ t A. Lemma 4.3.6 Let 0 < s A ≤ B ≤ t A for some scalars 0 < s ≤ t and λ ∈ [0, 1]. Then A∇λ B ≤ max{K (s) R , K (t) R }(Aλ B), where R = max{λ, 1 − λ}. Proof. Liao et. al. [129] proved that if x is a positive number and λ ∈ [0, 1], then (1 − λ) + λx ≤ K (x) R x λ . Thus for every strictly positive operator 0 < s I ≤ X ≤ t I , we have (1 − λ) + λX ≤ max K (x) R X λ . s≤x≤t

Substituting A− 2 B A− 2 for C, we get 1

1

(1 − λ) + λ(A− 2 B A− 2 ) ≤ max K (x) R (A− 2 B A− 2 )λ . 1

1

1

1

s≤x≤t

1

Multiplying A 2 to both sides in the above inequalities and using the fact that max K (x) = max{K (s), K (t)},

s≤x≤t

the desired inequality is obtained. We recall that K (h) is decreasing on (0, 1) and increasing on [1, ∞), K (h) = K ( h1 ) and K (h) ≥ 1 for every h > 0 [77].  Remark 4.3.7 We remark that Lemma 4.3.6 is a generalization of Lemma 4.3.5.     Since 0 ≤ m I ≤ A ≤ m I ≤ M I ≤ B ≤ M I or 0 ≤ m I ≤ B ≤ m I ≤ M I ≤ A, m M m then by easy calculation we have M A ≤ B ≤ m A. Now by letting s = M and t = M m in Lemma 4.3.6, Lemma 4.3.5 is obtained. We recall that f is operator monotone decreasing if − f is operator monotone. Lemma 4.3.8 Let g be a nonnegative operator monotone decreasing function on (0, ∞) and A > 0 . Then for every scalar λ ≥ 1 g(λA) ≥

1 g(A). λ

4.3 Operator Aczél Inequality Involving Operator Monotone Functions

101

Proof. First note that since g is analytic on (0, ∞), we may assume that g(x) > 0 for all x > 0. Otherwise, g is identically zero. Also, since g is an operator monotone decreasing on (0, ∞), so f = g1 is an operator monotone on (0, ∞) and hence for every λ ≥ 1 f (λA) ≤ λ f (A), which is equivalent to

(g(λA))−1 ≤ λ(g(A))−1 .

Thus g(λA) ≥

1 g(A).  λ

Ando and Hiai [12] proved that Theorem 4.3.9 Let g be a nonnegative operator monotone decreasing function on (0, ∞) and λ ∈ [0, 1]. Then g(A∇λ B) ≤ g(A)λ g(B). Corollary 4.3.10 Let g be a nonnegative operator monotone decreasing function on (0, ∞) and 0 < s A ≤ B ≤ t A for some constants 0 < s ≤ t. Then for all λ ∈ [0, 1] g(Aλ B) ≤ max{K (s) R , K (t) R }(g(A)λ g(B)), where R = max{λ, 1 − λ}. Proof. Put λ = max{K (s) R , K (t) R }. Since 0 < s A ≤ B ≤ t A, from Lemma 4.3.6 we have (4.3.3) A∇λ B ≤ λ(Aλ B). Notice that λ ≥ 1 and g is operator monotone decreasing. We have g(Aλ B) ≤ λg(λ(Aλ B)) (by Lemma 4.3.8) ≤ λg(A∇λ B) (by 4.3.3 and monotone decreasing og g) ≤ λ(g(A)λ g(B)), (by Theorem 4.3.9) and the desired inquality obtained.  Now, we can give the reverse of Aczél operator inequality as follows. In fact, we show upper bounds for the inequalities in Theorem 4.3.2. These reverse inequalities are proved for a nonnegative operator monotone decreasing function g and the condition of operator concavity has been omitted.

102

4 Operator Inequalities Involving Operator Monotone Functions

Theorem 4.3.11 Let g be a nonnegative operator monotone decreasing function on (0, ∞), 1p + q1 = 1 and 0 ≤ s A p ≤ B q ≤ t A p for some constants s, t. Then for all x ∈H g(A p  q1 B q ) ≤ max{K (s) R , K (t) R }(g(A p ) q1 g(B p )), 1 p

(4.3.4) 1 q

g(A p  q1 B q )x, x ≤ max{K (s) R , K (t) R } g(A p )x, x g(B q )x, x ,

(4.3.5)

where R = max{ 1p , q1 }. Proof. Letting λ = q1 and replacing A p and B q with A and B in Corollary 4.3.10, we reach the inequality (4.3.4). To prove the inequality (4.3.5), first note that under the condition 0 ≤ s A p ≤ B q ≤ t A p from Lemma 4.3.6, we have A p ∇λ B q ≤ max{K (s) R , K (t) R }(A p λ B q ). For convenience, set λ = max{K (s) R , K (t) R }. So, for the operator monotone decreasing function g (4.3.6) g(A p ∇λ B q ) ≥ g(λ(A p λ B q )). Now compute

g(A p  q1 B q )x, x ≤ λg(λ(A p  q1 B q ))x, x

( by Lemma 4.3.8)

= λ g(λ(A p  q1 B q ))x, x ≤ λ g(A p ∇ q1 B q )x, x ≤ λ g(A p ) q1 g(B q )x, x 1 p

( by (4.3.6)) ( by Theorem 4.3.9) 1

≤ λ g(A p )x, x g(B q )x, x q , where the last inequality follows from Lemma 4.2.4 that for every positive operators A, B and every x ∈ H

AλBx, x ≤ Ax, x 1−λ Bx, x λ . So we get 1

1

g(A p  q1 B q )x, x ≤ max{K (s) R , K (t) R } g(A p )x, x p g(B q )x, x q as desired. 

If A and B be commuting operators, the inequality (4.3.4) in Theorem 4.3.11 deduce the following corollary.

4.3 Operator Aczél Inequality Involving Operator Monotone Functions

103

Corollary 4.3.12 Let g be a nonnegative operator monotone decreasing function on (0, ∞), 1p + q1 = 1 and A, B be commuting positive operators such that 0 ≤ s A p ≤ B q ≤ t A p for some constants s, t. Then 1

1

g(AB) ≤ max{K (s) R , K (t) R }g(A p ) p g(B q ) q , where R = max{ 1p , q1 }. Notes and references. M. S. Moslehian proved Lemma 4.3.1, Theorem 4.3.2, Corollaries 4.3.3 and 4.3.4 in [172]. Theorem 4.3.6, Lemma 4.3.8, Corollary 4.3.10, Theorem 4.3.11, and Corollary 4.3.12 are due to [114].

4.4 Norm Inequalities Involving Operator Monotone Functions As we know, the celebrated Kantorovich constant [77] is defined by K (h) = (1+h) 4h for h > 0. We recall that K (h) is decreasing on (0, 1) and increasing on [1, ∞), K (h) = K ( h1 ) and K (h) ≥ 1. 2

Remark 4.4.1 Notice that Lemma 4.3.6 with λ =

1 2

gives the following inequality

A+B 1 ≤ K (h) 2 AB. 2

(4.4.1)

In the following, we get an analogue of the geometric concavity property f (A)λ f (B) ≤ f (Aλ B) for operator monotone functions involving the celebrated Kantorovich constant. Theorem 4.4.2 Let f : [0, ∞) → [0, ∞) be an operator monotone function and 0 < s A ≤ B ≤ t A, s, t > 0. Then for all λ ∈ [0, 1] f (A)λ f (B) ≤ max{K (s) R , K (t) R } f (Aλ B), where R = max{λ, 1 − λ}. Proof. First note that since f is analytic on (0, ∞), we may assume that f (x) > 0 for all x > 0. Otherwise, f is identically zero. Also, since f is an operator monotone function on [0, ∞), so it is operator concave [21]. That is (1 − λ) f (A) + λ f (B) ≤ f ((1 − λ)A + λB).

(4.4.2)

104

4 Operator Inequalities Involving Operator Monotone Functions

We notice that max{K (s) R , K (t) R } ≥ 1. Now compute f (A)λ f (B) ≤ (1 − λ) f (A) + λ f (B) (by Young inequality) ≤ f ((1 − λ)A + λB) (by (4.4.2))   ≤ f max{K (s) R , K (t) R }(Aλ B) (by Lemma 4.3.6 and monotonicity of f ) ≤ max{K (s) R , K (t) R } f (Aλ B). Note that the last inequality follows from the fact that for every nonnegative concave function f and every z ≥ 1, f (zx) ≤ z f (x).  As an immediate result we have the following corollary. Corollary 4.4.3 Let f : [0, ∞) → [0, ∞) be an operator monotone function and 0 < m I ≤ A, B ≤ M I . Then for all λ ∈ [0, 1] f (A)λ f (B) ≤ K (h) R f (Aλ B), where h =

M . m

In the remaining part of this section, we let H be a finite-dimensional Hilbert space. To get a norm inequality for convex operator function, the following Lemmas are needed. Lemma 4.4.4 ([39]) Let A, B ≥ 0 and g : [0, ∞) → [0, ∞) be a convex function with g(0) = 0. Then for every unitarily invariant norm . u g(A) + g(B) u ≤ g(A + B) u . Lemma 4.4.5 ([40]) Let f : [0, ∞) → [0, ∞) be a concave function. Let A ≥ 0 and let Z be expansive. Then for every unital invariant norm . u      f (Z ∗ AZ ) ≤  Z ∗ f (A)Z  . u u A norm inequality for an operator convex function g involving Specht’s ratio is obtained as follows. Theorem 4.4.6 ([85]) Let g : [0, ∞) → [0, ∞) be an operator convex function with g(0) = 0 and let 0 < m I ≤ A, B ≤ M I . Then for every unitarily invariant norm . u    g(AB)  g(A)g(B) u 2  ,  ≤ 2S (h)  AB u AB u where h =

M . m

4.4 Norm Inequalities Involving Operator Monotone Functions

105

Proof. Since g(t) is operator convex with g(0) = 0, then f (t) = g(t) is an operator t monotone function on (0, ∞) [21]. Thus, according to Lemma 4.1.3 we have f (A + B) ≤ f (2S(h)AB) , which is equivalent to

g (2S(h)AB) g(A + B) ≤ . A+B 2S(h)AB

(4.4.3)

Now compute 2 g(A)g(B) u A + B u g(A) + g(B) u ≤ (by Young inequality) A + B u g(A + B) u (by Lemma 4.4.4) ≤ A + B u    g(A + B)   ≤  A + B  (by submultiplicativity of unitarily invariant norm) u    g (2S(h)AB)   ≤  2S(h)AB  (by (4.4.3)) u = f (2S(h)AB) u   1 1  ≤ S(h) 2 f (2(AB)) S(h) 2  (by Lemma 4.4.5)  u  1 1  ≤ 2S(h) 2 f (AB) S(h) 2  u

= 2S(h) f (AB) u    g(AB)   .  = 2S(h)  AB  u

So we get

   g(AB)  g(A)g(B) u  ≤ S(h)   AB  . A + B u u

Thus, we have g(A)g(B) u AB u g(A)g(B) u ≤ 2S(h) (by Lemma 4.1.3) A + B u    g(AB)   ≤ 2S(h)2   AB  , (by (4.4.4)) u

(4.4.4)

106

4 Operator Inequalities Involving Operator Monotone Functions

and this completes the proof.



Next, we give a new sharp inequality of Theorem 4.4.6. Theorem 4.4.7 Let g : [0, ∞) → [0, ∞) be an operator convex function with g(0) = 0 and let 0 < m I ≤ A, B ≤ M I . Then for every unitarily invariant norm . u    g(AB)  g(A)g(B) u  ,  ≤ 2K (h)  AB u AB u where h =

M . m

Proof. Since g(t) is operator convex with g(0) = 0, then f (t) = g(t) is an operator t monotone function on (0, ∞) [21]. Thus, according to (4.4.1) we have   1 f (A + B) ≤ f 2K (h) 2 AB , which is equivalent to   1 g 2K (h) 2 AB g(A + B) ≤ . 1 A+B 2K (h) 2 AB

(4.4.5)

Now compute 2 g(A)g(B) u A + B u g(A) + g(B) u ≤ (by Young inequality) A + B u g(A + B) u (by Lemma 4.4.4) ≤ A + B u    g(A + B)   ≤  A + B  (by submultiplicativity of unitarily invariant norm) u     g 2K (h) 21 AB     (by (4.4.5)) ≤ 1    2K (h) 2 AB     u 1   =  f 2K (h) 2 AB  u   1 1  4 ≤ K (h) f (2(AB)) K (h) 4  (by Lemma 4.4.5)  u  1 1  ≤ 2K (h) 4 f (AB) K (h) 4  u

1 2

= 2K (h) f (AB) u

4.4 Norm Inequalities Involving Operator Monotone Functions

107

  1  g(AB)   = 2K (h) 2   AB  . u

So we get

  g(A)g(B) u 1  g(AB)   .  ≤ K (h) 2  A + B u AB u

(4.4.6)

Thus, we have g(A)g(B) u AB u 1 g(A)g(B) u ≤ 2K (h) 2 (by (4.4.1)) A + B u    g(AB)   ≤ 2K (h)   AB  , (by (4.4.6)) u

and this completes the proof.



Remark 4.4.8 Lin [138] proved that S(h) ≤ K (h) ≤ S 2 (h). That is, Theorem 4.4.7 implies Theorem 4.4.6. In the following, we present a norm inequality for operator monotone functions with the more general sandwich condition involving the λ-weighted operator geometric mean. Theorem 4.4.9 Let f : [0, ∞) → [0, ∞) be an operator monotone function and let 0 < s A ≤ B ≤ t A. Then for every unitarily invariant norm . u    f (Aλ B)  f (A)λ f (B) u  ≤ max{K (s) R , K (t) R }   A B  , Aλ B u λ u where R = max{λ, 1 − λ}. Proof. Compute f (A)λ f (B) u A B   λ u R max{K (s) , K (t) R } f (Aλ B) u ≤ (by Theorem 4.4.2) Aλ B u    max{K (s) R , K (t) R } f (Aλ B)   (by submultiplicativity property) ≤   Aλ B u    f (A B) λ  = max{K (s) R , K (t) R }   A B  .  λ u

108

4 Operator Inequalities Involving Operator Monotone Functions

Corollary 4.4.10 Let f : [0, ∞) → [0, ∞) be an operator monotone function and let 0 < m I ≤ A, B ≤ M I . Then for every unitarily invariant norm . u     f (A)λ f (B) u R  f (Aλ B)  ≤ K (h)  , Aλ B u Aλ B u where h =

M m

and R = max{λ, 1 − λ}.

Remark 4.4.11 If λ =

1 2

in Corollary 4.4.10, then

     f (AB)  f (A) f (B) u 1  f (AB)    ,   2 ≤ K (h)  ≤ S(h)  AB u AB u AB u by the fact that K (h) ≤ S 2 (h). Notes and references. All the results in this section are due to [87].

Chapter 5

Inequalities for Sector Matrices

More recent studies on inequalities are devoted to the study of matrices with the numerical range in a sector of the complex plane. In particular, this includes the study of accretive-dissipative matrices and positive definite matrices as special cases. Accretivedissipative matrices have attracted many researchers because of their rich applications in various areas of mathematics and related fields. Dissipative operators have found many applications in energy-conserving norms for the solution of hyperbolic systems of partial differential equations, where the term dissipative is used in the sense that energy is nonincreasing in time. Moreover, a matrix whose numerical range is contained in a sector region is called a sector matrix, and we extended some known inequalities to this new class of matrices. The readers can follow some new inequalities to extend them to this new class of matrices based on the ideas introduced in this chapter.

5.1 Haynsworth and Hartfiel Type Determinantal Inequality For A ∈ Mn , recall the Cartesian decomposition A = (A) + i(A) [21], where (A) =

A + A∗ , 2

(A) =

A − A∗ . 2i

The numerical range of A ∈ Mn , is defined by   W (A) = x ∗ Ax|x ∈ Cn , x ∗ x = 1 . For α ∈ [0, π2 ), let Sα be the sector in the complex plane given by

© Springer Nature Switzerland AG 2021 M. B. Ghaemi et al., Advances in Matrix Inequalities, Springer Optimization and Its Applications 176, https://doi.org/10.1007/978-3-030-76047-2_5

109

110

5 Inequalities for Sector Matrices

Sα = {z ∈ C|(z) > 0, |(z)| ≤ (z)(tan α)}   = r eiθ |r > 0, |θ| ≤ α . Clearly, if W (A) ⊂ S0 , then A is positive definite. As 0 ∈ Sα , if W (A) ⊂ Sα , then A is necessarily nonsingular. We consider A ∈ Mn to be partitioned as  A=

 A11 A12 , A21 A22

(5.1.1)

where diagonal blocks are square matrices. If A is nonsingular, then we partition A−1 conformally as A. If A11 is nonsingular, then we know that the Schur complement of A11 in A is defined by A = A22 − A21 A−1 11 A12 . A11 The term “Schur complement” and the notation were first brought in by Haynsworth. We refer the readers to [215] for a survey of this important notion and its far reaching applications in various branches of mathematics. Let A, B ∈ Mn be positive definite. It is well known that [103] det(A + B) ≥ det A + det B.

(5.1.2)

Haynsworth proved the following refinement of (5.1.2). Theorem 5.1.1 ([96]) Suppose A, B ∈ Mn are positive definite. Let Ak and Bk , k = 1, ..., n − 1, denote the k-th principal submatrices of A and B, respectively. Then   n−1  det Bk det(A + B) ≥ 1 + det A det Ak k=1   n−1  det Ak det B. (5.1.3) + 1+ det Bk k=1 Hartfiel obtained an improvement of Theorem 5.1.1 under the same condition as follows. Theorem 5.1.2 ([94]) Suppose A, B ∈ Mn are positive definite. Let Ak and Bk , k = 1, ..., n − 1, denote the k-th principal submatrices of A and B, respectively. Then

5.1 Haynsworth and Hartfiel Type Determinantal Inequality

111



 n−1  det Bk det(A + B) ≥ 1 + det A det Ak k=1   n−1  √ det Ak det B + (2n − 2n) det AB. + 1+ det Bk k=1

(5.1.4)

We recall that if A, B ∈ Mn be positive definite and be conformally partitioned as in (5.1.1), then [96] A+B A B ≥ + . (5.1.5) A11 + B11 A11 B11 Our purpose is to extend (5.1.5), then as an application, we obtain a generalization of Theorem 5.1.2 and so Theorem 5.1.1. We first present a few auxiliary lemmas. Lemma 5.1.3 Let A ∈ Mn be partitioned as in (5.1.1). If W (A) ⊂ Sα , then W ( AA11 ) ⊂ Sα . Proof. Clearly, if W (A) ⊂ Sα , then W (A∗ ) ⊂ Sα and W (A22 ) ⊂ Sα . Also, for any nonsingular X ∈ Mn , W (A) = W (X AX ∗ ). Therefore W (A−1 ) = W (A A−1 A∗ ) = W (A∗ ) ⊂ Sα . So W ((A−1 )22 ) ⊂ Sα . The desired result follows by observing that W ((

A −1 ) ) = W ((A−1 )22 ) ⊂ Sα .  A11

Lemma 5.1.4 ([57, 217]) Let A ∈ Mn with W (A) ⊂ Sα . Then A can be decomposed as A = X Z X ∗ for some invertible X ∈ Mn and Z = diag(eiθ1 , ..., eiθn ) with |θ j | ≤ α for all j. Lemma 5.1.5 Let A ∈ Mn with (A) positive definite. Then ((A))−1 ≥ (A−1 ). Proof. By [163], we have

−1 . (A−1 ) = (A) + ((A))((A))−1 ((A)) As ((A))((A))−1 ((A)) is positive semidefinite, so (A) + (A)((A))−1 (A) ≥ (A). Thus ((A))−1 ≥ (A−1 ).



112

5 Inequalities for Sector Matrices

Remark 5.1.6 See another proof of Lemma 5.1.5 in Lemma 5.20.2. Lemma 5.1.7 Let A ∈ Mn be partitioned as in (5.1.1). If (A) is positive definite, then (A) A )≥ ( . A11 (A11 ) (A) Proof. The notation (A makes sense as (A11 ) is the (5.1.1) block of (A). 11 ) Consider the Cartesian decomposition A = M + i N with M = (A) and N = (A) being conformally partitioned as A. Then we have the following equality relating the Schur complements [140]

−1 M N A −1 −1 ∗ = + i( ) + Y M11 − i N11 Y , A11 M11 N11

−1 −1 −1 −1 −1 is positive semidefinite, where Y = M21 M11 − N21 N11 . As R (M11 − i N11 )

−1 −1 −1 ∗ Y . This completes the proof.  so is R Y M11 − i N11 Lemma 5.1.8 Let A ∈ Mn with W (A) ⊂ Sα . Then secn (α) det((A)) ≥ | det A|. Proof. Consider the decomposition A = X Z X ∗ as in Lemma 5.1.4. Then after dividing by | det X |2 , it suffices to show secn (α) det((Z )) ≥ 1. But each diagonal entry of the diagonal matrix sec(α)(Z ) is no less than 1, implying the result.  Now, we can obtain an extension of (5.1.5). First of all, we remark that a direct extension of (5.1.5) is not valid. that is, assuming that A, B ∈ Mn with W (A), W (B) ⊂ Sα are conformally partitioned as in (5.1.1), it does not hold in general that 





A B A+B ≥R +R . R A11 + B11 A11 B11 Theorem 5.1.9 A, B ∈ Mn with W (A), W (B) ⊂ Sα be conformally partitioned as in (5.1.1). Then

sec2 (α)R

A+B A11 + B11



≥R

A A11



+R

B B11

 .

Proof. We consider the decomposition A = X ZX ∗ as in Lemma We further  5.1.4.  X 1 Z X 1∗ X 1 Z X 2∗ X1 , then A = . partition X as a 2-by-1 block matrix X = X2 X 2 Z X 1∗ X 2 Z X 2∗   X1 Let Y = (X ∗ )−1 = be conformally partitioned as X . X2

5.1 Haynsworth and Hartfiel Type Determinantal Inequality

 Then A−1 =

113

 Y1 Z −1 Y1∗ Y1 Z −1 Y2∗ . Clearly, Y2 Z −1 Y1∗ Y2 Z −1 Y2∗ cos2 (α)((Z ))−1 ≤ (Z −1 ),

it follows that

cos2 (α)Y2 ((Z ))−1 Y2∗ ≤ (Y2 Z −1 Y2∗ ).

Thus

cos2 (α) ((A))−1 22 ≤ ((A−1 )22 ),

which is equivalent to

(A) cos (α) (A11 )

−1

2

 A −1 , ≤ ( ) A11

taking the inverse of both sides yields

sec2 (α)

(A) (A11 )



Thus we get

 A −1 −1 ≥  ( ) A11 A ≥ ( ). ( by Lemma 5.1.5) A11

sec2 (α)

(A) (A11 )

 ≥ (

A ). A11

(5.1.6)

To finish the proof, we have



A+B A11 + B11



(A + B) (by Lemma 5.1.7) (A11 + B11 ) (A) (B) ≥ + (by 5.1.5) (A11 ) (B11 ) 



B A ≥ cos2 (α) R +R , (by 5.1.6) A11 B11 ≥

and this completes the proof.



As an application of Theorem 5.1.9, we present the following extension of Theorem 5.1.2. Apparently, the following theorem reduces to Theorem 5.1.2 when α = 0. Theorem 5.1.10 Suppose A, B ∈ Mn such that W (A), W (B) ⊂ Sα . Let Ak and Bk , k = 1, ..., n − 1, denote the k-th principal submatrices of A and B respectively. Then

114

5 Inequalities for Sector Matrices



3n−2

sec

n−1  det Bk (α)| det(A + B)| ≥ 1 + det Ak k=1

 + 1+

n−1  k=1

Proof. Clearly,

Ak+1 +Bk+1 Ak +Bk



det Ak det Bk

| det A| 

(5.1.7) √ | det B| + (2n − 2n) det AB.

∈ C, so

 

  Ak+1 + Bk+1    ≥ R Ak+1 + Bk+1 , k = 1, ..., n − 1.  A +B  Ak + Bk k k Now we put An = A and Bn = B. By Lemma 5.1.3, W ( AAk+1 ), W ( BBk+1 ) ⊂ Sα , then k k by Theorem 5.1.9 and Lemma 5.1.8

sec (α)R 2

Ak+1 + Bk+1 Ak + Bk





 Ak+1 Bk+1 ≥R +R Ak B   k 

  Ak+1   Bk+1  +  . ≥ cos(α)  A   B  k

k

       Ak+1 + Bk+1   Ak+1   Bk+1      ,  sec (α)  ≥ + Ak + Bk   Ak   Bk 

Hence

3

that is

       det(Ak+1 + Bk+1 )   det(Ak+1 )   det(Bk+1 )      ,  ≥ + sec (α)  det(A + B )   det(A )   det(B )  3

k

k

k

k

for k = 1, ..., n − 1. Taking the product for k from 1 to n − 1 yields sec3(n−1) (α)| det(A + B)|    n−1   det(Ak+1 )   det(Bk+1 )   +  ≥ |A1 + B1 |  det(A )   det(B )  . k k k=1 As |A1 + B1 | ≥ cos(α)(|A1 | + |B1 |), we therefore arrive at

5.1 Haynsworth and Hartfiel Type Determinantal Inequality

sec(3n−2) (α)| det(A + B)|    n−1   det(Ak+1 )   det(Bk+1 )      ≥ (|A1 | + |B1 |)  det(A )  +  det(B )  k k k=1    n    det(Ak )   det(Bk )   +  =  det(A )   det(B )  , k−1 k−1 k=1

115

(5.1.8)

where det A0 = det B0 = 1. Lin [136] Proved that for ak , bk > 0, k = 1, ..., n, a0 , b0 = 0, we have 

 n   ak bs bk ≥ an 1 + s + 1n−1 (5.1.9) + ak−1 bk−1 as k=1

   as + bn 1 + s + 1n−1 + (2n − 2n) an bn . bs Now by taking ak = | det Ak |, bk = | det Bk |, k = 0, 1, .., n − 1 in (5.1.9) and according to (5.1.8), the desired inequality is obtained.  We recall that a matrix A ∈ Mn is accretive-dissipative if both (A), (A) are π positive definite [81]. Note that if A is accretive-dissipative, then W (e−i 4 A) ⊂ S π4 . Thus, we have the following corollary. Corollary 5.1.11 Suppose A, B ∈ Mn are accretive-dissipative. Let Ak and Bk , k = 1, ..., n − 1, denote the k-th principal submatrices of A and B respectively. Then 

2

3 2 n−1

n−1  det Bk | det(A + B)| ≥ 1 + det Ak k=1

 + 1+



n−1  det Ak k=1

det Bk

| det A| 

√ | det B| + (2n − 2n) det AB.

Notes and references. We notice that Lemma 5.1.3, Lemma 5.1.5, Lemma 5.1.7, Lemma 5.1.8, Theorem 5.1.9, Theorem 5.1.10, and Corollary 5.1.11 also proved by [136].

116

5 Inequalities for Sector Matrices

5.2 Inequalities with Determinants of Perturbed Positive Matrices Let Mn , Rn , and N+ denote the set of n × n complex matrices, the set of real vectors of dimension n, and the set of positive integers, respectively. For A ∈ Mn , we denote by λ(A) the eigenvalues vector, i.e., λ(A) = (λ1 (A), λ2 (A), ..., λn (A)), where λi (A), i = 1, 2, ..., n, are all the eigenvalues of A. For x = (x1 , x2 , ..., xn ) ∈ Rn , we arrange ↓ ↓ ↓ the entries of x in the nonincreasing order of x1 ≥ x2 ≥ ... ≥ xn . Let x = (x1 , x2 , ..., xn ) ∈ Rn and y = (y1 , y2 , ..., yn ) ∈ Rn . As we know, We say k k   ↓ ↓ that x is weakly majorized by y, denoted by x ≺w y, if xj ≤ y j for all k : 1 ≤ j=1

j=1

k ≤ n. We recall that x is majorized by y, denoted by x ≺ y, if further

n  j=1

xj =

n 

yj.

j=1

In the following, we list some lemmas which are useful for the line of proof in Theorem 5.2.6.   A11 A12 be an n × n Hermitian matrix, where Lemma 5.2.1 ([203]) Let A = A21 A22 A11 , A22 are both square matrices. Then λ(diag(A11 , A22 )) ≺ λ(A), where diag(A11 , A22 ) is a block diagonal matrix with blocks A11 , A22 on the main diagonal. Lemma 5.2.2 ([206]) Let f (t) be a convex function, x = (x1 , x2 , ..., xn ) ∈ Rn and y = (y1 , y2 , ..., yn ) ∈ Rn . If x ≺ y, then ( f (x1 ), f (x2 ), ..., f (xn )) ≺w ( f (y1 ), f (y2 ), ..., f (yn )). Lemma 5.2.3 ([203]) Let A ∈ Mn . If (A) is positive definite, then det((A)) ≤ | det(A)|. 

A11 A12 Lemma 5.2.4 ([130]) Let A = A21 A22 W (A) ⊂ Sα and α ∈ [0, π2 ). Then

 ∈ Mn with A11 ∈ Mm such that m ≤ n2 ,

| det(A)| ≤ sec2m (α)| det(A11 ) det(A22 )|. Notice that the proof of Lemma 5.2.4 is in the next section. The following lemma is basically the content of the proof of Lemma 5.1.3.

5.2 Inequalities with Determinants of Perturbed Positive Matrices

117

 A11 A12 ∈ Mn . If W (A) ⊂ Sα , then W (I + A) ⊂ Sα , A21 A22 W (A11 ) ⊂ Sα , W (A22 ) ⊂ Sα , W (A−1 ) ⊂ Sα and W (X AX ∗ ) ⊂ Sα for any nonsingular X ∈ Mn . 

Lemma 5.2.5 Let A =

Now we are going to show the following first main theorem. Theorem 5.2.6 Assume that k ∈ N+ and that n 1 , ..., n k are positive integers. Assume k k and {Di }i=1 are two sequences of positive matrices such that for each that {Ci }i=1 i ∈ {1, 2, ..., k} the matrices Ci and Di are of format n i × n i . Assume that C is a positive matrix whose diagonal blocks are C1 , C2 , ..., Ck . The following inequality holds det (C + diag(D1 , ..., Dk )) det(C1 + D1 ) det(Ck + Dk ) ≥ ... . det C det C1 det Ck −1

−1

Proof. Note that the diagonal blocks of D − 2 C D − 2 are Di 2 Ci Di 2 , i = 1, 2, ..., k. By Lemma 5.2.1, we have 1

−1

−1

1

−1

−1

λ(diag(D1 2 C1 D1 2 , ..., Dk 2 Ck Dk 2 )) ≺ λ(D − 2 C D − 2 ). 1

1

As the function f (t) = log(1 + 1t ) is convex for t ∈ (0, ∞), it follows from Lemma 5.2.2 that n 

−1

−1

−1

−1

log(1 + λi (diag(D1 2 C1 D1 2 , ..., Dk 2 Ck Dk 2 ))−1 )

i=1



n 

log(1 + λi (D − 2 C D − 2 )−1 ), 1

1

i=1

which is equivalent to −1 −1 −1 −1 T r log(I + λi (diag(D1 2 C1 D1 2 , ..., Dk 2 Ck Dk 2 ))−1 ) 1 1 ≤ T r log(I + λi (D − 2 C D − 2 )−1 ) , by the fact that log det(A) = T r (log A) whenever A is positive definite. We get −1

−1

−1

−1

log det(I + λi (diag(D1 2 C1 D1 2 , ..., Dk 2 Ck Dk 2 ))−1 ) ≤ log det(I + λi (D − 2 C D − 2 )−1 ). 1

1

Now, taking exponential on both hand sides, we have

118

5 Inequalities for Sector Matrices −1

−1

−1

−1

det(I + λi (diag(D1 2 C1 D1 2 , ..., Dk 2 Ck Dk 2 ))−1 ) ≤ det(I + λi (D − 2 C D − 2 )−1 ). 1

1

The desired inequality follows by noting that det(I + D − 2 C D − 2 ) = det(I + DC −1 ) = 1

1

det(C + D) .  det C

In the following, we extend Theorem 5.2.6 to the class of matrices whose numerical ranges are contained in a sector. Theorem 5.2.7 Assume that k ∈ N+ and that n 1 , ..., n k are positive integers. Assume k k and {Di }i=1 are two sequences of positive matrices such that for each i ∈ that {Ci }i=1 {1, 2, ..., k} the matrices Ci and Di are of format n i × n i . Assume that the diagonal blocks of C are C1 , C2 , ..., Ck and D = diag(D1 , ..., Dk ) with W (C), W (D) ⊂ Sα , α ∈ [0, π2 ). The following inequality holds           det (C + diag(D1 , ..., Dk ))    ≥ cos2n (α)  det(C1 + D1 )  ...  det(Ck + Dk )  ,        det C det C1 det Ck where n =

k 

ni .

i=1

Proof. By Lemma 5.2.3 and Lemma 5.1.8, we have    det (C + diag(D1 , ..., Dk ))      det C det((C) + diag((D1 ), ..., (Dk ))) ≥ cosn (α) det (C) det((Ck ) + (Dk )) ) + (D det((C 1 1 )) ... ≥ cosn (α) det (C1 ) det (Ck )        det(C det(C + D ) + Dk )  1 1 k 2n    ≥ cos (α)   ...  . det C1 det Ck The second inequality above holds by Theorem 5.2.6 and the last inequality holds by Lemma 5.2.3 and Lemma 5.1.8 again.  Notice that when α = 0, Theorem 5.2.7 reduces to Theorem 5.2.6. Next, by the similar way as in Theorem 5.2.6 we obtain Theorem 5.2.8 Assume that k ∈ N+ and that n 1 , ..., n k are positive integers. Assume k k and {Di }i=1 are two sequences of positive matrices such that for each that {Ci }i=1 i ∈ {1, 2, ..., k} the matrices Ci and Di are of format n i × n i . Assume that C is a

5.2 Inequalities with Determinants of Perturbed Positive Matrices

119

positive matrix such that the diagonal blocks of C −1 are C1−1 , C2−1 , ..., Ck−1 . The following inequality holds det (C + diag(D1 , ..., Dk )) det(C1 + D1 ) det(Ck + Dk ) ... . ≤ det C det C1 det Ck Now we extend Theorem 5.2.8 to the class of matrices whose numerical range are contained in a sector as follows. Theorem 5.2.9 Assume that k ∈ N+ and that n 1 , ..., n k are positive integers. Assume k k and {Di }i=1 are two sequences of positive matrices such that for each that {Ci }i=1 i ∈ {1, 2, ..., k} the matrices Ci and Di are of the format n i × n i . Assume that the diagonal blocks of C −1 are C1−1 , C2−1 , ..., Ck−1 and D = diag(D1 , ..., Dk ) with W (C), W (D) ⊂ Sα , α ∈ [0, π2 ). The following inequality holds           det (C + diag(D1 , ..., Dk ))    ≤ sec2n (α)  det(C1 + D1 )  ...  det(Ck + Dk )  ,        det C det C1 det Ck where n =

k 

ni .

i=1

Proof. By Lemma 5.2.3 and Lemma 5.1.8, we have    det (C + diag(D1 , ..., Dk ))      det C det((C) + diag((D1 ), ..., (Dk ))) ≤ secn (α) det (C) det((Ck ) + (Dk )) ) + (D det((C 1 1 )) ... ≤ secn (α) det (C1 ) det (Ck )        det(C det(C + D ) + Dk )  1 1  k 2n   ≤ sec (α)   ...  . det C det C 1

k

The second inequality above holds by Theorem 5.2.8 and the last inequality holds by Lemma 5.2.3 and Lemma 5.1.8 again.  Remark 5.2.10 Notice that when α = 0, Theorem 5.2.9 reduces to Theorem 5.2.8. Notes and references. Theorem 5.2.6, Theorem 5.2.7, Theorem 5.2.8 and Theorem 5.2.9 are proved by [213].

120

5 Inequalities for Sector Matrices

5.3 Analogue of Fischer’s Inequality for Sector Matrices Suppose A ∈ Mn be partitioned as a 2 × 2 block matrix  A=

 A11 A12 , A21 A22

(5.3.1)

with A11 ∈ Mm and m ≤ n2 . In connection to the study of the growth factor in Gaussian elimination, researchers [57, 82, 83, 98, 109, 139] considered optimal (smallest) γ > 0 such that | det(A)| ≤ γ| det(A11 ) det(A22 )|. The well-known Fischer inequality [103] states that if A ≥ 0, then det A ≤ (det A11 )(det A22 ).

