Calculus, Vol. 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability [2 ed.] 9780471000075

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Calculus)))

M.

Torn

Apostol)

VOLUME

II)

and

Calculus

Multi-Variable

with

Algebra,

Differential

Applications

JOHN

EDITION)

New

WILEY &

York.

Chichester.

SONS) Brisbane

to

and Probability)

Equations

SECOND

Linear

\302\267 Toronto.

Singapore)))

EDITOR)

CONSULTING

George

Copyright

Springer, Indiana

@ 1969 All

by

rights

University)

John Wiley & Sons, Inc. reserved.

part of this publication

may be reproduced, stored in a retrieval system or transmitted fom1 or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as pennitted under Sections 107 or 108 of the 1976 United States No

in

any

either the prior written pennission of the Publisher, or fee to the Copyright payment of the appropriate per-copy Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax 750-4470. (978) Requests to the Publisher for permission should be addressed to the Pennissions Department, John Wiley & Sons, Inc., III River Street, Hoboken, NJ 07030, Copyright

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or for customer Printed

[email protected].)

service please,call

and bound by the

Hamilton

32 33 34 35 36 37 38 39

1 (800)-CALL-

Printing 40)))

WILEY

Company.)

(225-5945).

To

Jane

and

Stephen)))

PREFACE)

This book is a continuation present volume has beenwritten Sound

first.

in

training

is

to

As

in

in

divided

with

I,

philosophy

a

strong

of modern

I, historical

Volume

that

remarks

in the

prevailed

theoretical

mathematics

The

Edition.

Second

development. undue

without

are included to

give

the

of ideas.

evolution into three

the

Topics. The last two of Volume II so that all

chapters

Volume

Nonlinear parts, entitled Linear Analysis, I have of Volume been repeatedas the chapters the material on linear algebra will be complete

volume.

Part 1

an

contains

determinants,

matrices,

analysis,

in

particular

equations are

are

is combined the spirit convey

and Special

Analysis, first two

in one

author's Calculus, same underlying

the

with

technique

Every effort has been made on formalization. emphasis student a sense of participation

The secondvolume

the

of

by

proved

treated

Picard's

language of contraction

to linear

introduction

linear transformations, algebra, including are to and forms. given Applications eigenvalues, quadratic of to the study of linear differential differential Systems equations. with the help of matrix calculus. Existenceand uniquenesstheorems cast in the is also of successive approximations, which method operators.

of several variables. Differential calcul us is calculus of functions chain aid linear It includes rules for scalar and and with the of unified algebra. simplified and extremum to differential vector and fields, probJems. equations applications partial surface and calculus includes line integrals, multiple integrals, with integrals, Integral Here the treatment is along more or less classical lines and to vector analysis. applications forms. does not include a formal development of differential material The 3 are Probability and NumericalAnalysis. The special topics treated in Part with finite or countably infinite one dealing on probability is divided into two chapters, the other with uncountable spaces, random variables, and dissample spaces; sample in the study of both one- and The use of the calculus is illustrated functions. tribution

Part

2 discusses the

random

two-dimensional

The

last chapter

being on

different

variables.

contains kinds

an

introduction

of polynomial

to

numerical

approximation.

analysis,

Here again the

the chief ideas

a The book concludes with algebra. by the notation and terminology of linear and a discussion such as Simpson'srule, approximate integration formulas,

summation

emphasis

are

unified of

treatment

of Euler's

formula.)

vii)))

\\\"111)

Preface)

There

is ample

per week. It

material

in

this

volume

a knowledge

presupposes

The author recitation periodsper week,

calculus courses.

has

taught

allowing

for a full year's course meeting three or four times in most of one-variable calculusas covered first-year this material in a course with two lectures and two and about ten weeks for each part the omitting

starred sections.

This secondvolume

has been chapters planned so that many For example, the last chapter of each of the presentation. Part 1 by itself continuity

courses.

shorter

of

the

disrupting

bined coursein linear can choose topics to page

shows

which

Once again

the

algebra suit

logical

I acknowledge

In preparing the

his

and ordinary differential and preferences

needs

equations. by

can be

consulting

interdependence of the chapters. with pleasure the assistance of many

second edition

I

received

valuable

omitted for a variety

can be skipped without material for a comprovides instructor The individual

part

the

diagram

friends

and

help from Professors

on the next colleagues.

Herbert S.

of the of Washington, and Basil Gordon of the University of University a of Thanks Los Angeles, each of whom number are improvements. suggested for their assistance and cooperation. also due to the staff of Blaisdell Publishing Company to my wife for the many As it gives me special pleasure to expressmy gratitude before, I happily which has In dedicate this in she contributed. ways grateful acknowledgement

Zuckerman

California,

book

to her.)

T. M. Pasadena,

California

September

16,

1968)))

A.)

of the

Interdependence

Logical

Chapters)

IX)

1

LINEAR SP ACES I I

15

2

INTRODUCTION

LINEAR

TRANSFORM AND

TO

ATIONS

MATRICES

NUMERICAL ANAL

YSIS

3 DETERMINANTS

LINE

DIFFERENTIAL LINEAR

CALCULUS OF

DIFFERENTIAL EQUATIONS

I

I

7 SYSTEMS

INTEGRALS

SET

AND

FUNCTIONS

ELEMENTARY

PROBABILITY

FIELDS

VECTOR

4

-

AND

SCALAR

13

10

8

6

EIGENVALUES AND EIGENVECTORS

11

OF MULTIPLE

DIFFERENTIAL

EQUATIONS

INTEGRALS

5

14

EIGENV ALUES OF OPERATORS

ON

ACTING

EUCLIDEAN SPACES

I

CALCULUS

I

PROBABILITIES

12

9 APPLICATIONS DIFFERENTIAL CALCULUS)))

OF

SURFACE INTEGRALS

OF

CONTENTS)

1. LINEAR

PART

ANALYSIS)

1. LINEAR SPACES)

1.1

Introduction

1.2

The definition

1.3 1.4

3

of a linear space of linear spaces Examples consequences

Elementary

1.5

Exercises

1.6

Subspaces

1.7

Dependent

6

of the axioms

7

linear space

of a

8

sets

and independent

1.8 Bases and 1.9

3 4

in

a linear

9

space

12

dimension

13

Components

13

1.10 Exercises

1.11

Inner

Euclidean

products,

1.12 Orthogonality 1.13 Exercises

1.14

1.15Orthogonal 1.16 Best

space

20 of

Construction

14 18

Norms

spaces.

a Euclidean

in

sets.

orthogonal

complements.

approximation of

The Gram-Schmidt

22

process

26

Projections elements

in

a Euclidean

space

by

elements

in a

finite-

28

dimensional subspace

1.17Exercises

2.

30)

TRANSFORMATIONS

LINEAR

2.1

Linear

2.2

Null space and

2.3

Nullity

transformations

and

rank)

range

AND

MATRICES) 31

32 34) XI)))

Contents)

XlI)

35

2.4

Exercises

2.5

Algebraic operations

2.6

Inverses

transformations

36

38

2.7 One-to-one linear

41

transformations

2.8

Exercises

2.9

Linear transformations

2.10

Matrix representations of a matrix Construction

2.11

linear

on

42 with

prescribed

of linear

44

values

45

transformations

form

in diagonal

representation

2.12 Exercises

2.13

Linear

50

of matrices

spaces

51

2.14 Isomorphism between linear 2.15 Multiplication of matrices 2.16

matrices

Inverses

54

of linear equations

58

61 65 67

techniques

of square

matrices

2.20 Exercises 2.21

on matrices)

exercises

Miscellaneous

68)

DETERMINANTS)

3.

3.1

Introduction

3.2 Motivation for the choice of axioms for 3.3 A set of axioms for a determinant function 3.4

52

57

2.18 Computation 2.19

and

transformations

Exercises

2.17 Systems

48

a determinant

function

73

of determinants

Computation

71 72 76

3.5 The uniquenesstheorem

79

3.6

Exercises

79

3.7

The product formula for determinants of a nonsingular The determinant of the inverse

81

3.8

3.9 Determinantsand 3.10 The determinant 3.11 Exercises

3.12 Expansion

independence of

a block-diagonal

for determinant

3.14 The

determinant of a

3.15

cofactor

The

3.16 Cramer's

matrix

rule

3.17 Exercises)

matrix

83

83 84 85

formulas

3.13 Existenceof the

matrix

of vectors

determinants. function

transpose

Minors and cofactors

86

90

91

92

93 94)))

XIII)

Contents)

4.1

Linear

4.2

Eigenvectors

with

transformations and

eigenvalues

of 4.3 Linear independence

4.4

Exercises

4.5

The tinite-dimensional

4.6

Calculation

96

matrix representations diagonal of a linear transformation to distinct

corresponding

eigenvectors

97 eigenvalues

Matrices

102

case

finite-dimensional

107 same linear

the

representing

Similar matrices

transformation.

OF

EIGENVALUES

inner

and

Eigenvalues

ON

ACTING

OPERATORS

SPACES)

EUCLIDEAN

114

products

5.2 Hermitian and skew-Hermitiantransformations

115

and eigenvectors of Hermitian 5.3 Eigenvalues 5.4 Orthogonality of eigenvectors corresponding

and

skew-Hermitian

to distinct

operators

eigenvalues

118 set of

orthonormal

an

of

Existence

eigenvectors

for

and

Hermitian

5.10

acting on finite-dimensional spaces Matrix representations for Hermitian and skew-Hermitianoperators and skew-Hermitian The adjoint of a matrix matrices. Hermitian a or of Hermitian skew-Hermitian matrix Diagonalization Unitary matrices. Orthogonal matrices

5.11

Exercises

skew-Hermitian 5.7

5.8 5.9

operators

of a

Reduction

5.14 Applications Eigenvalues

quadratic

to

analytic

form

to a

of

130

geometry

a symmetric

transformation

obtained as

values

*5.18

properties

transformations

of

its

135

Extremal

5.20 Exercises

123

128

diagonal form

form

The finite-dimensional Unitary

122

122

134

*5.17

5.19

121

126

real quadratic

5.15 Exercises

*5.16

120

124

5.12 Quadratic forms 5.13

117

117

5.5 Exercises

5.6

108 112)

Exercises

5.1

103 106

matrix

4.10

5.

100 101

case. Characteristic polynomials of eigenvalues and eigenvectorsin the

4.7 Trace of a 4.8 Exercises

4.9

EIGENVECTORS)

AND

EIGENVALUES

4.

of eigenvalues

case

of a

symmetric

transformation

136

137 138 141)))

Contents)

XIV)

DIFFERENTIAL

LINEAR

6.

6.1

Historical

introduction

6.2

Review of

results concerning

6.3

Exercises

6.4

Linear

6.5

6.6 6.7 6.8

EQUATIONS) 142

linear

second orders

of first and

equations

of order n

differential equations

145

147

theorem

The existence-uniqueness of the solution dimension The

The algebra of of Determination

space of a homogeneouslinear

constant-coefficient

coefficients by

147

equation

148

operators

a basis of solutions

for

linear

equations

with

constant 150

of operators

factorization

6.9 Exercises

6.10 6.11

The

6.14

and nonhomogeneousequations

solution of the

a particular

of

of parameters matrix of Wronskian

the

of

homogeneous linear equation Special methods for determining a particular of first-order equation. Reduction to a system for

method

annihilator

The

determining

157 n

solutions

independent

of a 161

solution

linear

of the

nonhomogeneous

163

equations

a particular

solution

of

the

163

equation

nonhomogeneous

166

6.15 Exercises

6.16

6.17 Linear 6.18 The 6.19

with

analytic

coefficients

the Legendre

for

polynomials

182

equation

188)

7. SYSTEMS

7.2

Calculus

7.3

Infinite

7.4

Exercises

7.5

The

174 176

180

of Frobenius

6.24 Exercises

Introduction

169

177

Bessel

7.1

167 171

polynomials

formula

The method The

differential equations

Legendre equation

The Legendre

6.23

order

of second

equations

6.20 Rodrigues' 6.21 Exercises 6.22

on linear

exercises

Miscellaneous

154 156

nonhomogeneousequation.

of variation

method

6.12 Nonsingularity 6.13

the homogeneous

between

relation

Determination

The

143 144

OF DIFFERENTIAL EQUATIONS) 191

193

of matrix functions series

of matrices.

Norms of

matrices

194

195 exponential

matrix)

197)))

7.6

The

differential

satisfied

equation

Contents)

xv)

by etA

197

7.7 Uniqueness theorem for the matrix differential equation F'(t) = 7.8 The law of exponents for exponential matrices 7.9 Existence and uniqueness theorems for homogeneous linear with constant coefficients

7.10 The

problem

7.12

systems

200 201

theorem

203

Exercises

205

7.13 Putzer's method 7.14

199

etA

of calculating

7.11 The Cayley-Hamilton

198

AF(t)

for

calculating for

methods

Alternate

etA

205

etA in special

calculating

cases

208

211

7.15Exercises

7.16

linear

Nonhomogeneous

213

coefficients

with constant

syst6ms

215

7.17Exercises

7.18

The

7.19

A

= pet) yet) + Q(t) homogeneous linear systems

linear

217

system Y'(t) method for solving

general power-series

220

7.20 Exercises 7.21

Proof

7.22 The

221

of the existence

of successive

method

7.23 Proof

theorem by

of an

the

approximations theorem

ueness

existence-uniq

of successive

method

approximations nonlinear applied to first-order systems for first-order nonlinear systems

7.24 Exercises Successive

*7.26

Normed

*7.27

Contraction operators

*7.28

Fixed-point

and fixed points

approximations linear

233

of

Applications

the

234 235 237)

contraction

for

theorem

fixed-point

operators theorem

2. NONLINEAR

ANALYSIS)

8. DIFFERENTIAL CALCULUSOF 8.1 Functions from

Rn

8.2 Openballs

open

Exercises

8.4

Limits

8.5

Exercises

and

and

to Rm.

Scalar and vector fields

Exercises)

AND

243

244

sets

245 247

continuity

8.6 The derivative of a scalar field with 8.7 Directional derivatives and partial 8.8 Partial derivatives of higher order

SCALAR

FIELDS)

VECTOR

8.3

232

of operators

spaces

PART

8.9

227

229 230

*7.25

*7.29

222

respect

derivatives

to a

vector

251 252 254

255 255)))

Contents)

XVI)

8.10

Directional

8.11

The

derivatives and

8.12 The gradient 8.13 A sufficient 8.14 Exercises

8.15 A 8.16

a scalar

of

to geometry.

of scalar fields Level sets. Tangent

8.21

Matrix

8.22

Exercises

planes

fields

implies

271 272

continuity

for derivativesof vector form of the chain rule

fields

273

275

Sufficient conditions for the 8.24 Miscellaneousexercises)

*8.23

9.1

Partial

9.2

A

9.3

Exercises

9.4

The

differential

first-order

mixed

of

equality

derivatives

partial

277 281)

THE

OF

APPLICATIONS

9.

263 266 268 269

ses

Differentiability The chain rule

8.20

259

261

262

8.18 Derivativesof vector 8.19

field

for differentiability

for derivatives

Applications

8.1 7 Exerci

258

condition

rule

chain

257

continuity

derivative

total

CALCULUS)

DIFFERENTIAL

283

equations

partial

differential

equation

with

284

coefficients

constant

286 wave

one-dimensional

288

equation

9.5 Exercises

9.6

Derivatives

9.7

Worked

9.8

Exercises

292 of

functions

defined

294

implicitly

298

examples

302 303 308

9.9 Maxima, and saddle points minima, 9.10 Second-order formula for scalar fields Taylor 9.11 The nature of a stationary point determined by

the

eigenvalues

of the

test for extrema

Second-derivative

9.12

of functions

of

two

312

variables

313

9.13 Exercises

9.14

Extrema

with

constraints.

314

multipliers

Lagrange's

318

9.15 Exercises

9.]6

The

9.]

The

7

Hessian 310

matrix

extreme-value

small-span

for continuous

theorem

theorem

for continuous 10.

1 0.1

Introduction

10.2

Paths and line

LINE

319

scalar fields

scalar fields

(uniform

continuity)

321)

INTEGRALS) 323

integrals)

323)))

.. Contents)

10.3

Other

line

of

326

integrals

Exercises

328

10.6 The concept 10.7 Line integrals

as a line integral to arc length respect

of work with

10.8 Further applications of line 10.9

324

for line integrals

notations

10.4 Basicproperties 10.5

XVll)

328

329

330

integrals

331

Exercises

10.10 Open connected sets. 10.11 The second fundamental

10.12 Applications

to

of

Independence

the

332

path

of calculus

theorem

333

for line integrals

mechanics

335

10.13 Exercises

10.14 The first

336

of calculus

theorem

fundamental

10.15Necessary

for line integrals

vector field 10.16 Necessaryconditions for a vector field to be a gradient methods for constructing potential functions 10.17 Special 10.18

10.19

for a

conditions

sufficient

and

to

be a

337

339

gradient

340

342

345 346 349

Exercises

Applications

to exact

differential

of first

equations

order

10.20 Exercises

10.21Potential

sets)

on convex

functions

INTEG

MULTIPLE

11.

11.1

Introduction

11.2

Partitions

11.3

The double integral of a step function The definition of the double integral

11.4

350)

RALS) 353

of rectangles.