(5.3.2)

In [109], it was shown that if A is accretive and dissipative, then | det(A)| ≤ γ| det(A11 ) det(A22 )|, with γ = 3m . In [139], the bound was improved to γ=2 and

3m 2

n

γ = 2 2 if

if m ≤

n , 3

n n ≤m≤ . 3 2

As we know, for any α ∈ [0, π2 ) we have Sα = {z ∈ C|(z) > 0, |(z)| ≤ (z)(tan α)} . A subset C is a sector of half-angle α if it is of the form {eiϕ z : z ∈ Sα } for some ϕ ∈ [0, 2π). π ), then In [57], the author proved that if W (A) is a subset of half-angle α ∈ [0, 2m | det(A)| ≤ sec2 (mα)| det(A11 ) det(A22 )|. Suppose the eigenvalue problem  λ

   0 A12 A11 0 x x= A21 0 0 A22

(5.3.3)

5.3 Analogue of Fischer’s Inequality for Sector Matrices

121

for some nonzero x ∈ Cn . We will provide the optimal eigenvalue containment region for those λ satisfying (5.3.3), and the optimal eigenvalue containment region of the matrix −1 A−1 11 A12 A22 A21 ,

in case A11 and A22 are invertible. For this purpose, we begin with several Lemmas. We easily find that Lemma 5.3.1 For any φ ∈ [− π2 , π2 ], the function f : (0, π2 ) → R defined by f (θ) =

(cos(2φ) − cos(2θ)) sin2 (2θ)

is increasing. Lemma 5.3.2 Suppose z 1 = r1 eiθ1 ∈ C, z 2 = r2 eiθ2 ∈ C with r1 , r2 ∈ (0, ∞), 2 = eiψ . Let 2θ = θ1 − θ2 and 2φ = θ1 + θ2 − 2ψ. − π2 < θ2 ≤ θ1 < π2 and z1 +z 2 Then 2(cos(2φ) − cos(2θ)) . r1 r2 = sin2 (2θ) Proof. Consider the triangle T with vertices 0, z 1 , z 2 . Because eiψ is the midpoint of the side joining the vertices z 1 and z 2 , T can be divided into two triangles T1 and T2 with equal areas, where T1 has vertices 0, eiψ , r1 eiθ1 , and T2 has vertices 0, eiψ , r2 eiθ2 . Thus r1r2 sin(2θ) = r1 sin(θ1 − ψ) + r2 sin(ψ − θ2 ) r1 sin(θ1 − ψ) = r2 sin(ψ − θ2 ). It follows that (r1r2 sin(2θ))2 = (r1 sin(θ1 − ψ) + r2 sin(ψ − θ2 ))2 = 4r1r2 sin(θ1 − ψ) sin(ψ − θ2 ) = 2r1r2 (cos(2φ) − cos(2θ)). Hence r1r2 =

2(cos(2φ)−cos(2θ)) . sin2 (2θ)



We will also use some basic facts about the numerical range [105]. Lemma 5.3.3 Let L ∈ Mn . 1. If U ∈ Mn is unitary, then W (L) = W (U ∗ LU ). 2. The set W (L) is compact and convex. ˆ ⊆ W (L). 3. If Lˆ is a principal submatrix of L, then W ( L)

122

5 Inequalities for Sector Matrices

4. if L = L 1 ⊕ L 2 then W (L) = conv(W (L 1 ) ∪ W (L 2 )), where conv(S) denote the convex hull of the set S. 5. if L is normal, then W (L) is the convex hull of its eigenvalues. 6. If x ∈ Cn is a unit vector such that μ = x ∗ L x is a boundary point of W (L) with ¯ more than one support line, then L x = μx and L ∗ x = μx.  A11 A12 ∈ Mn with A11 ∈ Mm and m ≤ Theorem 5.3.4 Let A = A21 A22 be a subset of a sector of half-angle α ∈ [0, π2 ). 

n 2

and W (A)

(a) Suppose A11 ⊕ A22 is singular and x ∈ Cn is a nonzero vector in its kernel. Then Ax = 0 and (5.3.3) holds for every λ ∈ C with this nonzero vector x. (b) Suppose A11 and A22 are invertible. If λ ∈ C satisfies (5.3.3), then for α > 0 λ2 = {1 − r ei2φ : 0 ≤ r ≤

2(cos(2φ) − cos(2α)) , −α ≤ φ ≤ α} ∈ R, sin2 (2α)

and λ2 = [0, 1] ∈ R

if

α = 0.

−1 Moreover, for every eigenvalue μ of the matrix A−1 11 A12 A22 A21 , there is λ ∈ C 2 satisfying (5.3.3) so that μ = λ lies in the region R.

Proof. Without loss of generality, we may assume that W (A) ⊆ Sα . Let  B1 =

A11 0 0 A22



 and

B2 =

 0 A12 . A21 0

(a) Suppose B1 is singular and x ∈ Cn is nonzero such that B1 x = 0. By Lemma x , 5.3.3, for y = x 0 = y ∗ B1 y ∈ conv(W (A11 ⊕ A22 )) ⊆ W (A), so that 0 is a boundary point of W (A) with more than one support line. Thus, Ax = 0, and λB1 x = 0 = (A − B1 )x = B2 x for any λ ∈ C. (b) Assume A11 and A22 are invertible. Suppose λB1 x = B2 x for some nonzero unit vector x ∈ Cn . Let ξ1 = x ∗ B1 x and ξ2 = x ∗ B2 x. Then ξ1 ∈ W (B1 ) and ξ2 ∈ W (B2 ). We see that ξ1 + ξ2 ∈ W (B1 + B2 ) = W (A) and λξ1 = ξ2 . Note that A = B1 + B2 = Q ∗ (B1 − B2 )Q with Q = Im ⊕ (−In−m ), and hence W (A) = W (B1 − B2 ) contains X ∗ (B1 − B2 )x = ξ1 − ξ2 . so, ξ1 ± ξ2 ∈ W (A) which is a subset of Sα by our assumption. Observe that ξ1 = 0. Otherwise, by Lemma 5.3.3 0 ∈ W (B1 ) = conv(W (A11 ) ∪ W (A22 )) ⊆ W (A),

5.3 Analogue of Fischer’s Inequality for Sector Matrices

123

so that 0 is a boundary point of W (B1 ) with more than one support line implying that B1 is singular which contradicts our assumption. Without loss of generality, assume that ς(λ) = ς(1 + ξξ21 ) ≥ 0. Let z ± = r± eiθ± = 1 ±

ξ2 =1±λ ξ1

− − with r± ≥ 0, θ = θ+ −θ and φ = θ+ +θ . 2 2 If ξ1 = |ξ1 |eiω , then (ξ1 ± ξ2 ) = ξ1 (1 + ξξ21 ) has arguments θ± + ω ∈ [−α, α] as ξ1 ± 2 ξ2 ∈ W (A) ⊆ Sα . Note also that z1 +z = 1 has argument 0. It follows that −α − ω ≤ 2 θ− ≤ 0 ≤ θ+ ≤ α − ω. So 0 ≤ θ ≤ α. Applying Lemma 5.3.2 with (r1 eiθ1 , r2 eiθ2 ) = (r+ eiθ+ , r− eiθ− ) and ψ = 0. We have

2(cos(2φ) − cos(2θ)) sin2 (2θ) 2(cos(2φ) − cos(2α)) , ≤ sin2 (2α)

r+ r− =

where the inequality follows from Lemma 5.3.1. As a result 1 − λ2 = z + z − = r+r− ei(θ+ +θ− ) = r+r− ei2φ , lies in the region   2(cos(2φ) − cos(2α)) i2φ ˜ , −α ≤ φ ≤ α . R = re : 0 ≤ r ≤ sin2 (2α) Suppose B = B1−1 B2 . Then  B2 =

 −1 0 A−1 11 A12 A22 A21 . −1 0 A−1 22 A21 A11 A12

−1 −1 −1 If A−1 11 A12 A22 A21 has eigenvalues μ1 , ..., μm , then A22 A21 A11 A12 has eigenvalues μ1 , ..., μm together with n − m zeros. So, we may assume that B has eigenvalues λ1 , ..., λn such that λ2j = λ2m+ j = μ j for j = 1, ..., m and λl = 0 for l = 2m + 1, ..., n. Note that λ is an eigenvalue of B if and only if λ satisfies (5.3.3) for some nonzero x ∈ Cn . The second assertion of (b) follows. 

Remark 5.3.5 We notice that the containment region in section (b) of Theorem 5.3.4 is optimal [136]. Using the notation of Theorem 5.3.4, we see that if A11 and A22 are invertible, −1 i2φ : then all eigenvalues of the matrix C = Im − A−1 11 A12 A22 A21 lie in the set {r e

124

0≤r ≤ by

5 Inequalities for Sector Matrices 2(cos(2φ)−cos(2α)) , sin2 (2α)

−α ≤ φ ≤ α}. Thus, the spectral radius of C is bounded 

2(cos(2φ) − cos(2α)) max |φ|≤α sin2 (2α) 2(1 − cos(2α)) = sin2 (2α) 2 = sec (α),



and hence | det(A)| = | det(A11 ) det(A22 ) det(C)| ≤ sec2m (α)| det(A11 ) det(A22 )|. By continuity, one can remove the invertibility assumption on A11 and A22 .   A11 A12 Corollary 5.3.6 Let A = ∈ Mn with A11 ∈ Mm and m ≤ n2 and W (A) A21 A22 be a subset of a sector of half-angle α ∈ [0, π2 ). Then | det(A)| ≤ sec2m (α)| det(A11 ) det(A22 )|.

(5.3.4)

If A11 and A22 are invertible, then the eigenvalues of the matrix C = Im − A−1 11 A12 A−1 22 A21 ∈ Mm lie in the region   2(cos(2φ) − cos(2α)) , −α ≤ φ ≤ α . R˜ = r ei2φ : 0 ≤ r ≤ sin2 (2α) Suppose A is accretive-dissipative. Then W (e− 4 A) ⊆ S π4 . Applying Corollary 5.3.6 with α = π4 , we have the following result.   A11 A12 ∈ Mn be accretive-dissipative with A11 ∈ Corollary 5.3.7 Let A = A21 A22 n Mm such that m ≤ 2 . Then iπ

| det(A)| ≤ 2m | det(A11 ) det(A22 )|.

(5.3.5)

If A11 and A22 are invertible, then the eigenvalues of the matrix C = Im − A−1 11 A12 A−1 22 A21 ∈ Mm lie in the disk {z ∈ C : |z − 1| ≤ 1} . The author in [139] proposed the Corollary 5.3.6 and Corollary 5.3.7 as conjectures.

5.3 Analogue of Fischer’s Inequality for Sector Matrices

125

Notes and references. Lemma 5.3.2, Theorem 5.3.4, Corollary 5.3.6 and Corollary 5.3.7 are due to [130].

5.4 Analogues of Hadamard and Minkowski Inequality for Sector Matrices We recall that if W (A), W (B) ⊂ Sα , for some α ∈ [0, π2 ), then W (A + B) ⊂ Sα , A is nonsingular and (A) is positive definite. Moreover, W (A) ⊂ Sα implies that W (X ∗ A) ⊂ Sα for any nonzero n × m matrix X , and so W (A−1 ) ⊂ Sα . In the sequel, we always assume that α ∈ [0, π2 ). Sector matrices are a class of matrices whose numerical ranges are contained in Sα for some fixed α. The famous Hadamard’s inequality [103] states that if A = [ai j ] ∈ Mn is positive semidefinite, then (5.4.1) det A ≤ a11 a22 ...ann . There are many generalizations of (5.4.1) which is a consequence of Fischer’s inequality. Using a result of Lin [140], Choi [47] recently proved reverse Fischer’s type inequality as follows. Let Ai ∈ Mn , i = 1, 2, ..., n, be positive definite whose diagonal blocks are n j ( j) square matrices Ai , j = 1, ..., k and so n 1 + n 2 + ... + n k = n. Then det

 m  i=1

 Ai−1



  m  m   (k) (1) −1 −1 (Ai ) (Ai ) ≥ det ... det . i=1

(5.4.2)

i=1

Let A, B ∈ Mn be positive semidefinite. It is well known that det(A + B) ≥ det A + det B.

(5.4.3)

There are various extensions of (5.4.3) in the literature which is a consequence of the Minkowski inequality [103] 1

1

1

(det(A + B)) n ≥ (det A) n + (det B) n .

(5.4.4)

Another attractive extension of (5.4.3) is the following inequality [216] det(A + B + C) + det C ≥ det(A + C) + det(B + C),

(5.4.5)

for the n × n positive semidefinite matrix C. Setting C = 0, reduces to (5.4.3).

126

5 Inequalities for Sector Matrices

Recently, Lin [145] revealed one more connection between (5.4.3) and (5.4.5) as follows det(A + B + C) + det A + det B + det C ≥ det(A + B) + det(A + C) + det(B + C).

(5.4.6)

In the following, we present some analogues of the inequalities (5.4.1), (5.4.2) to (5.4.6) for sector matrices. Lemma 5.4.1 ([56]) If X ∈ Mn with W (X ) ⊂ Sα . Then sec2 (α)(X −1 ) ≥ ((X ))−1 . The following theorem is the analogue of Hadamard’s inequality (5.4.1) for sector matrices. Theorem 5.4.2 Let A = [ai j ] ∈ Mn with W (A) ⊂ Sα . Then | det A| ≤ secn (α)|a11 |.|a22 |...|ann |. Proof. Since W (A) ⊂ Sα , we see that (A) is positive definite. We have | det A| ≤ secn (α) det (A)

(by Lemma 5.1.8)

≤ sec (α)(a11 ).(a22 )...(ann ) (by 5.4.1) ≤ secn (α)|a11 |.|a22 |...|ann |. (by Lemma 5.2.3)  n

Remark 5.4.3 If α = 0 in Theorem 5.4.2, we get Hadamard’s inequality (5.4.1). We note that Corollary 5.3.6 is an analogue of Fischer’s inequality for sector matrices. If α = 0, Corollary 5.3.6 reduces to Fischer’s inequality (5.3.2). Theorem 5.4.4 Let Ai ∈ Mn , i = 1, 2, ..., n, be positive definite whose diagonal ( j) blocks are n j - square matrices Ai , j = 1, ..., k and so n 1 + n 2 + ... + n k = n. If W (Ai ) ⊂ Sα , i = 1, 2..., m, then   m   m    m        (1)  (k)   −1  3n −1 −1  Ai  ≥ cos (α) det (Ai ) (Ai ) ... det . det     i=1

i=1

i=1

Proof. Since W (Ai ) ⊂ Sα , i = 1, 2..., m, we see that (Ai ), i = 1, 2..., m, are positive definite. We have

5.4 Analogues of Hadamard and Minkowski Inequality for Sector Matrices

  m  m        −1  −1 Ai  ≥ det  Ai det   i=1 i=1   m  (Ai−1 ) = det

(by Lemma 5.2.3)

i=1



m  ≥ det cos (α) ((Ai ))−1





= cos (α) det 2n

127

i=1  m 

(by Lemma 5.4.1) 

((Ai ))

−1

.

(5.4.7)

i=1

Then det

 m 

 ((Ai ))−1

i=1

≥ det ≥ det

 m  i=1  m 

 ((Ai(1) ))−1

 m   (k) −1 ((Ai )) ... det

 ((Ai(1) )−1 )

... det

i=1  m 

i=1

 = det 

m 

(by 5.4.2)

 ((Ai(k) )−1 )

(by Lemma 5.1.5)

i=1

 (Ai(1) )−1

 ... det 

i=1

m 

 (Ai(1) )−1

i=1

  m   m     (1)  (1) n 1 +...+n k  −1 −1  ≥ (cos(α)) (Ai ) (Ai ) ... det  (by Lemma 5.1.8) det   i=1 i=1   m   m     (1)  (1)   (5.4.8) = cosn (α) det (Ai )−1 ... det (Ai )−1  .   i=1

i=1

Combining (5.4.7) and (5.4.8) deduce the desired inequality.



The following theorem is an analogue of the Minkowski inequality (5.4.4) for sector matrices. Theorem 5.4.5 Let A, B ∈ Mn with W (A), W (B) ⊂ Sα . Then 1

1

1

| det(A + B)| n ≥ cos(α)(| det A| n + | det B| n ). Proof. Since W (A), W (B) ⊂ Sα , we see that (A), (B) and (A + B) are positive definite. We have

128

5 Inequalities for Sector Matrices 1

1

| det(A + B)| n ≥ (det (A + B)) n (by Lemma 5.2.3) 1

1

≥ (det (A)) n + (det (B)) n (by 5.4.4) 1

1

≥ ((cosn (α)| det A|) n + ((cosn (α)| det B|) n (by Lemma 5.1.8) 1

1

= cos(α)(| det A| n + | det B| n ).  Notice that taking n’th power on both sides of Theorem 5.4.5, we have the following analogue of the inequality (5.4.3) for sector matrices. Corollary 5.4.6 Let A, B ∈ Mn with W (A), W (B) ⊂ Sα . Then | det(A + B)| ≥ cosn (α)(| det A| + | det B|). Similarly, we can get the following analogue of the determinantal inequalities (5.4.5) and (5.4.6), respectively. Theorem 5.4.7 Let A, B, C ∈ Mn with W (A), W (B), W (C) ⊂ Sα . Then | det(A + B + C)| + | det C| ≥ cosn (α)(| det(A + C)| + | det(B + C)|), and | det(A + B + C)| + | det A| + | det B| + | det C| ≥ cosn (α)(| det(A + B)| + | det(A + C)| + | det(B + C)|). Theorem 5.4.8 The function f (A) = log det (A) is a strictly concave function on the set of sector matrices in Mn . Proof. Let A ∈ Mn with W (A) ⊂ Sα . Then (A) > 0. So the assertion follows by the fact that the function f (A) = log det A is a strictly concave function on the set of positive definite matrices [103].  Corollary 5.4.9 Let A, B ∈ Mn with W (A), W (B ⊂ Sα and λ ∈ (0, 1). Then | det(λ A + (1 − λ)B)| ≥ cosn (α)| det A|λ | det B|1−λ . Proof. Since the function f (A) = log det (A) is strictly concave, we have log det(λ(A) + (1 − λ)(B)) ≥ λ log det (A) + (1 − λ) log det (B) = log(det (A))λ + log(det (B))1−λ = log(det (A))λ (det (B))1−λ .

(5.4.9)

5.4 Analogues of Hadamard and Minkowski Inequality for Sector Matrices

129

Taking exponential powers of both sides to the inequality above (5.4.9) yields the following inequality det(λ(A) + (1 − λ)(B)) ≥ (det (A))λ (det (B))1−λ .

(5.4.10)

Now we have | det(λA + (1 − λ)B)| ≥ det (λ A + (1 − λ)B) (by Lemma 5.2.3) = det(λ(A) + (1 − λ)(B)) ≥ (det (A))λ (det (B))1−λ (by 5.4.10) ≥ ((cosn (α)| det A|)λ ((cosn (α)| det B|)1−λ (by Lemma 5.1.8) = cosn (α)| det A|λ | det B|1−λ , as claimed.



In particular, when λ = 21 , we have the arithmetic-geometric mean determinantal inequality for sector matrices as follows. Corollary 5.4.10 Let A, B ∈ Mn with W (A), W (B ⊂ Sα . Then     det A + B  ≥ cosn (α)| det(AB)| 21 .   2

Notes and references. Theorem 5.4.2, Theorem 5.4.4, Theorem 5.4.5, Theorem 5.4.8, Corollary 5.4.6, Corollary 5.4.9, and Corollary 5.4.10 are due to [157].

5.5 Generalizations of the Brunn Minkowski Inequality The famous matrix form of the Burn Minkowski inequality stated in (5.4.4). In the last 50s, Ky Fan [61] gave a generalization of (5.4.4) in the following way. Theorem 5.5.1 Let A, B ∈ Mn be positive definite and let Ak , Bk denote the k-th leading principal submatrix of A, B, respectively. Then for each k = 1, ..., n,

det(A + B) det(Ak + Bk )

1  n−k



det A det Ak

1  n−k

+

det B det Bk

1  n−k

.

(5.5.1)

130

5 Inequalities for Sector Matrices

In [205], Yuan and Leng gave an extension for the inequality (5.5.1). They proved the following matrix form of the Brunn Minkowski inequality. Theorem 5.5.2 Let A, B ∈ Mn be positive definite and let Ak , Bk denote the k-th leading principal submatrix of A, B, respectively. If a and b are two nonnegative real numbers such that A > a In and B > bIn , then

1  n−k det(A + B) − det((a + b)In−k ) det(Ak + Bk ) 1 1

 n−k

 n−k det A det B ≥ − det(a In−k ) + − det(bIn−k ) . det Ak det Bk

Recently, Liu [158] obtained generalizations of Theorem 5.5.1 and 5.5.2 to a larger class of matrices, namely, matrices whose numerical range is contained in a sector as follows. Theorem 5.5.3 Let A, B ∈ Mn with W (A), W (B) ⊂ Sα and let Ak , Bk denote the k-th leading principal submatrix of A, B, respectively. Then   1  1  1     det A  n−k  det B  n−k  det(A + B)  n−k n+k       n−k .  det A  +  det B   det(A + B )  ≥ (cos α) k k k k Theorem 5.5.4 Let A, B ∈ Mn with W (A), W (B) ⊂ Sα and let Ak , Bk denote the k-th leading principal submatrix of A, B, respectively. If a and b are two nonnegative real numbers such that A > a In and B > bIn , then 1 

  n−k  det(A + B)  k   (sec α)  det(A + B )  − (cos α) det((a + b)In−k ) k k    1  1



  det A  det(a In k ) n−k  det B  det(bIn k ) n−k     − − ≥  +  . det Ak  (cos α)n det Bk  (cos α)n n+k n−k

The goal of this section is to improve and complement the previous two theorems. The next theorem provides a considerable improvement of theorem 5.5.3 as far as the coefficient is concerned. Theorem 5.5.5 Let A, B ∈ Mn with W (A), W (B) ⊂ Sα and let Ak , Bk denote the k-th leading principal submatrix of A, B, respectively. Then   1   1  1    det(A + B)  n−k  det A  n−k  det B  n−k n  ≥ (cos α) n−k   +   .  det(A + B )   det B   det A  k k k k

5.5 Generalizations of the Brunn Minkowski Inequality

131

Proof. Let C = A + B. Clearly, W (C) ⊂ Sα . By Lemma 5.1.3, W ( CCk ) ⊂ Sα , in particular,  CCK is positive definite. By Lemma 5.2.3, 

    det C  ≥ det  C .  C  C K

Moreover, since 



C CK





C Ck

(5.5.2)

K

by Lemma 5.1.7, we have

C det  Ck



 (C) ≥ det . (Ck )

(5.5.3)

Compute  1   det C  n−k    det C  k   1  C  n−k = det Ck  1

 n−k C ≥ det  (by (5.5.2)) Ck 1

 n−k (C) ≥ det (by 5.5.3) (Ck )

 1 det(C) n−k = det(Ck )  1  1

det(A) n−k det(B) n−k ≥ + (by (5.4.4)) det(Ak ) det(Bk )

 1  1

(cos α)n | det A| n−k (cos α)n | det B| n−k ≥ + (by Lemma 5.1.8) det(Ak ) det(Bk )   1  1 

| det A| n−k | det B| n−k n = (cos α) n−k + det(Ak ) det(Bk )   1  1    det A  n−k  det B  n−k n     n−k ≥ (cos α) , (by Lemma 5.2.3)  det A  +  det B  k k and this completes the proof.



Remark 5.5.6 From the proof of previous theorem, we have

132

5 Inequalities for Sector Matrices 1   1  n−k

 det C  n−k   ≥ det(C) .  det C  det(Ck ) k

(5.5.4)

However, in Liu’s proof of Theorem 5.5.3, he did not apply the Schur complement. So he only obtained a weaker result 1  1 

 n−k  det C  n−k  ≥ (cos α)k det(C)  .  det C  det(Ck ) k

Theorem 5.5.7 Let A, B ∈ Mn with W (A), W (B) ⊂ Sα and let Ak , Bk denote the k-th leading principal submatrix of A, B, respectively. Then   1   det(A + B)  n−k  3  det A    det(A + B )  ≥ (cos α)  det A k

k

k

 1   n−k  det B  +   det B

k

 1   n−k  . 

Proof. Let C = A + B. By Lemma 5.1.9, we have

C det  Ck





 A B 2 ≥ det (cos α)  + . Ak Bk

This gives 1 1

 n−k



 n−k C A B det  + ≥ (cos α)2 det  . Ck Ak Bk

Now compute   det C   det C

k

 1  n−k  

  1  C  n−k = det C  k

 1 n−k C (by (5.5.2)) Ck



 1 n−k A B + (by (5.5.5)) ≥ (cos α)2 det  Ak Bk 

 1

 1  n−k n−k A B det  ≥ (cos α)2 + det  (by (5.4.4)) Ak Bk    1    1   B  n−k A  n−k 2  + cos α det ≥ (cos α) cos α det (by Lemma 5.1.8) Ak  Bk     1  1   det A  n−k  det B  n−k   +  = (cos α)3  .   det Ak det Bk 

≥ det 

(5.5.5)

5.5 Generalizations of the Brunn Minkowski Inequality

133

Since 0 < cos α ≤ 1 for α ∈ [0, π2 ] when k ≥ 2n , and we have (cos α)3 ≥ 3 n (cos α) n−k . That is, the inequality in Theorem 5.5.7 is stronger than that in Theand weaker when 1 ≤ k < 2n . orem 5.5.5 when k ≥ 2n 3 3 The next theorem provides a considerable improvement of Theorem 5.5.4. Theorem 5.5.8 Let A, B ∈ Mn with W (A), W (B) ⊂ Sα and let Ak , Bk denote the k-th leading principal submatrix of A, B, respectively. If a and b are two nonnegative real numbers such that A > a In and B > bIn , then 1 

  n−k  det(A + B)    − det((a + b)In−k )  det(A + B )  k k  



  1  1  det A  det(a In k ) n−k  det B  det(bIn k ) n−k  − − ≥  + .  det B  det Ak  (cos α)n (cos α)n k n

(sec α) n−k

Proof. Let C = A + B. By (5.5.4),    det C  det(C)    det C  ≥ det(C ) . k k Now by compute 1 

  n−k  det C    − det((a + b)In−k )  det C  k 1

 n−k det C ≥ − det((a + b)In−k ) det Ck 1 1

 n−k

 n−k det A det B ≥ − det(a In−k ) + − det(a In−k ) det Ak det Bk

 1 | det A| det(a In−k ) n−k n ≥ (cos α) n−k − det Ak (cos α)n

 1 | det B| det(bIn−k ) n−k n + (cos α) n−k − det Bk (cos α)n 

  1  det A  det(a In k ) n−k n   n−k ≥ (cos α)  det A  − (cos α)n k 

  1  det B  det(bIn k ) n−k n   n−k + (cos α) ,  det B  − (cos α)n k

in which the second inequality is by Theorem 5.5.2, the third inequality is by Lemma 5.1.8, and the fourth inequality is by Lemma 5.2.3, respectively. 

134

5 Inequalities for Sector Matrices

Notes and references. The proof of Theorem 5.5.5, Theorem 5.5.7 and Theorem 5.5.8 are taken from [219].

5.6 A Lewent Type Determinantal Inequality The following interesting inequality was discovered first by Grothendieck [89] and reproved many times [131, 184, 193]. | det(I + A + B)| ≤ det(I + |A|) det(I + |B|).

(5.6.1)

A quick application of Bernoulli’s inequality, for contractions A, B and λ ∈ [0, 1], det(I + |A|)λ det(I + B)1−λ ≤ det(I + λ|A|) det(I + (1 − λ)|B|), (5.6.2) λ

det(I − |A|) det(I − |B|)

1−λ

≤ det(I − λ|A|) det(I − (1 − λ)|B|).

If A, B are positive contractions, then using a result of Ky Fan [103] gives det(I + A)λ det(I + |B|)1−λ ≤ det(I + λA + (1 − λ)B), (5.6.3) det(I − A)λ det(I − |B|)1−λ ≤ det(I − λA − (1 − λ)B). Notice that according to (5.6.1), (5.6.3) is stronger than (5.6.2) in the positive case. With this preliminaries, our purpose in this section is to present an analogue of the following inequality [112, 132] for determinantal functional. 1+ 1−

n  i=1 n 

λi xi ≤ λi xi

 n  1 + xi λi i=1

1 − xi

,

i=1

where xi ∈ [0, 1) and

n 

λi = 1, λi ≥ 0 for i = 1, ..., n.

i=1

First, we show a few auxiliary results. Lemma 5.6.1 Let A, B be positive trace class operators with A ≥ B and C be any self-adjoint trace class operator. Then T r (AC)2 ≥ T r (BC)2 .

5.6 A Lewent Type Determinantal Inequality

135

Proof. It is well known that if X, Y, Z are positive trace operators with X ≥ Y , then 1 1 1 1 Z 2 X Z 2 ≥ Z 2 Y Z 2 and so 1

1

1

1

Tr X Z = Tr Z 2 X Z 2 ≥ Tr Z 2 Y Z 2 = Tr Y Z. Thus T r X Z ≥ T r Y Z . We have T r (AC)2 = T r A(C AC) ≥ T r B(C AC) = T r A(C BC) ≥ T r B(C BC) = T r (BC)2 .  Lemma 5.6.2 Let Ai , i = 1, ..., n, be positive contraction and trace class. Then ⎛ ⎜ det ⎜ ⎝

I+ I−

n  i=1 n 

λi Ai λi Ai

⎞ n ⎟  ⎟≤ ⎠ i=1

I + Ai I − Ai

λi

,

i=1

where

n 

λi = 1, λi ≥ 0 for i = 1, ..., n.

i=1

Proof. It suffices to show the case where Ai , i = 1, ..., n, are positive (semi) definite matrices, the general case is via limiting arguments. The desired inequality is equivalent to showing that f (A) = ln(det(I + A)) − ln(det(I − A)) is convex over the set of positive (semi) definite contractions. The following claim is readily verified. Claim. A real-valued function g is convex over the convex set S if and only if ∀x, y ∈ S(x = y), 

x−y , φ(t) = g y + t x − y is convex over t ∈ [0, x − y]. Suppose U = U (t) and V = V (t) are matrices of the same size. We have the known formula d(U V ) dV dU =U + V, dt dt dt

136

5 Inequalities for Sector Matrices

and if U is invertible, then d(ln det U ) dU −1 = Tr U dt dt dU −1 dU −1 = −U −1 U . dt dt Now let X, Y (X = Y ) be positive definite contractions, denote by M= Then

X −Y . X − Y 

 t t Y + tM = X + 1− Y X − Y  X − Y 

is again a positive definite contraction for t ∈ [0, x − y]. Define φ(t) = f (Y + t M) = ln det(I + Y + t M) − ln(det(I − (Y + t M))), for t ∈ [0, x − y]. Then      φ (t) = T r (I + Y + t M)−1 M + T r (I − (Y + t M))−1 M , and  2  2  φ (t) = −T r (I + Y + t M)−1 M + T r (I − (Y + t M))−1 M . 

As (I + Y + t M)−1 ≤ (I − (Y + t M))−1 , by Lemma 5.6.1, we have φ (t) ≥ 0. Then the claim tell us f (A) is convex over the set of positive (semi) definite contractions.  Lemma 5.6.3 φ(t) = (1 − t)−1 is 2-positive over the contractions. Proof. The following identity is due to [151] for contractions A, B. (I − B ∗ B) − (I − B ∗ A)(I − A∗ A)−1 (I − A∗ B) = −(A − B)∗ (I − A A∗ )−1 (A − B).

An application of a result of Schur [103] reveals that 

(I − A∗ A)−1 (I − B ∗ A)−1 (I − A∗ B)−1 (I − B ∗ B)−1

Exchanging the role of A, B and their adjoints gives

 ≥ 0.

5.6 A Lewent Type Determinantal Inequality



137

(I − A A∗ )−1 (I − B A∗ )−1 (I − AB ∗ )−1 (I − B B ∗ )−1

 ≥ 0.

(5.6.4)

If A, B are contractions, so are A∗ B and B ∗ A. With (5.6.4), we have  (I − A∗ A)−1 (I − A∗ B)−1 (I − B ∗ A)−1 (I − B ∗ B)−1  I + A∗ (I − A A∗ )−1 A I + A∗ (I = I + B ∗ (I − AB ∗ )−1 A I + B ∗ (I    ∗ −1 I I ∗ (I − A A ) = + (A ⊕ B) (I − AB ∗ )−1 I I



− B A∗ )−1 B − B B ∗ )−1 B



 (I − B A∗ )−1 (A ⊕ B) ≥ 0. (I − B B ∗ )−1 

 A 0 . According to [8], 0 B  ∗  X X X ∗Y we identify any positive 2 × 2 operator matrix with the form , the Y ∗ X Y ∗Y conclusion follows.  Here we use A ⊕ B to denote the 2 × 2 operator matrix

Corollary 5.6.4 φ(t) = (1 + t)(1 − t)−1 is 2-positive over the contractions. Proof. From the proof of Lemma 5.6.3, we have (1 − t)−1 − 1 is also 2-positive over the contractions. The proof is completed by noting that φ(t) = 2(1 − t)−1 − 1.  Lemma 5.6.5 Let Ai , i = 1, ..., n, be contractive trace class operators. Then for any Liebian function f ,     f    where

n 

⎞ 2 ⎞ ⎛ ⎞ ⎛ n n     λi |Ai | λi |Ai∗ | 1+ 1+ ⎜ ⎟ ⎟ ⎜ ⎟ ⎜ i=1 i=1 i=1 ⎜ ⎟ ≤ f ⎜ ⎟f⎜ ⎟, n n n  ⎝ ⎠ ⎠ ⎠ ⎝ ⎝    ∗  1− λi Ai  1− λi |Ai | 1− λi |Ai | ⎛

1+

n 

λi Ai

i=1

i=1

i=1

λi = 1, λi ≥ 0 for i = 1, ..., n.

i=1

 Proof. An application of polar decomposition reveals ⎡

n 

λi |Ai∗ |

⎢ i=1 ⎢ n ⎣ 

i=1

λi Ai∗

 |Ai∗ | Ai ≥ 0 for any i, so Ai∗ |Ai |

⎤   ∗ λi Ai ⎥  n |Ai | Ai i=1 ⎥ ≥ 0. = λ i n  ⎦ Ai∗ |Ai | λi |Ai | i=1 n 

i=1

The desired inequality follows from Lemma 5.6.4.



138

5 Inequalities for Sector Matrices

Theorem 5.6.6 Let Ai , i = 1, ..., n, be contractive trace class operators. Then  ⎛ ⎞ n     I+ λi Ai   

n   ⎜ ⎟ I + |Ai | λi i=1 det ⎜ ⎟ ≤ det , n  ⎝ ⎠  I − |Ai |  i=1  I − λ A i i   i=1

where

n 

λi = 1, λi ≥ 0 for i = 1, ..., n.

i=1

Proof. Determinantal functional is a Liebian function, so by Lemma 5.6.4 we have  ⎛ ⎞ ⎛ ⎞ ⎞2 ⎛ n n n       1+ λi Ai  λi |Ai | λi |Ai∗ | I+ I+  ⎜ ⎟ ⎜ ⎟ ⎟ ⎜ i=1 i=1 i=1 det ⎜ ⎟ det ⎜ ⎟ ⎟ ≤ det ⎜ n n n  ⎝  ⎠ ⎝ ⎠ ⎠ ⎝    ∗   1− λi Ai  I− λi |Ai | I− λi |Ai |  i=1 i=1 i=1 λi  

n n  I + |Ai∗ | λi I + |Ai | ≤ det det I − |Ai | I − |Ai∗ | i=1 i=1 

n  I + |Ai | 2λi = det , I − |Ai | i=1 in which the second inequality is by Lemma 5.6.2 and the third equality is by the fact det(I + |A|) = det(I + |A∗ |), and the proof is completed.



Notice that Theorem 5.6.6 is essentially equivalent to the following theorem, though the latter looks stronger in the form. Theorem 5.6.7 Let Ai , i = 1, ..., n, be contractive trace class operators. Then ⎞ n   I +  λi Ai  

n ⎟  ⎜ I + |Ai | λi i=1 ⎟ ⎜ ⎠ ≤ n det ⎝ det ,   I − |Ai | i=1 I −  λi Ai  ⎛

i=1

where

n 

λi = 1, λi ≥ 0 for i = 1, ..., n.

i=1

Remark 5.6.8 When Ai in theorem 5.6.6 are moreover self-adjoint, a much more general result has been stated in [2].

5.6 A Lewent Type Determinantal Inequality

139

Notes and references. Lemma 5.6.1, Lemma 5.6.2, Lemma 5.6.3, Lemma 5.6.5, Corollary 5.6.4, and Theorem 5.6.6 are proved in [149].

5.7 Principal Powers of Matrices with Positive Definite Real Part For γ real, the principal power function z → z γ is the map from C \ (−∞, 0] to C defined by z γ = r γ eiγθ where z = r eiθ , r > 0 and −π < θ < π. The definition is extended to a complex matrix T with all its eigenvalue in C \ (−∞, 0] by 1 T = 2πi γ

# Γ

(z I − T )−1 z γ dz,

where Γ is a contour in the resolvent set of T which winds once about each eigenvalue and avoids (−∞, 0]. It is well known that if λ ∈ C \ (−∞, 0] and N is nilpotent, then (λI + N )γ = λγ (I + λ−1 N )γ , where the second factor on the right may be computed using the power series expansion γ(γ − 1) −2 2 λ N + ..., I + γλ−1 N + 2 which eventually terminate since N is nilpotent. Using this fact and Jordan canonical form, one has a way of calculating T γ with T as above. In the following, we are interested in establishing containment results for W (T γ ) in case γ ∈ (0, 1) and W (T ) is contained in the right half-space or in Sα . We note that there are some rather partial results in this direction in [81]. Containment results for the case γ, an integer, and many other interesting results can be found in [45]. We will need the following two integrals. Lemma 5.7.1 Let z ∈ C with (z) > 0 and γ ∈ (0, 1). Then

and

γπ 2 cos π 2

#

γπ 2 sin π 2

#

∞ 0

0



zt −γ dt = z −γ , z2 + t 2 t 1−γ dt = z −γ , + t2

z2

where z −γ denotes the principal powers of z. Proof. We use formula [88]

140

5 Inequalities for Sector Matrices

#

∞ 0

x μ−1 d x = β −μ B(μ, v − μ), (1 + βx)v

where |argβ| < π and (v) > (μ) > 0. The substitutions x = t 2 and β = z −2 , simple manipulations and the formulas B(x, y) = Γ (x)Γ (y)Γ (x + y)−1 , Γ (1) = 1, Γ (x)Γ (1 − x) = π csc(πx), 

to establish the desired result.

We immediately obtain the following corollary. Corollary 5.7.2 Let γ ∈ (0, 1) and T a complex square matrix with positive definite part. Then we have γπ 2 cos π 2 and

#

γπ 2 sin π 2



T (T 2 + t 2 I )−1 t −γ dt = T −γ ,

(5.7.1)

(T 2 + t 2 I )−1 t 1−γ dt = T −γ ,

(5.7.2)

0

#

∞ 0

where T −γ denotes the principal powers of T . Proof. If T is semisimple (diagonalizable), then the result is obvious. But semisimple matrices are dense in the space of all complex square matrices, so the result will follow by continuity provided that the integrand can be controlled. For (5.7.1), we do this for S semisimple and close to T by T (T 2 + t 2 I )−1 − S(S 2 + t 2 I )−1  = (T 2 + t 2 I )−1 (t 2 (T − S) − T (T − S)S)(S 2 + t 2 I )−1  ≤ T − S(T 2 + t 2 I )−1 (S 2 + t 2 I )−1 (t 2 + T S), and there is a similar estimate for (5.7.2). It follows that $# ∞ $ $ $

2 2 −1 2 2 −1 −γ $ $ T (T t + t I ) − S(S + t I ) dt $ $ ≤ CT − S 0

for some suitable constant C.



Theorem 5.7.3 Let γ ∈ (0, 1) and T a complex square matrix with positive definite part. Then T γ has positive definite real part and has its numerical range in S γπ2 .