Step functions

353 355 of

a function

defined and

bounded on a 357

rectangle

11.5 11.6

Upper and lower double integrals Evaluation of a double integral

11.7 Geometric interpretation 11.8

Worked

of

357 by

repeated

integral as

double

the

one-dimensional

a volume

examples

362

11.10

Integrability

of

continuous

11.11

Integrability

of

bounded

11.12 Doubleintegrals 11.14 Worked 11.15

359

360

11.9 Exercises

11.13Applications

integration

358

extended to

area

363

functions functions

with discontinuities

more

over

general regions

and volume

365 368 369

examples

Exercises

11.16 Further applications 11.17 Two theorems of 11.18 Exercises

364

of double

Pappus

integrals

371 373 376 377)))

Contents)

XVlll)

Green's theorem

11.19

11.21

the

in

378

plane

applications of Green'stheorem

11.20 Some

A necessary and

a two-dimensional

for

condition

sufficient

382

vector field to be a 383

gradient

11.22

385

Exercises

* 11.23 Green'stheorem * 11.24The winding * 11.25 Exercises

11.26

of

Change

11.27Special 11.28Exercises 11.29

for

connected

multiply

387

regions

389

number

391 in a

variables

392 396

double integral formula

transformation

of the

cases

399

of the transformation formula

Proof

11.30 Proof of

to

401

case

general case

higher

in

an n-fold

407

integral

409

examples

413)

11.34 Exercises

INTEGRALS)

12. SURFACE

12.1

Parametric

of a

representation

12.2 The fundamental

vector

12.3

The fundamental

12.4

Exercises

12.5

Area

of

surface

417

420

product

vector product as a normal

Surface

12.8

Change

12.9

Other

of parametric for

representation

surface

integrals

12.10 Exercises

12.11The

theorem

12.12 The curl and 12.13

of Stokes

divergenceof a vector

field

Exercises

12.14 Further properties 12.15 Exercises

* 12.16Reconstruction

of

of the a vector

curl

and

divergence

field from

its

* 12.17 Exercises

12.18Extensionsof 12.20

Stokes'

theorem

divergence theorem (Gauss'theorem) of the divergence theorem Applications

12.19 The

12.21Exercises)

the

surface

423

424

integrals

notations

to

424 surface

a parametric

12.6 Exercises

12.7

403 405

dimensions

11.32 Change of variables

11.33 Worked

in the

formula

transformation

the

11.31Extensions

a special

in

curl

429 430 432 434 436 438 440 442 443 447 448 452 453 457 460 462)))

Contents)

AND ELEMENTARY PROBABILITY)

13. SET FUNCTIONS

13.1

469

introduction

Historical

Finitely

additive set functions

13.3

Finitely

additive

13.4

Exercises

13.2

13.8

Worked

13.9

Exercises

470

471

measures

472

13.5 The definition 13.6 Specialterminology 13.7 Exercises

of

for finite sample

probability

to probability

peculiar

473

spaces

theory

475

477

477

examples

13.10 Some 13.11

TOPICS)

SPECIAL

3.

PART

XIX)

479

basic principlesof combinatorial

481

analysis

485

Exercises

13.12 Conditional probability

13.13 Independence

13.14

Exercises

13.15

experiments

Compound

13.16 Bernoulli trials

13.17 The

most

of successes

number

probable

in

n Bernoulli

trials

13.18 Exercises

13.19 Countable and 13.20

Exercises

13.21 The definition 13.22

sets

uncountable

of

infinite

sample

spaces

Exercises

14. The

Countability

14.3

Random

of

the

set of

507)

probability)

OF

CALCULUS of probability

definition

14.2

506

507

13.23 Miscellaneousexerciseson

14.1

for countably

probability

486 488 490 492 495 497 499 501 504

PROBABILITIES)

for uncountable

points

with

positive

variables

Distribution

14.6

Discontinuities

14.7

Discrete

14.8

Exercises

14.9

Continuous

510

probability

511

512

14.4 Exercises

14.5

sample spaces

513

514

functions

of

distribution

distributions.

functions

Probability mass functions

517 520 523

distributions. Density functions)

525)))

Contents)

xx)

14.1 0

Uniform

14. 12

over

distribution

14.11 Cauchy's

526

an interval

530

distribution

532

Exercises

533

14.13 Exponential distributions

14.14 Normal

distributions

14.15

Remarks

14.16

Exercises

general

539

distributions

540

of functions of random

14.17 Distributions 14.18

535

on more

541

variables

542

Exercises

of

14.19 Distributions

545

discrete distributions

14.20 Two-dimensional

14.21 Two-dimensional 14.22 Exercises

distributions.

continuous

Density

functions

546

548

14.23Distributions 14.24Exercises 14.25

543

variables

random

two-dimensional

of two

functions

of

random variables

550

553

556

and variance

Expectation

14.26 Expectation of

of a

a function

random variable

559

14.27Exercises 14.28

562

inequality

Chebyshev's

14.29 Laws 14.30

560 of large

The central

564

numbers

of the

theorem

limit

calculus of

probabilities

566

14.31Exercises

568

569)

References)

Suggested

TO NUMERICAL

15. INTRODUCTION

15.1

Historical

15.2

Approximations

571

introduction by

572

polynomials

approximation and normed Fundamental problems in polynomial 15.5 Exercises

15.3

Equally spaced

15.8

Error

15.9

Exercises

15.12

analysis

Factorial

577 579

interpolation points

in polynomial

582

583

interpolation

interpolation spaced

588

formula

interpolation

points.

The forward difference

A minimum

operator

590

592

polynomials

15.13 Exercises 15.14

575

approximation

585

Newton's

15.11Equally

574

spaces

polynomials

Interpolating

15.7

15.10

linear

Polynomial

15.4

15.6

ANALYSIS)

593

problem relative

to the

max

norm

595)))

. Contents)

15.15

596

polynomials

Chebyshev

15.16 A minimal 15.17

XXi)

property Chebyshev Application to the error formula of

598

polynomials for

interpolation

15.18 Exercises

15.19 Approximate integration. 15.20 Simpson's rule

The

The

610

summation

Euler

15.23 Exercises

Index)

formula

613

618

Suggested References Answers

rule

605

15.21 Exercises

15.22

trapezoidal

599 600 602

to

exercises

621

622 665)))

Calculus)))

PART

LINEAR

1

ANALYSIS)))

1)

LINEAR SPACES)

1.1

Introduction)

mathematics

Throughout

we encounter

added to each other and are such objects. themselves vectors numbers, infinite series, discuss a general mathematical can be

many examples

Other

mathematical

real-valued

are

examples

of

concept,

numbers

real

the complex chapter we

functions,

and vector-valuedfunctions. called a linear space, which

in n-space,

that

objects

numbers. First of all, the

by real

multiplied

this

In

all

includes

these

others as special cases. examples and many of any kind on which certain (called Briefly, a linear spaceis a set of elements operations a linear addition and multiplication can be performed. In defining by numbers) space, we do not specify the nature of the elements nor do we tell how the operations are to be on them. the have certain properties which Instead, we require that performed operations we take as axioms for a linear space. We turn now to a detailed description of these axioms.)

The

1.2

Let space

V

definition

of

a linear

space

a nonempty set of objects, calledelements. The set V is called satisfies the following ten axioms which we list in three groups.)

denote

if it

a

linear

Closure axioms) AXIOM

1.

CLOSURE

corresponds a unique AXIOM

2.

CLOSURE

every real number by

For

ADDITION.

UNDER

element

in V

called

the

sum

every pair of x and

BY UNDER MULTIPLICATION

a there correspondsan

element

NUMBERS.

REAL

in V

of elements x y, denotedby x

called

the

For

and

y

+ Y

in

V there

.

every x in V and and x, denoted

of a

product

ax.)

Axioms

for addition) LAW.

AXIOM

3.

COMMUTATIVE

AXIOM

4.

ASSOCIATIVE LAW.

For all

x and y

For all x,y,

and z

in

in

V,

we have

V, we

have (x

x + y

= y + x.

+ y) +

z

=

x +

(y +

z). 3)))

Linear

4)

AXIOM 5.

spaces)

OF ZERO ELEMENT.

EXISTENCE

for all

x+O=x) AXIOM 6.

x +

AxionlS

For etery

ASSOCIATIVE LAW.

7.

AXIOM

(- l)x

x

DISTRIBUTIVE

8.

AXIOM

by 0, such

V, denoted

that)

V.)

in

V, the

elenlent

has

(-l)x

the

property)

= o.)

LAW

V and

in

IN V.

ADDITION

FOR

all real

numbers a and b,

For all

x

have)

we

= (ab)x.)

a(bx)

}fe

in

x

in

by nunlbers)

nlultiplication

for

x

For etery

OF NEGATIVES.

EXISTENCE

elenlent

is an

There

and

y

in

V and

all real

a,

hate)

9.

AXIOM

a and b,

DISTRIBUTIVE

LAW

FOR

+

ay.)

all x

For

OF NUMBERS.

ADDITION

in

V and

all real

have)

H'e

(a + 10.

AXIOM

= ax

+ y)

a(x

OF IDENTITY.

EXISTENCE

b)x =

ax + bx .)

For every

x

in

V,

we have

Ix =

x .)

to emphasize above, are sometimescalled real linear spaces V If real real number is the elements of numbers. by multiplying in structure is called a Axioms and the number 2, 7, 8, 9, resulting replaced by complex vector or is to as a a linear referred linear linear Sometimes space complex space space. used as multipliers are also called scalars. A real linear simply a rector space; the numbers linear space has complex numbers as scalars. space has real numbers as scalars; a complex with linear spaces, all the theorems are deal of real we shall examples Although primarily further we use the term linear space without valid for complex linear spaces as well. When it is understood that the can be real or to be complex.) designation, space Linear

the fact

1.3

If

as defined

spaces,

are

we

that

Examples

of linear

we

the

specify

set

addition

EXAMPLE addition

and

Let

V

2.

Let

how to

satisfies

examples

V

=

V

add its elements and how to multiply The reader can easily space. the axioms for a real linear space.)

a linear

all

set of all real of real numbers.)

R, the

multiplication

of complex

tell

and

concrete example of

numbers, we get a each of the following EXAMPLE 1.

spaces

numbers,

and let

= C, the set of all numbers, complex numbers, and define ax to be multiplication

them verify

x + y and

ax be ordinary

x + y

to be ordinary

define of

the

complex

by that

number

x)))

of linear

Exalnples

a. Even

real number

the

by linear

space

V =

Let

3.

EXAMPLE

and multiplication

is a

it

for by

off reader

The

EXAMPLE

6.

EXAMPLE

7.

consider this

The

set of all

polynomials of

understood

9.

real-valued

V are

way:)

n is

not a

The set The

n is

where

n,

is also

polynomial

continuous on

a

interval.

given

(Whenever we

fixed.

included.)

linear spacebecause the closure of n degree need not polynomials

of two

sum

differentiable at a

set of all functions

The

of all

functions

we

f

at

defined

The

of

set

axioms are not have degree n.)

If the interval is

1 with

point.)

interval.)

= o.

f(l)

a nonzero number

0 by

replace

given

on a given

integrable

set of all functions

example. If

in this

the

space by C(a, b).)

this

11.

to

degree
1 . Yn on continuous 14. Let V be the set of an real functions [0, + 00) I

(b)

Prove that

(c)

Compute

2

f\037 e-tt

(t) dt

V

(x,

converges. Define ([, g)

= f;>

e-'l\"(t)g(t)

dt

.)))

and such

that

the

integral

Linear

22)

(a) Prove that in

the

Use the

(b) Prove that

(c)

15.

Cauchy-Schwarz

V is

a linear

pair of

each

for

(a) (ax, by)

16.

that

Prove

(a) (b)

ab(x, y). the following

+ y\\l2 \\Ix + yll2 \\Ix

(c) Prove

\\Ix

=

-

that the

-

IIx

if we

all

f and

functions

g

IIy\\l2

yl12 yl12

+

w

+

1.14 Constructionof

f:

+ bz)

= a(x,

Euclidean

space.

continuous

on

ay

general

theorem

space,

finite or

onalization

on [a,

in honor

of J.

y) +

b(x,z).

construction

P. Gram

an

interval

[a, b]

becomes

b].)

Gram-Schmidt

The

The

dimensional.

infinite

process,

for

dt,)

)g(t)

w(t)f(t

basis. linear space has a finite an orthogonal basis. This result will whose proof shows how to construct

construct

properties

formula)

finite-dimensional

Every

always

.

following

Ily\\l2.

continuous

sets.

orthogonal

. ..

the

x).

2(y,

2

functions complex-valued inner an product by the

function,

has

I/(t)g(t)\\ dt.]

+ (y, x).

define

fixed positive

is a

in every

valid

+

(f,g) = where

product

b. (b) (x,

(x, y)

= 2(x, y) = 2 IIx\\l2

inner

the

t

f \037 e-

a and

complex

identities are

space of all

space

unitary

-

Ilx

+

+ yl12

+

IIx\\l2

g) as an inner product. = tn, where n = 0, 1, 2,

(I,

g(t)

space, prove that

z, and

all

and

the integral

estimate

to

inequality

space with

= e- t

g) if I(t)

(I,

Compute

a complex Euclidean elements x, y and =

In

a

g) convergesabsolutely

V.)

[Hint:

17.

for (I,

integral

spaces)

(1850-1916)

process

space is Euclidean, we can as a consequence of a sets in any Euclidean orthogonal is called the Gram-Schmidtorthogand E. Schmidt (1845-1921).) If the

be deduced

Let Xl' X 2 , . . . , be a finite or infinite and let L(x l , . . . , xk ) denote the subs pace space V, sequence k these elements. Then there is the a of spannedby first corresponding sequence of elements . \302\267 in V which has the each \302\267 , for YI , Y2, following properties integer k: The is to in element element the subs \302\267 \302\267 , Yk-l). (a) Yk orthogonal every pace L(YI, \302\267 The subs . . is the same \302\267 \302\267 . as that , X k :) , Yk (b) pace spannedby YI, spanned by Xl' \302\267 THEOREM

1.13.

a

in

Euclidean

L(YI,

(c)

The sequence

YI,

Y2,

. . . , is

sequence of elementsin V there is a scalarC k such that Y\037 =

another

when

(1.14))

= k =

\302\267 \302\267 \302\267 ,

Yk)

unique, satisfying

= L(x l ,

\302\267 \302\267 \302\267 , X k) .)

exceptfor scalar factors. That is, ify\037, (a) and (b)for all k, then properties

Y\037,

. . . , is

for

each

k

CkYk.)

the elements YI, Y2, . . . , by induction. Now assume we have constructed YI, . . . , Y r so r. Then we defineYr+l by the equation)

We

Proof

take YI

THEOREM.

ORTHOGONALIZATION

of elements

construct

XI.

Yr+l

=

xr+l

-

r

!

i=l)))

aiYi

,

that

To start the process, (a) and (b) are satisfied

we

Construction

the

where

al , .

scalars

Y i is given

with

. . , ar

< r,

For j

be determined.

to

are

The Gram-Schmidtprocess)

sets.

of orthogonal

inner

the

23)

of

product

Yr+l

by)

(Yr+l'

Yi) =

(x r + l

, Yj)

r

-

!

=

Yj)

ai(Yi'

(x r+ 1 ,

-

Yj)

aj(Yi' Yj),

i=1)

since

(Yi' Yj)

= 0 if i

\037j.

If

Yi

\037

( 1.15))

make

we can

0,

=

ai

r+l

(X

,

Yr+l

i)

Y

to Yi by

orthogonal

taking)

\302\267

(Yi,Yj))

= 0,

If Yj

then

aj = o. Thus, elementsYl ,

to Yi for any choice of ai' and in this case we choose is to each well defined and is orthogonal of the earlier Yr+l it in is to the element Therefore, orthogonal every subspace)

is orthogonal

Yr+l

element

the . . . , Yr.

L(yl,...,Yr).)

This proves (a) when To prove (b) when given

that

L(Yl,

. . . ,

k = r + 1. k = r + 1, = L(x 1, Yr)

we must . . . , xr ).

that L(Yl, . . . , Yr+l) The first r elements Yl,

show

L(x l , hence

and

they are

in

the

differenceof This proves that) (1.14) is a

larger subspace in two elements

Y r+l)

(1.14)

Equation

argument gives

shows that xr+l the

inclusion

is the

in the

sum

proves

Finally

(b) when k = we prove (c) by

is true for k

r + I.

C L(x 1, of

we can

+l )

1 given

by

1 , . . . , x r + 1).

\302\267)

in L(Yl,

elements

two

. . . , Y r+l)

so a

similar

on k. element

r+ 1) .)

\302\267 \302\267 \302\267 , Y

(a) and

both

Therefore

The case k Y;+I.

L(Yl, so

\302\267 \302\267 \302\267 , xr

Yr+

other direction:)

induction

= r and considerthe

+ l ),

in)

are

\302\267 \302\267 \302\267 , x r ))

C L (yl, L (x1, \302\267 . \302\267 , X r+ 1) This

\302\267 \302\267 \302\267 , xr

L(x l ,

L(x l , . . . , xr + 1). The new element L(x l , . . . , X r + 1) so it, too, is in L(x

\302\267 \302\267 \302\267 ,

L(Yl,

=

. . . , Yr

Because

\302\267 \302\267 \302\267 ,

Y r+l)

(b) are

= 1 is trivial. of (b),

proved by induction this

element

on

assume

Therefore,

is

k.

(c)

in)

,)

write) r+l

Y;+l =

!

CiYi

=

Zr +

cr + 1Yr+l,

i=1)

. . . , Yr). We wish to prove that Zr = o. By property (a), both Y;+1 and are their difference, Zr, is orthogonal to Zr. In other orthogonal to Zr. Therefore, Cr+lYr+l = to so o. This itself, words,Zr is orthogonal Zr completesthe proof of the orthogonalization theorem.))) where

Zr E L(Yl,

Linear

24)

that

Xr+l

is

combination

a linear

elements Xl' . . . , Xr+l are independent, then

coefficientsa i in

= 0 for some r. Then have (1.14) Yr+l of Yl, . . . , Yr, and hence of Xl' . . . , X r , so the In other words, if the first k elements . . , Xk Xl' \302\267 elements Yl, . . . , Yk are nonzero. In this case the . . . 'Yk become) (1.15), and the formulas defining Yl,

construction, supposewe

In the foregoing shows

spaces)

(1.14)

are the

dependent. corresponding

are

given

by

r

(1.16)

Yl

=

Xl')

Y r+l

=

Xr+l _

\037

Yi)

(Xr+l'

\037

i=l

(Yi'

. Y\037

r =

for

1, 2,

...,k

-

1 .)

Yi))

formulas describe the Gram-Schmidt process for constructing an orthogonal set of the a . . . which same as set nonzero elements , Yk Yl, spans subspace given independent if X a for a X . In . . is basis finite-dimensional Euclidean . . . . , k , k Xl' Xl' particular, space, basis for the same space. We can also convert this to an then Yl, . . . , Yk is an orthogonal it by its norm. element Yi' that each orthonormal basis by normalizing is, by dividing we have the as a of Theorem 1.13 Therefore, following.) corollary These

1.14.

THEOREM

If

X

and

yare

Every finite-dimensional

elements

in a

Euclidean space has an

Euclidean space, with

Y -:;e 0,

basis.)

orthonormal

the element)

(X, y) Y

(y,

is called the projection of X the element Yr+l by subtracting

elements Yl, . . . , Yr. space

Figure

y))

In the Gram-Schmidt process (1.16),we construct y. each of the earlier from xr+l the projection of Xr+l along the construction geometrically in the vector 1.1 illustrates

along

Va.)

X3)

Y3 = X3

- a 1Y I -a

2Y2,

a.I _- (X3, Yi)

(\". . \\'y1, YI))

Y2

FIGURE

1.1

The Gram-Schmidt processin V s. constructed from a given independent

= X2 -

An orthogonal set set {Xl' X2' Xs} .)))

CY.,

C

{Yl' Y2'

__ (X2, Y.) (y., Y.))

Ys}

is

Construction

1. In

EXAMPLE

vectors

=

Xl

V4 ,

(1,

Solution.