5.7 Principal Powers of Matrices with Positive Definite Real Part

141

Proof. It will suffice to show that T −γ has its numerical range in S γπ2 . Therefore, we write T −γ = X + iY with X and Y Hermitian. We claim that γπ γπ X − cos Y (5.7.3) sin 2 2 is positive definite. Using (5.7.1) to estimate X and (5.7.2) to estimate Y , we see that it will suffice to show that # ∞

(T (T 2 + t 2 I )−1 ) − t((T 2 + t 2 I )−1 ) t −γ dt 0

is positive definite. In fact, we will show that the integrand is positive definite. A calculation shows that the integrand can be rewritten using

 T (T 2 + t 2 I )−1 ) − t((T 2 + t 2 I )−1



=  (T 2 + t 2 I )−1 T +  it (T 2 + t 2 I )−1

=  (T 2 + t 2 I )−1 (T + it I ) = (T − it I )−1 1 = (T ∗ + it I )−1 (T + T ∗ )(T − it I )−1 , 2 which is positive definite since T has positive definite real part. This completes the proof of the claim. A similar argument (or applying the above to T ∗ instead of T ) yields that sin

γπ 2

X + cos

γπ 2

Y

(5.7.4)

is positive definite. Adding (5.7.3) and (5.7.4), we see that X is positive definite. Now, let ξ be a unit vector and let z = ξ ∗ T −γ ξ. We must show z ∈ S γπ2 . We have (z) = ξ ∗ X ξ > 0 and (z) = ξ ∗ Y ξ. By (5.7.3) and (5.7.4) we have sin and the result follows.

γπ 2

(z) ± cos

γπ 2

(z) ≥ 0,



Corollary 5.7.4 Let α ∈ [0, π2 ), γ ∈ (0, 1) and T be a complex square matrix with W (T ) ⊆ Sα . Then W (T γ ) ⊆ Sαγ . Proof. Let β ∈ (0, π2 − α) and let R has positive definite real part. Applying Theorem 5.7.3 yields that W (R γ ) ⊆ S γπ2 . But R γ = eiβγ T γ , so we can deduce that % & π π W (T γ ) ⊆ z ∈ C; −γ( + β) ≤ arg(z) ≤ γ( − β) . 2 2

142

5 Inequalities for Sector Matrices

Rotating in the opposite direction and combining yields W (T γ ) ⊆ Sγ( π2 −β) . Taking the intersection of these sectors as β → π2 − α from below yields the required result.  As we know, for A and B positive definite matrices of the same size, the geometric mean G = AB is defined as the unique positive definite matrix such that G A−1 G = B. The reader may consult Bhatia [20] for an excellent discussion of this topic. The purpose of the following is to extend this theory to matrices A and B in this context. We define G by # dt 2 ∞ (5.7.5) (t A + t −1 B)−1 . G −1 = π 0 t The inversion t → t −1 preserves the measure t −1 dt, so this definition brings out the well known AB = BA immediately. It is immediate that G has positive definite real part. We also have Proposition 5.7.5 If W (A) ⊆ Sα and W (B) ⊆ Sα , then W (G) ⊆ Sα . Lemma 5.7.6 We have

2 π

# 0



z2

dt = z −1 + t2

(5.7.6)

for (z) > 0. Proof. Let left-hand side of (5.7.6) defines a function analytic in (z) > 0. For z > 0, (5.7.6) is correct. Hence, it is correct for (z) > 0.  Corollary 5.7.7 Let T be a complex square matrix with eigenvalues in (z) > 0. Then # 2 ∞ 2 (T + t 2 I )−1 dt = T −1 . π 0 Proof. This follows immediately from (5.7.6) as in the proof of Corollary 5.7.2.  1

Theorem 5.7.8 Let A and B have positive definite real part. Then G = A 2 1 21 1 1 21 1 1 1 1 A− 2 B A− 2 A 2 , G = B 2 B − 2 AB − 2 B 2 and G A−1 G = B.

Proof. Let λ be an eigenvalue of A− 2 B A− 2 . Then, there is a nonzero vector ξ 1 1 such that A− 2 B A− 2 ξ = λξ. It follows that Bη = λAη, where the nonzero vector 1 η = A− 2 ξ. Therefore, η ∗ Bη = λη ∗ Aη and λ does not lie on (−∞, 0]. Consequently 1 21 1 A− 2 B A− 2 has eigenvalues in (z) > 0. We then have 1

1

5.7 Principal Powers of Matrices with Positive Definite Real Part

143

# 2 ∞ 1 1 A 2 (B + t 2 A)−1 A 2 dt π 0 # −1 2 ∞ −1 1 = A 2 B A− 2 + t 2 I dt π 0 1 1 1 −2 = A− 2 B A− 2 .

A 2 G −1 B A 2 = 1

1

The first conclusion follows from this and the second follows by symmetry. Squar1 1 1 1 1 ing up now gives A 2 G −1 AG −1 A 2 = A 2 B −1 A 2 . Since A 2 is nonsingular, we get G A−1 G = B as required.  Note that a consequence of the above result is that A−1 B −1 = (AB)−1 . Proposition 5.7.9 Let A, B and H have positive definite real part and H A−1 H = B. Then H = G. Proof. We have H −1 AH −1 = B −1 and H −1 B H −1 = A−1 from which we deduce that H −1 (t A + t −1 B)H −1 = t B −1 + t −1 A−1 and hence that H (t A + t −1 B)−1 H = (t B −1 + t −1 A−1 )−1 . Substituting in (5.7.5) gives H G −1 H # 2 ∞ dt = (t B −1 + t −1 A−1 )−1 π 0 t = (B −1 A−1 )−1 = G. We deduce that (H G −1 )2 = I . But using an argument similar to one in the proof of Theorem 5.7.8, we see that H G −1 cannot have a negative eigenvalue. But the only possible eigenvalues are ±1, so 1 is the only eigenvalue. Again using (H G −1 )2 = I , we see that H G −1 = I or equivalently H = G.  Theorem 5.7.10 Suppose γ ∈ (0, 1), that T has positive definite real part and T  ≤ 1. Then we have T γ  ≤



Γ 21 γ Γ 21 (1 − γ) . √ 2 πΓ (γ)Γ (1 − γ)

Proof. For t > 0, we have t 2 I + t (T ∗ + T ) − t 2 T ∗ T ≥ 0,

144

or

5 Inequalities for Sector Matrices

(t I + T )∗ (t I + T ) ≥ (t 2 + 1)T ∗ T.

So

T (t I + T )−1  ≤ (t 2 + 1)− 2 . 1

Now, we have for γ ∈ (0, 1), Tγ =

sin(πγ) π

#



T (t I + T )−1 t −γ dt.

0

It follows that # sin(πγ) ∞ 2 1 (t + 1)− 2 t −γ π 0

Γ 21 γ Γ 21 (1 − γ) = √ , 2 πΓ (γ)Γ (1 − γ)

T  ≤



and this completes the proof.

Notes and references. All the results in this section are proved in [58].

5.8 Geometric Mean of Accretive Operators As we see in the previous section, generalizing the geometric mean, Drury defined a geometric mean for two accretive operators A, B ∈ B(H) via the formula

AB =

2 π

# 0



(t A + t

−1

B)

−1 dt

t

−1

,

(5.8.1)

in which we continue to use the standard notation AB for the geometric mean. This new geometric mean for accretive operators enjoys a lot of similar properties compared to the geometric mean for strictly positive operators. For example, AB = BA, (AB)−1 = A−1 B −1 . The purpose of the following note is to add a new property to the list. To give the main theorem, we need the following Lemma. Consider the function f (A) = ((A−1 ))−1 defined on the accretive part of B(H). Mathias [163] showed that f (A) is operator convex in the sense that f (λA + (1 − λ)B) ≤ λ f (A) + (1 − λ) f (B) for any accretive operators A, B ∈ B(H) and λ ∈ [0, 1].

5.8 Geometric Mean of Accretive Operators

145

As f (t A) = t f (A) for any t ∈ R, Mathias’s result can be equivalently stated in the following way. Lemma 5.8.1 For any accretive A, B ∈ B(H), f (A) + f (B) ≥ f (A + B). Proof. Let A = H1 + i S1 and B = H2 + i S2 be the Cartesian decomposition of A, B respectively. It could be verified that (A−1 ) = (H1 + S1 H1−1 S1 )−1 , so A−1 is again accretive. Using the Schur complement, the following block operator matrices, defined on H ⊕ H, 

S1 H1 S1 S1 H1−1 S1



 ,

 S2 H2 , S2 S2 H2−1 S2

are positive. So is their sum 

 H1 + H2 S1 + S2 , S1 + S2 S1 H1−1 S1 + S2 H2−1 S2

which it turn implies S1 H1−1 S1 + S2 H2−1 S2 ≥ (S1 + S2 )(H1 + H2 )−1 (S1 + S2 ). Now f (A) + f (B) = H1 + S1 H1−1 S1 + H2 + S2 H2−1 S2 ≥ H1 + H2 + (S1 + S2 )(H1 + H2 )−1 (S1 + S2 ) = f (A + B) as desired.



We are ready to present Theorem 5.8.2 Let A, B ∈ B(H) be accretive. Then (AB) ≥ (A)(B).

(5.8.2)

Proof. As A−1 , B −1 are again accretive. Applying Lemma 5.8.1 to t A−1 , t −1 B −1 for any t > 0 gives

−1 . t (A)−1 + t −1 (B)−1 ≥ ((t A−1 + t −1 B −1 )−1 ) It follows by the operator reverse monotonicity of the inverse

146

5 Inequalities for Sector Matrices

−1 ((t A−1 + t −1 B −1 )−1 ) ≥ t (A)−1 + t −1 (B)−1 .

(5.8.3)

Compute (AB)

=  (A−1 B −1 )−1 

# ∞ 2 −1 −1 −1 −1 dt (t A + t B ) = π 0 t # ∞

−1 dt 2  (t A + t −1 B −1 )−1 = π 0 t #

dt 2 ∞ −1 t (A)−1 + t −1 (B)−1 ≥ π 0 t

( by (5.8.3))



−1 = (A)−1 (B)−1 = (A)(B), and this completes the proof.



Recall that the harmonic mean of two accretive operators A, B is naturally defined as

A!B = 2(A−1 + B −1 )−1 .

As we have already observed that the set of accretive operators is closed under inversion, Lemma 5.8.1 is equivalent to f (A−1 ) + f (B −1 ) ≥ f (A−1 + B −1 ). Thus after taking inverses on both sides, we have a new formulation of Lemma 5.8.1 in terms of the harmonic mean. Proposition 5.8.3 Let A, B ∈ B(H) be accretive. Then (A!B) ≥ (A)!(B). The next corollary shows that Theorem 5.8.2 can be extended to more matrices. Corollary 5.8.4 Let Ak , Bk ∈ B(H), k = 1, ..., m, be accretive. Then 

 m  k=1

  Ak 

m  k=1

 Bk



m  (Ak )(Bk ). k=1

(5.8.4)

5.8 Geometric Mean of Accretive Operators

Proof. Clearly,

m 

Ak and

k=1



 m 

m 

147

Bk are again accretive. Compute that

k=1

  Ak 

k=1

m 



 ≥ 

Bk

k=1

m 

  Ak  

k=1

=

 m 

m 

 Bk

k=1

  m   Ak  Bk

k=1

k=1

m  ≥ (Ak )(Bk ), k=1

in which the first inequality is by Theorem 5.8.2 and this completes the proof.



A very useful property for the geometric mean of two strictly positive operators is the so-called extremal property, first proved byPusz and Woronowicz [183]. It says  A AB that if A, B ∈ B(H) are strictly positive, then , defined on H ⊕ H, is AB B positive and   A X ≥ 0 =⇒ AB ≥ X. (5.8.5) X B The situation turns out differently mean of accretive operators.

 for the geometric  A AB Indeed, it is not even true that  ≥ 0 for general accretive A, B ∈ AB B B(H). As this is

 0≤

A AB AB B





 A (AB) = , (AB) B

now (5.8.5) would imply (AB) ≤ (A)(B). Taking into account (5.8.2) one would conclude that (AB) = (A)(B), which is obviously false (for example, take A = B ∗ = z, where z is an arbitrary nonreal complex number with z > 0). One of the main motivations for the geometric mean is the comparison with the arithmetic mean. For strictly positive operators, it is well known that A+B ≥ AB. 2 A numerical example was given in Example 5.20.5 to show that

148

5 Inequalities for Sector Matrices



A+B 2

 ≥ (AB)

(5.8.6)

fails for general accretive A, B ∈ B(H). Here we give a theoretical explanation. Assume otherwise, we let B = A∗ , then (5.8.6) implies (AA∗ ) ≤ A = (A)(A∗ ), again contradicting (5.8.2). In contrast to (5.8.4), we can also show that it is in general not true that  

m 

  m   m    Ak  Bk Ak Bk , ≥

k=1

k=1

(5.8.7)

k=1

where Ak , Bk ∈ B(H), k = 1, ..., m, are accretive. For example, we may let m = 2, A1 = A∗2 = A, B1 = B2∗ = B and substitute into (5.8.7). Then one would get (A)(B) ≥ (AB), contradicting (5.8.2). Notes and references. The proof of Lemma 5.8.1, Theorem 5.8.2, and Corollary 5.8.4 are taken from [144].

5.9 Weighted Geometric Mean of Accretive Operators and Its Applications As we know, every A ∈ B(H) can be written in the so-called Cartesian decomposition form as follows A = A + iA (5.9.1) ∗



and A = A−A . We recall that if A, B ∈ B(H) are strictly positive with A = A+A 2 2i and λ ∈ (0, 1) is a real number, then the following quantities A∇λ B = (1 − λ)A + λB,

A!λ B = (1 − λ)A−1 + λB −1 , 1 1 1 1 λ Aλ B = A 2 A− 2 B A− 2 A 2 ,

(5.9.2)

are known as the λ-weighted arithmetic, λ-weighted harmonic and λ-weighted geometric operator means of A and B, respectively. As we know, we have A!λ B ≤ Aλ B ≤ A∇λ B.

(5.9.3)

5.9 Weighted Geometric Mean of Accretive Operators and Its Applications

149

Now, let A ∈ B(H) be as in (5.9.1). We say that A is accretive if its real part A is strictly positive. If A, B ∈ B(H) are accretive then so are A−1 and B −1 . Further, it is easy to see that the set of all accretive operators acting on H is a convex cone of B(H). Consequently, A∇λ B and A!λ B can be defined by the same formulas as previously whenever A, B ∈ B(H) are accretive. Clearly, the relationships

−1 are also valid foe A∇λ B = B∇1−λ A, A!λ B = B!1−λ A and A!λ B = A−1 ∇λ B −1 any accretive operators A, B ∈ B(H) and λ ∈ (0, 1). However, Aλ B cannot be defined by the same formula (5.9.2) when A, B ∈ B(H) are accretive, by Virtue of the presence of noninteger exponent for operators in (5.9.2). Recall that for the particular case λ = 21 , in section 5.7, we define AB via the following formula

AB =

2 π

# 0



(t A + t −1 B)−1

dt t

−1

.

(5.9.4)

It is proved that AB = BA and AB = (A−1 B −1 )−1 for any accretive operators A, B ∈ B(H). It follows that (5.9.4) is equivalent to # dt 2 ∞ (t A + t −1 B)−1 π 0 t # dt 2 ∞ A(t B + t −1 A)−1 B . = π 0 t

AB =

(5.9.5)

Now, we define Aλ B for the accretive operators A, B ∈ B(H). Definition 5.9.1 Let A, B ∈ B(H) be two accretive operators and let λ ∈ (0, 1). the λ-weighted geometric mean of A and B is defined by #

−1 sin(λπ) ∞ λ−1 −1 t dt A + t B −1 π o # sin(λπ) ∞ λ−1 t A (B + t A)−1 Bdt. = π o

Aλ B =

(5.9.6)

In the aim to justify our previous definition, we first state the following. Proposition 5.9.2 We have (i) If A, B ∈ B(H) are strictly positive, then (5.9.6) coincides with (5.9.2). (ii) If λ = 21 , then (5.9.6) coincides with (5.9.4). Proof. (i) Assume that A, B ∈ B(H) are strictly positive. From (5.9.6), it is easy to see that

# −1  1 sin(λπ) ∞ λ−1 1 1 −1 21 2 2 I +tA B A t dt A 2 . Aλ B = A π o

150

5 Inequalities for Sector Matrices

Since A 2 B −1 A 2 is self-adjoint positive, then it is sufficient by virtue of (5.9.2), to show that the following equality 1

1

a −λ =

sin(λπ) π

#



t λ−1 (1 + ta)−1 dt

o

holds for any real number a > 0. If we make the change of variables u = (1 + ta)−1 , the previous real integral becomes after simple manipulations as follows # sin(λπ) 1 (1 − u)λ−1 u −λ du aλπ o 1 sin(λπ) B(λ, 1 − λ) = λ a π 1 sin(λπ) = λ Γ (λ)Γ (1 − λ) a π 1 = λ, a and the proof of (i) is complete. (ii) Let A, B ∈ B(H) be accretive. If λ = 21 , then (5.9.6) yields AB =

1 π

#



0

dt (A−1 + t B −1 )−1 √ , t

which with the change of variables u =



t, becomes

# 2 ∞ −1 (A + u 2 B −1 )−1 du π 0 # 2 ∞ −1 −1 du = (u A + u B −1 )−1 . π 0 u

AB =

This with (5.9.5) and the fact that AB = BA deduces the desired result.



From a functional point of view, we are allowed to state another equivalent form of (5.9.6), which seems to be more convenient with a view to our objective in the sequel. Lemma 5.9.3 For any accretive operators A, B and λ ∈ (0, 1), there holds Aλ B =

sin(λπ) π

#

1 0

t λ−1 (A!λ B)dt. (1 − t)λ

Proof. If in (5.9.6) we make the change of variables t = the desired result. 

u 1−u

(5.9.7)

for u ∈ [0, 1), we obtain

5.9 Weighted Geometric Mean of Accretive Operators and Its Applications

151

Using the previous lemma, it is not hard to verify that the following formula Aλ B = B1−λ A,

(5.9.8)

persist for any accretive operators A, B and λ ∈ (0, 1). Moreover, it is clear that (A∇λ B) = (A)∇λ (B). About A!λ B, we state the following lemma. Lemma 5.9.4 For any accretive operators A, B and λ ∈ (0, 1), it holds that (A!λ B) = (A)!λ (B).

(5.9.9)

Proof. Let f (A) = ((A−1 ))−1 be defined on the convex cone of accretive operator A. For the operator convex f , we have f ((1 − λ)A + λB) ≤ (1 − λ) f (A) + λ f (B). This means that

−1

−1

−1 ≤ (1 − λ) (A−1 ) + λ (B −1 ) . ((1 − λ)A + λB)−1 Replacing in the latter inequality A and B by the accretive operators A−1 and B −1 , respectively, and using the fact that the map X → X −1 is operator monotone increasing for strictly positive operator X , we then deduce (5.9.9).  We now give a theorem which extends Theorem 5.8.2 as follows. Theorem 5.9.5 Let A, B be accretive operators and λ ∈ (0, 1). Then (Aλ B) ≥ (A)λ (B).

(5.9.10)

Proof. By (5.9.7) with (5.9.9) we can write # sin(λπ) 1 t λ−1 (A!t B)dt λ π 0 (1 − t) # sin(λπ) 1 t λ−1 ≥ (A)!t (B)dt, λ π 0 (1 − t)

(Aλ B) =

which, when combined with Proposition 5.9.2, implies the desired inequality.



Let A, B ∈ B(H) be strictly positive and λ ∈ (0, 1). The relative operator entropy S(A|B) and the Tsallis relative operator entropy Tλ (A|B) are defined by

152

5 Inequalities for Sector Matrices

1 1 1 1 S(A|B) = A 2 log A− 2 B A− 2 A 2 ,

(5.9.11)

Aλ B − A , λ

Tλ (A|B) =

(5.9.12)

see [70–72] for instance. The Tsallis relative operator entropy is a parametric extension in the sense that lim Tλ (A|B) = S(A|B). (5.9.13) λ→0

For more details about these operator entropies, we refer the reader to [78] and [173] and the related references cited therein. Here, our aim is to extend S(A|B) and Tλ (A|B) for accretive operators A, B. Following the previous study, we suggest that Tλ (A|B) can be defined by the same formula (5.9.12) whenever A, B ∈ B(H) are accretive and so Aλ B is given by (5.9.7). Precisely, we have the following Definition 5.9.6 Let A, B be accretive operators and λ ∈ (0, 1). The Tsallis relative operator entropy of A and B is defined by sin(λπ) Tλ (A|B) = λπ

#

1 0

t 1−t



 A!t B − A dt. t

(5.9.14)

This, with (5.9.12) and (5.9.10), immediately yields  (Tλ (A|B)) ≥ Tλ (A|B) for any accretive operators A, B and λ ∈ (0, 1). In view of (5.9.14), the extension of S(A|B) can be introduced via the following definition. Definition 5.9.7 Let A, B ∈ B(H) be accretive. The relative operator entropy of A and B is defined by # 1 A!t B − A dt. (5.9.15) S(A|B) = t 0 Proposition 5.9.8 If A, B ∈ B(H) are strictly positive, then (5.9.15) coincides with (5.9.11). Proof. Assume that A, B ∈ B(H) are strictly positive. By (5.9.15), with the definition of A!t B, it is easy to see that ⎛ 1 ⎜ S(A|B) = A 2 ⎝

# 0

1

1 1 −1 (1 − t)I + t A 2 B −1 A 2 −I t

⎞ ⎟ 1 dt ⎠ A 2 .

5.9 Weighted Geometric Mean of Accretive Operators and Its Applications

153

By similar arguments as those for the proof of proposition 5.9.2, it is sufficient to show that # 1 (1 − t + ta −1 )−1 log a = dt t 0 is valid for any a > 0. This follows from a simple computation of this latter real integral, so completing the proof.  Theorem 5.9.9 Let A, B ∈ B(H) be accretive. Then  (S(A|B)) ≥ S(A|B).

(5.9.16)

Proof. By (5.9.15) and (5.9.9), we have #

1

 (S(A|B)) = #

0 1

≥ 0

(A!t B) − A dt t (A)!t (A) − A dt. t

This, with Proposition 5.9.8, immediately yields (5.9.16).



In the following, we give a generalization of Proposition 5.9.8 Proposition 5.9.10 If A, B ∈ B(H) are strictly positive, then (5.9.14) coincides with (5.9.12). Proof. Putting 1 − t = Tλ (A|B) =

1 , s+1

(5.9.14) is calculated as

sin(λπ) λπ

#



  s λ−1 (A−1 + s B −1 )−1 − (1 + s)A ds.

0

For A, B > 0, we have sin(λπ) π and

#



s λ−1 (A−1 + s B −1 )−1 ds = Aλ B,

0

sin(λπ) π

which imply the assertion.

#



s λ−1 (1 + s)−1 ds = 1,

0



Now we give more about Aλ B where A, B ∈ B(H) are accretive operators. For any real number α, β > 0, we set αλ β = α1−λ β λ the real λ-weighted geometric mean of α and β.

154

5 Inequalities for Sector Matrices

Proposition 5.9.11 For any accretive A, B ∈ B(H) and λ ∈ (0, 1), the following equality (5.9.17) (α A)λ (β B) = (αλ β)(Aλ B) holds for every real number α, β > 0. Proof. Since Aλ B = B1−λ A, it is sufficient to prove that (α A)λ B = α1−λ (Aλ B). By equation (5.9.6), we have

−1 # sin(λπ) ∞ λ−1 A−1 −1 + tB t dt (α A)λ B = π α 0 # ∞

−1 sin(λπ) =α t λ−1 A−1 + αt B −1 dt. π 0 If we make the change of variables u = αt, and we use again (5.9.6), we immediately obtain the desired equality after simple manipulations.  Theorem 5.9.12 Let A, B ∈ B(H) be accretive and λ ∈ (0, 1). Then the following inequality n  ((Aλ B))−1 xk , xk  k=1

  n   n   −1 −1 (A) xk , xk  λ (B) xk , xk  ≤ k=1

(5.9.18)

k=1

holds for any family of vectors (xk )nk=1 ∈ H. Proof. By (5.9.10) with the left-hand side of (5.9.3), we have (Aλ B) ≥ (A)λ (B) ≥ (A)!λ (B), from which we deduce ((Aλ B))−1 ≤ (1 − λ)(A)−1 + λ(B)−1 . Replacing in this latter inequality A by t A, with t > 0 a real number, and using Proposition 5.9.11, we obtain t λ ((Aλ B))−1 ≤ (1 − λ)(A)−1 + tλ(B)−1 . This means that for any x ∈ H and t > 0, we have

5.9 Weighted Geometric Mean of Accretive Operators and Its Applications

155

t λ ((Aλ B))−1 x, x ≤ (1 − λ)(A)−1 x, x + tλ(B)−1 x, x, and so t

λ

n 

((Aλ B))−1 xk , xk 

k=1

≤ (1 − λ)

n 

(A)−1 xk , xk  + tλ

k=1

n 

(B)−1 xk , xk 

(5.9.19)

k=1

holds for any (xk )nk=1 ∈ H and t > 0. If xk = 0 for each k = 1, 2, .., n, then (5.9.18) is an equity. Assume that xk = 0 for some k = 1, 2, .., n. If we take ⎛ n

((Aλ B))−1 xk , xk 

⎜ k=1 t =⎜ n ⎝ 

(B)−1 xk , xk 

1 ⎞ 1−λ

⎟ ⎟ ⎠

> 0,

k=1

in (5.9.19) and then by easy mathematical computation, we immediately obtain the desired inequality.  As a consequence of Theorem 5.9.12, we have Corollary 5.9.13 Let A, B ∈ B(H) be accretive and λ ∈ (0, 1). Then $ $ $ $ $ $ $((Aλ B))−1 $ ≤ $(A)−1 $1−λ $(B)−1 $λ ,

(5.9.20)

where for any T ∈ B(H), T  = supx=1 T x is the usual norm of B(H). Proof. It follows from (5.9.18) with n = 1 and the fact that T  = sup T x = sup T x, x, x=1

whenever T ∈ B(H) is a positive operator.

x=1



It is interesting to see whether (5.9.20) holds for any unitarily invariant norm. Theorem 5.9.12 gives an inequality about ((Aλ B))−1 . The following inequality gives another inequality but involving (Aλ B). Theorem 5.9.14 Let A, B ∈ B(H) be accretive and λ ∈ (0, 1). Then

2 x ∗ , x



≤ (Aλ B)x ∗ , x ∗  (A)−1 x, xλ (B)−1 x, x

(5.9.21)

156

5 Inequalities for Sector Matrices

for all x, x ∗ ∈ H. Proof. Following [186], for any T ∈ B(H) strictly positive, the following inequality   T −1 x ∗ , x ∗  = sup 2x ∗ , x − T x, x x∈H

is valid for all x ∗ ∈ H. This, with (5.9.18) for n = 1, immediately implies that 2x ∗ , x ≤ (Aλ B)x ∗ , x ∗  + (A)−1 x, xλ (B)−1 x, x holds for all x, x ∗ ∈ H. In the latter inequality, we can, of course, replace x ∗ by t x ∗ for any real number t, for obtaining 2tx ∗ , x ≤ t 2 (Aλ B)x ∗ , x ∗  + (A)−1 x, xλ (B)−1 x, x.

(5.9.22)

If x ∗ = 0, the inequality (5.9.21) is obviously an equality. We then assume that x ∗ = 0. If in inequality (5.9.22), we take t=

x ∗ , x , (Aλ B)x ∗ , x ∗ 

then the desired inequality (5.9.21) is obtained by easy calculations. This completes the proof. 

Notes and references. The proof of Proposition 5.9.2, Proposition 5.9.8, Proposition 5.9.10, Proposition 5.9.11, Lemma 5.9.3, Lemma 5.9.4, Theorem 5.9.5, Theorem 5.9.9, Theorem 5.9.12, Corollary 5.9.13, and Theorem are taken from 5.9.14 [185].

5.10 Ficher Type Determinantal Inequalities for Accretive-Dissipative Matrices Suppose an accretive-dissipative complex matrix A ∈ Mn in the form of A = B + iC ∗ ∗ and C = A−A are both positive definite. Conformally partition where B = A+A 2 2i A, B, C as       B11 B12 C11 C12 A11 A12 , (5.10.1) = + i ∗ ∗ A21 A22 B12 B22 C12 C22 ,

5.10 Ficher Type Determinantal Inequalities for Accretive-Dissipative Matrices

157

such that all diagonal blocks are square. Say k and l (k, l > 0 and k + l = n) the order of A11 and A22 , respectively, and let m = min{k, l}. Determinantal inequalities for accretive-dissipative matrices were first investigated by [109] as follows Theorem 5.10.1 Let A ∈ Mn be a complex accretive-dissipative matrix and partitioned as in (5.10.1).Then | det A| ≤ 3m | det A11 |.| det A22 |.

(5.10.2)

A reverse direction to that of Theorem 5.10.1 has been given in [140]. We call this kind of inequalities the Fischer type determinantal inequality for accretive-dissipative matrices. In the following, we intend to give an improvement of (5.10.2). To do this, the following Lemmas are needed in the proof of our main result. Lemma 5.10.2 ([81]) Let A ∈ Mn be a complex accretive-dissipative matrix and partitioned as in (5.10.1). Then A = A22 − A21 A−1 11 A12 , A11 the Schur complement of A11 in A is also accretive-dissipative. Lemma 5.10.3 ([109]) Let A = B + iC be accretive-dissipative. Then A−1 = E − i F with E = (B + C B −1 C)−1 and F = (C + BC −1 B)−1 . Lemma 5.10.4 ([209]) Let B, C ∈ Mn be Hermition complex matrices and assume B is positive definite. Then B + C B −1 C ≥ 2C. Lemma 5.10.5 Let B, C ∈ Mn be positive semidefinite complex matrices. Then n

| det(B + iC)| ≤ det(B + C) ≤ 2 2 | det(B + iC)|. Proof. The first inequality follows from [211] while the second one follows from [24]. Here we provide a direct proof of Lemma 5.10.5 for the convenience of the readers. We may assume B is positive definite, the general case is by a continuity argument. 1 1 Let λ j , j = 1, ..., n be the eigenvalues of B − 2 C B − 2 . Then |1 + iλ j | ≤ 1 + λ j ≤ for j = 1, 2, ..., n. Compute

√ 2|1 + iλ j |,

158

5 Inequalities for Sector Matrices

| det(B + iC)| = det B.| det(I + i B − 2 C B − 2 )| 1

= det B.

n 

1

|1 + iλ j |

j=1

≤ det B.

n 

(1 + λ j )

j=1

= det B. det(I + B − 2 C B − 2 ) = det(B + C). 1

1

This proves the first inequality. To show the other, compute det(B + C) = det B. det(I + B − 2 C B − 2 ) 1

= det B.

n 

1

(1 + λ j )

j=1

≤ det B.

n  √

2|1 + iλ j |

j=1

= 2 2 det B.| det(I + i B − 2 C B − 2 )| n

1

1

n

= 2 2 | det(B + iC)|.  Theorem 5.10.6 Let A ∈ Mn be a complex accretive-dissipative matrix and partitioned as in (5.10.1). Then n

| det A| ≤ 2 2 | det A11 |.| det A22 |. Proof. Compute | det A| = | det(B + iC)| ≤ det(B + C) ( by Lemma 5.10.5) ≤ det(B11 + C11 ). det(B22 + C22 ) ( by (5.3.2)) n

n

≤ 2 2 | det(B11 + iC11 )|.2 2 | det(B22 + iC22 )| ( by Lemma 5.10.5) n

= 2 2 | det A11 |.| det A22 |. 

5.10 Ficher Type Determinantal Inequalities for Accretive-Dissipative Matrices

159

Theorem 5.10.7 Let A ∈ Mn be a complex accretive-dissipative matrix and partitioned as in (5.10.1). Then 3m

| det A| ≤ 2 2 | det A11 |.| det A22 |, where m = min{k, l} such that k and l are the order of A11 and A22 . Proof. By Lemma 5.10.3, we have A = A22 − A21 A−1 11 A12 A11 ∗ ∗ = B22 + iC22 − (B12 + iC12 )(B11 + iC11 )−1 (B12 + iC12 ) ∗ ∗ = B22 + iC22 − (B12 + iC12 )(E k − i Fk )(B12 + iC12 ) with −1 C11 )−1 , E K = (B11 + C11 B11 −1 B11 )−1 . Fk = (C11 + B11 C11

By Lemma 5.10.4 and the operator reverse monotonicity of the inverse, we get Ek ≤

1 −1 C , 2 11

Fk ≤

1 −1 B . 2 11

(5.10.3)

Setting AA11 = R + i S with R = R ∗ and S = S ∗ . By Lemma 5.10.2, we know R and S are positive definite. A calculation shows ∗ ∗ ∗ ∗ E k B12 + C12 E k C12 − B12 Fk C12 − C12 Fk B12 , R = B22 − B12 ∗ ∗ ∗ ∗ S = C22 − B12 Fk B12 + C12 Fk C12 − C12 E k B12 − B12 E k C12 .

It can be verified that ∗ ∗ ∗ ∗ Fk C12 + C12 Fk B12 ) ≤ B12 Fk B12 + C12 Fk C12 , ± (B12 ∗ ∗ ∗ ∗ ± (C12 E k B12 + B12 E k C12 ) ≤ B12 E k B12 + C12 E k C12 .

Thus,

∗ ∗ Fk B12 + C22 + 2C12 E k C12 . R + S ≤ B22 + 2B12

(5.10.4)

As B, C are positive definite, we also have −1 −1 ∗ ∗ B11 B12 and C22 > C12 C11 C12 . B22 > B12

Without loss of generality, we assume m = l. Compute

(5.10.5)

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5 Inequalities for Sector Matrices

    det A   A11  = | det(R + i S)| ≤ det(R + S) (by Lemma 5.10.5) ∗ ∗ ≤ det(B22 + 2B12 Fk B12 + C22 + 2C12 E k C12 ) (by (5.10.4)) −1 −1 ∗ ∗ B11 B12 + C22 + C12 C11 C12 ) (by (5.10.3)) ≤ det(B22 + B12

< det(2(B22 + C22 )) (by (5.10.5)) = 2m det(B22 + C22 )

m

≤ 2m .2 2 | det(B22 + iC22 )| (by Lemma 5.10.5) 3m

= 2 2 | det A22 |. The proof is completed by noting

det

A A11

 =

det A .  det A11

Notes and references. Lemma 5.10.5, Theorem 5.10.6 and Theorem 5.10.7 are due to [139].

5.11 Extensions of Fischer’s Inequality for Sector Matrices Let Mn,k denotes the set of n × k complex matrices. Throughout this section B(n 1 , ..., n k ) denotes the set of block matrices whose (i, j)-blocks are in Mni ,n j . If A ∈ B(n 1 , ..., n k ), the i-th main diagonal block of A is denoted by [A]i . We recall that Fischer’s inequality [103] states that if A is a positive semidefinite matrix whose main diagonal are Ai ∈ Mni , for i = 1, 2, ..., k, then det A ≤ det A1 det A2 ... det Ak .

(5.11.1)

Our purpose is to generalize Fischer’s inequality. To do this, we list some lemmas which are useful to establish our main result. Let [A] be the principal submatrix of the positive semidefinite matrix A corresponding to some fixed rows and columns. Lemma 5.11.1 For any A ≥ 0 and p ∈ [0, 1], [A p ] ≤ [A] p .

5.11 Extensions of Fischer’s Inequality for Sector Matrices

161

Proof. Since Φ(A) = [A] is a positive unital linear map on Mn , [A p ] ≤ [A] p follows from the fact that Φ(A p ) ≤ Φ p (A) which is due to [20].  Lemma 5.11.2 ([103]) If A, B be positive semidefinite with A ≤ B, then det A ≤ det B and X AX ∗ ≤ X B X ∗ for any X ∈ Mn . Lemma 5.11.3 Given positive definite matrices A and B, Let A : B = (A−1 + B −1 )−1 denote the parallel sum. Then we have [A : B] ≤ [A] : [B].

(5.11.2)

More generally, if Ai be positive definite for i = 1, 2, ...k and A1 : A2 : ... : Ak −1 −1 −1 denotes (A−1 1 + A2 + ... + Ak ) , then we have [A1 : A2 : ... : Ak ] ≤ [A1 ] : [A2 ] : ... : [Ak ]. Proof. Since Φ(A) = [A] is a positive unital linear map on Mn , (5.11.2) follows from Φ(A : B) ≤ Φ(A) : Φ(B) [20]. The general case can be easily proved by induction as follows. We have Φ(A1 : A2 : ... : Ak+1 ) = Φ((A1 : A2 : ... : Ak ) : Ak+1 ) ≤ Φ(A1 : A2 : ... : Ak ) : Φ(Ak+1 ) ≤ (Φ(A1 ) : ... : Φ(Ak )) : Φ(Ak+1 ) = Φ(A1 ) : ... : Φ(Ak ) : Φ(Ak+1 ), where the last inequality follows from the fact that A : B is monotonically increasing in the arguments A and B [20].  Lin [146] showed that Lemma 5.11.4 Let α ∈ (0, π2 ) and X ∈ Mn with W (X ) ⊆ Sα . Then (X −1 ) ≤ ((X ))−1 ≤ sec2 (α)(X −1 ). Lemma 5.11.5 ([217]) Let α ∈ (0, π2 ) and A, B ∈ Mn with W (A), W (B) ⊆ Sα . Then W (A + B) ⊆ Sα . Lemma 5.11.6 ([20, 48]) Let z be any nonzero complex number which is not on the negative real axis. Then for any γ ∈ (0, 1), we have z −γ =

sin(πγ) π

# 0



t −γ dt, t +z

where z −γ is the principal branch of the power function.