Y1

Y2

=

Xl =

=

X2 -

(1,

-1, 1, -1),

(x

, Yl)

2

=

Since

dimension

and

Yl

Yl

=

find)

(4 , 2, 0,

(x 3 ,

Yl)

(Yl,

Yl)

(x 3 , Y2)

-

Yl

=

Y2

(Y2,

X3

-

2'(1

_

_ 1 \" 1

\037

1)

The Legendre polynomials. In the linear dt , consider the infinite (x, y) = S\0371 x(t)y(t) product = tn. When the orthogonalization theorem is Y2, . . . ,

another sequence polynomials Yo, Yl, in work his matician A. M. Legendre(1752-1833) are easily calculated by the Gram-Schmidt polynomials of

1.

xo(t) =

=

l = i -1) dt

Yo) =

Yl

2

and)

( t) =

Xl ( t )

-

by

theory. First of

process.

The first all, we

few have

Yo) =

l i -1) t

=

dt

0,

Yo

Yo)

--

(t)

x1(t) =

=

t.

Yo))

relations)

we use the

(X2,

l i-1)

-

2 \"3,

Y2(t) =

X2(t)

t

2

d t

(X2,

=

Yl)

l 3 t dt i -1)

= 0,

(Yl ,

Yl)

=

l i -1)

2 t dt

= i,

obtain)

-

(x 2 ,

Yo)

Yo(t)

(Yo, Yo) Similarly,

the

with

. . , where

it yields sequence the French mathe-

this

to

applied

on potential

(Xl'

(Xl' Yo) (Yo,

to

of all polynomials, Xo, Xl' X2, . sequence

space

encountered

first

1).

1, 0,

(2,

6)

'1

that)

find

Next,

0) .

Since)

(Yo,

we

, 0 \" 0

1 /_

IIY211

EXAMPLE 2.

Yo(t)

Y2 = (0

-Y2 =

and)

IIYllI)

xn(t)

+

Yl

Y2))

=

-1:l.. -

inner

2) ,

since 0, the three vectors X2, X3 must be dependent. But Xl' Yl and Y2 are the vectors Xl and X 2 are independent. Therefore L(x l , x 2 , x 3) is a subspace of The set {Yl, Y2} is an orthogonal basis for this subspace. 2. each of Dividing basis consisting of the two vectors) Y2 by its norm we get an orthonormal

Y3

nonzero,

-

X3

process, we

-

X2

three

, Yl)

(Y1

Y3

=

Yl

the

by

and

(5,1,1,1),

Gram-Schmidt

the

Applying

=

X2

25)

for the subspace spanned X3 = (-3, -3,1, -3).)

basis

orthonormal

an

find

-1, 1, -1),

The Gram-Schmidtprocess)

sets.

of orthogonal

_ (x 2 , (Yl,

Yl)

Yl(t) =

2 t

_

!.

Yl))

we find that)

Y3(t) =

t

3

-

!t

,)

Y4(t)

=

t4

-

\037t2

+

-:l-s,)

Ys(t)

=

tS -

1./t3

+

t .))) -\037}l

26)

Linear

these polynomials

We shall encounter

spaces)

6

in Chapter

again

further

our

in

of differential

study

equations, and we shall prove that) n. ,

Yn ( t ) =

The

Pn

polynomials

as the

known

sequence ro,

qJo(t)

=

Legendre

\037t

,)

qJt(t)

qJ4(t) = 1.15.

for

=

(35t

l\037i

(2n)! 21l (n

by

given

formulas

the

(2n)! dt

2 (t -

n

l)

n

\302\267)

4

. . . , Ys

+

dt

t n (

2

-)

l)

n

The polynomialsin the corresponding orthonormal are called the normalized Yn/IIYnll Legendrepolygiven above, we

=

qJ2(t)

- 30t 2

2 n n!

=

f/Jn

Yo,

t ,)

\037i

1 dn

Yn( t ) =

!)2

polynomials.

\302\267 \302\267 \302\267 ,

r2,

CPt,

From

nomials.

n

by)

given

p n( t) =

are

d

-

2 t\037f

3),)

(3t

qJ5(t) =

that)

find

1),)

1\037 :1l'-

=

qJ3(t)

!\037t (5t

- 70t 3

(63tF)

+

3

-

3t),)

1St).)

Orthogonal complements. Projections

Let V be a Euclidean space and let 8 be a finite-dimensional We wish to subspace. Given an element x in V, to deterconsider the following type of approximation problem: as possible. mine an element in 8 whose distance from x is as small The distance between to be the norm II x - Y II . x and Y is defined two elements Before form, we consider a specialcase,illustrated discussing this problem in its general in Figure 1.2. Here V is the vector space V 3 and 8 is a two-dimensional subspace, a plane through

nearest

the

x

Given

origin.

in

V,

the

problem

is to

in

find,

the

8,

plane

that

point

s

to x.

If x is not in 8, then If x E 8, then clearly s = x is the solution. the nearest point s from x to the plane. This simple example suggests a perpendicular is obtained by dropping an approach to the general approximation problem and motivates the discussion that follows.)

Let S

DEFINITION.

orthogonal to S is denoted

S if

to

by

It is a simple In caseS is a If

E\037AMPLE. through

pretation

the

for

be a subset of a

exercise subspace, S is

to

An element

V.

space

The

set

a plane

the

through

to

this

origin,

as shown

plane.

This

in

example

Figure

in

V is

oj' all elements

that 81- is a subspace of V, whether 81- is called the orthogonalcomplement

verify

then

perpendicular the next theorem.)))

origin

Euclidean

it is orthogonal to every elementof S. 81- and is called \"8perpendicular.\

or not of

1.2,

said

to be

orthogonal

S itself

is one.

S.)

then

S1- is a line

also gives a geometric inter-

27)

Projections)

complements.

Orthogonal

s.l) .-- _--

s.l

1.15.

THEOREM let

and

(1.1

a sum

as

of

s+

s-L

2

( 1.18)) we prove

First

Proof

S is

IIxl1

finite-dimensional,

an

that

it has

a

s E

=

IIsll2

+

element

every

V be

x

is, we

V 3.)

a Euclidean space V can

in

in

be represented

have)

s1- E S1-.)

formula) Ils1-112.)

orthonormal

(1.17) actually exists. x, say {e1, . . . , en}.Given

basis,

sand s1- as follows:)

the elements

That

S-L.

and)

S)

theorem

Since

decomposition

orthogonal

finite

Let

THEOREM. Then

the Pythagorean

by

I I I , I I I I)

decomposition

and one in

where

,)

is given

the norm ofx

in S

one

elements,

x =

7))

Moreover,

two

subspace of V.

x=s+s.l)

orthogonal

DECOMPOSITION

ORTHOGONAL

a finite-dimensional

S be

uniquely

interpretation of the

Geometric

1.2

FIGURE

_-- ---.,

define

n S =

(1.19))

! i=l)

(x,

s1- =

ei)e i ,

X

-

s.)

s is the sum of the (x, ei)e i is the projection of x alongei . The element combination of the basis along each basis element. Sinces is a linear To prove that elements, s liesin S. The definition of s1- shows that Equation (1.17) holds. s1- lies in S1-, we consider the inner e j . We have) of s1- and any basis element product

Note

that

projections

each

term

of x

from

But

(1.19),

we

to every Next we prove that

is orthogonal has

(1.20))

two such

that

find

element the

= (x

ej)

(s1-,

x =

=

(x, ej )

- (s, ej)

s1-is orthogonal

(s,e j ) = (x,e j ), so in S, which means that

s-L E

say)

s +

s1-)

and)

x =

.)

to

ej .

Therefore

s1-

S1- .

(1.17) is

decomposition

orthogonal

representations,

- s, ej)

t + t1- ,)))

unique. Supposethat

x

Linear

28)

where sand

t

are

in S,

From (1.20), we have s - t E Sand t1. - s-L This shows that

the

Finally, we prove

-

=

-

is both

t

only

wish

orthogonal orthogonal

=

= t.l . But o. only prove to t1. - s-L and equal to t.l - s.l . to itself, we must have s - t = o. to

that s

prove

so we need

s.l,

element

that

t

=

t

the

by

We have)

formula.

Pythagorean

s.l, s + s.l)= (s,s) + (s.l,

s.l),)

since sand s.l are orthogonal.

This

(1.18).)

proves

.finite-dimensional subspace of a Euclidean space basis for s. If x E V, the element s defined by

V, and

an orthonormal

be

s.l

and

-

s

is unique. of x is given

= (s +

(x, x)

be a

S.l. We

t.l are in

decomposition that the norm

S

Let

{e1, . . . , en}

so S

being zero

terms

DEFINITION.

t-L

is the

2

remaining

=

t

E S1.

IIxl1

the

and

-

s

element

zero

the

Since

and s 1-

spaces)

the

let

equation)

n

! i=l)

S =

is called

the

on

of x

projection

We prove next that stated at the beginning

the of

x than

any

is,

all t

in

Proof

S;

By

the equality Theorem

for any t in S, we

sign holds

1.15

its norm

-

t

E S

is given

lis

s = t.

elements

by

problem

approximation

a finite-

in

S be a .finite-dimensional subspace of Then the projection of x on S is nearer to of x on S, we have) projection

sll


0, so we have Ilx This completesthe proof.)))

2 tl1

-

=

(x

=

Ilx 2

tl1

>

- s) + is an

Ilx

Sll2

(s

-

t).)

orthogonal decomposition of

+

lis

-

- s112,with

2 t1l

x

- t,

so

.)

equality

holding

if and

only

if

Best

1.

EXAMPLE

we exhibited

2n +

The

linear

prod uct

an orthonormal

CPo(x)=

(1.21))

1

1 elements

the

by

set of

CPo,

kx

cos

S of

!

of Ion

where

CPk)({Jk'

(f,

the

numbers

(I,

CPk)

can rewrite (1.22) in

are

called

the

form)

dimension 2n

(f,

+

1.

The ele-

Fourier

Then

we

have)

27T

=

CPk)

S.

subspace

Using the

coefficients off.

dx

f(X)CPk(X)

J.0

k=O)

The

1 .)

k >

for

TT

polynomials.

the projection

denote 2n

fn =

( 1.22))

kx ')

.J a subspace

CP2, . . . , where)

CPl,

sin

=

/P2k(X)

') TT

. . . , CP2n span

CPl,

trigonometric In

functions

trigonometric

.J

CPo,

29)

space)

[0, 217]by trigonometric polynomials. continuous on the interval [0, 217], 1.12 g) = S\0377TI(x )g(x) dx. I n Section

(I,

equation

/P2k-I(X) =

j- ' y217)

ments of S are called If IE C(O, 217),let

a Euclidean

in

of continuous functions on space of all real functions

Approximation

Let V = C(O, 217),the and define an inner

of elements

approximation

formulas

.)

in (1.21),

we

n

(1.23))

=

fn(x)

!

ta o +

(a

kx + b k

k cos

sin

kx),

k=1)

where) 27T

-1

ak =

17

1 0)

f(x) cos kx dx,

n.

The k = 0, 1, 2, . . . , nomial in (1.23) approximates the norm the sense that III

for

bk =

17 f 0)

In II

than

kx

the

that

dx

poly-

trigonometric

polynomial

trigonometric

in

S,

in

as possible.)

small

is as

other

any

f(x) sin

tells us

theorem

approximation

I better

27T

-1

on [-1, 1] by polynomials 2. Approximation of continuous functions of < n. Let V = C( -1, 1), the space of real continuous functions on [-1, 1], and let dx. The n + 1 normalized Legendre polynomials \302\267 . , CPn, (I, g) = S\037l I(x)g(x) CPo, CPl, \302\267 n + 1 consisting of all polyin Section 1.14, span a subspaceS of dimension introduced the projection of I on S. Then of degree we nomials < n. If IE C(-1, 1),let In denote EXAMPLE

degree

have) n

!

fn =

where

CPk)CPk'

(f,

(/,

=

/Pk)

k=O)

This is the

polynomial of degree
I

for

nt)

J\037

SA

dt,

x(t)y(t)

let

VS (6t 2 - 6t + 1))

set spanning the same subspaceas {xo,Xl' x2 }. linear space of all real functions f continuous on [0, + (0) and such that the = dt , and let Yo, Yl , Y2, . . . , be dt Define integral (f, g) S\037 e-fJ(t)g(t) converges. S \037 e-lj'2(t) n the set obtained by applying the Gram-Schmidt process to xo, Xl' X2' . . . , where xn(t) = t 3 2 2 = = = = > t for n o. Prove that yo(t) t t 4t + 2, Y3(t) 9t + I, YI (t) I, Y2(t)

form

5. Let

be

the

- 6.

18t

6. In

orthonormal

an

V

real

the

and

linear space C(I, 3) with the constant polynomial

inner

that

show

this g.

7.

In

the

show

real linear space that the constant

product

C(O,2) with

inner

polynomial

g nearest

=

(f, g)

g nearest to f is g

=!

this g.

8. In

the

real

find the

and

linear

linear

space C( -1, polynomial

1) with g nearest

inner

to f

3.

= S\037f(x)g(x) (f, g) to f is g = !(e2 - I).

product

product

(f,g)

Compute

Ilg

= -

f

dx,

S\037f(x)g(x) log

Compute dx,

S\037lf(x)g(x) 2 this 11 for

= S\0371Tf(x)g(x) inner linear space C(0,27T)with product (f, g) = I , Ul (x) = cos x, u 2 (x) = sin x, find spanned by uo(x) subspace polynomial nearest to f 10. In the linear space V of Exercise 5, let f(x) = e- X and find the linear polynomial 9.

In

the

In

the

to f.)))

real

Ilg

let f(x)

Compute

= I/x

let f(x)

Ilg

-

f

= -

f

2 11

for

e\037and 2 11

for

dx,

let f(x)

=

dx,

let f(x)

= x.

the

trigonometric

g.

that

e\037

is nearest

2)

2.1 Linear

MATRICES)

AND

TRANSFORMATIONS

LINEAR

transformations

One of the ultimate goals of domains and ranges are subsetsof

linear

study of functions

a comprehensive

is

analysis

spaces.

This chapter treats the simplest examples, branches of mathematics. Propertiesof more general tions, which them tions are often obtained by approximating by linear transformations. or operators. in all occur

mappings,

First we introduce somenotation

Vand

W

two

be

and

used to

each x

For

image of A

that

indicate

V, the

in

x onto

maps

element

T(x). If A T and

under

T is

a

T(x)

is any

Now

we

W is

in

of

is denoted

T, and

set of all

the

The

T(A).

by

whose valuesare in

the image of x under images T(x) for x in image of the domain V, T(

called

V,

is Vand

domain

that Vand Ware linear spaceshaving as follows.) transformation

assume

a linear

functions. Let

W)

whose

function

subset

--+

V

of T.

define

transforma-

The symbol)

sets.

T: will be

arbitrary

concerning

terminology

whose

are called transformations, called linear transforma-

Such functions

the

same

set of

T: V --+ W is If V and Ware linear spaces, a function two W it the V into has if of formation properties: following for all x and y in V, (a) T(x + y) = T(x) + T(y) for all in V and all scalars c.) (b) T(cx) = cT(x) DEFINITION.

W.

T we say that A is called the V), is the range

scalars, and

called a

linear

we

trans-

x

These properties are verbalized by saying The two properties can be combined scalars. T(ax for

all x,y

in Vand all

+

scalars a and b.

into

by) = By

aT(x) +

=

any

n elements

Xl' . . . , X n

in

V and

one

formula

aiT(x

multiplication

which states

by

that)

bT(y))

we also

induction,

have the more

general relation)

i ))

i\037l

TC\037aiXi) for

addition and

T preserves

that

any

n scalars

aI, .

. . , an

.)

31)))

Linear

32)

The reader can easily

The identity transfornlation. is called the identity

EXAMPLE 1. each

x in V,

EXAMPLE

2.

for

element of EXAMPLE

3. V.

T:

transformation

The

V

is denoted

and

transformation

transformations.)

---+ V,

where

I or by

by

The transformation T: V ---+ V which =ero transformation. and is denoted transformation the zero is called 0 by o.)

The

V onto

for all x in

examples are linear

the following

that

verify

and matrices)

transformations

by afixed

Multiplication When c =

1,

scalar c.

Herewe

identity

transformation.

is the

this

V ---+

T:

have

T(x) = Iv.)

maps each

T(x) = cx is the zero

V, where

When c

x

= 0, it

transformation.)

i = =

x

vector

Linear

4.

EXAMPLE

where

W =

Let V = Vn and 1, 2, . . . , n, define

equations.

1, 2, . . . , m and (Xl' . . . , xn ) in

k =

Vn

the

onto

y =

vector

Given

V m. Vn

T:

---+

Vm

(YI, . . . , Ym)

,

T maps each to the

V m according

in

a ik

numbers

real

mn

as follows:

equations) n

=

Yi

5. Inner

EXAMPLE

z in

element

fixed

of x

product

'with

T:

V

a fixed ---+ R

i =

for

k

aikx

1, 2, . . . , m

Let element. as follows: If

be

V

xE

.)

a real

V,

then

Euclidean space. T(x) = (x, z), the

For a inner

z.)

with

subspace. Let

Projection on a

EXAMPLE 6. dimensional

product V, define

! k=l)

subspace

T:

Define

V.

of

V

a Euclidean

be

V

as follows:

---+ S

space and let

If x E

then

V,

S be a finiteT(x)

is the

projection of x on S.) operator. Let V be the linear space of all real functions each The linear transformation which on an open interval (a, b). maps and is denoted by V onto its derivativef' is called the differentiation functionfin operator of D. we have D: V ---+ W, where D (I) = f' for each f in V. The space W consists Thus, all derivatives f'.) EXAMPLE

f

7.

The differentiation

differentiable

8.

EXAMPLE

The

on an

continuous

interval [a,

b]. Iff

g(x)

This transformation 2.2

Null

In

space

this

section,

THEOREM the

zero

2.1.

element

T is

and

Let

operator.

integration

called

=

E

V,

f' J(t)

V

define

be

if

dt)

the integration

linear

the

g = T(f) to

a
1. Prove that Tis linear and describe the null space and range of T. 29. Let V denote the linear space of all real functions continuous on the interval [- 7T,7T]. Let S be that subset of V consisting of all f satisfying the three equations) Let

27.

f(t)

f\"

dt = 0,)

f\" f(t)

that S is a subspaceof V. Prove that S contains the functions f(x) Prove that S is infinite-dimensional. Let T: V ---+ V be the linear transformation

dt

=

0,)

f\"

sin t dt

f(t)

=

O.)

Prove

(a)

(b) (c)

g(x)

30.

cost

=

f\" {1

= cos

nx

and

= sin

f(x)

as follows

defined

+ cos (x

-

t)}f(t)

: Iff

for

nx

E

V,g

each =

n =

2, 3, . . . .

T(f) means

that)

dt \302\267)

that T( V), the range of T, is finite-dimensional and find a basis for T( V). the null space of T. that such an f (f) Find all real c \037 0 and all nonzero f in V such that T(f) = cf. (Note lies in the range of T.) of a linear space V into a linear space W. If V is Let T: V -+ W be a linear transformation infinite-dimensional, prove that at least one of T( V) or N(T) is infinite-dimensional. (d)

Prove

(e)

Determine

dim N(T) = k, dim T( V) = r, let e 1 , . . . , e k be a basis for N(T) and , el'...' ek ek+l' . . . , ek + n be independent elements in V, where n > r. The elements Use this fact to obtain a T(ek+l)'...' T(ek+n) are dependent since n > r. Assume

[Hint:

let

contradiction.

2.5

])

operations

Algebraic

by

DEFINITION.

and

values

with

product

cT by

the

x

in

V.)))

in a given linear space W can be addedto eachother in Waccording to the following definition.)

Let

S: V ---+

in a

linear space

the

transformations

lie

scalars

T:

Wand W.

If

V

---+

be

W

c is any

two functions

scalar

in

W,

a

with

we define

T)(x) =

Sex) +

T(x),)

(cT)(x)

=

common

the sum S

equations)

(S +

(2.4))

for all

values

whose

Functions

be multiplied

on linear

cT(x))

and

can

domain

+ T and the

V

are especially

We

scalars as into W.

this

In

W.

on linear

operations

Algebraic

interested in the case we denote

also a linear set of all linear

where V is 2 (V, W) the

case by

37)

transformations)

the same

having transformations

space

of V

in 2( V, W), it is an easy exerciseto verify If Sand Tare two linear transformations that in 2(V, With T and cT are also linear transformations this is true. W). More than the operations just defined, the set 2( V, W) itself becomesa new linear The zero space. T transformation serves as the zero element of this space, and the transformation (-1) that all ten axioms for a linear is the negative of T. It is a straightforward matter to verify are satisfied. Therefore, we have the following.) space S +

space A

with

the

more

2(

operations

This

linear space

and can be defined

ST:

2.1

FIGURE

DEFINITION.

values

in

V,

and

Let U, let S:

the composition

V,

ST is the

be

W W

ST: =

(ST)(x)

to map is illustrated

Thus, This

x

by in

U

function \037

composition 2.1. Figure

or

is composition structure

first

transformations.)

U and

domain a function with V and values in

W.

domain

by the

for every x

ST, we

the

be

V

with

W defined

S[T(x)])

(2.4).)

W)

Let T: U \037

sets.

be another

function

-

in

as follows.)

the composition of two

Illustrating

V \037

U

a linear

W is

into

useof the algebraic

makes no

generally

quite

V

transformations

linear

on

operation

of

defined as

by scalars

multiplication

algebraic operation

transformations.

of

linear transformations

of all

W)

V,

of addition and

interesting

multiplication

of a

The set

2.4.

THEOREM

map

Then

equation)

in

x by

U.)

T and then map

T(x) by

S.

of real-valued functions has been encountered repeatedly in our study of Composition is, in general, not commutative. However, calculus, and we have seenthat the operation as in the case of real-valued functions, composition does satisfy an associative law.) THEOREM

2.5.

If T:

U \037

V,

S:

V

\037

W,

R(ST)

and

R: W

\037

= (RS)T.)))

X are

three functions,

then

we have)

Linear

38)

Both functions R(ST) and

Proof. in

= R[(ST)(x)] =

[R(ST)](x) which

that

proves

V \037

T:

powers of T inductively

R[S[T(x)]]

the

I is

Here

s:

Tn =

I,)

+ by)

W),

= S[T(ax

and let c be any

For any function

R

(S +

(b) For any

linear

with

+

=

by)]

itself.

We define integral

1

>

a and b, we S[aT(x)

the

algebraic

following.)

.)

that

law

associative

the

integers m

n.

and

transformations is again

linear.)

and if T: U \037 scalars, U \037 W is linear.)

same

composition

be linear spaces with

V

ST:

have)

+ bT(y)]

= aST(x)+

operations

of addition

same

the

scalars,

bST(y)

.)

and multiplica-

assume Sand

Tare

scalar. values

transformation

in V, we have)

TR)

and)

R: W \037

U , we

T) = RS + RT)

straightforward

n

the

with the

the

with

us

for

of linear

then

T)R = SR +

R(S +

The proof is a

= R[S[T(x)]],)

(RS) [T(x)]

V into

maps

TTn-l)

spaces

all scalars

U and

in

U, V, W

Let

2.7.

THEOREM

which

composition

transformations,

can be combined in 2(V, W) to give

of scalars

=

linear

Ware

V,

linear

For all x, y

Composition

2(V,

If U,

Ware

(ST)(ax

=

[(RS)T](x)

The reader may verify Tm+n for all nonnegative

transformation.

identity

2.6.

V \037

Proof.

(a)

each x

For

X.

in

as follows:)

implies the law of exponents TmTn The next theorem shows that the THEOREM

and

a function

V be

TO =

in

values

U and

domain

= (RS)T.)

R(ST)

Let

DEFINITION.

tion

(RS)Thave

we have)

U,

and

and matrices)

transformations

application

and)

of the

(cS)R

= c(SR).)

have)

R(cS)

= c(RS).)

definition of

composition and is left

as

an exerCIse.)

2.6

Inverses)

In our

study

of

real-valued

inversion of monotonic functions. more general class of functions.)))

functions Now

new we learned how to construct to extend the process of

we wish

functions inversion

to

by a

lnverses)

T, our goal is to find,

a function

Given

we have to distinguish which we call left

and right

and

ST

between

another

if possible,

Given two sets V and Wand a function T: a left inverse of T if S[T(x)]= xfor all x in V, ST =

\",here

Iv

of T if

is the

identity

T[R(y)] =

Y

all y

V.

on

transformation

for

in

that

is the

T(V),

=

and let W {O}. Define T: right inverses R: W \037 Vand =

V \037

R':

have a

follows:

W

V given

\037

=

S[T(l)] =

1=

This simpleexampleshows that be

1

S: T(V)

Afunction

\037

V

if)

is,

R: T(V) \037

V is

called a

inverse

right

with

two

T(l)

= T(2)

=

R'(O)

,)

right

Let

inverses.

= O.

V

has

function

This

=

{I, 2} two

by)

2 =

and)

S(O))

need not

inverses

left

W.

2.)

this would require)

S since

inverse

left

but

Was

R(O)

It cannot

V \037 that

IT(V),)

inverse

with no left

A function

EXAMPLE.

commutative, of inverses

kinds

on T(V).)

transformation

identity

not

Iv,)

Afunction is, if)

TR

where IT(v)

whosecomposition

inverses.)

DEFINITION. is called

function S

Since composition is in general TS. Therefore we introduce two

transformation.

the identity

T is

with

39)

S[T(2)] =

exist and that

S(O).) right

need not

inverses

unique.)

inveI see In fact, each y in T( V) has the = x , then select one such x and define R(y) = T(x) = y for eachy in T(V), so R is a right inverse. Nonuniqueness may occur T[R(y)] because there may be more than one x in V which maps onto a given y in T(V). We shall one x in V, 2.9) that if each y in T( V) is the image of exactly prove presently (in Theorem Every

'r: V

function

form y = T(x) for

then

right

at

\037

least

at least

W has

one x

in

are unique. if a left inverse

inverses

First we prove that

one

right

If we

V.

exists

it

is

unique

and,

at the same time, is

a

right

Inverse.)

A function

THEOREM 2.8.

inverse S,

S is

also a

T: right

V

\037

W

can

have at

most one left

Assume

S[T(x)]

If T

inverse.

has a left

inverse.)

S': T has two left inverses, S: T(V) \037 Vand = = We shall that Now T(x) for T(V). S'(y). y prove S(y)

Proof.

y in

then

=

x)

and)

S' [T(x)]

= x ,)))

T(V) some

\037

V.

Choose

x in V,

so we

any have)

Linear

40)

since both y in

all

for

=

T[S(y)]

=

x

T(x) =

we get

T,

Applying

Since y

y.

E

=

S'(y)

inverse.

Choose

, we

have y =

T(V)

so S(y) =

x ,

S'(y)

are unique.

left inverses

that

proves

and

any element y

in

T(x) for somex in

a left inverse, so)

S is

But

V.

that

which

x

inverse S is alsoa right

every left

that

prove

shall prove

We

Therefore S(y) =

inverses.

Therefore S = S'

T( V).

Now we T(V).

S' are left

Sand

and matrices)

transformations

= S(y).)

S[T(x)]

T[S(y)].

But

=

Y

T(x),

so

=

Y

which completes the

T[S(y)],

proof.)

The next

all functions

characterizes

theorem

A function T: 2.9. V elements onto distinct of

THEOREM elements

W has

--+

V

(2.5) is

Condition

Note:

that

x = y.

this

implies

r!=

(2.6)for

all

T(y).)

the statement) x =

implies)

T(y))

if, for

distinct

x and y in

V

y.) to be

is said

one-to-oneon

V.)

S, and assume that T(x) = T(y). We wish to prove = x and S[T(y)] = y, S, we find S[T(x)] = S[T(y)]. SinceS[T(x)] Applying = x with a left inverse is one-to-oneon its y. This proves that a function Thas

Assume

Proof.

(2.5) or

T satisfying

function

A

to

equivalent

T(x) =

(2.6))

is, if and only T(x)

only if T maps all x and y in V,)

if and

inverse

left

implies)

xr!=y)

(2.5))

a

of W; that

inverses.)

left

having

a left inverse

domain.

we prove the converse. Assume --+ V which is a left inverseof T.

Now S:

T(V)

there is exactly

(2.6),

we define

S on

one x in

V for

S(y) =

defined is a

S[T(x)] = x for

know

T-l

the

The linear

a inverse of is also

results

right

which y =

We

V.

then

=

y

T(x). Define S(y) to

a

exhibit

shall

T(x) for

function

some x in

be this

x.

the

function

V.

By

That

is,

x)

each

means

that)

ST = Iv.

V, so

x in

T(x)=y.)

Therefore,

S so

inverse)

be one-to-oneon is denoted

V.

by T-l.

The

unique

We say that

left inverse T is

invertible,

of T and

(which we

call

T.)

of this

transformations.)))

T(V),

T.)

T: V --+ W

Let

DEFINITION. we

of

inverse

left

If Y E

as follows:)

T(V)

Then we have

Tis one-to-oneon

section refer to

arbitrary

functions.

Now

we apply

these ideas to

linear

One-to-one

2.7 One-to-onelinear

41)

transformations)

transformations

the same section, V and W denote linear spaces with scalars, and T: V a linear transformation in .P(V, The us of T enables to express W). linearity one-to-one property in several forms.) equivalent

this

In

--+

denotes

THEOREM

statements

T: V --+

Let

2.10.

W be a linear

W). Then

in .P( V,

transformation

the

W

the

following

are equivalent.

(a) T is one-to-one

V.

on

and its inverse T-I: (b) T is invertible (c) For all x in V, T(x) = 0 implies the zero element of v.)

T(V) --+ V is x = o. That

linear. the null

is,

space N(T)

containsonly

and (c) implies (a). First Proof We shall prove that (a) implies (b), (b) implies (c), assume (a) holds. Then T has an inverse (by Theorem 2.9), and we must show that T-I is linear. Take any two elements u and v in T(V). Then u = T(x) and v = T(y) for some x and y in V. For any scalars a and b, we have)

+ bv =

au

T is

since

aT(x) + bT(y) =

Hence, applying T-I,

linear.

+ bv)

r-l(au

we

= ax

+

T(ax +

by),)

have)

=

by

+ bT-I(v),)

aT-leu)

Therefore so T-I is linear. (a) implies (b). Next assume that (b) holds. Take any x in V for which T(x) = o. Applying T-l is linear. find that x = T-l(O) = 0, since Therefore, (b) implies (c).

Take any two elements - v) = T(u) - T(v) = O,sou of the theorem is complete.)

(c) holds.

assume

Finally,

linearity,wehaveT(u

on

V,

and

the

V is

When

V is

T: V --+ W

be a

say dim

= n.

Let

2.11.

finite-dimensional,

(a) T is one-to-one (b) If el , . . . , ep elements (c)

dim T(V)

in

V

=

v

V with T(u) = T(v). By Therefore, Tis one-to-one

the

in

of

terms

next theorem.)

the

in .P( V,

transformation

Then

in

O.

property can be formulated

by

linear

v

we

following

statements

W) and assume that are equivalent.

V.

on are

elements

independent

in

V,

then

T(e l ),

...,

T(ep )

are

independent

T(V).

= n.

(d) If {el, . . . , en} Proof.

the one-to-one

finite-dimensional,

and dimensionality,as indicated

independence

THEOREM

proof

u and

-

T-I,

We shall

(a). Assume (a)

is a

prove

holds.

basis for that Let

(a) el,

V,

implies

{T(e l ),

then

(b), (b)

. . . , e1J be

...,

T(en )}

is a

basis for

implies (c), (c)implies

independent

elements of

(d), V

and

T(V).)

and

(d) implies

consider

the)))

42)

Linear

elements T(el), . . . , T(e

) in

p

and matrices)

transformations

T(V).

Suppose

that)

p

= 0

ciT(e i)

I i=l)

cl , .

scalars

certain

for

. . , cp

.

= 0,

i ciei ( t=l)

T

)

since T is

one-to-one.

(a) implies (b). (b)

T(el), . . . , T(e

independent, so C l

dim

Therefore

T(V). Then y = T(x) for

be

some

for

basis

dim

{T(e l ),...,

{T(el),

. . . , T(e

n )}

Finally, assume be a

basis for

=

y

=

T(x)

(d) holds. We

will

Therefore

Therefore x 2.8

then

Cl

=

Theorem

any

element

y in

I ciT(ei). dim

= o.

T(V)

Let

=

n,

{e l , . . . ,

so en}

write)

n

I ciei ,

\302\267 . = \302\267

(d).

implies

and

hence)

=

T(x)

i=l)

0,

Take

V.

= 0 impliesx

that T(x)

prove

, we may

(c)

n

=

by

implies (c).

But we are assuming

T(V).

spans

basis for T(V).

X =

If T(x)

But,

i=l)

If x E V

V.

the n elements

(b),

By

> n.

T(V)

o. Therefore

have)

and hence)

t'

T(en )}

is a

cp =

\302\267 . \302\267 =

n

\037 c.e. t \302\243.

i=l) Therefore

0

V.

n X =

t =

=

= n, so (b) a basis for

T(V)

and let {el , . . . , en} x in V, so we

(c) holds

assume

Next,

n.

dim T(V)


n

TmTn = Tm+n. If

of exponents:

(x +

=

(x, x

=

(x

T is invertible,

If ST

T commute,

(S +

(ST)n = prove that ST is

these formulas

T be the

Let Sand

(z, y, x)

and

T(x,

y,

(a) Determine the

ST -

of (x,

image

of

x

y, z)

(TS)2, (ST

T2, (ST)2,

S2,

TS,

+ y,

z) = (x, x

if ST

altered

be

must

transformations

linear

T)a =

(S +

and)

T2)

+ y

Va

\037

+ z), each

of

- TS)2.

(ST)-l

= T-1S-1. In other

reverse

order.

commute.

also

by

the

Sand

that

Prove

(c) Determine

Determine

29. Let Vand

xp(x). for

all n

31. Let

V be arbitrary

those

p in

V

for

those

p in

V

for

D be as in

Prove

30. Let Sand T an

+ Ta .)

formulas

S(x, y,

ST, TS,

under each

v, w)

z) =

of Va.

of the

transformations:

following

(d)

+ 3ST2

TS.

T are one-to-oneon Va and find the image of (u, S-l, T-l, (ST)-l, (TS)-l. (c) Find the image of (x, y, z) under (T - I)n for each n > 1 . Let D denote the 27. Let V be the linear space of all real polynomials p(x). which maps each polynomial and let T denote the integration operator = dt. Prove that DT = Iv but that TD \037 Iv. q given by q(x) S\037p(t) and range of TD. 28. Let Vbe the linear space of all real polynomials p(x). Let D denotethe that maps p(x) onto xp'(x). and let T be the linear transformation (a) Let p(x) = 2 + 3x - x2 + 4x a and determine the image of p under 2 transformations: D, T, DT, TD, DTTD, T2D2 - D T2. = in V for which . Determine those (b) p T(P) P (b)

In

V.

in

(x, y, z) is an arbitrary point the following transformations:

where

under

V and values

Sa + 3S2 T

Va defined

into

inductively by the for composition

> o.

that

T)2 = S2 + 2ST+

how

Indicate

26.

prove

+ z). + z).)

Tn is also invertible

that

domain

snTn for all integers n also invertible and that 23. If Sand T are invertible, of inverses, taken in words, the inverse of ST is the composition and commute, inverses 24. If Sand T are invertible prove that their that) 25. Let V be a linear space. If Sand T commute, prove

22. If Sand

- 1). z + 3).

1, Y + 2, + y, x + y + y,y + z, x

prove

z) =

(x, y,

1, z

+

Powers are defined the associative law

with T denote functions 25, Sand = TS, we say that Sand T commute.

through

TS.

+ l,y

=

= (T-1)n.)

(Tn)-l

In Exercises22

z) = (x

17. T(x,y,

=

given for one-to-

Tis

that be

DT

in !fJ(

Exercise

p(x)

real

DTn

polynomials

= Co +

operator

the polynomial the null space Describe

p onto

differentiation

operator

of the

following

that maps

p(x) onto

each

o.

which

- TD = I and that V, V) and assume that

> 1. the linear space of all polynomial

(DT - 2D)(p) = = which (DT - TD)n(p) Dn(p). 28 but let T be the linear transformation

differentiation

c1x +

ST

- TnD = nTn-1 - TS = I. Prove

. . .p(x). + cnx

for that

n >

2.

STn

- TnS

Let R, S, Tbe the functions n in V onto the polynomials

=

nTn-1

which

map

r(x),

s(x),)))

Linear

44)

and matrices)

transformations

and t(x), respectively,where) n r(x)

=

p(O)

=

s(x)

,)

I

Ck

Xk

-1

n

,

=

t(x)

k=l)

I

Ck

Xk + 1 .

k=O)

= 2 + 3x - x2 + x 3 and determine the image of p under each of the following 2 2 T S , S2T2, TRS, RST. R, S, T, ST, TS, (TS)2, (b) Prove that R, S, and T are linear and determine the null space and range of each. (c) Prove that T is one-to-one on Vand determine its inverse. (d) If n > 1 , express (Ts)n and sn Tn in terms of I and R. Refer to Exercise28 of Section 2.4. Determine whether T is one-to-oneon V. If it is, describe

Let p(x) transformations: (a)

32.

inverse.)

its

transformations

Linear

2.9

with

If V is tinite-dimensional, with prescribed values at the

be

n

el, .