(5.11.3)

162

5 Inequalities for Sector Matrices

Lemma 5.11.7 Let γ ∈ (0, 1) and T ∈ Mn whose spectrum does not intersect the negative real axis. Then we have Tγ = T −γ =

# #

∞ 0



T (t I + T )−1 dμ(t),

(5.11.4)

(t I + T )−1 dυ(t),

(5.11.5)

0

where μ and υ are positive measures on (0, ∞). Proof. Note that (5.11.5) follows from Lemma 5.11.6 with dυ(t) = Multiplying by z and replacing γ with 1 − γ in (5.11.3), we have zγ =

# 0

where dμ(t) = 



sin(πγ) −γ t dt. π

z dμ(t), t +z

sin(π(1−γ)) γ−1 t dt. Thus (5.11.4) results from the above representation. π

Lemma 5.11.8 Let α ∈ [0, π2 ), γ ∈ [0, 1] and T ∈ Mn with W (T ) ⊆ Sα . Then cos2γ (T γ ) ≤ ((T ))γ ≤ (T γ ). Proof. It is clear if γ = 0 or γ = 1. Using (5.11.4), we can show that the required inequalities hold if the following inequalities cos2 (α)[T (t I + T cos2 α)−1 ] ≤ ((T ))(t I + (T ))−1

(5.11.6)

−1

≤ (T (t I + T ) ) hold for t > 0. Using the following relation A(λI + A)−1 = I − λ(λI + A)−1 ,

(5.11.7)

we can show that the second inequality in (5.11.6) is equivalent to [(t I + T )−1 ] ≤ (t I + (T ))−1 , which is true by Lemma 5.2.5 and the first inequality in Lemma 5.11.4. Now we are going to prove the first inequality in (5.11.6). A simple calculation shows that cos2 (α)[T (t I + T cos2 α)−1 ] ≤ ((T ))(t I + (T ))−1 ,

5.11 Extensions of Fischer’s Inequality for Sector Matrices

163

if and only if [(t sec2 (α)T −1 + I )−1 ] ≤ ((T ))(t I + (T ))−1 .

(5.11.8)

By Lemma 5.2.5 and the first inequality in Lemma 5.11.4, we have [(t sec2 (α)T −1 + I )−1 ] ≤ (t sec2 (α)(T −1 ) + I )−1 . Thus (5.11.8) will follows from (t sec2 (α)(T −1 ) + I )−1 ≤ ((T ))(t I + (T ))−1 , or equivalently (t sec2 (α)(T −1 ) + I ) ≥ ((T ))−1 (t I + (T )) = t ((T ))−1 + I. The above follows directly from the second inequality in Lemma 5.11.4.



Lemma 5.11.9 Let α ∈ [0, π2 ), γ ∈ [−1, 0] and T ∈ Mn with W (T ) ⊆ Sα . Then (T γ ) ≤ ((T ))γ ≤ cos2γ (α)(T γ ). Proof. It is clear if γ = 0. By Lemma 5.11.4, it holds when γ = −1. Here we assume that γ ∈ (−1, 0). Using (5.11.5), we can show that the required inequalities hold if the following inequalities [(t I + T )−1 ] ≤ (t I + (T ))−1 ≤ [(t I + cos2 (α)T )−1 ] hold for t > 0. The first inequality follows from Lemma 5.2.5 and the first inequality of Lemma 5.11.4. Using (5.11.7), we have (t I + (T ))−1 ≤ [(t I + cos2 (α)T )−1 ] ⇐⇒t (t I + (T ))−1 ≤ [t sec2 (α)(t sec2 (α)I + T )−1 ] ⇐⇒I − ((T ))(t I + (T ))−1 ≤ [I − T (t sec2 (α)I + T )−1 ] ⇐⇒((T ))(t I + Re(T ))−1 ≥ [(t sec2 (α)T −1 + I )−1 ], and the last inequality above is (5.11.8).



164

5 Inequalities for Sector Matrices

Now we are going to obtain the generalizations of Fischer’s inequality (5.11.1) as follows. Theorem 5.11.10 Let Ai ∈ B(n 1 , ..., n k ) be positive semidefinite and Di ∈ B(n 1 , ..., n k ) be block diagonal matrices for 1 ≤ i ≤ m. Then  det

m 

 p Di Ai i

Di∗



i=1

k 

det

j=1

 m 

 p [Di ] j [Ai ] j i [Di ]∗j

i=1

for all 0 ≤ pi ≤ 1. Proof. By Fischer’s inequality (5.11.1), one gets det

 m 

 p Di Ai i

Di∗



i=1

k  j=1

=

k  j=1



k 

 m   pi ∗ det [Di Ai Di ] j i=1

 m   pi det [Di ] j [Ai ] j [Di∗ ] j i=1

 m   pi det [Di ] j [Ai ] j [Di ]∗j ,

j=1

i=1

which the last inequality follows from Lemma 5.11.1 and Lemma 5.11.2 and this completes the proof.  If Di in Theorem 5.11.10 are all identity matrices, we have the following result. Corollary 5.11.11 If Ai ∈ B(n 1 , ..., n k ) is positive semidefinite for 1 ≤ i ≤ m, then det

 m 

 p Ai i



i=1



k 

m  p det [Ai ] j i

j=1

i=1

 ,

for all 0 ≤ pi ≤ 1. Note that when m = 1, p1 = 1 and D1 = In 1 , Theorem 5.11.10 is reduced to Fischer’s inequality (5.11.1). We also provide a reverse of Theorem 5.11.10 as follows. Theorem 5.11.12 Let Ai ∈ B(n 1 , ..., n k ) be positive definite and Di ∈ B(n 1 , ..., n k ) be block diagonal nonsingular matrices for 1 ≤ i ≤ m. Then det

 m  i=1

for all −1 ≤ pi ≤ 0.

 p Di Ai i

Di∗



k  j=1

det

 m  i=1

 p [Di ] j [Ai ] j i [Di ]∗j

5.11 Extensions of Fischer’s Inequality for Sector Matrices

165

Proof. Let qi = − pi and E i = Di−1 for i = 1, ..., m. Under the same notation of Lemma 5.11.3, the desired inequality can be written equivalently as det(E 1∗ A11 E 1 : ... : E m∗ Aqmm E m ) q



k 

det([E 1 ]∗j [A1 ] j 1 [E 1 ] j : ... : [E m ]∗j [Am ] j m [E m ] j ). q

q

j=1

The above is shown by the following argument det(E 1∗ A11 E 1 : ... : E m∗ Aqmm E m ) q

k 



det([E 1∗ A11 E 1 : ... : E m∗ Aqmm E m ] j ) q

j=1 k 



det([E 1∗ A11 E 1 ] j : ... : [E m∗ Aqmm E m ] j ) q

j=1

=

k 

det([E 1∗ ] j [A11 ] j [E 1 ] j : ... : [E m∗ ] j [Aqmm ] j [E m ] j ) q

j=1



k 

det([E 1 ]∗j [A1 ] j 1 [E 1 ] j : ... : [E m ]∗j [Am ] j m [E m ] j ), q

q

j=1

where the first inequality follows from Fischer’s inequality (5.11.1), the second inequality from Lemma 5.11.3, and the third inequality from Lemma 5.11.1.  If pi = −1 and Di is the identity matrix for all i in the above theorem, we have the following corollary [47]. Corollary 5.11.13 If Ai > 0 for 1 ≤ i ≤ m, then det

 m  i=1

 Ai−1



k 



 m  −1 det [Ai ] j .

j=1

i=1

Now, we present some generalization of Theorem 5.11.10 and Theorem 5.11.12. Firstly, Theorem 5.11.10 is extended to the class of matrices whose numerical range is contained in a sector as follows. Theorem 5.11.14 Let Ai ∈ B(n 1 , ..., n k ) with W (Ai ) ⊆ Sα and Di ∈ B(n 1 , ..., n k ) be block diagonal nonsingular matrices for 1 ≤ i ≤ m. Then   m   m   k          pi ∗  pi n(1+2γ) ∗  Di Ai Di  ≤ sec (α) [Di ] j [Ai ] j [Di ] j  det det     i=1

j=1

i=1

166

5 Inequalities for Sector Matrices k 

for all 0 ≤ pi ≤ γ where γ ∈ (0, 1) and n =

ni .

i=1

Proof. By Corollary 5.7.4, Lemma 5.2.5 and Lemma 5.11.5, we have W(

m 

Di Ai i Di∗ ) ⊆ Sα , p

i=1

and W(

m 

[Di ] j [Ai ] j i [Di ]∗j ) ⊆ Sα . p

i=1

It follows from Lemma 5.1.8, Lemma 5.11.8 and Lemma 5.2.3 that   m      pi ∗  Di Ai Di  det   i=1  m   pi ∗ n ≤ sec (α) det (Di Ai Di ) i=1

= secn (α) det

 m 

 p Di (Ai i )Di∗

i=1

≤ sec (α) det n

 m  i=1

≤ sec

n(1+2γ)

(α) det

 2 pi

sec

 m 

(α)Di ((Ai ))

pi

Di∗

 Di ((Ai ))

pi

Di∗

i=1

≤ secn(1+2γ) (α)

k 

det

j=1

≤ sec

n(1+2γ)

(α)

k 



= sec

(α)

≤ secn(1+2γ) (α)

 p [Di ] j [(Ai )] j i [Di ]∗j

i=1

det (

j=1 n(1+2γ)

 m 



m 

 p [Di ] j [Ai ] j i [Di ]∗j )

i=1



k 

m  p det  [Di ] j [Ai ] j i [Di ]∗j

j=1

  m     pi ∗  [Di ] j [Ai ] j [Di ] j  . det  

i=1

k   j=1

i=1

The forth inequality above holds by Theorem 5.11.10. Thus the proof is completed. 

5.11 Extensions of Fischer’s Inequality for Sector Matrices

167

Taking γ = 1 in Theorem 5.11.14, the following corollary can easily be obtained. Corollary 5.11.15 Let Ai ∈ B(n 1 , ..., n k ) with W (Ai ) ⊆ Sα and Di ∈ B(n 1 , ..., n k ) be block diagonal nonsingular matrices for 1 ≤ i ≤ m. Then   m   m   k          pi ∗  pi 3n ∗  Di Ai Di  ≤ sec (α) [Di ] j [Ai ] j [Di ] j  det det     i=1

j=1

for all 0 ≤ pi ≤ 1, where n =

k 

i=1

ni .

i=1

Next, Theorem 5.11.12 is extended to the class of matrices whose numerical range is contained in a sector as follows. Theorem 5.11.16 Let Ai ∈ B(n 1 , ..., n k ) with W (Ai ) ⊆ Sα and Di ∈ B(n 1 , ..., n k ) be block diagonal nonsingular matrices for 1 ≤ i ≤ m. Then   m   m   k           pi ∗  p Di Ai Di  ≥ cos(1−2γ)n (α) [Di ] j [Ai ] j i [Di ]∗j  det det     i=1

j=1

for all γ ≤ pi ≤ 0 where γ ∈ (−1, 0) and n =

k 

i=1

ni .

i=1

Proof. It follows from Lemma 5.2.3, Lemma 5.11.9, Theorem 5.11.12 and Lemma 5.1.8 that   m      pi ∗  Di Ai Di  det   i=1   m  pi ∗ ≥ det ( Di Ai Di ) i=1

= det

 m 



p Di (Ai i )Di∗

i=1

≥ det

 m 

 cos

−2 pi

i=1

≥ sec

−2nγ

(α) det

(α)Di ((Ai ))

 m  i=1

pi

Di∗ 

Di ((Ai ))

pi

Di∗

168

5 Inequalities for Sector Matrices

≥ cos

−2nγ

(α)

k  j=1

= cos−2nγ (α)

k 

det

≥ cos−2nγ (α)

 p [Di ] j [(Ai )] j i [Di ]∗j

i=1

 m   pi ∗ det [Di ] j (([Ai ] j )) [Di ] j

j=1 k 

 m 

i=1

det

 m 

j=1

 p [Di ] j ([Ai ] j i )[Di ]∗j

i=1

  m  k      pi n(1−2γ) ∗  (α) [Di ] j [Ai ] j [Di ] j  , ≥ cos det   j=1

i=1



which gives the desired inequality.

Taking γ = −1 in Theorem 5.11.16, we drive the following corollary. Corollary 5.11.17 Let Ai ∈ B(n 1 , ..., n k ) with W (Ai ) ⊆ Sα and Di ∈ B(n 1 , ..., n k ) be block diagonal nonsingular matrices for 1 ≤ i ≤ m. Then   m   m   k          pi ∗  pi 3n ∗  Di Ai Di  ≥ cos (α) [Di ] j [Ai ] j [Di ] j  det det     i=1

for all −1 ≤ pi ≤ 0, where n =

j=1

k 

i=1

ni .

i=1

Notice that when α = 0, Corollary 5.11.17 reduces to Theorem 5.11.12. Notes and references. Lemma 5.11.1, Lemma 5.11.3, Lemma 5.11.7, Lemma 5.11.8, Lemma 5.11.9, Theorem 5.11.10, Theorem 5.11.12, Theorem 5.11.14, and Theorem 5.11.16 are proved by [48].

5.12 Singular Value Inequalities of Sector Matrices If the eigenvalue of A ∈ Mn are all real, then we denote its j-th largest eigenvalue by λ j (A). The j-th largest singular value of A is denoted by σ j (A). Note that σ j (A) = 1 (λ j (A∗ A)) 2 . Consider A ∈ Mn partitioned as  A=

 A11 A12 , A21 A22

(5.12.1)

5.12 Singular Value Inequalities of Sector Matrices

169

where A22 ∈ Mm such that m ≤ n2 . Suppose W (A) ⊂ Sα , it is clear that W (A11 ) ⊂ Sα and so A11 is invertible. As we know, the Schur complement of A11 in A is defined by A = A22 − A21 A−1 11 A12 . A11 Notice that (5.3.5) is equivalent to     det A  ≤ sec2m (α)| det A22 |.  A11  That is

m 

σj

j=1

A A11

 ≤ sec (α) 2m

m 

σ j (A22 ).

(5.12.2)

j=1

We aim to show a strengthening of (5.12.2). To do this, we start with some lemmas. Lemma 5.12.1 ([21]) For every A ∈ Mn , λ j (A) ≤ σ j (A), j = 1, ..., n.

(5.12.3)

The following lemma makes use of an explicit expression for the inverse of a 2 × 2 partitioned matrix [103]. Lemma 5.12.2 Let A ∈ Mn be invertible and partitioned as in (5.12.1). Also, we partition A−1 conformally as A and assume A11 is invertible. Then the (2, 2) block −1 . of A−1 , denoted by (A−1 )22 , is equal to AA11    Y1 X1 ∈ Mn and Y = ∈ Mn where X 2 and Y2 are X2 Y2 m × n such that Y X ∗ = In . Then 

Lemma 5.12.3 Let X =

λ j (X 2 X 2∗ )λm+1− j (Y2 Y2∗ ) ≥ 1

(5.12.4)

for j = 1, ..., m. Proof. As Y X ∗ = In , we have Y2 X 2∗ = Im . We notice that λm+1− j (Y2 Y2∗ ) =

1

, λ j (Y2 Y2∗ )−1

so (5.12.4) is the same as

λ j (Y2 Y2∗ )−1 ≤ λ j (X 2 X 2∗ ),

j = 1, ..., m.

(5.12.5)

170

5 Inequalities for Sector Matrices

We shall prove something stronger (Y2 Y2∗ )−1 ≤ X 2 X 2∗ .

(5.12.6)

As  0≤

Y2 X2



Y2 X2

∗

 Y2 Y2∗ Y2 X 2∗ X 2 Y2∗ X 2 X 2∗   Y2 Y2∗ Im , = Im X 2 X 2∗ 

=

using a well-known characterization of positivity in terms of the Schur complement [103], (5.12.6) follows.  Now we are in a position to prove our main theorem. Theorem 5.12.4 Let A ∈ Mn be partitioned as in (5.12.1) and W (A) ⊂ Sα . Then

σj

A A11

 ≤ sec2 (α)σ j (A22 ),

j = 1, ..., m.

(5.12.7)

Proof. By [57], we can write A in the following form A = X Z X ∗,

(5.12.8)

for some invertible X ∈ Mn and Z = diag(eiθ1 , ..., eiθn ) with |θ j | ≤ α for all j. Let Y = (X ∗ )−1 and partition X, Y as in Lemma 5.12.3. Then from (5.12.8), we have A22 = X 2 Z X 2∗ . By Lemma 5.12.1, σ j (X 2 Z X 2∗ ) ≥ λ j (X 2 (Z )X 2∗ ) ≥ λ j (X 2 X 2∗ ) cos(α), for j = 1, ..., m. Similarly, we have σ j (Y2 Z ∗ Y2∗ ) ≥ λ j (Y2 Y2∗ ) cos(α), for j = 1, ..., m. Combining these two inequalities and by Lemma 5.12.3, we get σ j (X 2 Z X 2∗ )σm+1− j (Y2 Z ∗ Y2∗ ) ≥ cos2 (α). Thus

5.12 Singular Value Inequalities of Sector Matrices

171

1 σm+1− j (Y2 Z ∗ Y2∗ )

= σ j (Y2 Z ∗ Y2∗ )−1 .

σ j (X 2 Z X 2∗ ) sec2 (α) ≥

But by Lemma 5.12.2, (Y2 Z ∗ Y2∗ )−1 =

A A11

and this completes the proof.



Next, we prove the following reverse inequality to (5.12.3). Theorem 5.12.5 Let A ∈ Mn be such that W (A) ⊂ Sα . Then σ j (A) ≤ sec2 (α)λ j (A),

(5.12.9)

for j = 1, ..., n. Proof. Since W (A−1 ) ⊂ Sα , it follows that (A−1 ) is positive definite. Hence, we may apply Lemma 5.12.1 to A−1 , deducing 1 λn+1− j (((A−1 ))−1 ) = λ j ((A−1 )) ≤ σ j (A−1 ) 1 = σn+1− j (A) for j = 1, ..., n. By replacing j with n + 1 − j, we have equivalently σ j (A) ≤ λ j (((A−1 ))−1 ). It remains to show that λ j (((A−1 ))−1 ) ≤ λ j (A) sec2 (α). To deduce this, let A = P + i Q with P positive definite and Q Hermitian. Then, (A−1 ) = (P + Q P −1 Q)−1 , and therefore

−1 = P + Qp −1 Q. (A−1 )

Again applying (5.12.8), we can write P = X diag(cos(θ1 ), ..., cos(θn ))X ∗ and Q = X diag(sin(θ1 ), ..., sin(θn ))X ∗ for some invertible X . But then

172

5 Inequalities for Sector Matrices



−1 (A−1 ) = X diag(sec(θ1 ), ..., sec(θn ))X ∗ ≤ sec2 (α)X diag(cos(θ1 ), ..., cos(θn ))X ∗ = sec2 (α)(A), and the desired inequality obtained.



Corollary 5.12.6 Let X, Y ∈ Mn be positive semidefinite. Then σ j (X + iY ) ≤

√ 2λ j (X + Y ),

for j = 1, .., n. Proof. By a continuity argument, we can assume without loss of generality that X and Y are positive definite. Then (1 − i)(X + iY ) = (X + Y ) + i(Y − X ) has its numerical range in S π4 . It follows that √ 2σ j (X + iY ) = σ j ((X + Y ) + i(Y − X )) ≤ 2λ j (X + Y ) as required.



The following result obtains immediately from Corollary 5.12.6. Corollary 5.12.7 Let X, Y, Z ∈ Mn such that 0 ≤ X ≤ Z and 0 ≤ Y ≤ Z . Then √ σ j (X + iY ) ≤ 2 2λ j (Z ) for j = 1, ..., n. The inequality (5.12.7) is strong enough to imply the following norm inequality. Theorem 5.12.8 Let A ∈ Mn be as in (5.12.1) and W (A) ⊂ Sα . Then $ $ $ A $ 2 $ $ $ A $ ≤ sec (α)A22  11 for any unitarily invariant norm .. In particular, when A is accretive-dissipative, we have $ $ $ A $ $ $ $ A $ ≤ 2A22 . 11

5.12 Singular Value Inequalities of Sector Matrices

173

Notes and references. We note that the proof of Theorem 5.12.4, Theorem 5.12.5, Theorem 5.12.8, Lemma 5.12.3, and Corollary 5.12.6 are taken from [56].

5.13 Extension of Rotfel’d Inequality for Sector Matrices Consider a partitioned matrix A ∈ Mn in the form (5.12.1) where A11 and A22 are square. Lee [126] proved the following theorem which is considered as an extension of the classical Rotfel’d theorem. Theorem 5.13.1 Let A ∈ Mn be positive semidefinite and partitioned as in (5.12.1). Let f : [0, ∞) → [0, ∞) be a concave function. Then  f (A) ≤  f (A11 ) +  f (A22 ).

(5.13.1)

As a further extension of the classic Rotfel’d theorem, Zhang [214] extended Theorem 5.13.1 to matrices with W (A) ⊂ Sα , for α ∈ [0, π2 ) as follows. Theorem 5.13.2 Let f : [0, ∞) → [0, ∞) be a concave function and let A with W (A) ⊂ Sα for α ∈ [0, π2 ) be partitioned as in (5.12.1). Then  f (|A|) ≤  f (|A11 |) +  f (|A22 |) + 2( f (tan(α)|A11 |) +  f (tan(α)|A22 |)).

(5.13.2)

Obviously, when α = 0 in (5.13.2), Theorem 5.13.2 is Theorem 5.13.1. If α = in (5.13.2), then A is an accretive-dissipative matrix and  f (|A|) ≤ 3( f (|A11 |) +  f (|A22 |)),

π 4

(5.13.3)

which is due to [214]. Our purpose is to provide another generalization of (5.13.1) in Theorem 5.13.6. To obtain this, the following Lemmas and notations are useful. We observe that if f : [0, ∞) → [0, ∞) is concave, then 0 ≤ A ≤ B =⇒  f (A) ≤  f (B).

(5.13.4)

We notice that (5.12.9) implies there exist a unitary matrix U ∈ Mn such that |A| ≤ sec2 (α)U (A)U ∗ . 

(5.13.5)

 A X ∈ Mm+n be positive semidefinite where A and X∗ B B are positive semidefinite. Then there exist unitary matrices U, V ∈ Mm+n such

Lemma 5.13.3 ([32]) Let

174

5 Inequalities for Sector Matrices



that

A X X∗ B



 =U

   0 0 A 0 V ∗. U∗ + V 0 B 0 0

(5.13.6)

Lemma 5.13.4 ([217]) Let A ∈ Mn be positive semidefinite and B ∈ Mn is Hermitian. Then (5.13.7) σ j (A) ≤ σ j (A + i B) for j = 1, 2, ..., n. The above lemma means that there exists a unitary matrix U ∈ Mn such that A ≤ U |A + i B|U ∗ .

(5.13.8)

Lemma 5.13.5 ([126]) Let A, B ∈ Mn be positive semidefinite matrices. Then σ j (A + B) ≤



2σ j (A + i B),

(5.13.9)

2U |A + i B|U ∗

(5.13.10)

for j = 1, 2, ..., n. The result states that (A + B) ≤



for some unitary matrix U ∈ Mn . Next, we present a new generalization of (5.13.1). Theorem 5.13.6 Let f : [0, ∞) → [0, ∞) be a concave function and let A with W (A) ⊂ Sα for α ∈ [0, π2 ) be partitioned as in (5.12.1). Then



 f (|A|) ≤  f sec2 (α)|A11 |  +  f sec2 (α)|A22 | .

(5.13.11)

Proof. We suppose that f (0) = 0. (It suffices to prove for the Ky Fan k-norms .k . This shows that we may assume f (0) = 0.) Consider the Cartesian decomposition A = R + i S, where R is positive semidefinite and S is Hermitian. By (5.13.5), there exists a unitary matrix U1 ∈ Mn such that |A| ≤ sec2 (α)U1 RU1∗ . Compute

5.13 Extension of Rotfel’d Inequality for Sector Matrices

175

|A| ≤ sec2 (α)U1 RU1∗ 

    R11 0 0 0 2 ∗ = sec (α)U1 U2 U2 + U3 U3∗ U1∗ 0 0 0 R22  

   |A11 | 0 0 0 2 ∗ ∗ ∗ ∗ ≤ sec (α)U1 U2 U4 U5 U3 U1∗ U4 U2 + U3 U5 0 0 0 |A22 |   2 sec (α)|A11 | 0 (U1 U2 U4 )∗ = U1 U2 U4 0 0   0 0 (U1 U3 U5 )∗ , + U1 U3 U5 0 sec2 (α)|A22 | where U1 , U2 , ..., U5 correspond to the unitary matrices in the inequalities (5.13.5)(5.13.10). The first equality is deduced by (5.13.6). The second inequality is followed by (5.13.8) and (5.13.10). By (5.13.4), the desired inequality follows  f (|A|) ≤  f (sec2 (α)|A11 |) +  f (sec2 (α)|A22 |), 

and the proof is complete.

Corollary 5.13.7 Let f (t) = t p with 0 < p ≤ 1 and let A with W (A) ⊆ Sα for α ∈ [0, π2 ) be partitioned as in (5.12.1). then |A| p  ≤ sec2 p (α)(|A11 | p  + |A22 | p ). π

Remark 5.13.8 A ∈ Mn is accretive-dissipative if and only if W (e−i 4 A) ⊆ S π4 . Let α = π4 in Theorem 5.13.6, we have  f (|A|) ≤  f (2|A11 |) +  f (2|A22 |). Since f is concave, we have f (α A) ≤ α f (A), α > 1. Thus  f (|A|) ≤  f (2|A11 |) +  f (2|A22 |) ≤ 2( f (|A11 |) +  f (|A22 |)). Remark 5.13.9 Notice that when α = 0 in Theorem 5.13.6, we get Theorem 5.13.1. Notes and references. Theorem 5.13.6 and Corollary 5.13.7 in thus section are proved in [64].

176

5 Inequalities for Sector Matrices

5.14 A Further Extension of Rotfel’d Inequality for Accretive-Dissipative Matrices Lemma 5.12.1 states that

A ≤ U |A|U ∗

(5.14.1)

for some unitary matrix U ∈ Mn . Consider a partitioned matrix A ∈ Mn in the form  A=

 A11 A12 , A21 A22

(5.14.2)

where A11 and A22 are square. Here, we extend Theorem 5.13.1 to accretivedissipative matrices and to matrices whose numerical ranges lie in a sector Sα , 0 ≤ α ≤ π4 . We list some lemmas which are useful for our main results. Lemma 5.13.5 means that U (A + B)U ∗ ≤

√ 2|A + i B|

(5.14.3)

for some unitary matrix U ∈ Mn . A reverse inequality was recently given in Corollary 5.12.6 which states that for positive semidefinite A and B, there exists a unitary matrix V ∈ Mn such that √ V |A + i B|V ∗ ≤ 2(A + B).  A X ∈ Mm+n be positive semidefinite, where A and X∗ B B are positive semidefinite. Then there exist unitary matrices U, V ∈ Mm+n such that       0 0 A 0 A X ∗ =U V ∗. U +V 0 B 0 0 X∗ B 

Lemma 5.14.1 ([32]) Let

Lemma 5.14.2 ([2]) Let f : [0, ∞) → [0, ∞) be a concave function and let R, S ∈ Mn be positive semidefinite. Then there exist unitary matrices U and V such that f (R + S) ≤ U f (R)U ∗ + V f (S)V ∗ . Lemma 5.14.3 ([21]) Let A, B be any two matrices. Then there exist unitary matrices U, V such that |A + B| ≤ U |A|U ∗ + V |B|V ∗ . Note that Theorem 5.13.1 easily follows from Lemma 5.14.1 and Lemma 5.14.2. Next, we will follow this approach for matrices with a sectorial assumption on

5.14 A Further Extension of Rotfel’d Inequality for Accretive-Dissipative Matrices

177

their numerical ranges. Firstly, Theorem 5.13.1 is extended to the class of accretivedissipative matrices as follows. Theorem 5.14.4 Let f : [0, ∞) → [0, ∞) be a concave function and let A ∈ Mn be accretive-dissipative partitioned as in (5.14.1). Then, $ $  √ $ $  √ $ $ $ $ 2 2 $ $ $ $  f (|A|) ≤ 2 $ f |A11 | $ + $ f |A22 | $ . $ $ $ $ 2 2

(5.14.4)

Proof. Here we suppose that f (0) = 0. (Otherwise, let g(t) = f (t) − f (0). Since  f (X )(k) = g(X )(k) + k f (0) for every positive semidefinite matrix X and Ky Fan k-norms .k , it suffices to prove (5.14.4) when f (0) = 0). Consider the Cartesian decomposition A = R + i S, where R, S are positive semidefinite. First, by Bourin and Ricard [42], we have |R + i S| ≤

1 {(R + S) + U1 (R + S)U1∗ } 2

for some unitary matrix U1 . From this, we have for some unitary matrices U j , V j , j = 2, 3, 4, 1 {(R + S) + U1 (R + S)U1∗ } 2      1 R11 + S11 0 0 0 ∗ V2∗ (by Lemma 5.14.1) U2 U2 + V2 = 0 0 0 R22 + S22 2       1 R11 + S11 0 0 0 ∗ + U1 U2 U2 + V2 V2∗ U1∗ 0 0 0 R22 + S22 2       1 |R11 + i S11 | 0 0 0 ∗ V ∗ (by (5.14.3)) U3 + V3 ≤ √ U3 0 0 0 |R22 + i S22 | 3 2       1 |R11 + i S11 | 0 0 0 ∗ + √ U1 U3 V ∗ U1∗ U3 + V3 0 0 0 |R22 + i S22 | 3 2       1 |A11 | 0 0 0 ∗ = √ U3 V∗ U3 + V3 0 0 0 |A22 | 3 2       1 |A11 | 0 0 0 ∗ + √ U4 V∗ . U4 + V4 0 0 0 |A22 | 4 2

|A| ≤

It follows from Ky Fan Dominance Theorem that, for some unitary matrices U j , V j , j = 5, 6,

178

5 Inequalities for Sector Matrices 

   0 0 0 U3∗ + V3 0 √1 |A | V3∗ 22 0 0 2   1   0 0 √ |A11 | 0 2 + U4 U4∗ + V4 0 √1 |A | V4∗ ) 22 0 0 2

 1

   0 0 √ |A11 | 0 2 ≤ U5 f V5∗ (by Lemma 5.14.2) U5∗ + V5 f 1 0 √ |A22 | 0 0 2

 1

   0 0 √ |A11 | 0 2 + U6 f U6∗ + V6 f V6∗ ) 1 0 √ |A22 | 0 0 2 $ $ $ $ $ $ $ $ 1 $ + 2 $ f √1 |A22 | $ , |A ≤ 2$ f | √ 11 $ $ $ $ 2 2

 f (|A|) ≤  f (U3

√1 |A11 | 2

which leads to the desired result.



The following example motivated by an example in Lin [139] shows that the equality case in Theorem 5.14.4 may happen. Example 5.14.5 Let f (t) = t and 

     1+i i −1 1 −1 1 1 A= = +i . i −1 i +1 −1 1 1 1 Obviously, f is concave, and A is accretive-dissipative. A calculation reveals that s1 (A) = s2 (A) = 2. Specify the √ norm in this example to the trace norm .tr . Since |A11 |tr = |A22 |tr = 2, |A|tr = 4, it follows that √  f (|A|)tr = |A|tr = 2(|A11 |tr + |A22 |tr ) $ $√ $  $ √ $ 2 $ $ 2 $ $ $ $ $ |A11 |$ + $ |A22 |$ =2 $ $ 2 $ $ 2 $ tr tr $ √ $  √ $ $  $ $ $ $ 2 2 $ $ $ $ =2 $f |A11 | $ + $ f |A22 | $ . $ $ $ $ 2 2 tr

tr

π

If W (A) ⊆ S π4 , then ei 4 A is an accretive-dissipative matrix. The following corollary is easily obtained. Corollary 5.14.6 Let f : [0, ∞) → [0, ∞) be concave and let A with W (A) ⊆ S π4 be partitioned as in (5.14.2). Then $ $  √ $ $  √ $ $ $ $ 2 2 $ $ $ $ |A11 | $ + $ f |A22 | $ .  f (|A|) ≤ 2 $ f $ $ $ $ 2 2

5.14 A Further Extension of Rotfel’d Inequality for Accretive-Dissipative Matrices

179

For a general A with W (A) ⊆ Sα for α ∈ [0, π2 ), Theorem 5.13.1 is extended as follows. Theorem 5.14.7 Let f : [0, ∞) → [0, ∞) be a concave function and let A with W (A) ⊆ Sα for α ∈ [0, π2 ) be partitioned as in (5.14.2). Then  f (|A|) ≤  f (|A11 |) +  f (|A22 |) + 2( f (tan(α)|A11 |) +  f (tan(α)|A22 |)). Proof. Here, we also suppose that f (0) = 0. Consider the Cartesian decomposition A = R + i S, where R is positive semidefinite and S is Hermitian. The condition W (A) ⊆ Sα tells that ±S ≤ tan(α)R. Then, there exists a unitary matrix U such that        S 0  0  = |S| 0 ≤ U tan(α)R  U ∗.  0 0  0 0 0 tan(α)R

(5.14.5)

By Lemma 5.14.3, we obtain, for unitary matrices U, V, U j , V j , j = 1, 2, 3, 

 |A| 0 0 0    A 0   =  0 0       R 0 S 0  +i =  0 0 0 0      R 0 |S| 0 ≤ U1 U1∗ + V1 V∗ 0 0 0 0 1     R 0 tan(α)R 0 ≤ U1 U1∗ + V1 U U ∗ V1∗ (by (5.14.5) 0 0 0 tan(α)R     R 0 tan(α)R 0 = U1 U1∗ + V1 U U ∗ V1∗ 0 0 0 0   tan(α)R 0 + V1 U V V ∗ U ∗ V1∗ 0 0     R11 0 0 0 U2∗ U1∗ + U1 V2 V2∗ U1∗ (by Lemma 5.14.1) = U1 U2 0 0 0 R22     tan(α)R11 0 0 0 U2∗ U ∗ V1∗ + V1 U V2 V2∗ U ∗ V1∗ + V1 UU2 0 0 0 tan(α)R22   tan(α)R11 0 + V1 U V U2 U2∗ V ∗ U ∗ V1∗ 0 0   0 0 V2∗ V ∗ U ∗ V1∗ + V1 U V V2 0 tan(α)R22     |R11 + i S11 | 0 0 0 V ∗ V ∗U ∗ U3∗ U2∗ U1∗ + U1 V2 V3 ≤ U1 U2 U3 0 0 0 |R22 + i S22 | 3 2 1

180

5 Inequalities for Sector Matrices

 tan(α)|R11 + i S11 | 0 U3∗ U2∗ U ∗ V1∗ (by (5.14.1)) 0 0   0 0 V ∗ V ∗ U ∗ V1∗ + V1 U V2 V3 0 tan(α)|R22 + i S22 | 3 2   tan(α)|R11 + i S11 | 0 + V1 U V U2 U3 U3∗ U2∗ V ∗ U ∗ V1∗ 0 0   0 0 + V1 U V V2 V3 V ∗ V ∗ V ∗ U ∗ V1∗ . 0 tan(α)|R22 + i S22 | 3 2 

+ V1 UU2 U3

(5.14.6)

It follows from Lemma 5.14.2 and (5.14.6) that 

 f (|A|) 0 0 0

  |A| 0 = f 0 0

  |R11 + i S11 | 0 U3∗ U2∗ U1∗ U4∗ ≤ U4 U1 U2 U3 f 0 0 

 0 0 V3∗ V2∗ U1∗ V4∗ + V4 U1 V2 V3 f 0 |R22 + i S22 |

  tan(α)|R11 + i S11 | 0 + U5 V1 UU2 U3 f U3∗ U2∗ U ∗ V1∗ U5∗ 0 0 

 0 0 V3∗ V2∗ U ∗ V1∗ V5∗ + V5 V1 U V2 V3 f 0 tan(α)|R22 + i S22 | 

 tan(α)|R11 + i S11 | 0 U3∗ U2∗ V ∗ U ∗ V1∗ U6∗ + U6 V1 U V U2 U3 f 0 0

  0 0 + V6 V1 U V V2 V3 f V3∗ V2∗ V ∗ U ∗ V1∗ V6∗ , 0 tan(α)|R22 + i S22 |

(5.14.7)

where U j , V j , j = 4, 5, 6, are unitary matrices. Taking the unitarily invariant norm on both sides of (5.14.7) can easily lead to the desired result.  If we take f (t) = t p , 0 < p ≤ 1 in Theorem 5.14.4 and Theorem 5.14.7, respectively, the following two corollaries can easily be derived. Corollary 5.14.8 Let 0 < p ≤ 1, and let A ∈ Mn be accretive-dissipative partitioned as in (5.14.2). Then |A| p  ≤ 2

1− p 2



|A11 | p  + |A22 | p  .

Corollary 5.14.9 Let 0 < p ≤ 1, and Let A with W (A) ⊆ Sα for α ∈ [0, π2 ) be partitioned as in (5.14.2). Then

5.14 A Further Extension of Rotfel’d Inequality for Accretive-Dissipative Matrices

181



|A| p  ≤ 1 + 2(tan α) p |A11 | p  + |A22 | p  . π

Remark 5.14.10 Matrix A is accretive-dissipative if and only if W e− 4 A ⊆ S π4 . Let α = π4 be in Theorem 5.14.7 to derive $ π $  f (|A|) = $ f |e− 4 A| $ $ π

$ $ π

$

≤ 3 $ f |e− 4 A11 | $ + $ f |e− 4 A22 | $ = 3 ( f (|A11 |) +  f (|A22 |)) .

(5.14.8)

The bound in (5.14.8) is weaker than that in Theorem 5.14.4. Notice that when α = 0, Theorem 5.14.7 reduces to Theorem 5.13.1. Notes and references. Theorem 5.14.4, Corollary 5.14.6, Theorem 5.14.7 and the following two last Corollaries 5.14.8, 5.14.9 are due to [214].

5.15 Hilbert-Schmidt Norm Inequalities for Accretive-Dissipative Operators Let B(H) stand for the C ∗ -algebra of all bounded linear operators on a complex Hilbert space H. For H = H  ⊕ H and T ∈ B(H), the operator T can be represented T11 T12 , with T jk ∈ B(H) for j, k = 1, 2. as an operator matrix T = T21 T22 For any T ∈ B(H), we can write T = A + i B, ∗

(5.15.1)



and B = T −T are Hermitian operators. This is the so-called in which A = T +T 2 2i Cartesian decomposition of T . In what follows, we will always represent the decomposition (5.15.1) as follows: 

T11 T12 T21 T22



 =

   A11 A12 B11 B12 +i , A21 A22 B21 B22

(5.15.2)

∗ in which T jk , B jk , A jk ∈ B(H), j, k = 1, 2. Then A12 = A∗21 and B12 = B21 . If T is a compact operator, we denote by σ1 (T ) ≥ σ2 (T ) ≥, ... the eigenvalues 1 of (T ∗ T ) 2 which are called the singular values of T . Thus, whenever we talk about singular values, the operators are necessarily compact.

182

5 Inequalities for Sector Matrices

A compact operator T is said to be in the Hilbert-Schmidt class, if

∞  j=1

σ 2j (T ) < ∞,

i.e., T ∗ T is the trace class. The Hilbert-Schmidt norm of operator T is defined by ' ( (∞ 2 σ j (T ), T 2 = ) j=1

√ and it is known that T 2 = T r (T ∗ T ), where T r denotes the usual trace functional. The Hilbert-Schmidt norm is of course a very special case of a much wider class of unitarily invariant norms T u , which satisfy T u = U T V u for all unitaries U, V ∈ B(H). Each of these unitarily invariant norms is defined on an ideal in B(H), and it will be implicitly understood that when we talk of T u , then the operator T is in this ideal. As we know, T is accretive (resp. dissipative) if in its Cartesian decomposition (5.15.1), A (resp. B) is positive and T is accretive-dissipative if both A and B are positive. We remark that T is dissipative if and only if −i T is accretive. Dissipative operators have found many applications, for example, Gunzburger and Plemmons [91] use results in [60] in their study of energy-conserving norms for the solution of hyperbolic systems of partial differential equations, where the term dissipative is used in the sense that energy is nonincreasing in time. Siegel [192] uses strictly dissipative matrices in the study of the analytic theory of abelian varieties. For the study of accretive-dissipative matrices in the numerical linear algebra context, we refer to [83, 98, 150]. Recent works devoted to studying this kind of operator or matrix are [81, 139, 140]. Our main consideration in this section is norm inequalities for the accretivedissipative operator matrix (5.15.2). The topic has been investigated in [27, 29], but from a different perspective. They are mainly concerned with inequalities between the norm of T and those of A and B, following or followed by [14, 49, 212]. We present inequalities between the norm of off diagonal blocks and that of diagonal blocks, the norm of the whole operator matrix, and that of its diagonal blocks. Unless otherwise specified, operators in this section are assumed to be in the Hilbert-Schmidt class. We can easily obtain that   X Y ∈ B(H) where X, Y, Z ∈ B(H) is positive, then Lemma 5.15.1 If Y∗ Z Y 22 ≤ X 2 Z 2 . Proposition 5.15.2 Let T ∈ B(H) be accretive-dissipative and partitioned as in (5.15.2). Then T12 22 ≤ 2T11 2 T22 2 .

5.15 Hilbert-Schmidt Norm Inequalities for Accretive-Dissipative Operators

183

Proof. Observe that ∗ ∗ ∗ T12 = A∗12 A12 + B12 B12 + i(A∗12 B12 − B12 A12 ), T12 ∗ ∗ ∗ ∗ T21 T21 = A21 A21 + B21 B21 − i(A21 B21 − B21 A∗21 ), ∗ ∗ = A∗12 A12 + B12 B12 − i(A∗12 B12 − B12 A12 ).

Compute T12 22 ≤ T12 22 + T21 22 ∗ = 2T r (A∗12 A12 + B12 B12 ) = 2(A12 22 + B12 22 ) ≤ 2(A11 2 A22 2 + B11 2 B22 2 ) ( by Lemma 5.15.1) * * ≤ 2 A11 22 + B121 22 A22 22 + B22 22 ) ( by Cauchy-Schwarz) = 2T11 2 T22 2 .  The following example illustrates that the factor 2 on the right-hand side of Proposition 5.15.2 is unavoidable. Example 5.15.3 Consider the matrix  T =

   1+ε 1 1 + ε −i +i , 1 ε i 1+ε

for ε > 0. A calculation reveals that T12 22 = 4 and T11 2 T22 2 = 2(1 + ε)2 , so T12 22 2 = . T11 2 T22 2 (1 + ε)2 As ε → 0+ , the ratio becomes 2. Proposition 5.15.4 Let T ∈ B(H) be accretive-dissipative and partitioned as in (5.15.2). Then (5.15.3) T12 2 T21 2 ≤ T11 2 T22 2 . Proof. According to the proof of the previous proposition, we have T12 22 + T21 22 ≤ 2T11 2 T22 2 .

(5.15.4)

Then (5.15.3) follows from an application of the arithmetic mean-geometric mean inequality to the left-hand side of (5.15.4) and this completes the proof. 

184

5 Inequalities for Sector Matrices

Remark 5.15.5 Proposition 5.15.4 is an extension of [98]. In the following, we give some estimates of T 2 in terms of the Hilbert-Schmidt norm of its diagonal blocks. Proposition 5.15.6 Let T ∈ B(H) be accretive-dissipative and partitioned as in (5.15.2). Then T 2 ≤ T11 2 + T22 2 . Proof. Compute T 22 = T11 22 + T12 22 + T21 22 + T22 22 ≤ T11 22 + 2T11 2 T22 2 + T22 22 = (T11 2 + T11 2 )2 , where the first inequality is by (5.15.4). The proof is completed by taking the square root.  Theorem 5.15.7 Let T ∈ B(H) be accretive-dissipative and partitioned as in (5.15.2). Then √ (5.15.5) T 2 ≤ 2T11 + T22 2 . Proof. We observe that (5.15.5) is equivalent to T11 22 + T12 22 + T21 22 + T22 22 ∗ ∗ ≤ 2(T11 22 + T22 22 + T r (T22 T11 + T11 T22 )),

that is, T12 22 + T21 22 ∗ ∗ ≤ T11 22 + T22 22 + 2T r (T22 T11 + T11 T22 )).

(5.15.6)

From (5.15.4), we know that T12 22 + T21 22 ≤ T11 22 + T22 22 .

(5.15.7)

∗ ∗ T11 + T11 T22 ) ≥ 0. Thus, to show (5.15.6), it suffices to show that T r (T22 1

1

1

1

2 2 2 2 As A11 , A22 , B11 , B22 are positive, so are A22 A11 A22 and B22 B11 B22 . A calculation gives

∗ ∗ T11 = (A∗22 − i B22 )(A11 + i B11 ) T22 = A22 A11 + B22 B11 − i(B22 A11 − A22 B11 ),

5.15 Hilbert-Schmidt Norm Inequalities for Accretive-Dissipative Operators

and so

185

∗ T22 = A11 A22 + B11 B22 + i(A11 B22 − B11 A22 ). T11

Thus ∗ ∗ T11 + T11 T22 ) = 2T r (A11 A22 + B11 B22 ) T r (T22 1

1

1

1

2 2 2 2 = 2T r (A22 A11 A22 + B22 B11 B22 )

≥ 0. This completes the proof.



We can give a unified improvement of Proposition 5.15.6 and (5.15.5) if the offdiagonal blocks T12 and T21 coincide. We need some lemmas to prove this assertion. Lemma 5.15.8 Let T ∈ B(H) be accretive-dissipative and partitioned as in (5.15.2). If T12 = T21 , then A12 = A21 and B12 = B21 , i.e., the off-diagonal blocks of A and B are Hermitian operators on H. Proof. Note that T12 = T21 means A12 + i B12 = A21 + i B21 .

(5.15.8)

Taking adjoint, we obtain ∗ ∗ = A∗21 − i B21 , A∗12 − i B12

i.e., A21 − i B21 = A12 − i B12 .

(5.15.9)

It follows from (5.15.8) and (5.15.9) that i(B12 − B21 ) = i(B21 − B12 ). Thus B12 = B21 . Then from (5.15.8), we get A12 = A21 .



Lemma 5.15.9 ([21]) If X, Y ∈ B(H) are such that X ∗ Y is normal, then X ∗ Y u ≤ Y X ∗ u , for any unitarily invariant norm .u .

186

5 Inequalities for Sector Matrices

 A11 A12 ∈ B(H) be positive with A12 = A21 , where A21 A22 ∈ B(H), j = 1, 2, are trace class. Then we have 

Lemma 5.15.10 Let A = A jk

T r A212 ≤ T r A11 A22 .

(5.15.10)

Proof. As A is positive, it has a (unique) positive square root B. Thus A = B ∗ B. Now one can select any isometry V from H to H and set [X Y ] = V B. Then A = (V B)∗ (V B)



X ∗ X X ∗Y Y ∗ X Y ∗Y

 .

That is, A11 = X ∗ X , A12 = A21 = X ∗ Y = Y ∗ X and A22 = Y ∗ Y . Applying Lemma 5.15.10 with respect to the Hilbert-Schmidt norm gives T r (X ∗ Y )2 ≤ T r X ∗ X Y ∗ Y, which is exactly (5.15.10). Theorem 5.15.11 Let T ∈ B(H) be accretive-dissipative and partitioned as in (5.15.2). If T12 = T21 , then (5.15.11) T 2 ≤ T11 + T22 2 . Proof. We observe that (5.15.11) is equivalent to ∗ ∗ T11 + T11 T22 ). 2T12 22 ≤ T r (T22 ∗ . Now Moreover, Lemma 5.15.8 tells us A12 = A∗12 and B12 = B12 ∗ B12 T12 22 = T r A∗12 A12 + T r B12 2 = T r A212 + T r B12 ,

and

∗ ∗ T11 + T11 T22 ) = 2T r (A11 A22 + B11 B22 ). T r (T22

Thus, by Lemma 5.15.10, the desired inequality(5.15.11) follows. In the following, we shall extend the previous results to an arbitrary unitarily invariant norm, with the sacrifice of the sharpness of inequalities. We start with two lemmas. Lemma 5.15.12 ([21]) Let x = (x1 , x2 , ...), y = (y1 , y2 , ...), α = (α1 , α2 , ...) be sequences of real numbers with entries arranged in decreasing order. Moreover, we

5.15 Hilbert-Schmidt Norm Inequalities for Accretive-Dissipative Operators

187

  assume the entries of α are nonnegative. If kj=1 x j ≤ kj=1 y j for all k = 1, 2, .... Then k k   αjxj ≤ αj yj j=1

j=1

for all k = 1, 2, .... Lemma 5.15.13 ([21]) If A ∈ B(H) is positive and B ∈ B(H) is Hermitian, then σ j (A) ≤ σ j (A + i B), for j = 1, 2, .... We shall also make use of the following facts without explicit referring, the details can be found in [21]. (i) For A, B ∈ B(H), k 

σ j (A + B) ≤

j=1

k 

σ j (A) + σ j (B),

j=1

k=   1, 2, .... X Y ∈ B(H), where X, Y, Z ∈ B(H) is positive, then (ii) If Y∗ Z k 

σ 2j (Y ) ≤

j=1

k 

σ j (X )σ j (Z ),

j=1

k = 1, 2, .... As a consequence k 

σ j (Y ) ≤

j=1

k 

1

1

σ j2 (X )σ j2 (Z ),

j=1

k = 1, 2, .... Theorem 5.15.14 Let T ∈ B(H) be accretive-dissipative and partitioned as in (5.15.2). Then T12 u T21 u ≤ max{T12 2u , T21 2u } ≤ 4T11 u T22 u for any unitarily invariant norm .u .

188

5 Inequalities for Sector Matrices

Proof. The first inequality is trivial. We only consider the second one. Let α = (α1 , α2 , ...) be a sequence with nonnegative entries and α1 ≥ α2 ≥ ..., α1 > 0. ∞  Define X α = α j σ j (X ) for X ∈ B(H) . k=1

Compute T12 α = = ≤ ≤

∞  k=1 ∞  k=1 ∞  k=1 ∞ 

α j σ j (T12 ) α j σ j (A12 + i B12 ) α j [σ j (A12 ) + σ j (B12 )] 1

1

1

1

α j [σ j2 (A11 )σ j2 (A22 ) + σ j2 (B11 )σ j2 (B22 )]

k=1



∞ 

1

1

α j [σ j (A11 ) + σ j (B11 )] 2 [σ j (A22 ) + σ j (B22 )] 2 (by Cauchy-Schwarz)

k=1



∞ 

1

1

α j [2σ j (A11 + i B11 )] 2 [2σ j (A22 + i B22 )] 2 (by Lemma 5.15.13)

k=1

=2

∞ 

1

1

α j σ j2 (T11 )σ j2 (T221 )

k=1

≤2

∞ 

α j σ j (T11 )

k=1

 21  ∞ 

 21 α j σ j (T22 )

(by Cauchy-Schwarz)

k=1 1

1

= 2T11 α2 T22 α2 , where throughout, Lemma 5.15.12 is used in the first and the second inequality. Similarly, one also has 1

1

T21 α ≤ 2T11 α2 T22 α2 . As α is arbitrary chosen, the alleged inequality follows from [105].



The following result is due to Bhatia and Kittaneh [24]. Proposition 5.15.15 If A, B ∈ B(H) are positive, then σ j (A + B) ≤



2σ j (A + i B)

(5.15.12)

5.15 Hilbert-Schmidt Norm Inequalities for Accretive-Dissipative Operators

189

for j = 1, 2, .... For our purpose, we need a result weaker than proposition 5.15.15. Lemma 5.15.16 If A, B ∈ B(H) are positive, then A + Bu ≤



2A + i Bu

for any unitarily invariant norm .u . We also need to invoke the following reverse relation. Proposition 5.15.17 ([211]) If A, B ∈ B(H) are positive, then k 

k 

σ j (A + B) ≥

j=1

σ j (A + i B)

(5.15.13)

j=1

for j = 1, 2, .... Indeed, something more is proved in [211]. That is, under the same hypothesis of Proposition 5.15.17, k 

σ j (A + |z|B) ≥

j=1

k 

σ j (A + z B),

j=1

for k = 1, 2, ... and for any complex number z. Also, we cannot expect further strengthening of (5.15.13) in the sense of (5.15.12), see [24]. Notice that we are informed by J. C. Bourin that a generalization of Proposition 5.15.17 to normal operator case can be found in [21, 34]. Corollary 5.15.18 If A, B ∈ B(H) are positive, then k 

σ j (A + B) ≥

j=1

k 

σ j (A + i B)

j=1

for k = 1, 2, .... Or equivalently, A + Bu ≥ A + i Bu for any unitarily invariant norm .u . Proof. This follows immediately from Proposition 5.15.17.



190

5 Inequalities for Sector Matrices

 Lemma 5.15.19 ([105]) Let A = j, k = 1, 2. Then

A11 A12 A21 A22

 ∈ B(H) be positive with A jk ∈ B(H),

Au ≤ A11 u + A22 u

for any unitarily invariant norm .u . Remark 5.15.20 It has been shown that if A12 = A21 , then Au ≤ A11 + A22 u for any unitarily invariant norm, see [35]. Theorem 5.15.21 Let T ∈ B(H) be accretive-dissipative and partitioned as in (5.15.2). Then √ (5.15.14) T u ≤ 2(T11 u + T22 u ) for any unitarily invariant norm .u . Furthermore, if T12 = T21 , then T u ≤



2T11 + T22 u .

(5.15.15)

Proof. Compute T u = A + i Bu ≤ A + Bu ( by Corollary 5.15.18) ≤ A11 + B11 u + A22 + B22 u ( by Lemma 5.15.19) √ ≤ 2(A11 + i B11 u + A22 + i B22 u ) ( by Lemma 5.15.16) √ = 2(T11 u + T22 u ), and this proves (5.15.14). Now we prove (5.15.15). By Lemma 5.15.8, we know that  A+B =

A11 + B11 A12 + B12 A21 + B21 A22 + B22



is positive with Hermitian off-diagonal blocks. Using a unitary similarity with   1 i I −I J=√ , 2 iI I we have    A11 + B11 A12 + B12 ∗ J = J A21 + B21 A22 + B22

A11 +B11 +A22 +B22 2





A11 +B11 +A22 +B22 2

 ,

5.15 Hilbert-Schmidt Norm Inequalities for Accretive-Dissipative Operators

191

where  stands for the unspecified operator. Then T u = A + i Bu $+ ,$ $ $ A11 +B11 +A22 +B22  $ $ 2 =$ $ A +B +A +B 11 11 22 22 $ $  2 u $ $ $ $ $ A11 + B11 + A22 + B22 $ $ A11 + B11 + A22 + B22 $ $ +$ $ (by Lemma 5.15.19) ≤$ $ $ $ $ 2 2 u u = A11 + B11 + A22 + B22 u √ ≤ 2A11 + A22 + i(B11 + B22 )u (by Lemma 5.15.16) √ = 2A11 + i B11 + A22 + i B22 u √ = 2T11 + T22 u

as desired.



Notes and references. Note that Lemma 5.15.1, Lemma 5.15.10, Lemma 5.15.8, Proposition 5.15.2, Proposition 5.15.4, Proposition 5.15.6, Theorem 5.15.7, Theorem 5.15.11, Theorem 5.15.14, Theorem 5.15.21, and Corollary 5.15.18 are due to [148].

5.16 Schatten p-Norm Inequalities for Accretive-Dissipative Matrices Suppose T ∈ Mn . For i, j = 1, 2, ..., n, let Ti j be square matrices of the same size such that the block matrix ⎡ ⎤ T11 T12 ... T1n ⎢ T21 T22 ... T2n ⎥ ⎢ ⎥ ⎢ . . ... . ⎥ ⎢ ⎥ (5.16.1) T =⎢ . ... . ⎥ ⎢ . ⎥ ⎣ . . ... . ⎦ Tn1 Tn2 ... Tnn is accretive-dissipative. Recently, some norm inequalities for 2 × 2 block accretive-dissipative matrices have been established using the Schatten p-norms. For p > 0 and A ∈ Mn , let

192

5 Inequalities for Sector Matrices

⎛ ⎞ 1p n  p A p = ⎝ σ j (A)⎠ , j=1

where σ1 (A), σ2 (A), ..., σn (A) are the singular values of A. So,

1 A p = T r |A| p p . For p ≥ 1, this is the Schatten p-norm of A. It is known that Schatten p-norms are important examples of unitarily invariant norms, and these include the spectral norm, the Hilbert-Schmidt norm, and the trace norm, which correspond to the cases p = ∞, p = 2 and p = 1, respectively. In the following, we will present some norm inequalities, which compare the Schatten p-norms of the off-diagonal blocks and those of the diagonal blocks. These norm inequalities are based on the following lemma. Lemma 5.16.1 ([92]) For  i, j = 1, 2, let Ti j be square matrices of the same size T11 T12 is accretive-dissipative. Then such that T = T21 T22 (i) For p ≥ 2, we have p

p

p

p

T12  pp + T21  pp ≤ 2 p−1 T11  p2 T22  p2 . (ii) For 0 < p ≤ 2, we have T12  pp + T21  pp ≤ 23− p T11  p2 T22  p2 . Lemma 5.16.2 ([194]) Let A, B ∈ Mn (C) be positive semidefinite. Then A pp + B pp ≤ A + B pp ≤ 2 p−1 (A pp + B pp ), for p ≥ 1. Lemma 5.16.3 ([24]) Let A, B ∈ Mn (C) be positive semidefinite. Then A + i B p ≤ A + B p ≤

√ 2A + i B p ,

for p ≥ 1. Theorem 5.16.4 Let T be represented as in (5.16.1). Then (i) For p ≥ 2, we have

5.16 Schatten p-Norm Inequalities for Accretive-Dissipative Matrices



Ti j  pp ≤ (n − 1)2 p−2

n 

i= j

193

Tii  pp .

i=1

(ii) For 0 < p ≤ 2, we have 

Ti j  pp ≤ (n − 1)22− p

n 

i= j

Tii  pp .

i=1

  T T Proof. (i) Suppose that p ≥ 2. Since T is accretive-dissipative, and since ii i j T ji T j j   T T is a principal submatrix of T , it follows that ii i j is accretive-dissipative. Now, T ji T j j   Tii Ti j applying (i) of Lemma 5.16.1 to , we get T ji T j j p p Ti j  pp + T ji  pp ≤ 2 p−1 Tii  p2 T j j  p2 for i = j. Consequently, using arithmetic-geometric mean inequality, we have

Ti j  pp + T ji  pp ≤ 2 p−2 Tii  pp + T j j  pp for i = j. Adding up the previous inequalities for i, j = 1, 2, ..., n, we get 

Ti j  pp ≤ (n − 1)2 p−2

i= j

n 

Tii  pp ,

i=1

which proves the first inequality. (ii) To prove thesecond inequality, let 0 < p ≤ 2. Now, applying (ii) of Lemma Tii Ti j 5.16.1 to , we get T ji T j j p p Ti j  pp + T ji  pp ≤ 23− p Tii  p2 T j j  p2 for i = j. Consequently, using arithmetic-geometric mean inequality, we have

Ti j  pp + T ji  pp ≤ 2 p−2 Tii  pp + T j j  pp , for i = j. Adding up the previous inequalities for i, j = 1, 2, ..., n, we get  i= j

Ti j  pp ≤ (n − 1)22− p

n  i=1

Tii  pp ,

194

5 Inequalities for Sector Matrices



as required.

For the Hilbert-Schmidt norm, we have the following corollary. Corollary 5.16.5 Let T be represented as in (5.16.1). Then T 22 ≤ n

n 

Tii 22 .

i=1

Proof. We have T 22 =

n 

Ti j 22

i, j=1

=

n 

Tii 22 +



Ti j 22 .

i= j

i=1

Using Theorem 5.16.4, for p = 2, we get T 22 ≤

n 

Tii 22 + (n − 1)

i=1

Thus T 22 ≤ n

n 

Tii 22 .

i=1

n 

Tii 22 .

i=1

Remark 5.16.6 It is possible to obtain Schatten p-norm inequalities related to the Hilbert-Schmidt norm inequality presented in Corollary 5.16.5. This can be achieved by employing various Schatten p-norm inequalities for partitioned operators given in [23]. We conclude here with the following Schatten p-norm inequalities involving sums of accretive-dissipative matrices. Theorem 5.16.7 Let S, T ∈ Mn (C) be accretive-dissipative. Then 2

−p 2

(S pp + T  pp ) ≤ S + T  pp ≤ 2

3( p−1) 2

(S pp + T  pp )

for p ≥ 1. Proof. Let S = A + i B and T = C + i D be the Cartesian decomposition of S and T . Then we have

5.16 Schatten p-Norm Inequalities for Accretive-Dissipative Matrices

195

S + T  pp = A + i B + C + i D pp = (A + C) + i(B + D) pp ≥2 =2 ≥2 ≥2 =2

−p 2 −p 2 −p 2 −p 2 −p 2

(A + C) + (B + D) pp (by Lemma 5.16.3) (A + B) + (C + D) pp (A + B pp + C + D pp ) (by Lemma 5.16.2) (A + i B pp + C + i D pp ) (by Lemma 5.16.3) (S pp + T  pp ),

which proves the first inequality. To prove the second inequality, note that S + T  pp = A + i B + C + i D pp = (A + C) + i(B + D) pp ≤ (A + C) + (B + D) pp (by Lemma 5.16.3) = (A + B) + (C + D) pp ≤ 2 p−1 (A + B pp + C + D pp ) (by Lemma 5.16.2) p

p

≤ 2 p−1 (2 2 A + i B pp + 2 2 C + i D pp ) (by Lemma 5.16.3) =2 as required.

3( p−1) 2

(S pp + T  pp )



Next, we introduce the accretive-dissipative versions of the Minkowski’s determinant inequality (5.4.4) and the following Young type determinantal inequality [104] (5.16.2) (det A)α (det A)1−α ≤ det(α A + (1 − α)B) for α ∈ (0, 1). Our results are based on the inequalities (5.4.4) and (5.16.2) and the following Lemma which we introduced it in section 5.10. Lemma 5.16.8 Let A, B ∈ Mn (C) be positive semidefinite. Then n

| det(A + i B)| ≤ det(A + B) ≤ 2 2 | det(A + i B)|. Theorem 5.16.9 Let S, T ∈ Mn (C) be accretive-dissipative. Then √

1

1

1

2| det(S + T )| n ≥ | det S| n + | det T | n .

Proof. Let S = A + i B and T = C + i D be the Cartesian decomposition of S and T . Then we have

196

5 Inequalities for Sector Matrices

| det(S + T )| = | det(A + i B + C + i D)| = | det((A + C) + i(B + D))| ≥2

−n 2

det((A + C) + (B + D)) (by Lemma 5.16.8)

=2

−n 2

det((A + B) + (C + D)).

By taking the nth roots of both sides of the previous inequality, we get 1

−1

1

| det(S + T )| n ≥ 2 2 (det((A + B) + (C + D))) n −1

1

1

≥ 2 2 ((det(A + B)) n + (det(C + D)) n ) (by (5.4.4)) −1

1

1

≥ 2 2 (| det(A + i B)| n + | det(C + i D)| n ) (by Lemma 5.16.8) −1

1

1

= 2 2 (| det S| n + | det T | n ), and so

√ 1 1 1 2| det(S + T )| n ≥ | det S| n + | det T | n , as required.



Theorem 5.16.10 Let S, T ∈ Mn (C) be accretive-dissipative. Then | det S|α | det T |1−α ≤ 2 2 | det(αS + (1 − α)T )| n

for α ∈ (0, 1). Proof. Let S = A + i B and T = C + i D be the Cartesian decomposition of S and T . Then we have | det S|α | det T |1−α = | det(A + i B)|α | det(C + i D)|1−α ≤ (det(A + B))α (det(C + D))1−α (by Lemma 5.16.8) ≤ det(α(A + B) + (1 − α)(C + D)) (by (5.16.2)) 1−α = αn det((A + B) + (C + D)) α 

1−α 1−α C + B+ D = αn det A+ α α 



   1−α 1−α n A+ C +i B + D  (by Lemma 5.16.8) ≤ αn 2 2 det α α

5.16 Schatten p-Norm Inequalities for Accretive-Dissipative Matrices

197

   1−α n  (C + i D)) = αn 2 2 det((A + i B) + α     1−α n = αn 2 2 det S + T  α n

= 2 2 | det(αS + (1 − α)T )|, which completes the proof.



Remark 5.16.11 It can be seen that the determinant inequalities in Theorems 5.16.9 and Theorems 5.16.10 are sharp. This can be demonstrated by considering the accretive-dissipative matrices 

1+i 1−i S= 1−i 1+i Note that

and







 1 + i −1 + i and T = . −1 + i 1 + i 1

1

1

2| det(S + T )| 2 = | det S| 2 + | det T | 2 ,

   S T   . + | det S| | det T | = 2 det 2 2  1 2

1 2

Notes and references. We notice that the proof of Theorem 5.16.4, Theorem 5.16.7, Theorem 5.16.9, Theorem 5.16.10, and Corollary 5.16.5 is due to [122].

5.17 Schatten p-Norm Inequalities for Sector Matrices A matrix T ∈ M2n (C) can be partitioned as a 2 × 2 block matrix  T =

 T11 T12 , T21 T22

(5.17.1)

where Ti j ∈ Mn (C), i, j = 1, 2. Let x = (x1 , ..., xn ), y = (y1 , ..., yn ) ∈ Rn . We write x ≤ y to mean x j ≤ y j for ↓ j = 1, ..., n. We rearrange the components of x and y in nonincreasing order x1 ≥ ↓ ↓ ↓ ... ≥ xn and y1 ≥ ... ≥ yn . It is well known that for A, B ∈ Mn (C), A ≤ B for all unitarily invariant norms . on Mn (C) if and only if σ(A) ≺ω σ(B). So, to some extent, the norm inequalities are essentially the same as the singular value majorization inequalities.

198

5 Inequalities for Sector Matrices

In this section, we will extend the results of Lemma 5.16.1 and Theorem 5.16.4 to a large class of matrices whose numerical ranges are contained in a sector called sector matrices. Firstly, we are interested in norm inequalities for sector block matrices in M2n (C) and Schatten p-norm inequalities involving sums of two sector matrices in Mn (C). For presenting and proving our results, we need the following several facts. The next lemma is known as the Ky Fan-Hoffman inequality in the literature. Lemma 5.17.1 ([21]) Let A ∈ Mn . Then A ≤ A

(5.17.2)

for all unitarily invariant norms . on Mn (C). Lemma 5.17.2 ([217]) Let A ∈ Mn (C) have W (A) ⊆ Sα for some α ∈ [0, π2 ). Then σ(A) ≺ω sec(α)λ(A). Equivalently, for all unitarily invariant norms . on Mn (C), A ≤ sec(α)A.

(5.17.3)

Lemma 5.17.3 ([217]) Let A be an n × n complex matrix such that W (A) ⊆ Sα for some α ∈ [0, π2 ) and let A = X Z X ∗ be a sectoral decomposition of A, where X is invertible and Z is unitary and diagonal. Then for any matrix R and all j = 1, ..., n, the following inequalities hold R R ∗ ≤ sec(α)(R(Z )R ∗ ) = sec(α)((R Z R ∗ )), and

σ 2j (R) ≤ sec(α)λ j (R(Z )R ∗ ) ≤ sec(α)σ j (R Z R ∗ ).

(5.17.4)

(5.17.5)

Lemma 5.17.4 ([217]) Let T ∈ M2n (C) be partitioned as in (5.17.1) and assume W (T ) ⊆ Sα for some α ∈ [0, π2 ). Then for any unitarily invariant norm . on Mn (C), max{T12 2 , T21 2 } ≤ sec2 (α)T11 T22 . Lemma 5.17.5 Let T ∈ M2n (C) be partitioned as in (5.17.1) and assume W (T ) ⊆ Sα for some α ∈ [0, π2 ). Then p

p

max{T12  pp , T21  pp } ≤ sec p (α)T11  p2 T22  p 2 for p > 0.

(5.17.6)

5.17 Schatten p-Norm Inequalities for Sector Matrices

199

Proof. For p ≥ 2, since the Schatten p-norms are the examples of unitarily invariant norms, then by Lemma 5.17.4, the result is clear. When  p < 2, let T11 be n × n.  0< C 1 , C1 ∈ Mn×2n (C). Then By Lemma 5.1.4, we write T = C ZC ∗ with C = C2 ∗ T12 = C1 ZC2 . By [21], we compute n 

p

σ j (T12 )

j=1

=

n 

σ j (C1 ZC2∗ ) p

j=1



n 

σ j (C1 )σ j (Z )σ j (C2∗ ) p

p

p

j=1

=

n 

σ j (C1 )σ j (C2∗ ) p

p

j=1

=

n 

p

p

σ j2 (C1 C1∗ )σ j2 (C2 C2∗ )

j=1



n 

p

p

sec p (α)σ j2 (C1 (Z )C1∗ )σ j2 (C2 (Z )C2∗ )

(by (5.17.4))

j=1



n 

p

p

sec p (α)σ j2 (C1 ZC1∗ )σ j2 (C2 ZC2∗ ).

(by (5.17.4))

j=1

These inequalities, together with the fact that weak log-majorization implies weak majorization, imply that n 

p

σ j (T12 ) ≤

j=1

n 

p

p

sec p (α)σ j2 (C1 ZC1∗ )σ j2 (C2 ZC2∗ )

j=1

=

n  j=1

p

p

sec p (α)σ j2 (T11 )σ j2 (T22 ) ⎛

≤ sec p (α) ⎝

n 

⎞ 21 ⎛ p σ j (T11 )⎠ ⎝

j=1

n 

⎞ 21 σ j (T22 )⎠ p

j=1 p 2

p 2

= sec p (α)T11  p T22  p , which the last inequality followed by Cauchy-Schwarz inequality. Thus

200

5 Inequalities for Sector Matrices p

p

T12  pp ≤ sec p (α)T11  p2 T22  p2 . Similarly, the inequality for T21 is proven. So p

p

max{T12  pp , T21  pp } ≤ sec p (α)T11  p2 T22  p 2 . This completes the proof.



With the above preparation, let us first present the generalization of Lemma 5.16.1 and Theorem 5.16.4 in the next two theorems. The following result is a generalization of Lemma 5.16.1. Theorem 5.17.6 Let T ∈ M2n (C) be partitioned as in (5.17.1) and assume W (T ) ⊆ Sα for some α ∈ [0, π2 ). Then p

p

T12  pp + T21  pp ≤ 2 sec p (α)T11  p2 T22  p2

(5.17.7)

for p > 0. Proof. By (5.17.6), we have T12  pp + T21  pp p

p

p

p

≤ sec p (α)T11  p2 T22  p2 + sec p (α)T11  p2 T22  p2 p

p

= 2 sec p (α)T11  p2 T22  p2 .  Remark 5.17.7 Set α = π4 in Theorem 5.17.6. That is T is an accretive-dissipative p matrix. Then sec p (α) = 2 2 . This means that p

p

p

T12  pp + T21  pp ≤ 2( 2 )+1 T11  p2 T22  p2 . When p > 4 or 0 < p < 43 , it is clear that Theorem 5.17.6 is stronger than the inequalities in Lemma 5.16.1. The next theorem is another generalization of Theorem 5.16.4. Theorem 5.17.8 For i, j = 1, 2, ..., n, Let Ti j be square matrices of the same size such that the bock matrix ⎤ ⎡ T11 T12 ... T1n ⎢ T21 T22 ... T2n ⎥ ⎥ ⎢ ⎢ . . ... . ⎥ ⎥ ⎢ T =⎢ . ... . ⎥ ⎥ ⎢ . ⎣ . . ... . ⎦ Tn1 Tn2 ... Tnn

5.17 Schatten p-Norm Inequalities for Sector Matrices

201

is a sector matrix. Then 

Ti j  pp

≤ (n − 1) sec (α) p

n 

i= j

Tii  pp

i=1

for p > 0. 

Tii Ti j Proof. It is easy to obtain that a principal submatrix T ji T j j   Tii Ti j matrix. Now, applying (5.17.7) to , we get T ji T j j p

 of T is also a sector

p

Ti j  pp + T ji  pp ≤ 2 sec p (α)Tii  p2 T j j  p2 for i = j and p > 0. Consequently, using the arithmetic-geometric mean inequality, we have Ti j  pp + T ji  pp ≤ sec p (α)(Tii  pp + T j j  pp ) for i = j and p > 0. Adding up the previous inequalities for i, j = 1, 2, ..., n, we get 

Ti j  pp ≤ (n − 1) sec p (α)

i= j

n 

Tii  pp ,

i=1

which proves the result.



Remark 5.17.9 Set α =

π 4

 i= j

p

in Theorem 5.17.8. Then sec p (α) = 2 2 . This means that Ti j  pp

≤ (n − 1)2

p 2

n 

Tii  pp ,

i=1

when T is an accretive-dissipative matrix. Obviously, if p > 4 or 0 < p < 43 , then our result is stronger than the inequalities in Theorem 5.16.4. We end this section with an extension of Schatten p-norm inequality for two positive semidefinite matrices to sector matrices. We extend Theorem 5.16.7 to sector matrices as follows. Theorem 5.17.10 Let S, T ∈ Mn (C) with W (S), W (T ) ⊆ Sα . Then for p ≥ 1 cos p (α)(S pp + T  pp ) ≤ S + T  pp ≤ 2 p−1 sec p (α)(S pp + T  pp ).

202

5 Inequalities for Sector Matrices

Proof. Let S = S + iS and T = T + iT be the Cartesian decomposition of S and T . Then we have S + T  pp = S + iS + T + iT  pp = S + T + i(S + T ) pp ≥ S + T  pp (by (5.17.2)) ≥ S pp + T  pp (by Lemma 5.16.2) ≥ cos p (α)S pp + cos p (α)T  pp (by 5.17.3) = cos p (α)(S pp + T  pp ), which proves the first inequality. To prove the second inequality, note that S + T  pp ≤ sec p (α)(S + T ) pp (by (5.17.3)) = sec p (α)S + T  pp ≤ 2 p−1 sec p (α)(S pp + T  pp ) (by Lemma 5.16.2) ≤ 2 p−1 sec p (α)(S pp + T  pp ), (by (5.17.2)) as required.



Notice that when α = 0, T and S are positive semidefinite and Theorem 5.17.10 implies Lemma 5.16.2. If S and T are accretive-dissipative, then α = π4 and Theorem 5.17.10 reduces to Theorem 5.16.7. Notes and references. In [156], We can find Lemma 5.17.5, Theorem 5.17.6, Theorem 5.17.8, and Theorem 5.17.10.

5.18 Schatten p-Norms and Determinantal Inequalities Involving Partial Traces The tensor product Mn ⊗ Mk is identified with the space Mn (Mk ), the set of n × n block matrices with each block in Mk . The complex matrices H ∈ Mn (Mk ) are partitioned as ⎤ ⎡ H11 H12 ... H1n ⎢ H21 H22 ... H2n ⎥ ⎥ ⎢ ⎢ . . ... . ⎥ ⎥ ⎢ H =⎢ . ... . ⎥ ⎥ ⎢ . ⎣ . . ... . ⎦ Hn1 Hn2 ... Hnn

5.18 Schatten p-Norms and Determinantal Inequalities Involving Partial Traces

203

with each block k × k. Here we introduce the definition of partial traces which is from quantum information theory. For any H ∈ Mn (Mk ), we write H=

m 

Ai ⊗ Bi

i=1

for some positive integer m. The partial trace T r1 H ∈ Mk and T r2 H ∈ Mn are defined [180] as follows T r1 H =

m 

(T r Ai )Bi , T r2 H =

i=1

m  (T r Bi )Ai , i=1

where T r stands for the usual trace, T r1 “trace out” the first factor and T r2 “trace out” the second factor. The actual forms of the partial traces are seen in [180] T r1 H =

n 

Hii , T r2 H = (T r Hi j )i,n j=1 .

i=1

Note that a density matrix on a bipartite system [180] is a positive semidefinite matrix in Mn ⊗ Mk with trace equal to one. Audenaert [3] proved an inequality for Schatten p-norms as follows. Theorem 5.18.1 Let A ∈ Mn (Mk ) be a density matrix. Then 1 + A p ≥ T r1 A p + T r2 A p .

(5.18.1)

Inequality (5.18.1) was used to prove the subadditivity of the so-called Tsallis entropies [3]. From the proof of Theorem 5.18.1, it is easy to see that if A is only a positive semidefinite matrix, then inequality (5.18.1) becomes T r A + A p ≥ T r1 A p + T r2 A p .

(5.18.2)

As an analogue result of Theorem 5.18.1, Lin [142] proved the following determinantal inequality. Theorem 5.18.2 Let A ∈ Mn (Mk ) be a density matrix. Then 1 + det A ≥ det(T r1 A)n + det(T r2 A)k .

(5.18.3)

Similarly, if A is only a positive semidefinite matrix, then inequality (5.18.3) becomes (T r A)nk + det A ≥ det(T r1 A)n + det(T r2 A)k .

(5.18.4)

204

5 Inequalities for Sector Matrices

In [63], Fiedler and Markham proved the following determinantal inequality for positive semidefinite matrices which was presented by partial traces in [143].

det(T r2 H ) k

k ≥ det H.

A careful examination of Lin’s proof in [143] actually revealed the following result

det(T r2 H ) kn

k ≥ det H.