Let

2.12.

THEOREM

u 1 , . . . , un

arbitrary V ---+ W

. . , en

an

n-dimensional linear space V. Let Then there is one and only one linear

k =

for

Uk)

x in V

1, 2,

. . . ,n

n

Ixke

k

T(x) =

then

,

k=l)

Every x in Xl'... multipliers

Proof the

x =

If

V can

,

X

n

.)

as follows:)

n

(2.8))

---+ W

theorem.)

next

the

in

V

that)

element

arbitrary

described

V, as

be a basis for an in a linear space W.

T(e k) =

T maps

of

T:

transformation

a linear

construct

always

basis elements

such

(2.7))

This

we can

elements

T:

transformation

values

prescribed

be expressed uniquely the components being

I XkU k=l)

a linear

as

of x

k .

of e1, . . . , en

combination the

to

relative

ordered

,

basis

matter to verify that T is (el , . . . , en). If we define T by (2.8), it is a straightforward linear. If x = ek for some k, then all components of x are 0 exceptthe kth, which is 1, so (2.8) gives T(e k ) = Uk' are required. To prove that there is only one linear transformation (2.7), let T' be another satisfying and compute We find that) T'(x).

n

n

T'(x)

Since T'(x)

(1,0)

and

= j

( k\037lxkek all

(0,

n

=

=

x in V,

the linear

Determine

EXAMPLE.

i=

= T(x) for

= T' )

k\037/kT'(ek)

we have

T' = T, which

transformation

T:

V 2

=

completes

---+ V 2 which

1) as follows:)

T(i) = i + j ,)

T(j)

=

T(x).)

k\037lXkUk

2i

- j .)))

the

maps

proof.) the basis

elements

Matrix

If x

Solution. =

T(x)

+

Xl;

=

of linear

representations

WI'

. . . , Wm

has

values

that a

shows

2.12

Theorem

linear space V is e1, . . . , en. Now,

linear

completely the

suppose

basis elements WI'

. . . , W m'

=

T(x) is given

, then

2x 2)i

+

(Xl

T:

transformation

V

\037

W

of

a given W is also finite-dimensional,say

space

basis for W. (The each element T(ek )

- j)

V 2

by its action on

nand be expressed

m mayor

dimensions can

+

(Xl

-

by)

x 2 )j.)

transformations

determined

be a

in W,

+ x2 (2i

+ j)

x l (;

45)

transformations)

element of

x 2j is an arbitrary

+ x 2 T(j)

xIT(i)

Matrix

2.10

=

of linear

representations

as

uniquely

a finite-dimensional of basis elements W = m, and let dim

set

may not be equal.) SinceT a linear combination of the

say)

m

T(e k) =

. . . , t mk are the components We shall display the m-tuple (t lk , . . .

! i=I)

of T(ek )

where t lk ,

tikw

i

,

to the

relative

ordered basis (WI'

. . . , W m ).

as follows:)

, t mk ) vertically,

t 1k)

t 2k) (2.9))

_t

This

one

is called a n elements

array

each of

the

of

pair

column

or a

vector

column matrix.

T(e 1), . . . , T(e n ). brackets to obtain the following

a matrix

mk)

We place

We

have

them side by

such side

a column and

enclose

vector for them

array:)

rectangular

t il

t I2

tIn)

t 21

t 22

t 2n)

t mi

t m2

t mn _)

We call it an m by n matrix, of m rows and n columns. The m X first row is the 1 x n matrix (t l1 , t 12 , . . . , tIn). matrix so the first subscript The scalars t ik are indexed displayed in (2.9) is the kth column. in which t ik occurs. i indicates the row, and the second subscriptk indicates the column We call t ik the ik-entry or the ik-element of the matrix. The more compact notation) This

is also

is called

array

or an

m

X

n

used to

matrix.

consisting

The

denote the

in

matrix

( (ik) ,)

or)

whose

ik-entry

( tik)F:k: I

is t ik

.)))

,)

1

Linear

46)

Thus, every

T of an n-dimensional space V into an m-dimensional matrix (t ik ) whose columns consist of the components of to the basis (WI' . . . , W m). We call this the matrix representation of ordered bases (el , . . . , en) for V and (WI' . . . , w m ) for choice of any element matrix to the T(x) relative (t ik ), the components can be determined as describedin the next theorem.) transformation

linear

space W gives rise to an T(e I ), . . . , T(en ) relative of T relativeto the given the W. Once we know

basis

. . . , W m)

(WI'

dim

=

W

and let

a linear

V = nand in 2( V, W), \302\273,heredim w and (WI' . . . , bases for Vand W, respectively, m ) be ordered whose entries are determined by the equations)

Let (el , . . . , en) x n matrix

m.

(t ik )

x n

m

T be

Let

2.13.

THEOREM

and matrices)

transformations

be the m

transforl11ation

m

(2.10))

Then

k)

T(e

! i=l)

=

tikw

i

1, 2, . . . , n

k =

for

,

.)

element)

an arbitrary

n

X =

(2.11))

k

!xke k=l)

in

V with

components (Xl'

. . . , xn )

. . . , en)

to (el ,

relative

by T onto

is mapped

the

element)

m

!

=

T(x)

(2.12))

YiWi

i=l)

in

W

with

(YI,

components

of X by

components

. . . , Ym)

the linear

relative to

I , . . . , w m ).

(HJ

related to

the

equations)

n

!

Yi =

(2.13))

Yi are

The

1, 2, . . . , m

i =

for

(ikXk

.)

k=l)

T to

Applying

Proof

each member of

where

each

Having

Yi

a pair

chosen

of bases(el, . . . , en)

every linear transformation we start ordered

with

any

bases

formation

elements of

T:

mn

T:

scalars

for Vand V

\037

W

=i\037

W,

having

V

\037

W

and

arranged as a then

it

is easy

this matrix

)

=i\037lYiWi')

proof.)

. . . ,

(WI'

W

m) for

Vand

rectangular that

representation.

there We

W, respectively,

if Conversely, choose a of and (t ik ) pair

representation matrix

to prove

the equations in (2.10). Then, T: V \037 W with these transformation linear one arbitrary point x in V is then given by Equations V by

( k\037/ikXk

a matrix

has

obtain)

m

n

Wi

This completesthe

by (2.13).

(2.10), we

using

m

=k\037lxkT(ek)=k\037lXki\037tikwi

is given

and

m

n

n

T(x)

(2.11)

(t

ik ).

is exactly one linear transT at the basis define simply

2.12, there is one and only of an The image values. T(x) prescribed by Theorem (2.12)

and (2.13).)))

Matrix representationsof linear EXAMPLE

of a

1. Construction 2 x 3 matrix)

linear

a given

from

transformation

matrix.

Suppose we

the

with

start

47)

transformations)

1 o)

G)

of unit

the usual bases

Choose

represents a linear in V3 onto the

T:

vector (YI, Y2)

in

V3 \037

Y2

of a

V 2

.

Then

maps an arbitrary the linear equations)

V 2 which

V 2 according

YI

V3 and

for

vectors

coordinate

transformation

-\037l)

to

=

3x I +

X 2

=

Xl +

OX 2

-

2x 3

+

4X3.)

the given matrix vector (Xl' X 2 , x 3)

,)

of a given linear transformation. of space p(x) polynomials degree < 3. This space has dimension 4, and we choose the basis (I, x, x 2 , x 3). Let D be the differentiation which operator each in V onto its derivative We can D as a linear maps polynomial p(x) p' (x). regard transformation of V into where W is the 3-dimensional of all real W, space polynomials of degree < 2. In W we choose the basis (I, x, X2). To find the matrix representation of D relative to this choice of bases, we transform (differentiate) each basiselement of Vand it a linear as combination of the basis elements of W. we find that) Thus, express V be

Construction

2.

EXAMPLE

Let

the linear

all

of

nlatrix

representation

real

2

D(I)=0=0+Ox+Ox D(x

coefficients

The

D.

2

2x =

of these the

Therefore,

) =

0 + 2x +

Ox

2

D(x

,)

polynomials determine the

o

1 =

1

3

) =

3x 2

the

by

Ox +

1 + =

0 +

of the

columns

is given

representation

required

=

D(x)

,)

following

Ox

2 ,)

Ox + 3x2 .) matrix representation 3 x 4 matrix:)

of

0

0

0020.

000 To also ordered

that the matrix representation emphasize on their order, let us reverse the order of the

dependsnot

basis

2 (x ,

nomials obtained basis

3)

2 (x ,

X,

x, I).

Then

above,

1) appear

but

basis

the

components

of these

order. Therefore, the

becomes)

000

3

0020.

o

1 0

0)))

only

elements

of Vare

elements

the

in reversed

basis

on the basis elements but in Wand use, instead, the

transformed into

the

same

polynomials relative to matrix

representation

the

polynew

of D now

Linear

48)

Let us compute x +

1 +

x

2

matrix

a third

3 + x ) for

V,

and

representation the basis (1, x,

for D,

x) 2

D(1 +

D(I)=O,)

matrix

the

x + x2 ,

x ,1+

basis elements of

W. The

for

(1, 1 +

basis

trans-

Vare

1

x 3)

+

0

1

1

Since it is

1 +

2x,)

+ 3x2 ,)

.

form

diagonal

of representations to try to choose the

matrix natural

simple form.

a particularly

=

3)

matrix representationin

different possible to obtain choices of bases,it is

tion by different matrix will have

x2 )

1

0 0 0 of a

x +

= 1 + 2x

0 0 2 2

Construction

+

D(1

,)

is

case

this

in

representation

=

x)

x + x2

D(1 +

2.11

the

using

as follows:)

formed

so

and matrices)

transformations

The next

a given shows

theorem

linear

bases so that that

transformathe

resulting

we can

make

from the upper left-hand corner 0 except possibly along the diagonal starting will a there be of ones of the matrix. this followed by zeros, the string Along \037.iagonal the to the of A number of ones rank transformation. matrix (t ik ) with all being equal matrix.) entries t ik = 0 when i \037 k is said to be a diagonal entries

the

all

THEOREM

dim

=

W

Let 2.14. m. Assume

a basis

exists

V

and

TE

!f

(e1, . . . , en)for

finite-dimensional linear spaces, with dim W) and let r = dim T( V) denote the rank of T. that) and a basis (WI' . . . , wm)for W such

W be (V, V

T(ei) =

(2.14))

Wi)

for

i =

1, 2,

V =

nand

Then

there

. . . , r ,)

and)

T(e

(2.15)) the

Therefore,

diagonal

matrix

(t ik )

=

i)

of T

0)

for

i = r

relative to these

+ 1, . . . , n

bases has all entries

First

Proof.

t 22 =

elementsform

for

the r

t rr = 1 \302\267)

dim T(V) = we construct a basis for W. Since T(V) is a subspaceof Wwith w r . Theorem has a basis ofr elements in W, say WI'..., 1.7, these By a subset of some basis for W. Therefore we can adjoin elements w r + 1 , . . . ,

that) \302\267 \302\267 , W r' (WI' \302\267

(2.16)) is a

except

\302\267 \302\267 \302\267 =

r, the space T(V) m so

zero

entries)

t 11 =

W

.)

basis for

W.)))

W r+1,

. \302\267 \302\267 , W m))

matrix representation

of a

Construction

in

diagonal

49)

form)

first r elements Wi in (2.16) is the image of at in V and call it e i . Then T(e i ) = Wi for i = 1,2, . . . , r so (2.14) Now let k be the dimension of the null space N(T). is satisfied. By Theorem 2.3 we have n = k + r. Sincedim N(T) = k, the space N(T) has a basis of k elements in V which we designate as er + l , . . . , e r+ k . For each of these consisting is satisfied. elements, Equation Therefore, to complete the proof, we must show (2.15) that the ordered set) element

one

in V. Choose

for

a basis

V.

one such element

V =

dim

Since

independent. Suppose

that

some

of the

Each

V.

er ,

,...,

(e l

7))

(2.1

is

a basis for

we construct

Now

least

n

=

linear

er + k ))

e r + 1 ,...,

r + k, we need only that show of them is zero, combination

these

elements are

say)

r+k \037

k

(2.18))

0 .

= c.e. \037\037

i=l)

T and

Applying

using

Equations

and (2.15), we

(2.14)

r

r+k \037 c.T \302\243.,l

i=l

that)

find

( e.l )

=

\037 c.w. \037l \302\243.,

=

0

.

i=l)

and hence WI' . . . , W r are independent, to) reduces terms in (2.18) are zero, so (2.18) But

C1

=

cr =

\302\267 \302\267 \302\267 =

o. Therefore,

the

r

first

r+k \037 i=r-t-l)

e r+ l

But

,.

. . , er + k are independent o. Therefore, all the

c r+k =

\302\267 \302\267 \302\267 =

basis for

V.

This

completes

since C i in

cie i

they

(2.18)

=

0 .

a basis

form

are zero,

for N(T), and

so the

elements

in

hence cr + l (2.17)

=

form a

the proof.)

D is the differentiation to Example 2 of Section 2.10,where operator V of the space W of polynomials of of 3 into maps space degree polynomials the method the range T( V) = W, so T has rank 3. Applying < 2. In this degree example, the for basis basis for used to prove Theorem 2.14, we choose W, (1, 2 ). example any 3 We A set of polynomials in V which is onto these elements given by (x, !X2, lx ). map is a basis the constant polynomial 1, which extend this set to get a basisfor V by adjoining for the null space of D. Therefore, if we use the basis (x, !X2, lx 3 , 1) for Vand the basis We

EXAMPLE.

refer


1 the

affected,

involve

A 2

of each minor we have)

If the

each kth

multiplied

affectedbut row.)))

row

A 2

by t, some

so fk row

is

homogeneous

of Ail

gets

in by

Ail-)

= tfi(Al'

t, where

in the multiplied

I , - -.,

tfl(A

A 2

\302\267 \302\267 , An) , \302\267

by

is not

All

t, so

we

An).)

by t and the

multiplied

first

the

=

tall detAIl gets

Ail

all is multiplied

coefficient

The

row.

first

\302\267 \302\267 , \302\267 , An)

fi is homogeneous of A is multiplied

det

1 and 2 the same will for f. be true first row of A by a scalar t. The minor

An) =

,...,

fi(tAl' Therefore

the

row

first

so again

= (-I)i+lail

\302\267 \302\267 \302\267 , An)

we verify that eachh satisfiesAxioms Consider the effect of multiplying the

affected have)

kth

I,

coefficient ail

is not

\302\267)

row.

minor

k > I , the row.

kth

by

t.

If j

Hence

\037

Akl

k,

is not

the

every fi is

affected

coefficient

but a kl is ail is not

homogeneous in

the

The

A similar

2.

shows

argument

We prove next

that

transpose) in every

is additive

eachh

Axiom 3', the

I satisfies

that

of a

determinant

weak version

91)

row, soI

1 and

Axioms

satisfies

3.

Axiom

of

From Theorem

3. that I satisfiesAxiom Axiom 3', assume two adjacent rows of A are equal, I satisfies say Ak = has two equal rows so A k+l . Then, except for minors Akl and A k+1,1, each minor Ail det Ail = o. Therefore the sum in (3.26) consists only of the two terms corresponding to

3.1, it

follows

then

verify that

To

j = k andj = k

+

1,)

I(A l ,

(3.27))

=

\302\267 \302\267 \302\267 , An)

Akl +

(-1 )k+lak 1 det

(-1

k + l , 1 det Ak+1

)k+2a

,)1

\302\267

ak + l 1 since Ak = A k + l . Therefore the two terms in (3.27) so I(A l , . . . , An) = o. Thus, I satisfies Axiom 3'. = 1 and a,l = 0 for = Iwehaveall Axiom 4. A When Finally, weverifythat/satisfies in each term I. is the matrix n so of order > Also, All 1, (3.26) is zero except identity j 4.) the first, which is equal to 1. Hence1(11, . . . , In) = 1 so I satisfiesAxiom and akl =

A k1 = Ak+l 1

But

only in sign,

differ

have In the foregoing proof we could just as well of the first-column kth-column minors Aik instead

defined

a functionl

used

minors Ajl. In

in

if we

fact,

of the

terms

let)

n

(3.28))

= !(-l)i+k

, An)

f(Al'...

A ik

det

aik

,

i=l)

this I satisfies all four same type of proof shows that functions are unique, the expansion determinant

the

exactly

those in The but

also

are

(3.21)

expansion reveal

3.14 The

all equal to det A. formulas (3.28) not

a new

propertiesand

aspect of the

with

are

definition may

Although primarily not

alter

another

I

2

3

[4

5

6)

,

be given

as

whose

i, j

square

transpose

of

A

and

denoted

if)

1

4)

2

5

3

6)

.)

follows.)

can be applied to any We prove next matrices.

its determinant.)))

At =

then

the

example,

]

The transposeof an is au .) entry

transposition with

called

matrix

of A. For

OF TRANSPOSE.

DEFINITION

At

connection

columns

the

A-_ A formal

existence

of determinants-a

theory

A is

matrix

each

and

a transpose

of

rows of At

in (3.28)

formulas

of determinant functions connection between rowis discussed further in the next section.) the

establish

only

This

column-properties.

determinant

Associated by At. The

matrix

axioms for a determinant

Since

function.

m

X n

matrix

rectangular that

transposition

A

matrix

=

(aii);:j\037l

we shall of a

is the

n

X

m

be concerned

square matrix

does

92)

Determinants) any n

For

3.11.

THEOREM

A we

n matrix

X

A =

det

B =

proof is that the

The

Proof

then,

Assume,

=

At

we

minors

det

For

on n.

induction

by

have)

n

At.)

=

(hi').

= 2 the

1 and n

result is easily verified.

- 1. Let A = minors and det B by

theorem is true for matrices of Expanding det A by its first-column

n

order

) and

first-row

have)

n

n

det

A =

'+1

I\037 ( \302\243.,

-1)'

j=l)

a jl det

A

det

j1 ,

B =

\037 ( \302\243., j=l)

-I)'

'+1

Blj .

h lj det

of transpose we have b lj = ail and B1i= (AjI)t. Since n - 1 we have det Bli = assuming the theorem is true for matrices of order the foregoing sums are equal term Hence so det A = det B.) by term, the definition

from

But

3.15 The

let

(a ij its

we det

are Ail.

matrix

cofactor

Theorem 3.5 showed that if A is nonsingular then det A \037 o. The next theorem proves the converse. That is, if det A \037 0, then A-I exists. Moreover, it gives an explicit formula for expressing A-I in terms of a matrix formed from the cofactors of the entries of A. In Theorem 3.9 we proved that the cofactor of is to i, j aij equal (-1 )i+i det A ij , where is the minor of Let us A. denote this cofactor i, Ai' j by cof aij. Thus, by definition,)

=

cof aii

OF THE COFACTOR

DEFINITION

cofactor matrixt

cof

The apart

next theorem from a scalar

THEOREM

A =

shows that

the

factor, the

identity

For any n

3.12.

X

if det

In particular,

the

much of the older

\037 a.l1,) 't,''-

literature

an entirely

different

A

matrix

\037

0 the

literature

calls object,

it

I =

of product matrix

A

n matrix

A (cof

(3.29))

t In

( cof

by

A.

cof

inverse

of

A-I

=

A

with

n >

2 we

aij is

called the

?2

-) 1 . \"_ 't,.) \037,

transpose

of its

cofactor

matrix

is,

have)

(det A)/.)

exists

1

det A)

the transpose of the

the

and

is given

hy)

(cof A)t.

cofactor

adjoint of A. However, current discussed in Section 5.8.)))

the

i+i det A,

with

is cof

entry

have)

I.)

A)t = A

we

Thus,

(( _l )

ii .)

whose i, j

matrix

The

MATRIX.

is denoted

of A and

(-1 )i+i det A

matrix

is called the

nomenclature

of A. Some of the term adjoint for

adjugate

reserves

rule)

Cramers'

3.9 we

Using Theorem

Proof.

of its kth-row cofactors by

A in terms

det

express

93)

the

formula) n

(3.30

!

A =

det

))

a

k;

cof a k ;

.

;=1)

k fixed

Keep

row of

A

and apply i

some

for

\037

the ith det B = 0 because ith-row cofactors we have)

whose

and

B whose ith

new matrix

to a

relation

this

k,

rows are

remaining

rows of B are equal.

and kth

to the kth of A. Then in terms of its

is equal

row

same

the

as those B

det

Expressing

n

=\037b..cofb.. k 0

detB

(3.31))

;=1)

since

But

the ith row boo'l:1

Hence (3.31)

of B is equal to =

a k1).

cof

and)

boo t1

=

o.

of A we have)

row

kth

the

0=

cof a..t1)

for

every

j.)

states that) n \037

k

(3.32))

a k . cof

j=l)

Equations

(3.30) and

1

a..'l:1 =

0

k

if

(3.32) together can be written

(3.33))

But fore

\037 j=l)

appearing on the

the sum (3.33)

implies

3.16

3.13.

Cramer's

left

of (3.33)

det

A

{0

if

i =

k

if

i

k.)

is the k, i entry

\037

of the

product

A(cof A)t.

There-

(3.29).)

As a direct corollary sufficient condition for

THEOREM

a..t1 =

a k 1' cof

i.)

follows:)

as

n \037

\037

of

3.5 and

Theorems

a squarematrix

A square

matrix

to

A

3.12 we have

the

necessary

following

and

be nonsingular.)

is nonsingular

if and

only

if det

A

\037

O.)

rule

Theorem 3.12can alsobe used to give explicit formulas for the solutions of a system of with are a coefficient The formulas called matrix. Cramer's nonsingular equations rule, in honor of the Swiss mathematician Gabriel Cramer (1704-1752).) linear

THEOREM

Xl'

3.14.

If a system

RULE.

CRAMER'S

of

n

linear

\302\267 \302\267 \302\267 , x n ,)

n \037

k

;=1)

a..x. t1

1

=

b.t

(i =

1, 2,

. . . , n))))

equations

in n unknowns)

Determinants)

94)

has a

gil'en

coefficient-matrix

nonsingular

A =

(aii)'

k cofa

k J';'

is a

there

then

by

\037b

A

det

The

Proof

can be

system

L

the system

for

j =

for

1, 2, . . . , n

as

written

a matrix

=

equation,)

B,) b l)

Xl)

X and

matrices, X =)

B are column

B=)

.

x n)

is a

unique solution X

det

It should

in

follow

(3.34)

be noted

that

the

A is

nonsingular)

b n)

1

A-IB =

X =

The formulas

Since

by

given

(3.-35))

.)

k=l)

AX

there

solution

n

1

x,;J =

(3.34))

where

unique

the formulas)

components

by equating for x j

formula

(cof AYB.

A)

in

in

be expressed

can

(3.34)

(3.35).)

as the quotient of

two

determinants,)

x.1 = where

C j is

the matrix

obtained from

A

det C j

')

det A

of A

the jth column

by replacing

by

column

the

matrix B.)

3.17 Exercises

1.

Determine

the

cofactor

[:

:l

matrices:

following

3

0

1

1

-1 -2

0

2

1

3

2 -1 (b)

(a)

each of the

matrix of

0 5 1 -1 -2

2 (c)

-2

2

3

matrices in Exercise the inverse of each of the nonsingular Determine 3. Find all values of the scalar A for which the matrix AI - A is singular,

2.

1

(b)

(\037) [\037

-\037l)

0

0

-1

2 -2)

2

-2 o)

4

(c))

11

-2

19

-3

-8)

2

8

14 . -5)))

1

6 3)

1. if A

is equal

to)

95)

Exercises)

of its cofactor matrix: 2, prove each of the following properties (cof A)tA = (det A)/. with the transpose of its cofactor matrix). (A commutes (c) A(cof 5. Use Cramer's rule to solve each of the following systems: 2x - y + 4z = 7, (a) x + 2y + 3z = 8, -y + z = 1 . = = z 3 2x + 5 Y + 3z = 4. , (b) x + y + 2z Y for a straight line in the 6. (a) Explain why each of the following is a Cartesian equation xy-plane

4. If A is an (a)

matrix

x n

n

with

two

passing through

distinct

[

and

State

-

I Xl

-

Y

Y2

-

YI)

I

and

(X 2 , Y2).)

= 0;

det

YI) y]

prove corresponding relations for

a

X

Y

1

Xl

YI

1

X2

Y2

in

plane

= o.

1)

3-space

passing

through three

points.

(c) State noncolinear

and prove correspondingrelations functions

X in

(a, b).

n

a circle in

for

the xy-plane passing

points.

2

7. Given each

X2

(Xl'

points

X-X

det

distinct

(b)

3x -

0,

(b)

n >

= (cof A)t. A)t = (cof A)t A

cof (At)

fij,

each differentiable on an interval p' (x) is a sum the derivative

(a,

Prove that

of n

b), define

through

F(x) = det

three

[fij(x)] for

determinants,)

n

p' (x)

=

! i=l)

det

Ai(x)

,

obtained in the ith row of [fij(x)]. the functions where Ai(x) is the matrix by differentiating of the form W(x) = in which n x n matrix of functions each row after the first [U\037i-I)(X)], matrix in honor of the Polish of the previous row, is called a Wronskian matheis the derivative Prove that the derivative of the determinant of W(x) matician J. M. H. Wronski (1778-1853). of the matrix obtained by differentiating each entry in the last row of W(x). is the determinant

8. An

[Hint: Use Exercise

7.])))

4)

4.1 Linear

with

transformations

EIGENVECTORS)

AND

EIGENVALUES

matrix

diagonal

representations

linear V. Those Let T: V \037 V be a linear transformation on a finite-dimensional space V are called T which are of coordinate for of (basis) independent any system properties all a basis are shared the matrix of T. If intrinsic properties of T. They by representations can be chosenso that the resulting matrix has a particularly simple form it may be possible to detect some of the intrinsic properties directly from the matrix representation. ask the simplest types of matrices are the diagonal matrices. Therefore we might Among 2 In has matrix whether linear transformation a Chapter every diagonal representation. we treated the problem of finding a diagonal matrix a linear transforof representation that T: V \037 W, where dim V = n and dim W = m. In Theorem 2.14we proved mation the a w that there always exists a basis(el , . . for Vand basis . . . for W such , (WI' en) m) = bases is a In if W V matrix of T relative to this of matrix. pair diagonal particular, will be a square diagonal we want matrix. The new feature now is that to use the the matrix it is not always possible to find a diagonal same basis for both Vand W. With this restriction for T. We turn, then, to the problem of determining matrix representation which transdo have a diagonal matrix formations representation.

.,

If A =

Notation:

(aij) is a diagonal

easy to give a necessary diagonal matrix representation.) It is

matrix

diagonal

in V and

a correspondingset of scalarsAI, T(u k )

Conversely,

. . . , An

is an

if there

satisfying

=

set

(4.1),

the

then

is a 96)))

representation

of T relative

to

V

exists an

U 1 , . . . , Un

V,

diag

the basis

(AI'

. \302\267 \302\267 , ann) .

to have

set

dim V = n. of elements

a

If T has a UI

, . . . , Un

that)

1,2,...

,n.)

in V and

a correspondingset of scalars

. . . , An))

(uI , .

a22 ,

transformation

where

independent

such

k =

for

\037

matrix)

A =

= diag (all'

condition for a linear

. . . , An

Aku k)

independent

we write A

T:

then there

representation,

(4.1 ))

AI,

and sufficient

linear transformation

Given a

THEOREM 4.1.

matrix,

. . , un).)

and

Eigenvectors

first

Assume

Proof.

(el , . . . , en).

some basis

97)

transformation)

a diagonal matrix representation A = (a ik ) relative to the action of T on the basis elements is given formula) by has

T

that

of a linear

eigenvalues

The

n

!

=

1\"(e k )

=

i

aike

akke k

i=l)

a ik =

since

Now

0 for i

(4.1).

Since

u

1, . . .

0 for i

\302\245= k,

with

Uk

=

e k and

=

Ak

a kk

.

and scalars AI' . . . , An exist satisfying a basis for V. If we define a kk = Ak and form , Un are independent they T then the matrix A = (a ik ) is a diagonal matrix which represents

to the basis (u1,

relative

(4.1)

proves

elements

independent

suppose

a ik

=

This

k. \302\245=

U

1 , . . . , Un

. . . , un).)

matrix representation the problem a diagonal of a linear of finding transformation of finding been transformed to another problem,that elements U 1 , . . . , Un independent and scalars AI' . . . , An to satisfy (4.1). Elements Uk and scalars Ak satisfying (4.1) are called of T, respectively. In the next section we study and eigenvalues eigenvectors eigenvectors Thus

has

and eigenvalues

t

a more

in

4.2 Eigenvectorsand In

and

not

are

Let T:

S -+

eigenvalue of T if there

space and S denotesa subspace

transformation of S element x in S such that)

a linear

be

V

T(x) =

(4.2)) x is

element

The

transformation of

V.

S

The spaces

dimensional.)

finite

nonzero

is a

a linear

a linear

to be

required

DEFINITION.

of

eigenvalues

V denotes

discussion

this V

setting.)

general

AX

called an eigenvectorof T belonging

A scalar

V.

into

A

is called

an

.)

to

The scalar

A.

A

is called

an eigenvalue

corresponding to x.)

one eigenvalue

is exactly

There

have T(x) = Note:

AX

and

Although

excludes0 as an A

eigenvalue The

following

EXAMPLE

defined every

by

1. the

nonzero

T(x)

=

flX

Equation

eigenvector. with associated

Multiplication equation

some

(4.2) One

\302\245= 0,

AX

x.

eigenvector

given

then

eigenvector x.

the meaning

=

0 and

tlX so any

against

A

=

In fact, if we

fl.

A, the definition 0 is to have exactly one scalar

of these concepts.)

scalar. Let T: S -+ V be the linear transformation each x in S, where c is a fixed scalar. In this example

by a fixed

T(x) = cx for

element of

x

= always holds for x reason for this prejudice

a given

illustrate

examples

corresponding to a

for

S is an

eigenvector

belonging

to the

scalar

c.)

are partial translations of the German words Eigenvektor and words eigenvector and eigenvalue use the terms characteristic vector, or proper vector as synonyms for Eigenwert, respectively. Someauthors roots.))) values, proper values, or latent Eigenvalues are also called characteristic eigenvector.

t The

98)

and eigenvectors)

Eigenvalues

be a

that It

A.

is

T(x) = AX. easy to prove

to

corresponding

general,

4.

EXAMPLE

becauseit scalars.

Al

The

that

plane

in

form,

z =

Thus, each z \037 0 not

real unless Now

of

the

plane

of

through

V =

space,

ajixed

angle

of eigenvectorsmay as a linear space V 2 (R),

with

This

rx.

depend

in two

two basis

if

rx

is

not

eigenvectors. Thus of scalars for V.)

If

T rotates

z

through

an

angle A = e irl .

is an

eigenvector with eigenvalue multiple of 1T. the plane as a real linear space,

\037is

consider

re i8 .

be a =

T(i)

in

reflection

T(j) = j,

i,

is of special example on the underlying

different ways:

rx

of

then

Note

field of

(1) As a 2-

elements (1, 0) and linear

interest

(0, 1),and

V =

space,

VI

(C)

,

can be expressed eirlz. the eigenvalue eirl is

VI (C) T(z)

that

= rei(8+rl) =

an integer

numbers the rotation words,

let T

1. xy-plane is an eigenvectorwith eigenvalue the form where has c each of them \037 0; ck,

as scalars;

numbers

polar

O. In

eigenvalue

.)

in the

vector

the existence can be regarded

linear

real

real

the

with

eigenvector

be zero, by eigenvalue for null space of T

if 0 is an

Conversely, if

or (2) as a I-dimensional complex with one basis element 1, and complex numbers as scalars. Each element z \037 0 Consider the second interpretation first.

with

by))

it cannot

exists In fact,

Let S = V = V 3(R) and basis vectors i, j, k as follows:

eigenvectors are those

shows

dimensional

+

.)

5. Rotation

EXAMPLE

yare

subspace. The spaceE(A) is infinite-dimensional. If E(A) at least one nonzero element x

contains

null space of T. each of these is an

xy-plane. on the

T act

nonzero

Every

-1

eigenvalue

of T -

the

in

Reflection

remaining

belonging in E(A) we have)

eigenvectors

so E(A) is a may be finite- or

the

then

null space

That is, let

T(k) = -k.

It

all

and

= A(ax

bAY

E E(A)

> 1 , since E(A)

in

is

elements

is the

the xy-plane.

The

0, so x

nonzero

E(A)

+

aAx

of zero eigenvalues.If an eigenvector zero scalar can be an eigenvalue.

the

Ox =

=

T: S \037 V be the set of all elements x in

A.)

However,

x then T(x) contains any

aT(x) + bT(y) =

by) =

3. Existence

EXAMPLE definition.

Let E(A)

= AX. Let

T(x)

the zero element 0 and subspace of S, becauseif x

is a

E(A)

then dim E(A)

is finite-dimensional

that

set contains

This

Hence (ax + by) corresponding to A.

the eigenspace

all x such A.

eigenvalue

a and b.

all scalars

for

an

having

that

+

T(ax

called

E(A) consisting of

eigenspace

linear transformation

S such

to

The

2.

EXAMPLE

T has real

eigenvalues

only

an integer multiple of 1T then the existence of eigenvectors

V 2 (R).

if

rx

Since is an

the scalars

integer

of

V 2 (R)

multiple of

1T.

are In

real other

T has no real eigenvaluesand hence no and eigenvaluesmay depend on the choice

The differentiation Let V be the linear space of all real functions operator. f derivatives of order on a transforhaving every given open interval. Let D be the linear = I'. mation which maps eachlonto its derivative, The eigenvectorsof D are those D(f) nonzero functions f satisfying an equation of the form))) EXAMPLE 6.

and

Eigenvectors

=

I' some

for

a

an

c is

where

functions

I(x)

examples

like

= ceAX

c

o. \302\245=

The

a

function

V is

where

one

this

with

continuous

=

g(x)

The eigenfunctions

T (if any

of

space

I: f(t)

=

relationf(x)

the

only X/ A

with

eigenfunction exists we

real A. If an Af'(x), for

candidates

c

from A

0 and \302\245=

we

which

eigenfunctions o. \302\245=

However,

o =

e a/ A is never

zero f, so T has

zero we no

see that

the

g =

define

a
o. Therefore the the same ellipse in the to c = 9.)

if c

ellipse

c, represents corresponding

ellipse

The reduction of a quadratic form all points (x, y) in the plane which

ax2

YlY2-system.) XAXt

equation

= c,

coordinate

original

written

system.

geometry

analytic

(5.10))

the

in

+

diagonal form

to a

a Cartesian

satisfy

bxy +

to

the set

identify

of

of the form)

= o.)

+f

dx + ey

+

cy2

can be used

equation

set is always a conicsection,that is, an ellipse, hyperbola, parabola, cases (the empty set, a singlepoint, or one or two straight lines). form the The type of conic is governed that the is, terms, by quadratic by second-degree ax2 + bxy + cy2. To conform with the notation used earlier, we write Xl for x, X 2 for y, and express this quadratic form as a matrix product,) We

shall

'or one of

find

that this

the

degenerate

XAX

a where

X

=

[Xl' X2] and

a diagonal form of eigenvectors

AlY\037 UI

+

A

=

A2Y\037'

t

=

ax\037

b/2 .

[ b/2

c

where

AI'

, U 2 determines

a new

]

+

bXIX2 +

By a

A2 are

cx\037,)

rotation

the

Y

=

XC we

of A.

eigenvalues

set of coordinate axes,relative

reduce An

to

this

form

set

orthonormal

which

to

the Cartesian

equation (5.10) becomes) (5.11))

AIY\037

with

new

term

YIY2,

coefficients d' so the type

+

A2Y\037

+

and e' in the linear of conic is easily

d'YI + e'Y2 terms. identified

In

this

+ f

=

0,)

equation

by examining

there is no mixed the eigenvalues Al

product and

A 2 .)))

to analytic

Applications

131)

geometry)

If the

if AI' A 2 have conic is not degenerate, Equation (5.11) representsan ellipse if either if A a and have a Al or A 2 is 2 AI' parabola signs, opposite sign, hyperbola three cases correspond to AIA2 > 0, AIA2 < 0, and AIA2 = o. We illustrate specific

2 4xy + 5y

2X2 +

1.

( 5.12)) form

quadratic

U2

.

C = t

+ 4X1X2

2x\037

has

=

2), where the

reduces

This

[ determine the effect on the form X = yct and obtain) [YI,

_ 1

Y2]

.J5

[

2]

1)

4

the

completing

13

in

squares

6y\037

system. and We

of an

equation

The positive

ellipse

is the

Geometrically,this the

(-

-

YIY2 axes

with

but

equation of the

YI

)5

ellipse

the

Yl

its

with

the

-

+

all

three

YI

+

this

! = o.)

as

follows:)

6.J5

at the

point (lJs,

XC in the

+

YI

2Y2)')

Y2')

6z; =

systems

in

the

)'1)'2-

eigenvectors

U1

writing)

I)S,)

9,)

-!)5) by the

determined

are

axes

Z2

=

Y2 +

!)S.)

to a new system of coordinate axesparallel In the zlz2-systerTl the center of the ellipse.

is simply)

coordinate

(-

To

= 9.)

lJ-S)2

center Y2

-

2 Z2

2

The ellipse and

1

6y\037.