(5.18.5)

Now, we extend inequality (5.18.5) to sector matrices. For presenting and proving our results, we need the following several lemmas. Lemma 5.18.3 ([187]) Let X and Y be generalized Pauli matrices on Ck , these 2πi j operators act as X e j = e j+1 , Y e j = e k e j , where i is the imaginary unit, e j is the j-th column of Ik and ek+1 = e1 . Then k 1  (In ⊗ X l Y j )H (In ⊗ X l Y j )∗ = (T r2 H ) ⊗ Ik . k l, j=1

(5.18.6)

According to Lemma 5.1.8, we have Lemma 5.18.4 If H ∈ Mn (Mk ) and W (H ) ⊂ Sα , then | det H | ≤ secnk (α) det(H ).

(5.18.7)

Lemma 5.18.5 ([207]) If H ∈ Mn (Mk ) and W (H ) ⊂ Sα , then λ j (H ) ≤ σ j (H ) j = 1, ..., nk.

(5.18.8)

Lemma 5.18.6 ([217]) If H ∈ Mn (Mk ) and W (H ) ⊂ Sα , then for all unitarily invariant norms . on Mn (Mk ), H  ≤ sec(α)H .

(5.18.9)

With the above preparation, we can present the main theorem which extends (5.18.5) to sector matrices. Theorem 5.18.7 Let H = (Hi j )i,n j=1 ∈ Mn (Mk ) and W (H ) ⊂ Sα . Then    det(T r H ) k  2   | det H | ≤ secnk (α)  .   kn

5.18 Schatten p-Norms and Determinantal Inequalities Involving Partial Traces

205

Proof. As the determinant functional is log concave over the cone of positive semidefinite matrices [103], we have | det H | =

k 

1

|det H | k 2

l, j=1

=

k  1  det[(In ⊗ X l Y j )H (In ⊗ X l Y j )∗ ] k 2 l, j=1



k  1  nk sec (α) det[(In ⊗ X l Y j )H (In ⊗ X l Y j )∗ ] k 2 ( by (5.18.7)) l, j=1



⎞ k  1 [(In ⊗ X l Y j )H (In ⊗ X l Y j )∗ ]⎠ ( by (5.18.6)) ≤ secnk (α) det ⎝ 2 k l, j=1

 T r2 (H ) ⊗ Ik = secnk (α) det , (5.18.10) k where X , Y and U are unitary matrices. Notice that we use the fact that (U AU ∗ ) = U (A)U ∗ in the first inequality. Since (T r2 H ) = T r2 (H ), the following result is obtained.

det

T r2 (H ) ⊗ Ik k



Thus, | det H | is bounded by

 Ik n = (det(T r2 (H )))k det k n   I k  k = |det((T r2 H ))| det  k n    I k ≤ |det(T r2 H )|k det  (by (5.18.8)) k    Ik  = det T r2 H ⊗ k      T r2 H = det ⊗ Ik  . k

(5.18.11)

206

5 Inequalities for Sector Matrices



T r2 (H ) ⊗ Ik (by (5.18.10)) | det H | ≤ secnk (α) det k     T r2 H ⊗ Ik  (by 5.18.11) ≤ secnk (α) det k     det(T r2 H ) n  = secnk (α)  , kn and the proof is completed.



Now, let us present our generalizations of (5.18.1) and (5.18.3) in the next two theorems. The following theorem is a generalization of (5.18.1). Theorem 5.18.8 Let A ∈ Mn (Mk ) and W (H ) ⊂ Sα . Then T r |A| + A p ≥ cos(α)T r1 A p + cos(α)T r2 A p .

(5.18.12)

Proof. By (5.18.8), it is easy to obtain T r (|A|) ≥ T r (A) and A p ≥ A p . Compute T r (|A|) + A p ≥ T r (A) + A p ≥ T r1 (A) p + T r2 (A) p (by (5.18.2)) = (T r1 A) p + (T r2 A) p ≥ cos(α)T r1 A p + cos(α)T r2 A p , (by (5.18.9)) where the first equality is followed by (T r1 A) = T r1 (A) and (T r2 A) = T r2 (A).  Remark 5.18.9 When α = 0, our result (5.18.12) reduces to (5.18.2). Moreover, if A is a density matrix, then the inequality (5.18.12) is (5.18.1). So, Theorem 5.18.8 is a generalization of Theorem 5.18.1. Finally, we give a generalization of the result (5.18.3). Theorem 5.18.10 Let A ∈ Mn (Mk ) and W (H ) ⊂ Sα . Then (T r |A|)nk + det |A| ≥ cosnk (α)| det(T r1 A)|n + cosnk (α)| det(T r2 A)|k . (5.18.13) Proof. Firstly, we prove (T r (A))nk ≤ (T r |A|)nk . By (5.18.8), we have

5.18 Schatten p-Norms and Determinantal Inequalities Involving Partial Traces

(T r (A))nk

207

⎛ ⎞nk nk  =⎝ λ j (A)⎠ ⎛ ≤⎝

j=1 nk 

⎞nk σ j (A)⎠

j=1

= (T r |A|)nk . Next, we prove (5.18.13) as follows. (T r |A|)nk + det |A| ≥ (T r (A))nk + det |A| ≥ det((T r1 A))n + det((T r2 A))k (by (5.18.4))  

| det(T r1 A)| n | det(T r2 A)| k ≥ + (by (5.18.7)) seck (α) secn (α) | det(T r2 A)|k | det(T r1 A)|n + = secnk (α) secnk (α) nk = cos (α)| det(T r1 A)|n + cosnk (α)| det(T r2 A)|k , and the desired inequality obtained.



Remark 5.18.11 When α = 0, our result (5.18.13) reduces to (5.18.4). Moreover, if A is a density matrix, then the inequality (5.18.13) is (5.18.3). Thus, Theorem 5.18.10 is a generalization of Theorem 5.18.2.

Notes and references. All Theorems 5.18.7, 5.18.8 and 5.18.10 are due to [204].

5.19 Ando-Choi Type Inequalities for Sector Matrices As we introduced in Theorem 3.4.1, if A, B ∈ Mn are positive definite, then for any positive linear map Φ and λ ∈ [0, 1], we have Φ(Aλ B) ≤ Φ(A)λ Φ(B).

(5.19.1)

Φ(A!λ B) ≤ Φ(A)!λ Φ(B).

(5.19.2)

Ando [11] also proved that

208

5 Inequalities for Sector Matrices

We recall that Choi’s inequality says that for any positive unital linear map Φ, it holds (5.19.3) Φ −1 (A) ≤ Φ(A−1 ), where A ∈ Mn is positive definite. For a survey on the positive linear maps, we refer to [20]. In this section, we intend to extend (5.19.1), (5.19.2), and (5.19.3) to sector matrices. Then, we introduce this class of matrices and the recently defined weighted geometric mean for accretive matrices. We recall the geometric mean of two accretive matrices (5.7.5) and the weighted geometric mean for accretive matrices (5.9.6). The following first lemma gives the closure property of sector matrices under the positive linear map. Lemma 5.19.1 Let Φ be a positive linear map. If A ∈ Mn with W (A) ⊂ Sα , then W (Φ(A)) ⊂ Sα . In particular, if A ∈ Mn is accretive, then so is Φ(A). Proof. First of all, note that for any T ∈ Mn , (Φ(T ) + Φ(T )∗ ) 2 (Φ(T ) + Φ(T ∗ )) = 2 

(T + T ∗ ) =Φ 2 = Φ(T ),

Φ(T ) =

in which the second inequality is by a Lemma in [20]. That is, Φ(T ) = Φ(T ). Similarly, we have Φ(T ) = Φ(T ). Now since W (A) ⊂ Sα , by definition we have ±A ≤ (tan α)A. Applying the map Φ to the previous inequality gives ±Φ(A) ≤ (tan α)Φ(A). Equivalently ±Φ(A) ≤ (tan α)Φ(A), that is, W (Φ(A)) ⊂ Sα , as required.



5.19 Ando-Choi Type Inequalities for Sector Matrices

209

A reverse of Lemma 5.1.5 is as follows. Lemma 5.19.2 ([146]) Let A ∈ Mn with W (A) ⊂ Sα . Then (A)−1 ≤ (sec α)2 A−1 . The remarkable property about the weighted geometric mean of accretive matrices was proved in Theorem 5.9.10 which the next lemma complements it. Lemma 5.19.3 Let A, B ∈ Mn such that W (A), W (B) ⊂ Sα and let λ ∈ [0, 1]. Then (cos α)2 (Aλ B) ≤ (A)λ (B). Proof. By Lemma 5.1.5, we have (A−1 + λB −1 )−1 ≤ (A−1 + λB −1 )−1 . On the other hand, by Lemma 5.19.2 we have A−1 + λB −1 ≥ (cos α)2 ((A)−1 + λ(B)−1 ). Thus,

(A−1 + λB −1 )−1 ≤ (sec α)2 ((A)−1 + λ(B)−1 )−1 .

Combining previous two inequalities gives # sin λπ ∞ t−1 λ (A−1 + λB −1 )−1 dt π #0 sin λπ ∞ t−1 ≤ λ (sec α)2 ((A)−1 + λ(B)−1 )−1 dt π 0 = (sec α)2 ((A)λ (B)).

(Aλ B) =

The desired inequality follows.



According to Lemma 5.17.2, we have Lemma 5.19.4 Let A ∈ Mn such that W (A) ⊂ Sα . Then for any unitarily invariant norm ., cos αA ≤ A. Now we start with an extension of (5.19.1). Theorem 5.19.5 Let A, B ∈ Mn such that W (A), W (B) ⊂ Sα . Then for any positive linear map Φ, it holds

210

5 Inequalities for Sector Matrices

(cos α)2 Φ(Aλ B) ≤ (Φ(A)λ Φ(B)),

(5.19.4)

where λ ∈ [0, 1]. Proof. Lemma 5.19.3 tells us that (cos α)2 (Aλ B) ≤ (A)λ (B). Applying the positive linear map Φ to the previous inequality and by Lemma 5.19.1, we obtain (cos α)2 Φ(Aλ B) ≤ Φ((A)λ (B)). Now we estimate Φ((A)λ (B)) ≤ Φ(A)λ Φ(B) (by (5.19.1)) = (Φ(A))λ (Φ(B)) ≤ (Φ(A)λ Φ(B)), (by Theorem 5.9.10) and the proof is completed.



Corollary 5.19.6 Let A, B ∈ Mn such that W (A), W (B) ⊂ Sα . Then for any positive linear map Φ and unitarily invariant norm ., it holds (cos α)3 Φ(Aλ B) ≤ Φ(A)λ Φ(B),

(5.19.5)

where λ ∈ [0, 1]. Proof. By Lemma 5.19.4 and Lemma 5.19.1, we have cos αΦ(Aλ B) ≤ Φ(Aλ B). Then by the inequality (5.19.4) and Lemma 5.17.1, (cos α)2 Φ(Aλ B) ≤ (Φ(A)λ Φ(B)) ≤ Φ(A)λ Φ(B).  Corollary 5.19.7 Let A, B ∈ Mn be accretive-dissipative. Then for any positive linear map Φ and unitarily invariant norm ., it holds √ 2 Φ(Aλ B) ≤ Φ(A)λ Φ(B), 4 where λ ∈ [0, 1].

(5.19.6)

5.19 Ando-Choi Type Inequalities for Sector Matrices

211 π

Proof. It is easy to observe that W (e− 4 A) ⊂ S iπ4 , W (e− 4 B) ⊂ S π4 . Moreover, by (5.9.6), iπ

(e− 4 A)λ (e− 4 B) = e− 4 iπ





sin πλ π

#



t λ−1 (A−1 + t B −1 )−1 dt

0

= e− 4 (Aλ B). iπ

One readily finds that (5.19.6) follows from (5.19.5) by specifying α to be equal to π .  4 Using exactly the same approach, one could state analogous results for the weighted harmonic mean, we leave the details of the proof for the interested reader. Theorem 5.19.8 Let A, B ∈ Mn such that W (A), W (B) ⊂ Sα . Then for any positive linear map Φ, it holds (cos α)2 Φ(A!λ B) ≤ (Φ(A)!λ Φ(B)), where λ ∈ [0, 1]. Corollary 5.19.9 Let A, B ∈ Mn such that W (A), W (B) ⊂ Sα . Then for any positive linear map Φ and unitarily invariant norm ., it holds (cos α)3 Φ(A!λ B) ≤ Φ(A)!λ Φ(B), where λ ∈ [0, 1]. Corollary 5.19.10 Let A, B ∈ Mn be accretive-dissipative. Then for any positive linear map Φ and unitarily invariant norm ., it holds √ 2 Φ(A!λ B) ≤ Φ(A)!λ Φ(B), 4 where λ ∈ [0, 1]. Next, we present an extension of (5.19.3) and related norm inequalities. Theorem 5.19.11 Let A ∈ Mn such that W (A) ⊂ Sα . Then for any positive linear map Φ, it holds (cos α)2 Φ −1 (A) ≤ Φ(A−1 ). Proof. By Lemma 5.19.1 and Lemma 5.1.5, we have Φ −1 (A) ≤ (Φ(A))−1 = (Φ(A))−1 .

212

5 Inequalities for Sector Matrices

Now by Choi’s inequality (5.19.3), (Φ(A))−1 ≤ Φ((A)−1 ). Finally, by Lemma 5.19.2, we have Φ((A)−1 ) ≤ (sec α)2 Φ(A−1 ) = (sec α)2 Φ(A−1 ). So the desired inequality follows.



Similar to the proof of Corollary 5.19.6 and Corollary 5.19.7, we could present the following results. Corollary 5.19.12 Let A ∈ Mn such that W (A) ⊂ Sα . Then for any positive linear map Φ and unitarily invariant norm ., it holds (cos α)3 Φ −1 (A) ≤ Φ(A−1 ). Corollary 5.19.13 Let A ∈ Mn be accretive-dissipative. Then for any positive linear map Φ and unitarily invariant norm ., it holds √

2 −1 Φ (A) ≤ Φ(A−1 ). 4

Notes and references. The proof of Lemma 5.19.1, Lemma 5.19.3, Theorem 5.19.5, Theorem 5.19.11, Corollary 5.19.6, and Corollary 5.19.7 are taken from [198]. Note that the stated last corollaries are also given in [198].

5.20 Geometric Mean Inequality for Sector Matrices Consider A ∈ Mn partitioned as  A=

 A11 A12 , A21 A22

(5.20.1)

where A22 ∈ Mm such that m ≤ n2 . Assume that A11 is invertible. We recall that the Schur complement of A11 in A is defined as AA11 = A22 − A21 A−1 11 A12 . It is clear that A is invertible whenever W (A) ⊂ Sα . If W (A) ⊂ Sα , then W (A11 ) ⊂ Sα , that AA11 is well defined.

5.20 Geometric Mean Inequality for Sector Matrices

213

 X 11 X 12 is invertible, then we also partition X −1 conformally as X so X 21 X 22 that (X −1 )22 means the (2, 2) block of X −1 . We need two simple lemmas.   X 11 X 12 Lemma 5.20.1 If X = is positive definite, then X 21 X 22 

If X =

(X 22 )−1 ≤ (X −1 )22 . Proof. Note that (X −1 )22 =

X X 11

−1

−1 = (X 22 − X 21 X 11 X 12 )−1

≥ (X 22 )−1 . A generalization of this lemma can be found in [210].



Lemma 5.20.2 If X ∈ Mn has a positive definite real part, then (X −1 ) ≤ (X )−1 . Proof. Consider the Cartesian decomposition X = Y + i Z . Then (X −1 ) = (Y + Z Y −1 Z )−1 ≤ Y −1 = (X )−1 .  Inequality (5.12.7) can be equivalently written as   A  A

11

   ≤ sec2 (α)U ∗ |A22 |U 

(5.20.2)

for some unitary matrix U ∈ Mm . We present the following result, which says that concerning the real parts of AA11 , A22 , an analogue of (5.20.2) is valid without bringing in a unitary matrix. Theorem 5.20.3 Let A ∈ Mn be partitioned as in (5.20.1) and W (A) ⊂ Sα . Then



A A11

 ≤ sec2 (α)A22 .

(5.20.3)

214

5 Inequalities for Sector Matrices

Proof. Consider the Cartesian decomposition A = B + iC. The condition W (A) ⊂ Sα implies that ±C ≤ tan(α)B and so −1

±B 2 C B −1

−1 2

≤ tan(α).

−1

This yields (B 2 C B 2 )2 ≤ tan2 (α), i.e., C B −1 C ≤ tan2 (α)B. In particular,

(C B −1 C)22 ≤ tan2 (α)B22 .

(5.20.4)

Note that sec2 (α) = 1 + tan2 (α), so (5.20.4) is equivalent to cos2 (α)(B + C B −1 C)22 ≤ B22 .

(5.20.5)

With (5.20.5), we can find upper bounds for (B22 )−1 , (A22 )−1 = (B22 )−1

−1 ≤ sec2 (α) (B + C B −1 C)22

≤ sec2 (α) (B + C B −1 C)−1 22 (by Lemma 5.20.1)

= sec2 (α) (A−1 ) 22 = sec2 (α)(A−1 )22    A −1 2 = sec (α) A11 −1

A 2 (by Lemma 5.20.2). ≤ sec (α)  A11

Therefore 



A A11



≤ sec2 (α)A22 as desired.



Remark 5.20.4 Note that (5.20.3) can be written as (tan2 (α)A22 + A21 A−1 11 A12 ) ≥ 0. , which yields On the other hand, if W (A) ⊂ Sα , then W (A A−1 A∗ ) = W (A∗ ) ⊂ Sα −1 −1  A A ⊂ W (A−1 ) ⊂ Sα . As A11 is a principal submatrix of A−1 , we have W A11 Sα . In particular,

(A22 + A21 A−1 11 A12 ) ≥ 0.

However, under the assumption W (A) ⊂ Sα , it is in general not true that

5.20 Geometric Mean Inequality for Sector Matrices

215

(A22 + A21 A−1 11 A12 ) ≥ 0. As we know, the geometric mean for two sector matrices A, B ∈ Mn is #



AB =

(t A + t −1 B)−1

0

dt . t

(5.20.6)

We recall the following noncommutative AM-GM inequality is known for positive definite matrices A, B ∈ Mn A+B . (5.20.7) AB ≤ 2 Is there an analogue for sector matrices? The first thought is whether it holds

(AB) ≤ 

A+B 2

 (5.20.8)

for sector matrices A, B ∈ Mn . The answer is no as the following example shows. Example 5.20.5 Let 

10 3+i A= 3 + i 2 + 4i





 2 − 4i −1 − 4i ,B = . −1 − 4i 2−i

It is easy to verify that A, B have positive definite real part. Using the Matlab, one computes that   6.2830 2.0747 (AB) = . 2.0747 3.2251 However, det((

A+B ) − (AB)) = −0.8083 < 0, 2

violating (5.20.8). Now, our main result is a correct extension of (5.20.7). We need a lemma which can be regarded as a complement of Lemma 5.20.2. Lemma 5.20.6 If X ∈ Mn with W (A) ⊂ Sα , then (X )−1 ≤ sec2 (α)(X −1 ). Proof. The inequality is implicit in the proof of obtaining inequality (5.12.9). We omit the details. 

216

5 Inequalities for Sector Matrices

Theorem 5.20.7 Let A, B ∈ Mn be such that W (A), W (B) ⊂ Sα . Then (AB) ≤

sec2 (α) (A + B). 2

Proof. Compute (AB) = (A−1 B −1 )−1 

# ∞ dt 2 = (t A−1 + t −1 B −1 )−1 π 0 t # dt 2 ∞ (t A−1 + t −1 B −1 )−1 = π 0 t # dt 2 ∞ (by Lemma 5.20.2) ≤ (tA−1 + t −1 B −1 )−1 π 0 t # ∞ dt 2 ≤ sec2 (α) (by Lemma 5.20.6) (t (A)−1 + t −1 (B)−1 )−1 π 0 t = sec2 (α)((A)−1 (B)−1 )−1 = sec2 (α)(A)(B) sec2 (α) (A + B) 2 sec2 (α) (A + B).  = 2 ≤

Next, we present some implications of Theorem 5.20.7. We recall that for a Hermitian matrix X ∈ Mn , λ j (X ) means the j-th largest eigenvalue of X . We need an auxiliary lemma. Lemma 5.20.8 ([21]) Let A ∈ Mn be such that W (A) ⊂ Sα . Then λ j (A) ≤ σ j (A). Theorem 5.20.9 Let A, B ∈ Mn be such that W (A), W (B) ⊂ Sα . Then σ j (AB) ≤ for j = 1, ..., n. Proof. Compute

sec4 (α) σ j (A + B), 2

(5.20.9)

5.20 Geometric Mean Inequality for Sector Matrices

217

σ j (AB) ≤ sec2 (α)σ j ((AB)) (by 5.12.9) sec4 (α) σ j ((A + B)) (by Theorem 5.20.7) 2 sec4 (α) ≤ σ j (A + B), (by 5.20.9) 2 ≤

as claimed.



Theorem 5.20.10 Let A, B ∈ Mn be such that W (A), W (B) ⊂ Sα . Then AB ≤

sec3 (α) A + B 2

for any unitarily invariant norm. Proof. The claimed result follows from the following chain of inequalities AB ≤ sec(α)(AB) sec3 (α) (A + B) 2 sec3 (α) A + B. ≤ 2 ≤

The argument in each step is the same as in the proof of Theorem 5.20.9 except for the first inequality, where we used a result of Zhang [217]. 

Notes and references. In this section, we take the proof of Lemma 5.20.1, Lemma 5.20.2, Lemma 5.20.6, Theorem 5.20.3, Theorem 5.20.7, Theorem 5.20.9, and Theorem 5.20.10 from [146].

5.21 Weighted Geometric Mean Inequality for Sector Matrices The main topic of the present section is an analogue of (5.9.3) for sector matrices. Liu and Wang [159] presented a complement of Theorem 5.20.7. More precisely, under the same condition of Theorem 5.20.7, it holds (AB) ≥ cos2 (α)(A!B).

(5.21.1)

218

5 Inequalities for Sector Matrices

The main aim of this section is to extend Theorem 5.20.7 and (5.21.1). Moreover, we use the established result to give some new determinantal inequalities. Next, we extend (5.9.3) to sector matrices. Theorem 5.21.1 Let A, B ∈ Mn be such that W (A), W (B) ⊂ Sα . Then cos2 (α)(A!λ B) ≤ (Aλ B) ≤ sec2 (α)(A∇λ B),

(5.21.2)

where λ ∈ [0, 1]. Proof. We estimate

−1 (A!λ B) =  (1 − λ)A−1 + λB −1

−1 ≤  (1 − λ)A−1 + λB −1 (by Lemma 5.1.5)

−1 −1 −1 = (1 − λ)A + λB

−1 ≤ cos2 (α)(1 − λ)(A)−1 + cos2 (α)λ(B)−1 (by Lemma 5.20.6)

2 −1 −1 −1 = sec (α) (1 − λ)(A) + λ(B) ≤ sec2 (α)((A)λ (B)) (by (5.9.3)) ≤ sec2 (α)(Aλ B) (by (5.9.10)). This proves the first inequality of (5.21.2). To show the second one, we notice that by Lemma 5.1.5, for any t ≥ 0

−1 . (A−1 + t B −1 )−1 ≤ A−1 + tB −1 On the other hand, by Lemma 5.20.6 we have

A−1 + tB −1 ≥ cos2 (α) (A)−1 + t (B)−1 . By the operator reverse monotonicity of the inverse, it follows Hence,

Therefore,

A−1 + tB −1

−1



−1 ≤ sec2 (α) (A)−1 + t (B)−1 .

−1

−1 ≤ sec2 (α) (A)−1 + t (B)−1 .  A−1 + t B −1

5.21 Weighted Geometric Mean Inequality for Sector Matrices

219

#

−1 sin λπ ∞ λ−1 −1 t  A + t B −1 dt π 0 #

−1 sin λπ ∞ λ−1 2 ≤ t sec (α) (A)−1 + t (B)−1 dt π 0 = sec2 (α)((A)λ (B))

(Aλ B) =

≤ sec2 (α) ((1 − λ)(A) + λ(B)) = sec2 (α)(A∇λ B). This completes the proof.



In the following, we present some determinant inequalities that make use of the previous results. As we know, the following lemma [104] is known as the OstrowskiTaussky inequality. Lemma 5.21.2 If A ∈ Mn is accretive, then it holds det(A) ≤ | det A|. In lemma 5.1.8, the reverse of the Ostrowski-Taussky inequality introduced. Now, we can give our main result. Theorem 5.21.3 Let A, B ∈ Mn such that W (A), W (B) ⊂ Sα . Then cos3n (α)| det(A!λ B)| ≤ | det(Aλ B)| ≤ sec3n (α)| det(A∇λ B)|,

(5.21.3)

where λ ∈ [0, 1]. Proof. We give the proof of the first inequality of (5.21.3). The proof of second inequality is similar and we omit it. By Lemma 5.1.8, | det(A!λ B)| ≤ secn (α) det (A!λ B). Note that if M ≥ N are positive semidefinite, then det M ≥ det N ≥ 0. Hence, by the first inequality of (5.21.2), we obtain det (A!λ B) ≤ sec2n (α) det (Aλ B). So we have reached | det(A!λ B)| ≤ sec3n (α) det (Aλ B). It follows by Lemma 5.21.2 that

220

5 Inequalities for Sector Matrices

| det(A!λ B)| ≤ sec3n (α)| det(Aλ B)|, and so we complete the proof of the first inequality.



Notes and references. The main results of this section stated in Theorem 5.21.1, and Theorem 5.21.3 are taken from [199].

Chapter 6

Positive Partial Transpose Matrix Inequalities

In this chapter, we present some inequalities related to 2 × 2 block PPT matrices and positive matrices partitioned in blocks. When these positive block matrices are PPT, some nice results have been obtained, motivated by the Quantum Information Science. Our study follows a natural thought that conclusions drawn under the PPT assumption should be stronger than those drawn under only the usual positivity assumption. Moreover, we believe the new result presented in this work is of interest in its own right and may serve to understand better the intrinsic properties of PPT matrices in quantum sciences and information theory. Note that multipartite quantum states that have a positive partial transpose concerning all bipartitions of the particles can outperform the separable state in linear interferometers.

6.1 Singular Value Inequalities Related to PPT Matrices Positive semidefinite matrices partitioned into 2 × 2 blocks play an important role in matrix analysis. A recent monograph [20] contains an excellent exposition of this point. It is well known that the transpose map is not 2-positive, that is, assuming each block is square,     A X∗ A X ≥0 ≥ 0. X B X∗ B Motivated by the theory of quantum information [106, 179], there is a need to introduce a stronger class of positive matrices, that is, matrices whose partial transpose are also positive.   A X In 2 × 2 block case, we say is positive partial transpose (PPT for short) X∗ B if © Springer Nature Switzerland AG 2021 M. B. Ghaemi et al., Advances in Matrix Inequalities, Springer Optimization and Its Applications 176, https://doi.org/10.1007/978-3-030-76047-2_6

221

222

6 Positive Partial Transpose Matrix Inequalities



A X X∗ B





 ≥ 0 and

A X X∗ B

A X∗ X B

 ≥ 0.



Thus, whenever we say is PPT, the off-diagonal blocks are necessarily square. In [18], Besenyei formulated the following remarkable trace inequality which A X ≥ 0 with X square, then says that if X∗ B T r AB − T r X ∗ X ≤ (T r A)(T r B) − |T r X |2 .

(6.1.1)

Inequality (6.1.1) arises from the subadditivity of q-entropies, see [3] and references therein  details.  for more A X ≥ 0 with X square, then it is clear that If X∗ B (T r A)(T r B) − |T r X |2 ≥ 0, since trace functional is Liebian [195]. However, T r AB − T r X ∗ X may take a negative value. For example, taking 

A X X∗ B





1 ⎢0 =⎢ ⎣0 1

0 0 0 0

0 0 0 0

⎤ 1 0⎥ ⎥ ≥ 0, 0⎦ 1

(6.1.2)

then T r AB − T r X ∗ X = −1. The following result proposes a condition for the positivity of T r AB − T r X ∗ X . We remark that it is an extension of Lemma 2.10 in [148].   A X Theorem 6.1.1 Let ∈ M2n be PPT. Then X∗ B T r X ∗ X ≤ T r AB.

(6.1.3)

Proof. Without loss of generality, we may assume A to be positive definite. As   A X is PPT, we have B ≥ X A−1 X ∗ and B ≥ X ∗ A−1 X , thus X∗ B A 2 B A 2 ≥ (A 2 X A− 2 )(A 2 X A− 2 )∗ , 1

and

1

1

1

1

1

A 2 B A 2 ≥ (A− 2 X A− 2 )∗ (A− 2 X A 2 ). 1

1

1

1

1

1

6.1 Singular Value Inequalities Related to PPT Matrices

223

Taking trace and adding up gives 2(T r AB) ≥ A 2 X A− 2 2F + A− 2 X A 2 2F , 1

1

1

1

where . F means the Frobenius norm. Thus, it suffices to show A 2 X A− 2 2F + A− 2 X A 2 2F ≥ 2X 2F = 2(T r X ∗ X ). 1

1

1

1

Let A = U DU ∗ be the spectral decomposition of A with U unitary and D = diag(d1 , ..., dn ) a diagonal matrix. Let also Y = [yi j ]i,n j=1 = U ∗ XU . Then A 2 X A− 2 2F + A− 2 X A 2 2F = D 2 Y D − 2 2F + D − 2 Y D 2 2F 1

1

1

1

1

1

1

1

n n



dj di |yi j |2 + |yi j |2 d d j i i, j=1 i, j=1 n

dj di |yi j |2 = + d d j i i, j=1

=

≥2

n

|yi j |2

i, j=1

= 2Y 2F = 2X 2F = 2(T r X ∗ X ), from which (6.1.3) follows.



Recall that A ∈ Mm×n is contractive if A∗ A ≤ In and strict contractive if In − A A is further invertible. In studying the theory of functions of several complex variables, L.K. Hua [108] discovered an intriguing positive matrix which now carries his name [8]. The Hua matrix is given by ∗

⎡ H =⎣

(In − A∗ A)−1 (In − B ∗ A)−1 ∗

(In − A B)

−1



(In − B B)

−1

⎤ ⎦,

where A, B ∈ Mm×n are strictly contractive. It was only recently observed that H is PPT [8]. Thus, thanks to Theorem 6.1.1, we have the following corollary. Corollary 6.1.2 Let A, B ∈ Mm×n be strictly contractive. Then T r (In − A∗ B)−1 (In − B ∗ A)−1 ≤ T r (In − A∗ A)−1 (In − B ∗ B)−1 . The following result is a weak log-majorization version of Theorem 6.1.1 which is due to Bourin.

224

6 Positive Partial Transpose Matrix Inequalities

 Theorem 6.1.3 Let

A X X∗ B

k

 ∈ M2n be PPT. Then

σ j (X ) ≤

j=1

k

1

1

σ j (A 2 B 2 ), k = 1, ..., n.

j=1

Equivalently, k

λ j (X ∗ X ) ≤

k

j=1

λ j (AB), k = 1, ..., n.

j=1

Proof. We have X = A 2 C B 2 for some contraction C ∈ Mn . Similarly, X ∗ = 1 1 A 2 D B 2 for some contraction D ∈ Mn . Therefore, 1

k

1

λ j (X ∗ X ) =

j=1

k

1

1

1

1

σ j (A 2 D B 2 A 2 C B 2 )

j=1



k

1

1

1

1

σ j (D B 2 A 2 C B 2 A 2 )

j=1



k

1

1

1

1

σ j (B 2 A 2 )σ j (B 2 A 2 )

j=1

=

k

λ j (AB),

j=1

in which the first inequality is by the fact [21] that k = 1, 2, ..., whenever P Q is normal. 

k j=1

σ j (P Q) ≤

k j=1

σ j (Q P),

In this connection, we shall show that Theorem 6.1.3 can be self-improved. Recall that the geometric mean of two positive definite matrices A, B ∈ Mn , denoted by AB, is the positive definite solution of the Ricatti equation X B −1 X = A. The notion of the geometric mean can be uniquely extended to all A, B ≥ 0 by a limit from above [20] AB = lim (A + εIn )(B + εIn ). ε→0

 Lemma 6.1.4 If

A X X∗ B



 ∈ M2n is PPT, then so is

 AB X . X ∗ AB

Proof. This follows immediately from Lemma 3.1 in [9].



Here, we give a generalization of Theorem 3.3 in [9] as follows.

6.1 Singular Value Inequalities Related to PPT Matrices

 Theorem 6.1.5 Let

A X X∗ B

k

225

 ∈ M2n be PPT, then

σ j (X ) ≤

k

j=1

σ j (AB) ≤

j=1

k

1

1

σ j (A 2 B 2 ),

j=1

for k = 1, ..., n. 

AB X X ∗ AB



is PPT. Now applying Theorem 6.1.3 to Proof. By Lemma 6.1.4,   AB X gives the first inequality. The second inequality is well known and it X ∗ AB has various generalizations, see, for example, [28]. It is also apparent that n

σ j (AB) =

j=1

This completes the proof.

n

1

1

σ j (A 2 B 2 ).

j=1



Next, we study some singular value inequalities on PPT matrices.   A X Proposition 6.1.6 Let ∈ M2n be PPT. Then X∗ B 2X  ≤ A + B

(6.1.4)

for any unitarily invariant norm. Again, the example in (6.1.2) shows that (6.1.4) may fail without the PPT assumption. As the Hua matrix is PPT, Proposition 6.1.6 entails Corollary 6.1.7 ([137]) Let A, B ∈ Mm×n be strictly contractive, then 2(In − A∗ B)−1  ≤ (In − A∗ A)−1 + (In − B ∗ B)−1  for any unitarily invariant norm. Under the same assumption as in Proposition 6.1.6, one may wonder whether a stronger level inequality (in the sense of Bhatia and Kittaneh [25]) 2σ j (X ) ≤ σ j (A + B), j = 1, ..., n  P 2 + Q2 P Q + Q P ∈ M2n is PPT, P Q + Q P P 2 + Q2 whenever P, Q ∈ Mn are Hermitian. But it is known that σ j (P Q + Q P) ≤ σ j (P 2 + 

is true! The answer is no. For example,

226

6 Positive Partial Transpose Matrix Inequalities

Q 2 ) fails in general [25]. Nevertheless, for the Hua matrix, the answer is affirmative. This looks surprising. We need some lemmas to give the main result in Theorem 6.1.11. We recall the first one which is due to Ky Fan and Hoffman. Lemma 6.1.8 For every A ∈ Mn with A ≥ 0, σ j (A) ≤ σ j (A), j = 1, ..., n. With Lemma 6.1.8, we can present the following result, which can be regarded as a complement of (6.1.5). Lemma 6.1.9 Let A, B ∈ Mm×n be contractive. Then for j = 1, ..., n   2σ j (In − A∗ B) ≥ σ j (In − A∗ A) + (In − B ∗ B) . Proof. It is clear that A∗ A + B ∗ B ≥ A∗ B + B ∗ A = 2(A∗ B), thus

1 (In − A∗ B) = In − (A∗ B) ≥ In − (A∗ A + B ∗ B) ≥ 0. 2

Hence, by Lemma 6.1.8, we have σj

(In − A∗ A) + (In − B ∗ B) 2

so the required result follows.



  ≤ σ j (In − A∗ B)   ≤ σ j In − A∗ B , j = 1, ..., n,



We need to invoke the following powerful tool, which was recently established by Drury [59] as a solution to the question raised in [25]. Lemma 6.1.10 Let A, B ∈ Mn be positive. Then for j = 1, ..., n  2 σ j (AB) ≤ σ j (A + B). Now we are in a position to present Theorem 6.1.11 Let A, B ∈ Mm×n be contractive. Then for j = 1, ..., n.     2σ j (In − A∗ B)−1 ≤ σ j (In − A∗ A)−1 + (In − B ∗ B)−1 .

(6.1.5)

6.1 Singular Value Inequalities Related to PPT Matrices

227

Proof. For any j = 1, ..., n, by Lemma 6.1.10, it suffices to show      σ j (In − A∗ A)−1 (In − B ∗ B)−1 ≥ σ j (In − A∗ B)−1 , which is equivalent to 

σ j ((In − B ∗ B)(In − A∗ A)) ≤ σ j (In − A∗ B),

(6.1.6)

1 for every invertible X ∈ Mn . Again by Lemma 6.1.10, since σ j (X −1 ) = sn− j+1 (X ) (6.1.6) would follow if

σj

(In − B ∗ B) + (In − A∗ A) 2



  ≤ σ j In − A ∗ B ,

but this is the content of Lemma 6.1.9. Therefore, Theorem 6.1.11 is proved.



As a bodyproduct of the proof of Theorem 6.1.11, we have the following proposition. Proposition 6.1.12 Let A, B ∈ Mm×n be strictly contractive. Then for j = 1, ..., n     σ j (In − A∗ B)−1 (In − B ∗ A)−1 ≤ σ j (In − A∗ A)−1 (In − B ∗ B)−1 , or equivalently,     σ j (In − A∗ B)(In − B ∗ A) ≥ σ j (In − A∗ A)(In − B ∗ B) . To finish this section, we show that Proposition 6.1.12 implies the following result of Marcus [161]. If A ∈ Mn , we let the eigenvalues of A be so arranged such that |λ1 (A)| ≥ |λ2 (A)| ≥ ... ≥ |λn (A)|. Proposition 6.1.13 Let A, B ∈ Mm×n be contractive. Then for each k satisfying 1 ≤ k ≤ n, k

|λn− j+1 (In − A∗ B)|2 ≥

j=1

k

(1 − λ j (A∗ A))(1 − λ j (B ∗ B)).

j=1

Proof. It is well known [21] that for any X, Y ∈ Mn , k

j=1

σ j (X ) ≥

k

j=1

|λ j (X )| and

k

j=1

σ j (X )σ j (Y ) ≥

k

σ j (X Y ),

j=1

for each k satisfying 1 ≤ k ≤ n − 1 with equality at k = n. This implies

228

6 Positive Partial Transpose Matrix Inequalities k

sn− j+1 (X ) ≤

j=1

and

k

k

|λn− j+1 (X )|,

j=1

sn− j+1 (X )sn− j+1 (Y ) ≥

j=1

k

sn− j+1 (X Y ),

j=1

for each k satisfying 1 ≤ k ≤ n. Compute k

|λn− j+1 (In − A∗ B)|2 ≥

j=1

k

 2 sn− j+1 (In − A∗ B) j=1



k

  sn− j+1 (In − A∗ A)(In − B ∗ B)

j=1



k

sn− j+1 (In − A∗ A)sn− j+1 (In − B ∗ B)

j=1

=

k

λn− j+1 (In − A∗ A)λn− j+1 (In − B ∗ B)

j=1

=

k

(1 − λ j (A∗ A))(1 − λ j (B ∗ B)),

j=1

in which Proposition 6.1.12 plays a role in the second inequality.