--.J5

Y2

+

6(Y2

Yl and

by

6)5

rewrite

same as introducing new origin at the

Z\037+

+

Y =

rotation

X2 =

Y2),)

y\037

to)

2Y2) =

+

Yl

Y2 we

in Figure 5.2. U2 , as indicated the equation further can simplify Zl =

form

the

is

matrix

.J5

transformed

- t)S)2 +

directions of

(5.12) to

(2YI +

eigenvectors

becomes)

and

Yl

0'1

This is the

1

the foregoing

diagonalizing

.J5

equation Y\037+

By

Xl =

2 of

Example

orthonormal set of

equation of

.J5

Cartesian

in

o.)

orthogonal

part of

, 2

.J5

transformed

1/\0375.

we write the

Y2) +

+

(2YI

An

part

13xis

part 4x1 +

Therefore the linear

!=

-

13x2

this as)

rewrite

We

O.

6, and an

=

A2

quadratic

linear

2

1

=

=

t

=

one treated

the

is

5x\037

!

4x 1 +

+

5x\037

Al = 1 ,

2]

-1

[Xl'X2]

+

eigenvalues

t(l,

13y -

4x +

+

4X1X2 +

+

2x\037

section. Its matrix Ul = t(2, -1), 2 1

The

some

with

examples.

EXAMPLE

The

same The

the zero.

or)

ZI

+

9

are shown

=

1.

3/2) in

Figure

5.2.)))

132)

acting on Euclidean

of operators

Eigenvalues

/

/

/

,

spaces)

Y2

/)

- - -...

--

-

----

Xl)

\"

\"

,

,

\" Yl)

2 1)

is 2.

EXAMPLE

and

Rotation

5.2

FIGURE

by the translation

followed

2X2

4xy -

-

-

2x\037

The Al

=

Zl

Yl

4X1X 2 -

x\037

-

3,

where

t

A2

=

of rotation

=

-2.

-2

1/)5.

X =

set

An orthogonal YCt

1

=

(2YI +

the

-

=

2y\037

-

2

3YI

(2YI +

+

Y2)

squares

2]

2Y2)')

+ 2Y2) -13

(-YI \037\037

-

2 2Y2

-

16

18 YI

+

../5 the

-1),

2 1

or)

completing

t(2,

C = t

(-YI +

the eigenvalues

has

/S

;5

By

as)

equation becomes)

transformed

3y\037

this

[ -1) 1

X2 =

Y2),)

../5 Therefore

UI

us

gves

Xl

is

is

!\0375.)

matrix

This

[ -2 ] of eigenvectors

diagonalizing matrix

in

Yl

3(yl

and

Y2 we

- 1)5)2-

Y2

-

13 =

../5

obtain the equation)

2(y2

-

XC

13 = O.)

-

.

-1

Y =

rotation

rewrite

We

10X2

=

A

orthonormal

An

!\0375,

4x I + 2

where

is XAXt,

part

quadratic

= Z2 Y2 +

13= o.

4x + lOy -

-

y2

=

-

The

axes.

coordinate

of

translation

tJ-S)2 =

12,)))

o.)

=

0,)

.

U2

The

=

t(l, 2),

equation

to analytic

Applications

.

133)

geometr.v)

X2

I I I I I I I I I I)

Xl)

, \"

,

,

\"

\"

Zl

\"Yl)

3ZI

(a) Hyperbola:

-

which

Zl

=

- tJs,

Yl

(b) Parabola:

curves

in

with its center at

a hyperbola = Z2

represents

= 12

The

5.3

FIGURE

lation

2z\037

Y2

- tv

5

(!Js, tJ-S) in this

simplifies

2

3z;

The

is shown

hyperbola

tions of the

positiveYl 3.

EXAMPLE

9x

2

+

= 12,)

- 2z:

in

2

4

A2

=

O.

24xy + 16y2

The eigenvectors

+

- 20x +

24x1X2 +

for the quadratic

set of

An orthonormal

15y

16x\037

-

part is A

eigenvectors

3 onal

is C

matrix

diagonalizing

=! [

Xl = Therefore

the

25y\037

This simplifies

to y\037 +

parabolais shown

in

Y2

-

!(3Yl

Cartesian

transformed

-

\037j(3Yl

= 0, the

Figure

.

1

6)

=

U

1 and

We rewrite

O.

20x 1 +

15x2=

9 matrix

symmetric

The trans-

to)

U2 determine

the

direc-

Y2 axes.)

and

9x\037

The

0

YlY2-system.

further

Z2

Zl ---=

or)

5.3(a).

Figure

the

equation

+ Y2 =

3.)

2 and

Examples

YT

5.3(b).)))

4 4Y2)

12

=

[

12 16]

is U 1 =

!(3, 4),

this

O.)

. Its eigenvalues U2

=

!(

-4

3'] ,)

The

X2

of rotation

equation

=

!(4Yl

+

as)

are

-4, 3). X =

An

yct

Al =

25,

orthog-

gives

us)

3Y2).)

equation becomes) -

4Y2) +

.lj(4Yl +

3Y2)

equation of a parabola.with

=

its

o.) vertex

at the

origin. The

134)

4.

EXAMPLE

three

x2 +

equations

eigenvalues; the first (x, y) = (0, 0), and

represents third

the

alone does not reveal conic section. For example,the 2 2 2 2 the same 1, x + 2y = 0, and x + 2y = -1 all have a nondegenerate ellipse, the second is satisfied only by The as set. last two can be the regarded empty represents

degenerate cases of the ellipse. The graph of the equation y2 = two lines y = 1 and y = parallel The

parabola.

if either x -

2y

x2 equation = 0 or x +

hyperbola.

polynomial A

- a

-b/2

det

A

[ -b/2 the

Therefore

-1.

=

2y

o.

the

eigenvalues

x-axis. The equation

1=

0 representsthe cases of the

as degenerate

be regarded

can

These

-

y2

lines since 0 representstwo intersecting This can be regardedas a degenerate

satisfied

it is

of the

case

bxy + cy2 + dx + ey + f = 0 represents section, then the type of conic can be determined quite easily. ax 2 + bxy + cy2 is) of the matrix of the quadratic form

conic

characteristic

0 is the

4y2 =

However, if the Cartesian a nondegenerate

spaces)

a degenerate

represents

=

2y2

of

A knowledge

cases.

Degenerate

whether the Cartesian equation

acting on Euclidean

of operators

Eigenvalues

-

ax 2 +

equation

=

A2 -

(a +

]

- ib2 )

!(4ac

- b2 ).)

=

-

(A

-

A1)(A

A 2 ).

C)

of the eigenvalues

product

(ac

+

C)A

The

AIA2

=

ac

is)

- !b2

=

Since the type of conic is determined by the algebraic sign of the product Al A 2 , we see that the conic is an ellipse,hyperbola, or parabola, as 4ac - b 2 is positive, negative, or according 2 2 ax 4ac zero. The number b is called the discriminant of the quadratic form + bxy + has the values 34, -24, and 0, respectively.) cy2. In Examples 1, 2 and 3 the discriminant

Exercises)

5.15

In each of Exercises of A ; (c) an

eigenvalues

1.

XIX2.

3.

x\037 +

4.

34x\037

In each the

4X 1X2 +

+

4x\037

2.

2X 1X2 -

-

1 through orthonormal

find

(a)

set of

a symmetric matrix A for eigenvectors; (d) an orthogonal

x\037.

x\037.

24x1X2 +

18,

identify

y2

-

what

value

o be a pair 20. If the equation bounds

7.

3x\037

+

X2X3.

4X1X3 + x\037 x;. + 4X 1X 2 + 8X1X3 +

4X2X3

+

3x;.

a sketch of the

section

conic

represented

by

equation.

-

19. For

2x\037 +

and make

- 5 = O. 2xy + 2X2 = 9. y2 O. 2xy + 5x 10.y2 - 2xy + X2 - 5x = O. 11. 5x2 - 4xy + 2y2 - 6 = O. 12. 19x2 + 4xy + 16y2 - 212x+ 104y = 356. 13. 9X2 + 24xy + 16y2 - 52x + 14y = 6.

8.

x\037

C.

matrix

diagonalizing

+ X 1X 2 + X 1X3

5.

6.

form; (b) the

the quadratic

41x\037.

of Exercises 8 through

Cartesian

7,

of

(or values) lines?

is 21T/vi

ax 2 + 4ac

of c will

the

graph

14.

5x

15.

2

x

16. 2X2

17. x2 18. xy of the

2

+ 5y2 + 2xy + y2 + 4xy + 5y2 +

6xy

+ Y

-12

_2y2

+4xy

- 2x

= O. 2x + 2y + 3 = o. - 2x - Y - 4 = O. - 2

-

=

2

=0.))) O.

Cartesian equation 2xy

an ellipse, bxy + cy2 = 1 represents - b2 . This a gives geometric meaning

prove that to

the

the

-

area

discriminant

4x +

7y

of the

4ac

+

region

c =

- b2 .)

it

Eigenvalues of a symmetric Eigenvalues of a symmetric

* 5.16t

the requirement

we drop

Now

the eigenvalues

Suppose x is

of a symmetric an

operator

belonging to

between

a relation

form.

and

with norm 1

eigenvector

and we find

finite-dimensional its quadratic

V be

that

135)

of its quadratic form

as values

obtained

transformation

of its quadratic form

as values

obtained

transformation

an

T(x) =

Then

A.

eigenvalue

Ax

we have)

so

since

x) =

values

T: V \037

Let

Q(x) = (T(x), x). that Q takes on the V =

Let

EXAMPLE.

V be the

Then unit

2

I

i=l

of T are Al = 4, A 2 the minimum and maximum

eigenvalues

respectively, = 1. In fact, x\037

on this

Q(X) = has

its smallest

points satisfying circle

are

x\037)

=

Q(x)

eigenvectors

ellipse are eigenvectors

on

unit sphere

the

called

Euclidean to be found

a real

are

in

space among

V.

V,

the

the

and

with

matrix

A

as inner

dot product

usual

4

0

[0

8)

.

=

Then

the

]

2

IaijXiXj =

j=l)

=

4xi +

It is easy

8.

which

values

8x\037.

to see that

these

Q takes on the

eigenvalues

are,

circle

x\037 +

unit

+

4x\037

=

4 +

4x\037,)

where

-1




Then if

not change sign,

the

of

V. Choose

in

element

inner

any

and

product,

t 2Q(y)

= at +

t

ty)

t 2(T(y), bt

on V we

all real

for

+

x)

t(T(y),

is nonnegative

0)

x+

+ tT(y),

(T(x)

2

y)

,)

have the

inequality)

.)

2 polynomial p(t) = at + bt has its minimum at t = o. = a 2(T(x),y), so (T(x),y) = O. Since y was arbitrary,

(T(x),

at +

bt

T(x\302\273 2




(5.14)) A =

Let

or minimum) at a point is Q(u), the

all x

for

have

Q(u)

(u, u)

= 1.

value

of Q on the

Then

u is

an eigen-

unit

sphere.)

we have)

Then

u.

at

Q(u))

x) = 1 we

If (x,

Q(u).

u with extreme

eigenvalue

a minimum

137)

case)

finite-dimensional

with (x,

=

= A(X, x)

x) = 1.) x)

(AX,

(5.14) can

so inequality

be

as)

written

(5.15))

Then x =

ay,

where

(T(x), x) But (T(y),

y) >

Ily

=

(Ay,

II

=

that

(5.15)

y)

since

(y, y)

= ay. get (5.15) for x Since(T(x),x) - (AX, x) = (T(x) - AX, x) > 0, or)

a 2(T(y),

=

ay)

(T(ay),

= 1.

(T(x) -

(S(x), x) >

(5.16)) x =

have equality

u we

x))

(Ax,

for all x in

is valid

Ilxll =

Suppose

both

Multiplying

AX,

S =

where

0,)

= a 2 (AY,

members

of this

rewrite

we can

x),

(Ax, x)

and)

y))

T

-

and hence also in (5.16). states that the quadratic (5.16) V. When x = u we have Ql(U) =

S is

V.

a.

Hence)

1.

a 2 we

When

x} >

= 1. Now we prove

(x, x)

provided

(T(x),

in

(5.14)

y).)

(5.15)

inequality

by

inequality

in the

form

AI.)

The

linear

transformation

form Ql given Ql(X) = symmetric. Inequality by is on o. Theorem Therefore, by (S(x), x) nonnegative 5.13 we must have S(u) = o. In other words, T(u) = AU, so u is an eigenvector for T, and A = if Q has a minimum Q(u) is the corresponding eigenvalue. This completesthe proof u.

at

If there

*

is a maximum

at

5.13

to the

apply

Theorem

5.18

The

order,

increasing

in

the

foregoing

form

proof

are reversed

and we

Ql.)

case

dim

V

=

n.

Then T

has n

Al
0

and AX

(b),

write

be an

exist

there

matrix

\302\273'here Ak

eigenvectors

of T relative to is the eigenvalue

AX) = =

T(x)

to A. Then

belonging

eigenvector

x

0 \302\245=

and T(x)

=

Ax.

(x, x))

or)

x) =

AX( x,

implies IAI = T(y) = I\"y and

IAI2, this = AX,

(x, x) .)

1.

the inner

compute

product (T(x),

T()'))

We have)

ways.

T is

The

(5.11) we get)

(T(x), T(y)) since

then

complex

an orthonormal

form

(a), let x

x in Equation (Ax,

Since

n, and

Uk.)

To prove

Proof

=

V

which

diagonal matrix

is the

basis

this

dim

of T

\302\267 \302\267 \302\267 , un

unitary .

We

also

= (x, y))

have)

T(y)) =:

(T(x),

=

(Ax,I\"Y)

Afi(x,

y))

= 1. Afi yare eigenvectors. Therefore Afi(x, y) = (x, }'), so (x, }') = 0 unless (a), so if we had Afi = 1 we would also have AX = Afi, X = fi, A = 1\", which 1 and Therefore contradicts the assumption that A and I\" are distinct. \302\245= (x, .r) = O. Afi Part (c) is proved by induction on n in much the same way that we proved Theorem 5.4, is in that part The only change the correspondingresult for Hermitian operators. required of the proof which shows that T maps S.l into itself, where) x and

since

But

AX

=

1 by

s1- =

Here

Ut

is an

eigenvector of

T with

{x I

x E

AI.

eigenvalue

u1 =

-1 At

(x,

V,

T(u

1)

=

u

t)

From

= O}.)

the equation

Al

T(u

1))))

T(u1) =

)\"lli

1 \\\\:e find)

140)

since Al'\0371 =

=

I Al12

Hence T(x) E

u l)

=

= (T(x), ):IT(u l)

note

S1. and

x in

any

= AI(X, u l ) = o.)

rest of the

The

itself.

spaces)

that)

T(u l )

AI(T(x),

if x E S1. , so T maps S1.into 5.4, so we shall not repeat the

S1.

of Theorem

that

choose

1 . Now

(T(x),

acting on Euclidean

of operators

Eigenvalues

proof is identicalwith

details.)

transformations next two theorems describe properties of unitary The dimensional space. We give only a brief outline of the proofs.)

a linear

dim V =

Assume

5.17.

THEOREM Then

T:

transformation

orthonormal

if E is

In particular,

T is

then

if and

is unitary

= (ei, ej))

(T(ei), T(ej)

( 5.18))

(el , . . . , en)

E =

let

and

n

---+ V

V

all i

if and

unitary

a fixed

a finite-

basis for

v.

if)

only

for

be

on

and j

ifT

only

.)

maps

E onto an

orthonormal

basis.)

Sketch of proof.)

x =

Write

\037

Xiei,

( \037lxiei'

and) T(y)) =

(x,y)

A =

(au) be

this basis.

the

T is

Then

jej

(T(x),

= n

V

matrix

representation

unitary

if and

n

=

n T(e;)).)

i\037

t:lXiYiT(ei)'

and let (el , . . . , en) be of a linear transformation if A is

only

e j ) is the

(ei,

unitary,

that

is,

an

orthonormal V

T:

if and only

---+

the

of

ij-entry

n

(e i

Since

A is

the matrix of

T we

(T( e i ), T( e j)

=

, e j)

have

identity

matrix,

compare

this

with

( k':?lakiek, (5.20)

\037

k=l

T(ei)

=

Qkiaki

and

\037 akiQki. k=l)

Lk=l akiek' n )

use Theorem

T(ej) =

\037\037=l arie

n

=

ar;e r I:l

to

Equation (5.19)

n =

n

n

N ow

=

V.

if)

implies)

( 5.20))

basis for V relative

= I.)

A* A Since

e j ),)

t:/iYiei'

T(y).)

( 5.19))

Sketchof proof.

i\037

)

( i\037lxiT(e;),t:/jT(ej) )

Assume dim

THEOREM 5.18. Let

with

n

=

n

n

Now compare

n

t:/

we have)

Then

\037 Yiej.

n

n

=

(x, y)

(T(x),

=

Y

n akiori

k\037l r\037l

5.17.)))

r , so)

e k , er)

= k\037lakiok;')

141)

Exercises) 5.19.

THEOREM

(a)

Every

is nonsingular

A

matrix

unitary and A-I

= is a

has

A

the following

matrix. Each of At, A, and A* unitary The eigenvalues of A are complex numbers of Idet AI = 1,. if A is real, then det A = ::l: 1 .)

(b) (c) (d)

The

of Theorem

proof

properties:

.

A*

5.19 is left as an

absolute

for

exercise

1.

value

the reader.)

5.20 Exercises)

1. (a) Let

= cx, where c is a fixed scalar. Prove given by T(x) = T is unitary if and only if lei 1. that on V are those des(b) If V is one-dimensional, prove that the only unitary transformations cribed in (a). In particular, if V is a real one-dimensionalspace,there are only two orthogonal = x and T(x) = -x. transformations, T(x) n x n matrix A. 2. Prove each of the following statements about a real orthogonal

(a) If

4.

(

the transformation

A

is a

A

is a

linear and norm-preserving, prove that T is unitary. = I. V is both unitary and Hermitian, prove that T2 bases for Let (e l , . . . , en) and (Ul, . . . , un) be two orthonormal that there is a unitary T which maps one of these transformation a real a such that matrix is unitary:) Find the following T is

6. If T:

8.

V be

- l)k .

5. If

7.