Notes and references. Theorem 6.1.1, Theorem 6.1.3, Theorem 6.1.5, Theorem 6.1.11, Corollary 6.1.2, Lemma 6.1.4, Lemma 6.1.8, and Proposition 6.1.13 are proved by [147].

6.2 Matrix Inequalities and Completely PPT Maps Consider



A11 A12 ⎢ A21 A22 ⎢ ⎢ . . A=⎢ ⎢ . . ⎢ ⎣ . . Am1 Am2

⎤ ... A1m ... A2m ⎥ ⎥ ... . ⎥ ⎥, ... . ⎥ ⎥ ... . ⎦ ... Amm

(6.2.1)

6.2 Matrix Inequalities and Completely PPT Maps

229

with each block in Mn . The usual transpose of A is ⎡

T T A11 A21 ⎢ AT AT ⎢ 21 22 ⎢ . . T A =⎢ ⎢ . . ⎢ ⎣ . . T T A2m A1m

⎤ T ... Am1 T ⎥ ... A2m ⎥ ... . ⎥ ⎥. ... . ⎥ ⎥ ... . ⎦ T ... Amm

When looking for the definition of partial transpose, two natural versions would come into mind. They are ⎡

A11 A21 ⎢ AT AT ⎢ 12 22 ⎢ . . τ A =⎢ ⎢ . . ⎢ ⎣ . . A1m A2m

⎤ ... Am1 T ⎥ ... Am2 ⎥ ... . ⎥ ⎥, ... . ⎥ ⎥ ... . ⎦ ... Amm



T T A11 A12 ⎢ A21 A22 ⎢ ⎢ . . AΓ = ⎢ ⎢ . . ⎢ ⎣ . . T T Am2 Am1

⎤ T ... A1m ... A2m ⎥ ⎥ ... . ⎥ ⎥. ... . ⎥ ⎥ ... . ⎦ T ... Amm

One observes that AΓ = (Aτ )T . Generally speaking, if a property is valid for Aτ , the same property is also valid for AΓ . For simplicity, here whenever we say partial transpose of A, we refer to Aτ (other authors may use AΓ for partial transpose of A). The primary use of partial transpose is materialized in the so-called PPT criterion in quantum information theory [99, 106], where “PPT” stands for positive partial transpose. For an explanation of the meaning of the partial transpose, we recommend [43]. For notational convenience, we use  m A = Ai j i, j=1 ∈ Mm (Mn ), for (6.2.1). A is PPT if both A and Aτ are positive. A linear map Φ on Mn is said to be m-positive if  m  m (6.2.2) Ai j i, j=1 ≥ 0 ⇒ Φ(Ai j ) i, j=1 ≥ 0. It is said to be completely any integer m ≥ 1. Φ is m if (6.2.2) is true for  positive m completely PPT if further Φ(Ai j ) i, j=1 is PPT Φ(Ai j ) i, j=1 ≥ 0). A comprehensive survey of the standard results on completely positive maps can be found in [20, 178]. Let X ∈ Mn . It is well known that the trace map is completely positive. We shall prove Φ(X ) = X + (T r X )I is completely PPT. Part of the motivation for our result comes from an ingenious observation due to Choi [46]: The map Ψ (X ) = (n − 1)(T r X )I X , from Mn to Mn , is (n − 1)-positive, but is not n-positive. Before Choi’s result the only positive, but not completely positive, maps failed to be even 2-positive. Similarly, concrete examples of PPT maps are rare in the literature. Apparently, the map Φ(X ) is completely positive, thus the main effort is made to show that it is a PPT map. Some interesting consequences of this result are also

230

6 Positive Partial Transpose Matrix Inequalities

included. Though we confine to matrices with matrix entries, extensions to matrices with entries from bounded operators on Hilbert spaces can be similarly stated. Two-by-two block case is most appealing and it contains the essential idea, so we treat this case separately. Lemma 6.2.1 Let x, y ∈ Cn . Then 

x x ∗ + x ∗ x I yx ∗ + x ∗ y I x y ∗ + y ∗ x I yy ∗ + y ∗ y I

 ≥ 0.

Proof. Without loss of generality, we may assume x, y to be unit vectors. Noting that ((x ∗ y)x − y)((x ∗ y)x − y)∗ ≥ 0, we have

|y ∗ x|2 x x ∗ + yy ∗ ≥ (y ∗ x)yx ∗ + (x ∗ y)x y ∗ .

(6.2.3)

Consider the Schur complement yy ∗ + I − (x y ∗ + y ∗ x I )(x x ∗ + I )−1 (yx ∗ + x ∗ y I ) x x∗ (yx ∗ + x ∗ y I ) = yy ∗ + I − (x y ∗ + y ∗ x I ) I − 2 = yy ∗ + (1 − |y ∗ x|2 )I − [(1 − 2|y ∗ x|2 )x x ∗ + (y ∗ x)yx ∗ + (x ∗ y)x y ∗ ] ≥ yy ∗ + (1 − |y ∗ x|2 )I − [(1 − 2|y ∗ x|2 )x x ∗ + |y ∗ x|2 x x ∗ + yy ∗ ] (by 6.2.3) = (1 − |y ∗ x|2 )(I − x x ∗ ) ≥ 0. 

The required result follows. 

A B Proposition 6.2.2 If B∗ C with Φ(X ) = X + (T r X )I . Proof. We may write





 ∈ M2 (Mn ) is positive, then so is

A B B∗ C



 =

X Y



X Y

∗

Φ(A) Φ(B ∗ ) Φ(B) Φ(C)

,

for some X, Y ∈ Mn×2n . Thus, we need to show  H :=

X X ∗ + (T r X X ∗ )I Y X ∗ + (T r Y X ∗ )I X Y ∗ + (T r X Y ∗ )I Y Y ∗ + (T r Y Y ∗ )I

Let X = [x1 X 2 ... x2n ] and Y = [y1 y2 ... y2n ]. Then

 ≥ 0.



6.2 Matrix Inequalities and Completely PPT Maps



2n 

⎢ ⎢ j=1 ⎢ H =⎢ ⎢ ⎣ 2n j=1



2n 

⎢ ⎢ j=1 ⎢ =⎢ ⎢ ⎣ 2n j=1

=

2n

j=1

⎡ ⎣

x j x ∗j

+ (T r

j=1

x j y ∗j + (T r

x j x ∗j + x j y ∗j + x j x ∗j

2n 

+

2n  j=1 2n  j=1

2n  j=1

x j x ∗j )I x j y ∗j )I

j=1 2n  j=1

y j x ∗j

+ (T r

2n  j=1

y j y ∗j + (T r

2n  j=1

⎤ y j x ∗j )I y j y ∗j )I

2n 

y j x ∗j +

⎥ ⎥ ⎥ ⎥ ⎥ ⎦



(T r y j x ∗j )I ⎥ ⎥ j=1 j=1 ⎥ ⎥ ⎥ 2n 2n   ⎦ ∗ ∗ y j y j + (T r y j y j )I

(T r x j y ∗j )I y j x ∗j

2n 

2n 

(T r x j x ∗j )I

x ∗j x j I

231

j=1

+

x ∗j y j I

x j y ∗j + y ∗j x j I y j y ∗j + y ∗j y j I

j=1

⎤ ⎦.



The positivity of H now follows from Lemma 6.2.1.

Using a similar argument as in [137], we have the following proposition.   A B ∈ M2 (Mn ) is positive, then Proposition 6.2.3 If B∗ C 2B + (T r B)I  ≤ A + C + (T r (A + C))I  for any unitarily invariant norm.   A B ∈ M2 (Mn ) is positive, then Proposition 6.2.4 If B∗ C ⎡ ⎣

(T r C)A + B B ∗ (T r B ∗ )B + AC ∗



(T r B)B + C A (T r A)C + B B

⎤ ⎦ ≥ 0.

(6.2.4)

Consequently, (T r A)(T r C) + (T r B ∗ B) ≥ (T r AC) + |T r B|2 . Proof. Again, we write



A B B∗ C





 =

Y ∗Y Y ∗ X for some X, Y ∈ Mn×2n . As X ∗Y X ∗ X

(6.2.5)

 X X∗ XY ∗ , Y X∗ Y Y ∗  ≥ 0, by Proposition 6.2.2, we have

232

6 Positive Partial Transpose Matrix Inequalities

⎡ ⎣

(T r Y ∗ Y )I + Y ∗ Y ∗



(T r X ∗ Y )I + X ∗ Y







(T r Y X )I + Y X (T r X X )I + X X

⎦ ≥ 0.

Now (6.2.4) follows by observing that ⎡ ⎣

(T r C)A + B B ∗ (T r B ∗ )B + AC



⎦ (T r B)B ∗ + C A (T r A)C + B ∗ B ⎡ ⎤ (T r Y Y ∗ )X X ∗ + X Y ∗ Y X ∗ (T r Y X ∗ )X Y ∗ + X X ∗ Y Y ∗ ⎦ =⎣ (T r X Y ∗ )Y X ∗ + Y Y ∗ X X ∗ (T r X X ∗ )Y Y ∗ + Y X ∗ X Y ∗

 =

⎡ ⎤  (T r Y ∗ Y )I + Y ∗ Y (T r X ∗ Y )I + X ∗ Y  ∗ X 0 ⎣ ⎦ X 0 ≥ 0. 0 Y 0 Y (T r Y ∗ X )I + Y ∗ X (T r X ∗ X )I + X ∗ X

In view of (6.2.4), we have (T r C)A + B B ∗ + (T r A)C + B ∗ B ≥ (T r B ∗ )B + AC + C A + (T r B)B ∗ . 

Taking trace on both sides yields (6.2.5).

The Schur product (or the Hadamard product) [103] of two matrices X, Y ∈ Mn is denoted by X ◦ Y . The variants of the following inequality are more or less known. Lemma 6.2.5 Let X, Y ∈ Mn . Then (X ∗ Y ) ◦ (Y ∗ X ) + (X ∗ X ) ◦ (Y ∗ Y ) ≥ 0. Proof. Notice that 

X ∗ X X ∗Y Y ∗ X Y ∗Y



 ≥ 0,

Y ∗Y Y ∗ X X ∗Y X ∗ X

 ≥ 0,

by the Schur product theorem [103], we have ⎡ ⎣

(X ∗ X ) ◦ (Y ∗ Y ) (X ∗ Y ) ◦ (Y ∗ X ) (X ∗ Y ) ◦ (Y ∗ X ) (X ∗ X ) ◦ (Y ∗ Y )

⎤ ⎦=

from which the required result follows. Now we are in a position to present





X ∗ X X ∗Y Y ∗ X Y ∗Y

  ∗  Y Y Y∗X ◦ ≥ 0, X ∗Y X ∗ X

6.2 Matrix Inequalities and Completely PPT Maps

233

Theorem 6.2.6 The map Φ(X ) = X + (T r X )I is completely PPT. Proof. Using a similar idea as to the proof of Proposition 6.2.2, it suffices to show H := [x j xi∗ + xi∗ x j I ]i,m j=1 ≥ 0 for any⎡x j ∈⎤Cn , j = 1, ..., m. y1 ⎢ .. ⎥ Let ⎣ . ⎦ ∈ Cmn with y j ∈ Cn , j = 1, ..., m. Consider the quadratic form assoym ciated with H , ∗

y Hy =

m

m

(yi∗ x j xi∗ y j + xi∗ x j yi∗ y j ).

i=1 j=1

Let X = [x1 ... xm ], Y = [y1 ... ym ]. By Lemma 6.2.5, (X ∗ Y ) ◦ (Y ∗ X ) + (X ∗ X ) ◦ (Y ∗ Y ) ≥ 0. Hence, y ∗ H y = e∗ ((X ∗ Y ) ◦ (Y ∗ X ) + (X ∗ X ) ◦ (Y ∗ Y ))e ≥ 0, where e ∈ Cm is a vector with all entries equal to one. This completes the proof.  The following result is essentially due to Hiroshima [99] and Horodecki et. al. [106], see also [137].  m Theorem 6.2.7 Let A = Ai j i, j=1 ∈ Mm (Mn ) be PPT. Then   

  m   A A ≤  jj   j=1  for any unitarily invariant norm. Thus, thanks to Theorem 6.2.6, we have the following proposition. Proposition 6.2.8 Let A j ∈ Mm×n , j = 1, ..., l. Then   

 l   ∗ ∗  [(T r Ai∗ A j )I + Ai∗ A j ]li, j=1  ≤  ((T r A A )I + A A ) j j j j    j=1  for any unitarily invariant norm. The following theorem is an extension of Proposition 6.2.4 to higher number of blocks.

234

6 Positive Partial Transpose Matrix Inequalities

 m Theorem 6.2.9 Let A = Ai j i, j=1 ∈ Mm (Mn ) be positive. Then [(T r A(m−i+1)(m− j+1) )Ai j + Ai(m− j+1) A(m−i+1) j ]i,m j=1 ≥ 0. Proof. We may write A as



⎤⎡ ⎤∗ X1 X1 ⎢ .. ⎥ ⎢ .. ⎥ ⎣ . ⎦⎣ . ⎦ , Xm Xm

for some X j ∈ Mn×mn , j = 1, ..., m, so Ai j = X i X ∗j , i, j = 1, ..., m. Define ⎤ ⎡ ∗ ⎤∗ Xm X m∗ ⎢ .. ⎥ ⎢ .. ⎥ = ⎣ . ⎦⎣ . ⎦ . ⎡

B = [Bi j ]i,m j=1

X 1∗

X 1∗

∗ Thus B = [Bi j ]i,m j=1 = [X m−i+1 X m− j+1 ]i,m j=1 ≥ 0. By Theorem 6.2.6, we have

[(T r B ji )I + B ji ]i,m j=1 ≥ 0. That is m  ∗ ∗ C = (T r X m−i+1 X m− j+1 )I + X m− j+1 X m−i+1 i, j=1 m  ∗ ∗ = (T r X m− j+1 X m−i+1 )I + X m− j+1 X m−i+1 i, j=1 ≥ 0. ⎡ ⎢ ⎢ The conclusion follows from PC P ≥ 0 with P = ⎢ ⎢ ⎣ ∗

X1

⎤ .

.

⎥ ⎥ ⎥. ⎥ ⎦

.



Xm

Notes and references. The proofs of Lemma 6.2.1, Lemma 6.2.5, Proposition 6.2.2, Proposition 6.2.3, Proposition 6.2.4, Theorem 6.2.6, and Theorem 6.2.9 are taken from [134].

6.3 Hiroshima’s Type Inequalities for Positive Semidefinite Block Matrices

235

6.3 Hiroshima’s Type Inequalities for Positive Semidefinite Block Matrices In 2003, Hiroshima proved a very beautiful result on majorization relations which has useful applications in quantum physics [90, 107] and references therein. However, this result seems not widely known in the field of matrix analysis. Theorem 6.3.1 ([99]) Let H = [Hi j ] ∈ Mm (Mn ). If Im ⊗ T r1 H ≥ H ≥ 0, then   

  m   H  ≤ T r1 H  =  H jj ,   j=1 

(6.3.1)

for any unitarily invariant norm. In particular, if H is positive partial transpose, then (6.3.1) holds. Indeed, independent of the Hiroshima paper, we have the special case of Theorem 6.3.1 as follows:   A X H= ≥ 0 ⇒ H  ≤ A + B (6.3.2) X B for any unitarily invariant norm [141]. We remark that a sharper observation that entails (6.3.2) is the following theorem [37, 38]. Theorem 6.3.2 We have   1 A X H= ≥ 0 ⇒ H = (U (A + B)U ∗ + V (A + B)V ∗ ), X B 2 for some isometries U, V . Here, we fix some notation. The tensor product Mm ⊗ Mn is canonically identified with Mm (Mn ). Here Mm (Mn ) is the space of m × m block matrices with entries in Mn . Let A j ∈ Mm , B j ∈ Mn , j = 1, ..., p, and consider the tensor product H = p  A j ⊗ B j . As H ∈ Mm (Mn ), we can write H = [Hi j ] with Hi j ∈ Mn . The partial j=1

trace of H is defined as [181]. p

T r1 H = (T r A j )B j . j=1

236

6 Positive Partial Transpose Matrix Inequalities

It is readily verified that T r1 H =

m 

H j j . The partial transpose map H → H τ is

j=1

defined on Mm ⊗ Mn as H τ =

P  j=1

A Tj ⊗ B j . If H and H τ are both positive, then we

say H is positive partial transpose. We remark that the observation H is positive partial transpose ⇒ Im ⊗ T r1 H ≥ H is due to Horodecki et. al. [106]. In this section, we make use of Theorem 6.3.1 to prove some new results on matrix norm inequalities. It is expected that Hiroshima’s theorem will become a practical tool in the field of matrix analysis. Our first result is the following norm inequality of commuting type. It is an immediate consequence of Hiroshima’s result. Proposition 6.3.3 Let X 1 , X 2 , ..., X k ∈ Mm×n such that X i∗ X j is Hermitian for all 1 ≤ i, j ≤ k. Then     

 

 k   k   ∗ ∗   (6.3.3) X j X j ≤  X j X j    j=1   j=1  for any unitarily invariant norm. Proof. Denote H by

⎤ X 1∗ ⎥ ⎢ H = [Hi j ] = ⎣ ... ⎦ [X 1 ... X k ]. ⎡

X k∗

Then H ∈ Mk (Mn ) is positive and Hi j = H ji , therefore H equals its partial transpose. Thus, by Theorem 6.3.1, we have     

 

  k   k  ∗    = H  ≤  H X X jj j j .    j=1   j=1  On the other hand, ⎡ ∗ ⎤   X1    ⎢ .. ⎥  H  = ⎣ . ⎦ [X 1 ... X k ]    X∗  k  ⎡ ∗ ⎤  X1     ⎢ .. ⎥ = [X 1 ... X k ] ⎣ . ⎦    X k∗    

  k  ∗  = X j X j.  j=1 

6.3 Hiroshima’s Type Inequalities for Positive Semidefinite Block Matrices

The desired result follows.

237



The special case of Proposition 6.3.3 when k = 2 has been observed in [141]. The inequality (6.3.3) is elegant as an inequality of commuting type. To the authors’ best knowledge, another example of commuting-type norm inequality is the following, and there are no others. Proposition 6.3.4 ([21]) Let X, Y ∈ Mm×n such that X ∗ Y is Hermitian. Then X ∗ Y + Y ∗ X  ≤ X Y ∗ + Y X ∗ 

(6.3.4)

for any unitarily invariant norm. The usefulness of inequality (6.3.4) has been demonstrated in matrix analysis [21, 121, 195]. We now present an application of Proposition 6.3.3 following the spirit of [21]. Theorem 6.3.5 Let A, B, X ∈ Mn . Then (A A∗ + X B ∗ B X ∗ ) ⊕ (B ∗ B + X ∗ A A∗ X ) ≤ (A∗ A + A∗ X X ∗ A) ⊕ (B B ∗ + B X ∗ X B ∗ )

(6.3.5)

for any unitarily invariant norm. Proof. We first prove the case where A, B are Hermitian. Consider  T =

   A 0 0 X . , S= 0 B X∗ 0

Take X 1 = T , X 2 = ST , as X 1∗ X 2 = T ST is Hermitian. Then by Proposition 6.3.3 for k = 2, we have T 2 + ST 2 S ≤ T 2 + T S 2 T , i.e., (A2 + X B 2 X ∗ ) ⊕ (B 2 + X ∗ A2 X ) ≤ (A2 + AX X ∗ A) ⊕ (B 2 + B X ∗ X B).

(6.3.6)

For the general case, consider the polar decompositions A = |A|U , B = |B|V . Then (6.3.6) yields (|A|2 + Y |B|2 Y ∗ ) ⊕ (|B|2 + Y ∗ |A|2 Y ) ≤ (|A|2 + |A|Y Y ∗ |A|) ⊕ (|B|2 + |B|Y ∗ Y |B|) for any Y ∈ Mn .

(6.3.7)

238

6 Positive Partial Transpose Matrix Inequalities

As |A| = AU ∗ = U A∗ , |B| = BV ∗ = V B ∗ , we have |A|2 = A A∗ = U A∗ AU ∗ , |B|2 = B B ∗ = V B ∗ BV ∗ . Substituting these into (6.3.7) and setting Y = X V ∗ gives (6.3.5).  Remark 6.3.6 In particular, for the Schatten- p norm, (6.3.5) leads to A A∗ + X B ∗ B X ∗  pp + B ∗ B + X ∗ A A∗ X  pp ≤ A∗ A + A∗ X X ∗ A pp + B B ∗ + B X ∗ X B ∗  pp . For the trace norm, (6.3.5) becomes an equality, and so we have the following determinantal inequality: det(A A∗ + X B ∗ B X ∗ ) det(B ∗ B + X ∗ A A∗ X ) ≥ det(A∗ A + A∗ X X ∗ A) det(B B ∗ + B X ∗ X B ∗ ). In the following and the subsequent results in this section, we are mainly concerned with the application of Theorem 6.3.1 when H ∈ M2 (Mn ). We restate Theorem 6.3.1 here as a lemma.   A X ∈ M2 (Mn ) be positive. Furthermore, if H is PPT, Lemma 6.3.7 Let H = X∗ B then H  ≤ A + B for any unitarily invariant norm. The next lemma plays an important role in our analysis.   A X Lemma 6.3.8 ([200]) Let H = ∈ Mm+n be positive with A ∈ Mm and X∗ B B ∈ Mn . Then 2σ j (X ) ≤ σ j (H ), j = 1, ..., min{m, n}. It should be mentioned that Lemma 6.3.8 has several variants. We refer to [200] and references therein for equivalent forms. Ky Fan’s dominance theorem [21] reveals an important relation between singular value inequalities and norm inequalities. More precisely, let A, B ∈ Mn . Then the following statements are equivalent: (i)

k  j=1

σ j (A) ≥

k 

σ j (B) for k = 1, ..., n;

j=1

(ii) A ≥ B for any unitarily invariant norm .. Thus, Lemma 6.3.8 is strong enough to entail  H=

A X X∗ B

 ≥ 0 =⇒ 2X  ≤ H 

(6.3.8)

6.3 Hiroshima’s Type Inequalities for Positive Semidefinite Block Matrices

239

for any unitarily invariant norm. The following theorem states a relation between the norm of diagonal blocks of H and the norm of its off-diagonal block. Theorem 6.3.9 Let A, B ∈ Mm×n be strictly contractive. Then 2(I − A∗ B)−1  ≤ (I − A∗ A)−1 + (I − B ∗ B)−1 

(6.3.9)

for any unitarily invariant norm. Proof. As H is PPT, (6.3.9) follows immediately from Lemma 6.3.7 and (6.3.8).  From the proof of Theorem 6.3.9, we find that in proving certain norm inequalities, it suffices to show the corresponding (block) matrix is PPT. One may suspect that Theorem 6.3.9 is a special case of a much more general result. In particular, it is tempted to ask whether it is true 

A X X B

 ≥ 0 =⇒ 2X  ≤ A + B?

(6.3.10)

The answer is no, as the following example shows. Example 6.3.10 Take  A=

     1 1 1 −1 1 −1 , B= , X= . 1 1 −1 1 1 −1

 A X is positive. A simple calculation gives It is easy to check in this case X∗ B σ1 (X ) = 2, σ2 (X ) = 0 and σ1 (A + B) = σ2 (A + B) = 2. Thus, in this case, by Ky Fan’s dominance theorem, 2X  ≥ A + B holds for every unitarily invariant norm. 

Remark6.3.11 In spite of the failure of (6.3.10), we have the following. A X Assume ∈ M2 (Mn ) is positive. Then X∗ B X + X ∗  ≤ A + B

(6.3.11)

for any unitarily invariant norm. Indeed, 

A X X∗ B



 ≥ 0 =⇒ =⇒

A + B X + X∗ X + X∗ A + B

k

j=1

σ j (X + X ∗ ) ≤

 ≥0 k

j=1

σ j (A + B), k = 1, ..., n,

240

6 Positive Partial Transpose Matrix Inequalities

which is stronger than (6.3.11). Proposition 6.3.12 Let A, B ∈ Mn be positive and U ∈ Mn be unitary. Then A + U B ≤ A + B + U BU ∗ 

(6.3.12)

for any unitarily invariant norm. Proof. First, we observe that  A+UB A + B + U BU ∗ A + B + U BU ∗ A + BU ∗       A A B 0 U BU ∗ U B ≥ 0. = + + BU ∗ B A A 0 U BU ∗ 

H=

Second,  A + BU ∗ A + B + U BU ∗ H = A+UB A + B + U BU ∗       U BU ∗ 0 A A B BU ∗ ≥ 0. + = + U B U BU ∗ 0 B A A τ



Thus, H is PPT. The desired result then follows by Lemma 6.3.7 and (6.3.8).



On the other hand, Lee [127] proved the following. Proposition 6.3.13 Let A, B ∈ Mn . Then A + B ≤

√ 2|A| + |B|

for any unitarily invariant norm. Equivalently, if A, B ≥ 0, then for any unitary matrix U ∈ Mn , √ (6.3.13) A + U B ≤ 2A + B. There is no obvious relation between (6.3.12) and (6.3.13). However, when A ≥ 0 and B = I , we have a stronger inequality A + U  ≤ A + I , which is due to Ky Fan and Hoffman [62]. Notes and references. The proofs of Proposition 6.3.3, Theorem 6.3.5, Theorem 6.3.9, and Proposition 6.3.12 in this section are due to [137].

6.4 Geometric Mean and Norm Schwarz Inequality

241

6.4 Geometric Mean and Norm Schwarz Inequality   A B ≥ 0 implies It is well known that positivity of a 2 × 2 operator matrix B∗ C √ B ≤ A.C for operator norm .. This can be considered as an operator version of the Schwarz inequality. In fact, since for {X 1 , ..., X n } and {Y1 , ..., Yn }, ⎡ n ⎢ j=1 ⎢ ⎢ ⎢ n ⎣ j=1

X j X ∗j Y j X ∗j

n  j=1 n  j=1

X j Y j∗ Y j Y j∗

⎤ ⎥

  ∗ n  ⎥ Xj Xj ⎥= . ≥ 0, ⎥ Y Yj j ⎦ j=1

the following Schwarz inequality follows:        

   

 n n n 

       ∗ ∗  ∗   ≤ . X Y X X Y Y j j j j  j j .    j=1   j=1   j=1  For other versions of the operator Schwarz-type inequalities, see [1] and [67] and references therein. On the other hand, there is a well-established notion of geometric √ mean AC for operators A, C ≥ 0 for which AC ≤ A.C.  section, we study under what conditions on A, B, and C or on B alone  In this A B ≥ 0 implies B ≤ AC. We say that the norm Schwarz inequality holds B∗ C for A, C ≥ 0 and B if B ≤ AC. In particular, we are interested in establishing conditions on B such that 

A B B∗ C

 ≥ 0 =⇒ B ≤ AC.

(6.4.1)

Here at first, we summarize known properties of a 2 × 2 operator matrix and also those of geometric mean. Lemma 6.4.1 ([20]) The following statements are mutually equivalent for operators A, B, C.   A B ≥ 0; (1) B∗ C   C B∗ (2) ≥ 0; B A 1 1 (3) A, C ≥ 0 and B = A 2 W C 2 for some W with W  ≤ 1; (4) A ≥ 0 and C ≥ B ∗ (A + εI )−1 B for all ε > 0. When A > 0, B ∗ (A + εI )−1 B can be replaced simply by B ∗ A−1 B.

242

6 Positive Partial Transpose Matrix Inequalities

For the convenience of the readers, we recall that Lemma 6.4.2 ([20]) geometric mean for A, C > 0 has the following properties: AC = CA; 1 AC = (AC) 2 when AC = C A; −1 ; A−1 C −1 = (AC) √ (α A)(β B) = αβ(AC); A → AC inmonotone increasing;   A AC A X 6. ≥ 0 and AC = max{X ≥ 0; ≥ 0}; AC C X C ∗ ∗ ∗ 7. (X AX )(X C X ) ≥ X (AC)X for all X .

1. 2. 3. 4. 5.

In view of monotonicity, the notion of the geometric mean is uniquely extended to all A, C ≥ 0 as the limit in the strong operator topology AC = lim(A + εI )(C + εI ). ε↓0

Since ∀ε > 0,



A B B∗ C



 ≥ 0 ⇐⇒

 A + εI B , B ∗ C + εI

and AC = lim (A + εI )(C + εI ), ε↓0

throughout our discussions on the norm Schwarzinequality, we may always assume A B that A > 0 and C > 0 in the inequality ≥ 0. B∗ C In the following, we study a necessary condition for (6.4.1). Let r (X ) denote the spectral radius of X. That is, r (X ) = max{|λ|; λI − X is not invertible}. Since it is known [93] that r (X ) is described by using norms of iterates of X as 1

r (X ) = lim X n  n , n→∞

we can see r (X ) ≤ X  and r (X Y ) = r (Y X ), ∀X, Y.

(6.4.2)

An operator B is called normalized if r (B) = B. A normal operator B, in particular, a self-adjoint operator, is normalized [93]. Since, by Lemma 6.4.1, 

A B B∗ C





≥0⇒C ≥B A

−1

 B and

A B B ∗ B ∗ A−1 B

 ≥ 0,

(6.4.3)

6.4 Geometric Mean and Norm Schwarz Inequality

243

and by the definition of geometric mean, we have A(B ∗ A−1 B) = A 2 .|A− 2 B A− 2 |.A 2 , 1

1

1

1

(6.4.4)

where |X | = (X ∗ X ) 2 is the modulus of X , property (6.4.1) for B is equivalent to the following: 1 1 1 1 B ≤ A 2 .|A− 2 B A− 2 |.A 2  (6.4.5) 1

for all A > 0. Lemma 6.4.3 For all A > 0 and B, we have A 2 |A− 2 B A− 2 |A 2  ≤ A− 2 B A− 2 . 1

1

1

1

1

1

Proof. Since both sides are positive homogeneous of order 1 with respect to B, it suffices to prove that A− 2 B A− 2  = 1 ⇒ A 2 |A− 2 B A− 2 |A 2 ≤ I. 1

1

1

1

1

1

Now A− 2 B A− 2  = 1 implies A 2 B ∗ A−1 A 2 , and hence 1

1

1

1

|A− 2 B A− 2 |2 ≤ A−2 . 1

1

In view of the Lowner theorem [20], this implies |A− 2 B A− 2 | ≤ A−1 . 1

Hence

1

A 2 |A− 2 B A− 2 |A 2 ≤ I, 1

and this completes the proof.

1

1

1



Theorem 6.4.4 If (6.4.1), equivalently (6.4.5) holds for B, then B is a normalized. Proof. By Lemma 6.4.3 and (6.4.5), we have B ≤ A 2 .|A− 2 B A− 2 |.A 2  ≤ A− 2 B A− 2 . 1

1

1

1

1

1

Now the assertion follows from the following characterization of the spectral radius [93] 1 1 r (B) = inf{A− 2 B A− 2  : A > 0}.  

244

6 Positive Partial Transpose Matrix Inequalities

In the converse direction, we have the following theorem.   1 1 A B Theorem 6.4.5 If B is normalized, then ≥ 0 implies B ≤ A 2 C 2 . ∗ B C Proof. In view of Lemma 6.4.1, we have by (6.4.2) 1

1

B = r (B) = r (A 2 W A 2 ) 1

1

1

1

1

1

= (W C 2 A 2 ) ≤ C 2 A 2  = A 2 C 2 .  

 A AC Finally, since by Lemma 6.4.2, ≥ 0 and AC is self-adjoint and AC C hence normalized, we have from Theorem 6.4.5, 1

1

AC ≤ A 2 C 2 . A little sharper inequality holds [13] for A, C ≥ 0 as follows: 1

1

1

1

1

AC ≤ A 4 C 2 A 4  ≤ A 2 C 2 . Now, we present the sufficient conditions. Lemma 6.4.6 We have     Aj B B A1 A2 ≥ 0 ( j = 1, 2) ⇒ ≥ 0. B∗ C j B ∗ C1 C2 Proof. In view of Lemma 6.4.1, the assumption means that B ∗ A−1 j B ≤ C j , which implies by Lemma 6.4.2, −1 ∗ −1 ∗ −1 B ∗ (A1 A2 )−1 B = B ∗ (A−1 1 A2 )B ≤ (B A1 B)(B A2 B) ≤ C 1 C 2 .

 Again, by Lemma 6.4.1, this implies

B A1 A2 B ∗ C1 C2

 ≥ 0.



Theorem 6.4.7 For A > 0 and B, we have  B ≤ A(B ∗ A−1 B).A(B A−1 B ∗ ). Proof. This follows from Lemma 6.4.6 via 

A B B ∗ B ∗ A−1 B



 ≥ 0 and

B A−1 B ∗ B A B∗

 ≥ 0. 

6.4 Geometric Mean and Norm Schwarz Inequality

245

 Theorem 6.4.8 If a positive operator matrix 

C B Proof. Since by Lemma 6.4.1, we have B∗ A   AC B that ≥ 0, which implies B ∗ CA B ≤



 A B is PPT, then B ≤ AC. B∗ C  ≥ 0. It follows from Lemma 6.4.6

AC.CA = AC. 

Theorem 6.4.8 can be automatically applied to the case of self-adjoint B. But a little more can be said as an extension of Lemma 6.4.2.   A B Theorem 6.4.9 If B is self-adjoint and ≥ 0, then AC ≥ ±B. B C  Proof. In the proof of Theorem 6.4.8, it is shown that implies immediately that AC ≥ ±B.



AC B B AC

 ≥ 0 which

Theorem 6.4.10 If B is a scalar multiple of a unitary operator, then 

A B B∗ C

 ≥ 0 ⇒ B ≤ AC.

Proof. We may assume that B is unitary, and we will prove via (6.4.3) that A(B ∗ A−1 B) ≥ 1 = B. Suppose by contradiction, that 1 , 1+ε

∃ε > 0,

A−1 (B ∗ AB) ≥ (1 + ε)I,

∃ε > 0.

A(B ∗ A−1 B) ≤ or equivalently by Lemma 6.4.2,

Since A−1 (B ∗ AB) = A− 2 |A 2 B A 2 |A− 2 by (6.4.4), this leads to 1

1

1

1

1

1

|A 2 B A 2 | ≥ (1 + ε)A, and hence

246

6 Positive Partial Transpose Matrix Inequalities

A ≥ (1 + ε)A, 

which is a contradiction.

 Theorem 6.4.11 Positivity

A B B∗ C

 ≥ 0 implies B ≤ AC if one of the fol-

lowing conditions is satisfied. (1) AB = B A; (2) B ∗ A−1 B = B A−1 B ∗ ; (3) ∃α > 0 such that C = α A.  Proof. (1) Commutativity implies B ∗ A−1 B = |B|A−1 |B|, so that

 A |B| ≥0 |B| C

by Lemma 6.4.1. Now appeal to Lemma 6.4.2. (2) Appeal to Theorem 6.4.7. (3) Appeal to Lemma 6.4.2.  Theorems 6.4.9 and 6.4.10 suggest that (6.4.1), equivalently (6.4.5), will hold for all normal B. At present, we can settle (6.4.5) for all normal B only when dim(H) = n = 2. The proof, as seen below, is quite specialized to the case n = 2. Theorem 6.4.12 For positive definite operator A ∈ M2 and normal B ∈ M2 , B ≤ A 2 .|A− 2 B A− 2 |.A 2 . 1

1

1

1

Proof. We may assume that B = 0. Since both sides of the desired inequality are positive homogeneous of order 1 with respect to B, it suffices to show that D = A 2 .|A− 2 B A− 2 |.A 2 ≤ I ⇒ B ≤ 1. 1

1

1

1

(6.4.6)

Now we have, by definition of modulus, A− 2 D A−1 D A− 2 = A− 2 B ∗ A−1 B A− 2 , 1

1

1

1

and hence, with S = A−1 > 0, B ∗ S B = DS D.

(6.4.7)

Since B ∈ M2 is normal, we may assume that it is a diagonal matrix [93] as follows:  λ1 0 . B= 0 λ2 

(6.4.8)

6.4 Geometric Mean and Norm Schwarz Inequality



Write D=

d1d d12 d21 d22



247

 and S =

 s11 s12 . s21 s22

(6.4.9)

Then it follows from D ≥ 0 and S > 0 that d11 , d22 ≥ 0, d¯12 = d21 and s11 , s22 ≥ 0, s12 ¯ = s21 .

(6.4.10)

If min(d11 , d22 ) = 0, then D ≥ 0 implies d12 = 0 = d21 . Then it follows from (6.4.7) and (6.4.8) that |λ j | = d j j ( j = 1, 2), so that B = max(|λ1 |, |λ2 |) = max(d11 , d22 ) ≤ D ≤ 1. Therefore, we may assume that d j j > 0 ( j = 1, 2). Taking determinants of both sides of (6.4.7) we have by (6.4.8) |λ1 λ2 | = det(D) ≥ 0.

(6.4.11)

Computing the (1, 1)-entry and the (2, 2)-entry of each side of (6.4.7), we have by (6.4.8), (6.4.9), and (6.4.10), 2 + d11 (s12 d21 + s21 d12 ) + s22 |d12 |2 , s11 |λ1 |2 = s11 d11

and 2 . s22 |λ2 |2 = s11 |d12 |2 + d22 (s12 d21 + s21 d12 ) + s22 d22

Then it follows from (6.4.11) that d22 s11 |λ1 |2 − d11 s22 |λ2 |2 2 2 = d22 s11 d11 − d11 s11 |d12 |2 + d22 s22 |d12 |2 − d11 s22 d22

= d11 s11 (d11 d22 − |d12 |2 ) + d22 s22 (|d12 |2 − d11 d22 ) = (d11 s11 − d22 s22 )|λ1 |.|λ2 |. Therefore, we have (s11 |λ1 | + s22 |λ2 |)(d22 |λ1 | − d11 |λ2 |) = 0.

(6.4.12)

Since s11 , s22 > 0 and max(|λ1 |, |λ2 |) > 0, (6.4.12) implies d22 |λ1 | = d11 |λ2 |. Then it follows from (6.4.11) and (6.4.13) that

(6.4.13)

248

6 Positive Partial Transpose Matrix Inequalities

|d12 |2 = d11 d22 − |λ1 |.|λ2 | d22 2 − |λ1 |2 ) = (d11 d11 d11 2 = (d22 − |λ2 |2 ) , d22 and hence by (6.4.11), |λ1 |2 =

d11 d11 (d11 d22 − |d12 |2 ) = det(D), d22 d22

and correspondingly, |λ2 |2 =

d22 det(D). d11

Since det(D) ≤ d11 d22 , we can conclude that |λ1 | ≤ d11 and |λ2 | ≤ d22 , and finally B = max(|λ1 |, |λ2 |) = max(d11 , d22 ) ≤ D ≤ 1. This completes the proof of (6.4.6).



Notes and references. Lemmas 6.4.3 and 6.4.6 are due to [9]. Also, the proofs of Theorem 6.4.4, Theorem 6.4.5, Theorem 6.4.7, Theorem 6.4.8, Theorem 6.4.9, Theorem 6.4.10, Theorem 6.4.11, and Theorem 6.4.12 are taken from [9].

6.5 Inequalities Involving the Off-Diagonal Block of a PPT Matrix The classical study of positive matrices partitioned in blocks has numerous applications. When these matrices have positive partial transpose, some nice results have been obtained, motivated by the Quantum Information Science, as the NielsenKempe majorization [176] criterion (also [17]) and its extension by Hiroshima [99] (also [152]). Let     A X∗ A X τ ≥ 0 and M = ≥ 0. M= X B X∗ B

6.5 Inequalities Involving the Off-Diagonal Block of a PPT Matrix

249

Then M is PPT. For such matrices, the Nielsen-Kempe-Hiroshima majorization holds: For all k = 1, ..., 2n, k

 λj

j=1

A X X∗ B

 ≤

k

λ j (A + B),

(6.5.1)

j=1

with equality for k = 2n. Here, given a Hermitian m × m matrix Z , we use the notation λ j (Z ) for the eigenvalues of Z arranged in decreasing order, 1 ≤ j ≤ m, and we merely set λ j (Z ) = 0 for j > m. When X = X ∗ , a stronger result holds as in Theorem 6.3.2. In Theorem 6.3.2 and the majorization (6.5.1), the full matrix is connected to the sum of its diagonal blocks. The main result of this section connects the off-diagonal part X of the full matrix with the geometric mean of its diagonal blocks, where the usual limit from the above process is used in case of noninvertible matrices. That is, in this section, we give an operator inequality involving the off-diagonal block of a positive partial transpose matrix and the geometric mean of its diagonal blocks. Some new eigenvalue inequalities are derived, nicely completing the existing ones.   A X ∈ M2n is PPT, then for some unitary matrix V ∈ Mn , Theorem 6.5.1 If X∗ B AB + V ∗ (AB)V . 2

|X | ≤

Proof. From the PPT assumption, we infer that the matrix ⎡

A ⎢0 ⎢ ⎣0 X

0 A X∗ 0

0 X B 0

⎤ X∗ 0 ⎥ ⎥ ≥ 0. 0 ⎦ B

By the well-known maximal characterization of the geometric mean, we then infer  ±

so that

0 X X∗ 0







   A 0 B 0  0 A 0 B   AB 0 = , 0 AB



AB X X ∗ AB

 ≥ 0.

(6.5.2)

Mimicking the technique of the proof of [32, Proposition 2.11], we observe from (6.5.2) that

250

6 Positive Partial Transpose Matrix Inequalities

[−V ∗ I ]



AB X X ∗ AB



−V I

 ≥ 0,

for any matrix V . Letting V be the unitary factor in the polar decomposition X = V |X | yields the result.  By the min-max principle, this nice operator inequality is equivalent to a remarkable eigenvalue inequality.   A X ∈ M2n is PPT, then for all j = 1, ..., n, Corollary 6.5.2 If X∗ B λ j (2|X | − AB) ≤ λ j (AB). The following special case is already nontrivial.   A X ∈ M2n is PPT where A is invertible, then X  ≤ 1. Corollary 6.5.3 If X ∗ A−1 In the course of the proof of Theorem 6.5.1, we gave an original and very simple proof of a recent and remarkable result of Ando in the proof of Theorem 6.4.8, the positivity of the matrix (6.5.2). This is the main step of the proof. As we know, for 0 ≤ S, T ∈ Mn , the weak log-majorization S ≺w log T means k

λ j (S) ≤

j=1

k

λ j (T ),

j=1

for all k = 1, ..., n.   A X ∈ M2n is PPT, then Theorem 6.5.4 If X∗ B |X | ≺w log AB. Proof. From (6.5.2), we get 1

1

X = (AB) 2 C(AB) 2 , for some contraction C. This implies, thanks to Horn inequalities, the log-majorization of the theorem.  From Theorem 6.5.1 follows an eigenvalue inequality which nicely complements the log-majorization of Theorem 6.5.4. The proof is a simple application of Weyl’s inequality

6.5 Inequalities Involving the Off-Diagonal Block of a PPT Matrix

251

λ j+k+1 (H ) ≤ λ j+1 (H ) + λk+1 (H ), for any Hermitian H ∈ Mn with the convention λ1 (H ) = 0 if j > n.   A X ∈ M2n is PPT, then for all j, k = 0, 1, 2, ..., n, Corollary 6.5.5 If X∗ B 2λ j+k+1 (|X |) ≤ λ j+1 (AB) + λk+1 (AB). In particular, λ2 j−1 (|X |) ≤ λ j (AB), j ≥ 1. For instance, with A, B, X ∈ Mn , n ≥ 5, we have λ5 (|X |) ≤ λ3 (AB). This combined with Theorem 6.5.4 shows that the eigenvalues of the blocks of a PPT matrix have strong constraints. As another application of Theorem 6.5.1, we recapture a sharper form of the trace inequality T r (X ∗ X ) ≤ T r (AB) given (6.1.3).   A X ∈ M2n is PPT, then for some unitary matrices U, V ∈ Corollary 6.5.6 If X∗ B Mn , U ∗ (AB)2 U + V ∗ (AB)2 V . X∗X ≤ 2 Therefore

T r (X ∗ X ) ≤ T r (AB)2 ≤ T r (AB).

Proof. Since 0 ≤ S ≤ T implies 0 ≤ S ≤ L ∗ T L for some unitary matrix L, the operator inequality follows from Theorem 6.5.1 combined with the operator convexity of t → t 2 . The first trace inequality is an immediate consequence. The second one is 1 1 well known and easy to prove. Let W be the unitary part in AB = A 2 W B 2 . Then 1

1

1

1

1

1

T r (AB)2 = T r (B 2 A 2 W B 2 A 2 W ) ≤ T r (A 2 B A 2 ), by the Cauchy-Schwarz inequality |T r Z 2 | ≤ T r (Z ∗ Z ).



Remark 6.5.7 In fact, the first trace inequality in Corollary 6.5.6 can be improved as a weak log-majorization in Theorem 6.5.4, meanwhile the last trace inequality can be strengthened as a well-known log-majorization, that is, Theorem 6.5.4 can be inserted as a new weak log-majorization in the series 1

1

|X | ≺w log (AB) ≺log |A 2 B 2 |, equivalently, X ∗ X ≺w log (AB) ≺log A 2 B A 2 . 1

1

On the other hand, we have a special case of the Ando-Hiai log-majorization [13],

252

6 Positive Partial Transpose Matrix Inequalities

(A2 B 2 ) ≺log (AB)2 , so it would be valuable to give an example, if any, of a PPT matrix whose blocks satisfy T r (X ∗ X ) ≥ T r (A2 B 2 ). Remark 6.5.8 Without the PPT assumption, the trace inequality in Corollary 6.5.6, T r (X ∗ X ) ≤ T r (AB), can be reversed. Here is a generic example. Let 

A X X∗ B



⎡ =⎣

1

f (C) 1 2



1

f 2 (C)V g 2 (C) 1 2

g (C)V f (C)

⎤ ⎦,

g(C)

where 0 ≤ C ∈ Mn , V is unitary in Mn , and f (t), g(t) are nonnegative functions defined on [0, ∞) such that f (t) is nondecreasing and g(t) is nonincreasing. Then we have T r (X ∗ X ) ≥ T r (AB) [33].

Notes and references. Note that Theorem 6.5.1 and Theorem 6.5.4, Corollaries 6.5.2, 6.5.3, 6.5.5 and 6.5.6 are due to [128].

6.6 Unitarily Invariant Norm Inequalities of PPT Matrices Hayajneh and Kittaneh conjectured the following inequality in [95]. Theorem 6.6.1 Let A1 , A2 , B1 , B2 ∈ Mn be positive semidefinite with A1 A2 = A2 A1 and B1 B2 = B2 B1 . Then 1

1

A1 A2 + B1 B2  ≤ (A1 + B1 ) 2 (A2 + B2 )(A1 + B1 ) 2  holds for any unitarily invariant norm. The following lemma is a variation of Theorem 6.1.3.   A X Lemma 6.6.2 Let be PPT, where A, B, X ∈ Mn . Then X∗ B X ∗ X  ≤ A 2 B A 2  1

holds for any unitarily invariant norm.

1

6.6 Unitarily Invariant Norm Inequalities of PPT Matrices

253

 1 1 A X ≥ 0, we have X = A 2 C B 2 for some contraction C ∈ Mn X∗ B 1 1 [20]. Similarly, X ∗ = A 2 D B 2 for some contraction D ∈ Mn . Hence for k = 1, ..., n, 

Proof. Since

k

σi (X ∗ X ) =

i=1

k

1

1

1

1

σi (A 2 D B 2 A 2 C B 2 )

i=1



k

1

1

1

1

1

1

σi (D B 2 A 2 C B 2 A 2 )

i=1



k

1

1

σi (B 2 A 2 )σi (B 2 A 2 )

i=1

=

k

1

1

σi (A 2 B A 2 ),

i=1

where the first inequality is by the fact [21] that k

σ j (X Y ) ≤

j=1

k

σ j (Y X ),

k = 1, 2, ...,

j=1

whenever X Y is normal. The second one is due to the fact that σ(AB) ≺log σ(A) ◦ 1 1 1 1 σ(B) [216]. Therefore, we have σ(X ∗ X ) ≺w log σ(A 2 B A 2 ), then σ(X ∗ X ) ≺w σ(A 2 B A 2 ). Consequently, the assertion follows.  Now, we are in a position to present the following theorem. Theorem 6.6.3 For i = 1, 2, ..., k, let 0 ≤ Ai , Bi ∈ Mn such that Ai commutes with Bi for each i. Then   k    k 2   21  k  k  21   

 k 

 



1 1        2 2    ≤ ≤ A B A B A B A  i i i i i i i           i=1

i=1

i=1

i=1

i=1

holds for any unitarily invariant norm. Proof. The first inequality from the result of Audenaert [4]. We prove the second 1

1

inequality. Since Ai commutes with Bi , then Ai2 also commutes with Bi2 . Hence 1

1

Ai Bi and Ai2 Bi2 are both positive semidefinite. It is easy to see that ⎡ ⎢ ⎣

1

1 2

1

Ai2 Bi2

Ai 1 2

Ai Bi

Bi

⎤ ⎥ ⎦,

i = 1, ..., k

254

6 Positive Partial Transpose Matrix Inequalities

are positive semidefinite, by using the Schur complement. So they are all PPT. Consequently, ⎡ k ⎤ k 1 1   2 2 Ai Bi ⎥ ⎢ i=1 Ai i=1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ k k 1 1 ⎣ ⎦  2 2 Ai Bi Bi i=1

i=1

is also PPT. By Lemma 6.6.2, we get   2   21  k  k  21   k   k 



 21 21  

    Ai Bi Ai Bi Ai   ≤ ,  i=1   i=1  i=1 i=1 and this completes the proof.



Immediately, we deduce Theorem 6.6.1 from Theorem 6.6.3. Notes and references. Lemma 6.6.2, Theorem 6.6.3, and consequent corollary are taken from [160].

6.7 On Symmetric Norm Inequalities for Positive Block Matrices The numerical range [103] is a convex set on the complex plane. By the width of a numerical range, we mean the smallest possible ω such that the numerical range is contained in a strip of width ω. It is useful to notice that if the width of W (A) is ω, then one can find a θ ∈ [0, π] such that r I ≤ (eiθ A) ≤ (r + ω)I for some r ∈ R. We refer the reader to [103] for basic properties. Bourin and Mhanna [41] recently obtained a novel norm inequality for positive block matrices as follows.   M11 M12 be a positive matrix with each block square. Theorem 6.7.1 Let M = M21 M22 Then for all symmetric norms M ≤ M11 + M22 + ω I , where ω is the width of the numerical range of M12 .

6.7 On Symmetric Norm Inequalities for Positive Block Matrices

255

In particular, if the numerical range of M12 is a line segment, then the previous theorem gives [164] (6.7.1) M ≤ M11 + M22 . To the author’s best knowledge, Mhanna’s study [164] provides the first example for (6.7.1) to be true without the PPT condition. We refer to [37, 99] for some motivational background. Bourin and Mhanna’s proof of Theorem 6.7.1 makes use of a useful decomposition for 2 × 2 positive block matrices [32]. Their approach seems difficult for an extension to a higher number of blocks, as remarked in their paper [41]. It is the purpose of the present section to provide such an extension. Our extension of Theorem 6.7.1 to a higher number of blocks is as follows. Theorem 6.7.2 Let M = [Mi, j ]i,m j=1 be a positive matrix with each block Mi, j ∈ Mn . Then for all symmetric norms, M ≤ 

m

i=1

Mi,i +

m−1

ωi I ,

i=1

where ωi (i = 1, ..., m − 1) is the average of the widths of W (Mi,i+1 ± Mi,i+2 ± ... ± Mi,m ). Proof. By Ky Fan’s dominance theorem [103], it suffices to show that the inequality is true for the Ky Fan norms .k , k = 1, ..., n. The proof is by induction. The base case m = 2, i.e., Theorem 6.7.1 was treated in [41]. We include a proof for completeness. The presentation is slightly different from that in [41]. As M is positive, we may write  ∗ X [X Y ] M= Y∗ for some X, Y ∈ M2n×n so that M11 = X ∗ X , M12 = X ∗ Y , and M22 = Y ∗ Y . Clearly, Mk = X X ∗ + Y Y ∗ k . As the norm of M is invariant if we replace Y with eiθ Y , we may assume that r I ≤ (X ∗ Y ) ≤ (r + ω)I for some r ∈ R and that ω is the width of W (M12 ). Compute

256

6 Positive Partial Transpose Matrix Inequalities

 1 (X + Y )(X + Y )∗ + (X − Y )(X − Y )∗  k 2      1  (X + Y )(X + Y )∗ k + (X − Y )(X − Y )∗ k ≤ 2     1  (X + Y )∗ (X + Y ) + (X − Y )∗ (X − Y ) = k k 2     1   X ∗ X + Y ∗ Y + 2(r + ω)I  +  X ∗ X + Y ∗ Y − 2r I  ≤ k k 2  =  X ∗ X + Y ∗ Y + ω I k   =  M1,1 + M2,2 + ω I k .

Mk =

This completes the proof of the base case. Suppose the asserted inequality is true for m = l for some l > 2. Then we consider the m = l + 1 case. In this case, M could be written in the form ⎤ ⎡ ∗ ⎤ ⎡ ∗ X1 X 1 X 1 ... X 1∗ X l X 1∗ X l+1 ⎥ ⎢ .. ⎥ ⎢ .. .. .. ⎥ ⎢ . ⎥ ⎢ . . . ⎥=⎢ ⎥ [X 1 ... X l X l+1 ], ⎢ ⎣ X ∗ X 1 ... X ∗ X l X ∗ X l+1 ⎦ ⎣ X ∗ ⎦ l l l l ∗ ∗ ∗ ∗ X l+1 X 1 ... X l+1 X l X l+1 X l+1 X l+1 for some X, Y ∈ M(l+1)n×n . Again, we assume by multiplying X l+1 with a rotation unit that (6.7.2) s I ≤ (X l∗ X l+1 ) ≤ (s + ωl )I for some s ∈ R. Consider the following l × l block positive matrices: ⎡ ⎢ ⎢ ⎢ M1 = ⎢ ⎢ ⎣

X 1∗ .. .

∗ X l−1

⎤ ⎥ ⎥ ⎥ ⎥ [X 1 ... X l−1 ⎥ ⎦

X l + X l+1 ],

(X l + X l+1 )∗ and

⎡ ⎢ ⎢ ⎢ M2 = ⎢ ⎢ ⎣

X 1∗ .. .

∗ X l−1

⎤ ⎥ ⎥ ⎥ ⎥ [X 1 ... X l−1 ⎥ ⎦

X l − X l+1 ].

(X l − X l+1 )∗ Let αi (i = 1, ..., l − 2) be the average of the widths of W (X i∗ X i+1 ± ... ± X i∗ X l−1 ± X i∗ (X l + X l+1 ),

6.7 On Symmetric Norm Inequalities for Positive Block Matrices

257

∗ and let αl−1 be the width of W (X l−1 (X l + X l+1 )). Similarly, let βi (i = 1, ..., l − 2) be the average of the widths of

W (X i∗ X i+1 ± ... ± X i∗ X l−1 ± X i∗ (X l − X l+1 ), ∗ (X l − X l+1 )). We observe that and let βl−1 be the width of W (X l−1

ωi =

αi + βi , i = 1, ..., l − 1. 2

By the inductive hypothesis,  l−1  l−1 



  M1 k ≤  X i∗ X i + (X l + X l+1 )∗ (X l + X l+1 ) + αi I  ,   i=1 i=1 k  l−1  l−1 



  X i∗ X i + (X l − X l+1 )∗ (X l − X l+1 ) + βi I  . M2 k ≤    i=1

i=1

k

Furthermore, by (6.7.2), we have  l+1  l−1 



  ∗ M1 k ≤  X i X i + 2(s + ωl )I + αi I  ,   i=1 i=1 k  l+1  l−1 



  X i∗ X i − 2s I + βi I  . M2 k ≤    i=1

i=1

k

Now we proceed to estimating the norm of M, ∗ ∗ + X l X l∗ + X l+1 X l+1 k Mk = X 1 X 1∗ + ... + X l−1 X l−1 1 ∗ = X 1 X 1∗ + ... + X l−1 X l−1 + (X l + X l+1 )(X l + X l+1 )∗ 2 1 + (X l − X l+1 )(X l − X l+1 )∗ k 2 1 ∗ ≤ X 1 X 1∗ + ... + X l−1 X l−1 + (X l + X l+1 )(X l + X l+1 )∗ k 2

258

6 Positive Partial Transpose Matrix Inequalities

1 ∗ + X 1 X 1∗ + ... + X l−1 X l−1 + (X l − X l+1 )(X l − X l+1 )∗ k 2 1 = (M1 k + M2 k ) 2  l+1 l−1 

1  ∗  ≤  X i X i + 2(s + ωl )I + αi I   2  i=1 i=1 k  l+1  l−1  



1  +  X ∗ X i − 2s I + βi I   2  i=1 i i=1 k  l+1  l 



  = X i∗ X i + ωi I    i=1 i=1 k  l+1  l 



  = Mi,i + ωi I  .   i=1

i=1

k

Thus, the asserted inequality is true for m = l + 1, so the proof of induction step is complete.  The previous theorem contains an important special case of Hiroshima’s theorem when all off-diagonal blocks are Hermitian [37]. Corollary 6.7.3 Let M = [Mi, j ]i,m j=1 be a positive matrix with each block Mi, j ∈ Mn . If all off-diagonal blocks are Hermitian or all off-diagonal blocks are skew Hermitian, then for all symmetric norms,   m  

  Mi,i  . M ≤    i=1

As we know, for general X, Y ∈ Mn , it is clear that W (X + Y ) ⊂ W (X ) + W (Y ), but there is no simple comparison relation between the width of W (X + Y ) and the widths of W (X ) and W (Y ), for example, when X is Hermitian and Y is skew Hermitian. Therefore, it seems interesting to capture a special case of Theorem 6.7.2 for block tridiagonal matrices. ⎤ ⎡ A1 X 1 0 · · · 0 ⎢ ∗ .. ⎥ ⎢ X 1 A2 X 2 . ⎥ ⎥ ⎢ ⎥ ⎢ Corollary 6.7.4 Let A = ⎢ 0 X ∗ . . . . . . ⎥ be a positive matrix with 0 2 ⎥ ⎢ ⎥ ⎢ . .. ⎣ .. . Am−1 X m−1 ⎦ ∗ 0 · · · 0 X m−1 Am each block in Mn . Then for all symmetric norms

6.7 On Symmetric Norm Inequalities for Positive Block Matrices

259

 m  m−1 



  A ≤  Ai + ωi I  ,   i=1

i=1

where ωi are widths of W (X i ), i = 1, ..., m − 1.

Notes and references. We note that Theorem 6.4.6 and two following Corollaries are due to [154].

6.8 Matrix Norm Inequalities and Majorization Relation for Singular Values According to Weyl’s Theorem [21, 208], we have |λ(A)| ≺log σ(A). As x ≺ω,log y implies x r ≺ω y r for any r > 0, Weyl’s theorem can in slightly weaker form be stated as for all r > 0. (6.8.1) |λ(A)|r ≺ω σr (A), We recall that the sum of the k largest singular values of a matrix defines a norm, known as the k-th Ky Fan norm. The convexity of the Ky Fan norms can be expressed as a majorization relation as follows: σ( p A + (1 − p)B) ≺ω pσ(A) + (1 − p)σ(B), for any p such that 0 ≤ p ≤ 1. When A and B are positive semidefinite, their singular values coincide with their eigenvalues and we have λ( p A + (1 − p)B) ≺ pλ(A) + (1 − p)λ(B).

(6.8.2)

For positive semidefinite matrices A and B, the eigenvalues of AB are real and nonnegative. Furthermore, λ(AB) ≺log λ(A) ◦ λ(B), see [208]. Hence, we also have λ(AB) ≺ω λ(A) ◦ λ(B).

(6.8.3)

Now, we start with a rather technical result concerning a majorization relation for singular values. Lemma 6.8.1 Let S be an n × m complex matrix and let L and M be diagonal, positive semidefinite m × m matrices. Then

260

6 Positive Partial Transpose Matrix Inequalities

σ(S L diag(S ∗ S)M S ∗ ) ≺ω σ((S(L M) 2 S ∗ )2 ) ≺ω σ(S L S ∗ S M S ∗ ). 1

(6.8.4)

Proof. Let us begin with the first majorization inequality. Since L, M, and diag(S ∗ S) are diagonal, they commute and we can write S L diag(S ∗ S)M S ∗ = S(L M) 2 diag(S ∗ S)(L M) 2 S ∗ . 1

1

This is a positive semidefinite matrix, hence its singular values are equal to its eigen1 1 values. The same is true for (S(L M) 2 S ∗ )2 . Let us introduce X = S(L M) 4 . Then we have to show that λ(X diag(X ∗ X )X ∗ ) ≺ λ(X X ∗ X X ∗ ). In terms of the matrix T = X ∗ X ≥ 0, this is equivalent to λ(T diag(T )) ≺ λ(T 2 ). Note that there exist some number m of unitary matrices U j such that diag(T ) =

m

(U j T U ∗j ) j=1

m

.

Exploiting inequalities (6.8.2) and (6.8.3) in turn, we obtain  1  1 λ(T diag(T )) = λ T 2 diag(T )T 2 ⎛ ⎞ m

1 1 1 (U j T U ∗j )T 2 ⎠ = λ ⎝T 2 m j=1 ≺

m 

1  1 1 λ T 2 U j T U ∗j T 2 m j=1

m

 1  λ T U j T U ∗j = m j=1

≺ω =

m

  1 λ(T )λ U j T U ∗j m j=1

m

1 2 λ (T ) m j=1

= λ2 (T ),

6.8 Matrix Norm Inequalities and Majorization Relation for Singular Values

261

which proves the first inequality of (6.8.4). 1 For the second inequality, note that since (L M) 2 and S ∗ S are both positive semidefinite, their product has real and nonnegative eigenvalues. Thus, by Weyl’s Theorem (6.8.1) with r = 2, λ2 ((L M) 2 S ∗ S) = |λ(L 2 S ∗ S M 2 )|2 ≺ω σ 2 (L 2 S ∗ S M 2 ). 1

1

1

1

1

This implies that σ



1 2

S(L M) S



2

  1 1 = λ (L M) 2 S ∗ S(L M) 2 S ∗ S   1 = λ2 (L M) 2 S ∗ S  1  1 ≺ω σ 2 L 2 S ∗ S M 2   21 1 1 2 ∗ ∗ 2 2 M S SLS SM =λ   1 1 = λ M 2 S∗ S L S∗ S M 2 = λ(S L S ∗ S M S ∗ ) = |λ(S L S ∗ S M S ∗ )| ≺ω σ(S L S ∗ S M S ∗ ),

where in the last line we again exploit Weyl’s Theorem (6.8.1) with r = 1. This proves the second inequality of (6.8.4).  We can now state and prove the main result. Theorem 6.8.2 For i = 1, ..., k, let Ai and Bi be positive semidefinite n × n matrices such that for each i, Ai commutes with Bi . Then for all unitarily invariant norms,  k    k  k 2   k    

  



1 1         2 2 ≤ Ai Bi  ≤  Ai Bi Ai Bi  .         i=1

i=1

i=1

(6.8.5)

i=1

Proof. Let Ai and Bi have eigenvalue decompositions Ai = Ui L i Ui∗ ,

Bi = Ui Mi Ui∗ ,

where the Ui are unitary matrices and L i , Mi are positive semidefinite diagonal matrices. Let L=

k " i=1

Li ,

M=

k " i=1

Mi , S = (U1 |U2 |...|Uk ).

262

6 Positive Partial Transpose Matrix Inequalities

Then

k

k

Ai = S L S ∗ ,

i=1

Bi = S M S ∗ ,

k

i=1

Ai Bi = S L M S ∗ .

i=1

In addition, the diagonal elements of S ∗ S are 1 since all columns of S are normalized. Hence, diag(S ∗ S) = I . By Lemma 6.8.1, we then have  σ

k



 Ai Bi

≺ω σ (

i=1

k



 1 2

1 2

Ai Bi )2

≺ω σ (

i=1

which is equivalent to (6.8.5).

k

Ai )(

i=1

k

 Bi ) ,

i=1



Remark 6.8.3 The case k = 2 is an inequality recently conjectured by Hayajneh and Kittaneh in [95] and proven by them for the trace norm and the Hilbert-Schmidt norm. A simple consequence of Theorem 6.8.2 is that for any set of k positive semidefinite matrices Ai , all positive functions f and g and all unitarily invariant norms,  k   k  k   

 

    f (Ai )g(Ai ) ≤  f (Ai ) g(Ai )  .      i=1

i=1

i=1

Setting k = 2, f (x) = x p , and g(x) = x q yields the inequality     p+q A + B p+q  ≤ (A p + B p )(Aq + B q ) , which was conjectured by Bourin [34]. The way that Audenaert proved Theorem 6.8.2 is original. Considering that alternative proofs may provide new perspectives to these elegant results, we take the chance to do it in the following. To do this, we need two lemmas. The following lemma is a special case of [30, Theorem 5] by taking f (t) = t r , t ∈ [0, ∞). Lemma 6.8.4 For i = 1, ..., k, let Ai ∈ Mn be positive semidefinite. Then  k   k r    

    r Ai  ≤  Ai  ,      i=1

r > 1.

i=1

Notice that the inequality reverses for 0 ≤ r ≤ 1.   A X Lemma 6.8.5 Let where A, B, X ∈ Mn , be PPT. Then X∗ B X ∗ X  ≤ AB.

(6.8.6)

6.8 Matrix Norm Inequalities and Majorization Relation for Singular Values

263

Proof. Inequality (6.8.6) is a quick consequences of Theorem 6.1.3, which is attributed to Bourin.  Now we are ready to present the new proof of Theorem 6.8.2 as follows. 1

1

1

1

Proof. As Ai commutes with Bi , Ai2 commutes with Bi2 . Thus, Ai Bi and Ai2 Bi2 are positive semidefinite. Now  k   k  

   1 1 2      Ai2 Bi2  Ai Bi  =       i=1 i=1  2  

 1 1  k  2 2 , ≤ A B i i    i=1 

(by Lemma 6.8.4)

yielding the first inequality in Theorem 6.8.2. Using the Schur complement, it is easy to see that ⎡ 1 1 ⎤ Ai Ai2 Bi2 ⎢ ⎥ ⎣ ⎦ 1 1 2 2 Bi Ai Bi are positive semidefinite, so they are PPT. Hence ⎡  1 1 ⎤ k k 2 2 i=1 Ai i=1 Ai Bi ⎥ ⎢ ⎦ ⎣ 1 1 k k 2 2 i=1 Bi Ai i=1 Bi is PPT. By (6.8.6), we get  2   k    k  k 



 21 21     ≤  Ai Bi Ai Bi  ,       i=1 i=1 i=1 completing the proof of the second inequality in Theorem 6.8.2.



Under the same condition as in Theorem 6.8.2, it is easy to see that for two positive semidefinite A, B ∈ Mn , ⎡ ⎣

Ai2

Ai Bi

Bi Ai

Bi2

⎤ ⎦,

i = 1, ..., k

are positive semidefinite with off-diagonal blocks Hermitian, so is

264

6 Positive Partial Transpose Matrix Inequalities



k 

Ai2

⎢ i=1 ⎢ ⎢ ⎢ k ⎣ Bi Ai i=1

k 



Ai Bi ⎥ ⎥ ⎥. ⎥ k ⎦  2 Bi

i=1

i=1

By an extremal property of the matrix geometric mean [20], we have k

i=1

Ai Bi ≤

 k

i=1

  k 

Ai2  Bi2 , i=1

where the inequality is in the sense of Loewner partial order. It follows   k  k   k    



   2 2  Ai Bi  ≤  Ai  Bi       i=1 i=1 i=1  1   k  21  

  k 2 2

 2  , ≤ Ai Bi   i=1  i=1

(6.8.7)

in which the second inequality is by a known log-majorization relation [28]. Inequality (6.8.7) is clearly a complement of the inequality in Theorem 6.8.2. Lemma 6.8.6 ([15, 208]) Let A, B ∈ Mn be positive semidefinite and 0 ≤ q ≤ 1. Then j = 1, ..., n. σ j (Aq B 1−q ) ≤ σ j (q A + (1 − q)B), Consequently, Aq B 1−q  ≤ (q A + (1 − q)B. Recall that the Ky Fan k-norm, a special class of unitarily invariant norm, is k  defined as .k = σ j (A), 1 ≤ k ≤ n. If the eigenvalues of A are all real, then the j=1

largest one is denoted by λmax (A). Audenaert [5] used a similar technique from [15] to prove the following theorem. It would be clear to the reader that the proof idea below is quite different from Audenaert’s original proof. Our proof relies on Lemma 6.8.6. Theorem 6.8.7 For all X, Y ∈ Mn and all q ∈ [0, 1], X Y ∗ 2 ≤ q X ∗ X + (1 − q)Y ∗ Y  ≤ (1 − q)X ∗ X + qY ∗ Y . Proof. By Lemma 6.8.6,

6.8 Matrix Norm Inequalities and Majorization Relation for Singular Values

265

q X ∗ X + (1 − q)Y ∗ Y  ≥ (X ∗ X )q (Y ∗ Y )1−q , and similarly

(1 − q)X ∗ X + qY ∗ Y  ≥ (Y ∗ Y )q (X ∗ X )1−q .

Therefore, it suffices to show (X ∗ X )q (Y ∗ Y )1−q (Y ∗ Y )q (X ∗ X )1−q  ≥ X Y ∗ 2 .

(6.8.8)

By a generalization of Ky Fan’s dominance theorem [155], (6.8.8) is equivalent to (X ∗ X )q (Y ∗ Y )1−q k (Y ∗ Y )q (X ∗ X )1−q k ≥ X Y ∗ 2k ,

(6.8.9)

for all k = 1, ..., n. First of all, let us show (6.8.9) for Ky Fan 1-norm which is equivalent to spectral norm. (X ∗ X )q (Y ∗ Y )1−q 1 (Y ∗ Y )q (X ∗ X )1−q 1 ≥ (X ∗ X )q (Y ∗ Y )(X ∗ X )1−q 1   ≥ λmax (X ∗ X )q (Y ∗ Y )(X ∗ X )1−q   = λmax (X ∗ X )(Y ∗ Y )   = λmax (Y X ∗ )(X Y ∗ ) = X Y ∗ 21 . That is,       σ1 (X ∗ X )q (Y ∗ Y )1−q σ1 (Y ∗ Y )q (X ∗ X )1−q ≥ σ12 X Y ∗ .

(6.8.10)

Secondly, using a standard argument via the anti-symmetric product [21], (6.8.10) yields   k k k       σ j (X ∗ X )q (Y ∗ Y )1−q σ j (Y ∗ Y )q (X ∗ X )1−q ≥ σ j (X Y ∗ ), j=1

j=1

j=1

for k = 1, ..., n. This gives [105] k  k

   

σ j (X ∗ X )q (Y ∗ Y )1−q σ j (Y ∗ Y )q (X ∗ X )1−q ≥ σ j (X Y ∗ ), j=1

j=1

for k = 1, ..., n. The right-hand side is just X Y ∗ k . By the Cauchy-Schwarz inequality, the left-hand side is clearly bounded above by

266

6 Positive Partial Transpose Matrix Inequalities

⎞⎛ ⎞ ⎛ k k



 ∗ q ∗ 1−q   ∗ q ∗ 1−q  ⎠⎝ ⎠ ⎝ σ j (X X ) (Y Y ) σ j (Y Y ) (X X ) j=1

j=1 ∗



= (X X ) (Y Y ) q

1−q



k (Y Y )q (X ∗ X )1−q k .

Therefore, (6.8.9) and so the conclusion follows.



Notes and references. We take the proof of Theorem 6.8.2 and Lemma 6.8.1 from [4]. The proofs of Lemma 6.8.5 and Theorem 6.8.7 are due to [153].

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Index

A Accretive, 149 Accretive-dissipative, 115, 181 Aczél inequality, 98 Adjoint of the operator, 9 Algebra, 5 Ando’s inequality, 79 Ando-Hiai log-majorization, 251 Arithmetic-geometric mean determinantal inequality, 129 Arithmetic-geometric mean inequality, 19

Doubly stochastic, 4

B Block diagonal, 7 Block diagonal matrix, 7 Bounded linear operators, 5 Burn Minkowski inequality, 129

G Generalized Kantorovich constant, 82, 93 Gram-Schmidt procedure, 2

C Canonical decomposition, 11 Cartesian decomposition, 109 Cauchy-Schwarz inequality, 68 Choi-Davis-Jensen inequality, 66 Choi’s inequality, 62 Classical Rotfel’d theorem, 173 Closure property of sector matrices, 208 Compact, 10 Concavity property of geometric mean, 84 Contraction, 8

E Extremal property, 147

F Fischer’s inequality, 160 Frobenius norm, 13 Functional Calculus, 10, 32

H Hadamard’s inequality, 125 Heinz inequality, 20 Heinz means, 19 Hermitian, 9 Hilbert space, 2 Hilbert-Schmidt class, 182 Hilbert-Schmidt norm, 12, 182 Hilbert-Schmidt space, 12 Holder-McCarthy inequality, 99 Hua matrix, 223

I Inner product, 1 D Determinant, 6 Dissipative, 182 Doubly concave, 92

K Kantorovich constant, 61

© Springer Nature Switzerland AG 2021 M. B. Ghaemi et al., Advances in Matrix Inequalities, Springer Optimization and Its Applications 176, https://doi.org/10.1007/978-3-030-76047-2

275

276 Kantorovich inequality, 61 Ky Fan Dominance Theorem, 14 Ky Fan-Hoffman inequality, 198 Ky Fan k-norm, 13

L Linear mapping, 3

M Majorization of vectors, 4 Majorized, 116 Matrix Casel, 79 Matrix vector norm, 13 Maximal characterization, 65 Minimax Principle, 92 Minkowski inequality, 125

N Nonsingular, 110 Normalized, 242 Numerical radius, 18 Numerical range, 121 Numerical range and radius, 17

O Olson order, 96 Operator arithmetic mean, 32 Operator Cassels inequality, 79 Operator convex, 89 Operator Diaz-Metacalf inequality, 79, 80 Operator geometric mean, 32 Operator Kantorovich inequality, 63 Operator Klamkin-Mclenaghan inequality, 79 Operator matrix, 181 Operator means, 16 Operator monotone, 89 Operator monotone and convex functions, 16 Operator norm, 8 Operator Polya-Szeg ´ o¨ inequality, 80 Operator Shisha-Mond inequality, 81 Orthogonal, 2, 75 Orthonormal basis, 2 Ostrowski-Taussky inequality, 219

P Partial isometry, 10 Partial traces, 203 Polar decomposition, 10

Index Positive definite, 10 Positive linear map, 17 Positive operator, 10 Positive partial transpose, 221 Positive semidefinite, 10 Positive unital linear map, 62 Principal power function, 139 Principal submatrix, 6 Projection theorem, 8 R Ratio type reverse Ando’s operator inequality, 82 Representing function, 17 Reverse Ando’s operator inequality, 81 Reverse Cauchy-Schwarz inequality, 69 Reverse Fischer’s type inequality, 125 Reverse of Aczél operator inequality, 99 Reverse Ostrowski-Taussky inequality, 219 Reverse Young’s inequality, 38 Ricatti equation, 224 Riesz-Fischer theorem, 3 S Sandwich condition, 100 Schatten p-class, 11 Schatten p-norm, 13, 192 Schur complement, 7, 8, 110 Schwarz inequality, 1, 241 Self-adjoint, 9 Singular value, 168 Specht’s ratio, 20, 24, 89 Spectral norm, 13 Spectral radius, 18, 242 Spectrum of the operator, The, 9 Squared Pólya -Szegó operator inequality, 85 Strictly positive, 10 Supplemental Young inequality, 39 T Trace, 6, 12 Trace class operators, 11 Transpose map, 221 U Unitary invariant norm, 13 W Weak log-majorization, 5

Index Weak majorization, 4 Weakly majorized, 116 Weighted arithmetic-geometric, The, 19 Weighted operator arithmetic mean, 17, 32 Weighted operator geometric mean, 17, 32 Weyl’s inequality, 250

277 Width of a numerical range, 254 Wielandt inequality, 75

Y Young’s inequality, 19