\037

real eigenvalue of A, then A = 1 or A = -1. (b) complex eigenvalue of A, then the complex conjugate \037is also an eigenvalue of A. In other words, the nonreal eigenvalues of A occur in conjugate pairs. (c) If n is odd, then A has at least one real eigenvalue. n. An orthogonal transformation T: V \037 V V be a real Eucljdean space of dimension Let 1 is an eigenvalue for T. This 1 is called a rotation. If n is odd, prove that with determinant shows that every rotation of an odd-dimensional space has a fixed axis. [Hint: Use Exercise 2.] k. Prove that det A = Given a real orthogonal matrix A with -1 as an eigenvalue of multiplicity If

3.

V

T:

V \037

1.

a ia

1(1

a 9. If A

(I

10. If

A

is a

skew-Hermitian

+

A)(I

is a

A)-l

unitary

matrix,

A

matrix and

13.

A

square

matrices

prove that

is Hermitian,

12. Prove

that

any

matrix are

- 1)

+ i)

la(l

- i)

1

la(2 -

-\"2)

prove

that both

1-

A

unitary

matrix

is called

normal

1 +

if

A is

Prove

the other.

i)

and

prove

nonsingular,

onto

bases

1 + A

is unitary.

Hermitian.

11. If

ia(2i

2:1

a Euclidean space V.

that

(I

are

and

nonsingular

- A)(I +

A)-l

that

is skew-

- il is is and that (A - iI)-l(A + unitary. nonsingular can be diagonalized by a unitary matrix. * = A *A. Determine which of the following types if AA

il)

A

normal.

Hermitian matrices. matrices. (b) Skew-Hermitian (c) Symmetric matrices. 14.If A is a normal matrix (AA* = A* A) and

matrices. (d) Skew-symmetric matrices. (e) Unitary (f) Orthogonal matrices.

(a)

if

U is

a

unitary

matrix,

prove

that U* A

U is normal.)))

of

6)

DIFFERENTIAL

LINEAR

EQUATIONS)

6.1 Historical introduction The

of differential equations began in the 17th century when Newton, Leibniz, and solved some simple differential of the first and second orderarising equations in geometry and mechanics. These early about 1690, discoveries, beginning

history

Bernoullis

the

from

problems

seemed to suggest that

calculus.Therefore, for solving

traction,

at

means,

elementary

by

and composition, of calculus.

division,

multiplication,

was aimed

early work

equations

of times to the

number

expressed in

of the

much

differential

of all differential equations of the familiar terms

solutions

the

could be

problems

physical

based on

and

geometric

elementary functions of ingenious techniques say, by addition, sub-

developing is to

that

applied

integration,

a

only

finite

functions

familiar

Special methods such as separation more or less haphazardly

devised

variables

of

end

the

before

and the

use of integrating

of the 17th century.

factors

were

During the 18th

and were developed, primarily by Euler, Lagrange, few differential equations could be solved relatively means. Little by little, mathematicians it was hopeless by elementary began to realize that to try to discover methods for solving all differential Instead, they found it more equations. or not fruitful to ask whether a given differential has any solution at all and, equation when it has, to try to deduce properties of the solution from the differential itself. equation Within this framework, mathematicians began to think of differential as new equations more

century,

procedures

systematic

Laplace. It soon becameapparent

that

sources of functions. An

in the theory

phase

important

general trend toward a the

obtained

more

developed early in approach

rigorous

first \"existence theorem\" of the form)

for

19th

the

century,

paralleling

the

calculus. In the 1820's, Cauchy He proved that every equations.

to the

differential

first-order equation

y' =f(x,y))

has a

the

whenever

solution

example is the

One important

right Ricatti

member, f(x,

y' = P(X)y2 where

of

the

142)))

P,

Q, and

Ricatti

R are

equation

functions.

given in

any

open

y),

certain

satisfies

general

conditions.

equation)

+ Q(x)y +

Cauchy's

interval

(-r,

work

R(x),) implies the existence

r) about

the

origin,

provided

of a solution P, Q,

and)

R have

cases

Experience of solutions

differential

linear

expansions

power-series

in some

that

solution

this

has shown

differential

A

differential

linear

of

equation

Q are given

and

P

theorem for

this

the

satisfies given

by

the

P and

Assume

6.1.

THEOREM

in J and let b

be

f(x) =

( 6.2)) \037'here

A(x)

Linear

=

S\037pet)

equations

dt

be-A(x)

on

are

the so-called

problems. linear of

Some

equations

the

principal

orders

form)

r

an existence-uniqueness

intert:al J.

only

condition

initial

e-A(x)

proved here.)

an open

is one and

the

+

we

restate

we

Then there

equation (6.1) and explicit formula)

of the

I

Volume

which

differential

and second

first

about

generality

= Q(x),)

Q are continuous

number.

real

any

In

8.3)

(Theorem

equation

is one

+ P(x)y

functions.

much

scientific

of

variety

of

equations

order

first

y'

(6.1))

where

The

coefficients.

types.

showed

(1809-1882)

these

Among

143)

means.

equations offirst orderand next section gives a review

I-linear

concerning linear

of results

Review

6.2

occur

in Volume

simple types of second order with constant results concerning these equations.)

for a few in a great

except

which

results of

to obtain

is difficult

equations,

discussed

r). In 1841 Joseph Liouville be obtained by elementary

in (-r, cannot

it

that

equations were

offirst and secondorders

concerning linear equations

of results

Review

Chooseany

one function

f(a)

= b.

point

a

y =

j'(x) u'hich This junction is

Q(t)l0(t) dt,)

.)

of second

order are those of the

Po(x)y\"

+ P1(x)y'

form)

+ P 2 (x)y

=

R(x).)

If the coefficients Po, PI' P2 and the right-hand member R are continuous on some interval if on and is never zero an theorem existence J, J, (discussed in Section 6.5) guarantees Po J. Nevertheless, there is no general formula that solutions always exist over the interval to (6.2) for expressing these solutions in terms of Po, PI' P 2 , and R. Thus, even analogous is far from complete, except in in this of the (6.1), theory relatively simple generalization if If R is and cases. the coefficients are constants zero, all the solutions can be special and in of functions determined terms explicitly polynomials, exponential trigonometric by I in was Volume the following theorem which (Theorem 8.7).) proved

THEOREM

(6.3)

6.2.

Consider

the differential y\" +

equation)

ay' +

by

=

0,)))

Linear

144)

- 00, (

interval

constants. Let d = a2

given real

b are

a and

where

+ (0) has

equations)

differential

-

y = e-aX/2[CIUl(X) +

(6.4))

where C1 and C2 are algebraic sign of d

(a) If d = (b) If d > (c) If d


=

f(x)

for

convergent

-

Ix

Xo)n,

of a homogeneouslinear

If the coefficients

r.




- xo)n-l

nan(x

ex:>

y\"

=

\037 n=2

n(n

must

P 2 in

satisfy

given

ex:>

y'

series for PI and

the coefficients an

-

l)an(x

-

=

\037 (n

+

l)a

n+

l (x

n=O)

the power

series for

- xo)n,

ex:> xo)n-2

=

\037 (n n=O)))

+

2)(n +

1)a n + 2(x - xo)n.

y

term

170)

The products Pl(x)y' and

Linear

differential

are

given by the

P2(X)Y

=

P I (x)y'

equations)

power seriest)

-

l)a HI b n - k )(x

+ n\037oct(k

xot)

and)

- xot.

Plx)y

=n\037ottakCn-k)(X series

these

When

(n +

are substituted

2)(n +

1)an + 2

kt[(k

they

differential

be satisfied if

will

equation

-

k +

find)

=

xo)n

O.

akcn-kJ}

0, M2

O.

>

recursion

formula

k

M




for

series 2

-

Ix

.

An

n+

t

Replacing

n

k

k=O)

-

- 1 in

n

by

that (n +

2)(n +

and

(6.34)

-

n+2

1)A

A

l)nA

+ 2)Mt ') 2)(n + l)t

1

n-t

(6.34)

l)n + (n

+

(n

--

[-1 times the resulting equation from = M(n + 2)An+l. Therefore) n+ l

subtracting

[-l(n +

(n + we

I)A kt

+

1 L\037(k

is dominated xo)n An for all n > 0, so the series 2 an(x by the - xoln converges if Now we use the ratio test to show that 2 An Ix

A n+2 and

n+l

t.


0 we have)

Pn(-X)

This

satisfies the Legendre

which

polynomial

only

of degree

nll

n

2 =

[Pn(x)]2 I -1

2

dx =

+

2n

. 1)

as a linear combination be expressed is a iff polynomial of degree n we

can

Po, PI' . . . , Pn. In fact,

of

the

Legendre

have)

n

= ICkPk(X),

f(x)

k=O)

where) Ck

=

2k

relation

the orthogonality

it

,

every

polynomial

g of

polynomial P n

Legendre

degree lessthan

has

dx

This

n.

real zeros

n distinct

\302\267

that)

follows

t1 g(x)P n(x) for

dx

f(X)Pk(X)

I -1)

2

From

I

+ 1

=

0)

property

and

that

can be they

used to

all lie in

prove

the

that

the open interval

(-1,1).)

6.21 1.

Exercises)

ex = 0 has the solution The Legendre equation (6.35) with polynomial in the series not a (6.41). Equation U2, polynomial, given by (a) Show that the sum of the seriesfor U2 is given by)

1 U2(X)

(b) ex

Verify =0.)))

directly that

the

function

=

1 +

x

-

x)

log 1 2:

U 2 in

for

part (a) is

Ul (x)

= 1 and

a solution

Ixl < 1 .)

a solution

of the

Legendre

equation

when

Linear

178)

2.

Show

that the

f defined

function

differential the

by

equation

3. The

Legendre

(a) If a, b, C

are

with a

constants

x)

-

(6.38)

-

b)y']'

and

=

cy

(x a Legendre two

-

2

x)y\" +

equation.

independent

power-series

on an interval

nonnegative

of the

form

0)

valid

a power-series

of

solution

the

valid on an two

-

l)y'

2xy' +

Hermite

=

2exy

one

=

2y

x =

At +

B,

0)

equation)

0)

solutions is a polynomial

of these

differential

when

ex

is a

interval

functions

of the

(3 +

r)y' + 3x

2

+ x 2y'

, r) .

B analytic

on

-

+

(exx

an interval

(x o

=

!

the C has

0)

!

-

r , Xo

-

xo)n,

B(x)

=

!

+

r) ,

for

all

x

r ,

Xo

\037 o.

-

bn(x

C(x) = A(x)B(x) is also analytic the power-series expansion)

!

n=O)

say)

xo)n.

n=O)

on

product

=

valid

= 0)

(x

o

n

00

C(x)

n anx

00 an(x

n=O)

It can be shown that exercise shows that

2

2)y

00

A(x)

=

y

of the

form ( -r

A and

equation)

solution of the form y = xdifferential equation)

for all x. Find a second a power-seriessolution

x 2y\"

7. Given

form

integer.

xy\" +

6. Find

-

, r). Show that

( -r

-

(2x

solutions of the y\"

Find

equation of

a differential

that

to a Legendreequation by a change of variable of the Determine A and B in terms of a and b. Use the method suggested in part (a) to transform the equation)

Find

linear

(6.39).

> o.

A

(b)

as a

function

= o.)

l)y

1 > 0, show

> band 4c + - a)(x

+

ex(ex

Express this

1.

be transformed

with

5.

-

- l)y']'

2

type)

can

to

1

(6.35) with ex = Legendre equation solutions and in Equations Ul U2 given in the can be form) written (6.35) equation

[(x

4.

log

the

[(x

the

2\"

x

1 + -

x

f(x) = 1 for Ixl < 1 satisfies combination of the

equations)

cn(x

-

xo)n,

where

Cn

=

!

akbn-k k=O)))

.

-

+

r).

This

179)

Exercises)

(a) Use gi ven

Leibniz's rule

the nth

for

derivative of a product

of Cis

derivative

nth

Now

use the fact

that

O)

A(k)(X

n

i (k) ) A(k) (x)B(n-k) (x) . k=O

=

c(n)(x)

(b)

that the

to show

by)

= k! ak

= (n

and B(n-k)(xo )

- k)!bn-k

to

obtain)

n C(\037) (x

c(n) (xo) =

Since

of the

14, Pn(x) denotesthe properties of the Legendre

proofs

8. (a) UseRodrigues'

to show

formula

1

(b)

If m

\037 n,

orthogonality

the equation

in

10.

(a)

= (x2

Let f(x)

-

Use

l)n.

formula

dx

2(2n)!

(b)

The substitution

the

relation)

x =

cost

that

J;

TT /2

t dt

=

1o and

Rodrigues'

formula

to

tl

the

(1

- m(m

on the

integral

-

x2)n

I [Pn(x)]

that)

+

l)]PnPm.)

(x) dx.)

left

is equal

S\037(1

- x2)n dx to

- 2) . . . 2 (2n + 1)(2n - 1) . . . 3 2n(2n

2

dx =

2

2n

of the

to)

dx.)

obtain)

I -1

n)

that)

(x)f(n-l)

f(nH)

=

O.)

to show

the integral

transforms

sin2n+1

-

=

show

ex

(with

to 1 to give an alternate proof

-I

parts

by

integration

to deduce

repeatedly

1)

Pn(x)Pm(x)dx =

t/(n)(x)f(n)(x)

Apply this

+

[n(n

from

(a)

tl

I)Qn(x),)

Pm to

and

Pn

by

x2)(PnP'm -P\037Pm)]' =

integrate relation)

-

of Legendre's equation

solution

satisfied

equations

of degree n. These exercises in Section 6.20.

= (-I)n.

= 1.

x

-

[(1

+ (x

+ l)n

(x

2n

is a polynomial. = 1 and that Pn( -1) Pn(l) that P n(x) is the only polynomial

having the value 1 when 9. (a) Use the differential

described

polynomials

Qn(x) Prove

polynomial

Legendre

(b) Provethat (c)

.

that)

=

Pn(x) where

.

akbn_k

\037 k=O)

proves the required formula for Cn

, this

8 through

In Exercises outline

n! Cn

= n!

o)

+)))

1

\302\267

. 1)

SO/2

sin2n+It

dt.

Use

Linear

180)

11. (a) Show

equations)

differential

that)

(2n) !

=

Pn(x)

xn +

2n (n!)2

Qn(x),)

is a polynomial of degreelessthan n. Qn(x) Express the polynomialf(x) = X4 as a linear combination of Po, PI' n can be expressed as a linear (c) Show that every polynomial! of degree P . . . \302\267 , , n PI' Po Legendre polynomials where

(b)

12.

If! is

(a)

a

of degree

polynomial

n,

P , P 3 , and 2

P4 . of the

combination

write)

n

!

=

f(x)

\302\267

CkPk(X)

k=O)

because of Exercise11(c).]Fora fixed m, 0 < m < n, multiply -1 to 1. Use Exercises from and integrate 9(b)and 10(b)

is possible

[This

this

by P m(x)

equation

sides

both

of

the

to deduce

relation)

13. UseExercises9 less

show

11 to

and

=

2

that

l

+ I

2m

Cm

dx.) f _l!(X)Pm(x) dx =

J:.Ig(x)Pn(x)

0 for every

g

polynomial

of degree

n.

than

Use Rolle's theorem to show that P n cannot have any multiple zeros in the open (-1 , 1). In other words, any zeros of P n which lie in (-1 , 1) must be simple zeros. (b) Assume Pn has m zeros in the interval (-1 , 1). If m = 0, let Qo(x) = 1. If m

14. (a)

Xl' X2, .

where

Qm(x) has

(c) Use part

. . , X m are

(b), This

tradiction.

of Pn

the m zeros

. . .

X2)

in (-1

(x

let)

- x m ),)

Show

, 1).

> 1,

at each

that,

point x

( -1

, 1),

to a

con-

in

sign as

same

the

- xl)(x -

= (x

Qm(X)

interval

along shows

Pn(x). with Exercise 13, to that Pn has n distinct

the inequality zeros, all of which that

show

real

m < lie in

n

leads

the

interval

open

(-1,1).

15. (a) Show

6.22 In

that

Evaluate

(b)

the value

the

The method

of

6.17 we

Section

of

the

J\037l

integral

J\037l P n(x)P\037+l(x) integral X Pn(x)Pn_l(x) dx.)

learned how to find

about a analytic near

an interval is not

example, supposewe

point Xo Xo,

try

2

x)'

( 6.44))

nearXo

=

o.

If we assumethat

differential equation

a solution

we are ledto the a n +l

y =

!

n2

P2

are

Xo

mayor

-

n

k akx

-

exists

+

1)

may

differential

and substitute

1)

an \302\267))) n

analytic.

If either not exist.

PI or For

equation)

0

formula)

recursion

=)

PI and

-y=)

-y

equation)

0)

solution of the

a power-series \",

=

valid near

solutions

power-series

to find

+ P 2 (x)y

coefficients

the

where

of the differential

solutions

power-series

+ PI(x)y'

y\"

in

of n.

independent

Frobenius

(6.43))

P2

dx is

this

series

in the

this

Although

test showsthat of

solution

us a

gives this

power

we

P2 are given

PI and

x =

for

about

Xo

(6.44)

Equation

put

satisfies (6.44), the ratio formally o. Thus, there is no power-series = o. This example does not violate in the form (6.43) we find that the

which

akx

only

valid in any open interval

(6.44)

coefficients

k \037

series converges

6.13 because when

Theorem

=

power seriesy

181)

of Frobenius)

method

The

by)

- -1

PI(X) =

and)

1

-

=

P2(X)

2.

x)

X2)

do not have power-seriesexpansionsabout the coefficient of y\" in (6.44) has the value 0 when

functions

These

is that

the x

origin. = 0;

The difficulty here in other words, the

point at x = o. of a complexvariable of is needed functions to appreciate the knowledge theory difficulties in the investigation of differential equations near a singular encountered point. some important special casesof equations with can be treated by However, singular points For example, suppose the differential equation in (6.43) methods. is equivalent elementary to an equation of the form) differential

has

equation

where

-

(x

( 6.45))

this

a singular

of the

A

P and

case

of (6.45)

Q have

we say (x

by

-

Xo is a

the

XO)2

x

\037

Xo.

If P(xo)

efficient of y' or the so Theorem 6.13

\037

0

regular

not

will

P(x)

x

Q(xo) of y

coefficient

0,)

in some

Xo

+

r).

both

In sides

becomes)

+

or

=

Q(x)y

- r, (xo open interval of the If we divide singular point equation.

equation

y\"

for

- xo)P(x)y'+

(x

expansions

power-series

that

+

XO)2y\"

-

(x

Xo

\037

0,

not

wiJI

Q(x)

y' +

be applicable.

-

2

y

=

0

xo))

if Q(x o ) = 0 and Q' (xo) \037 0, either have a power-series expansion about the

the co-

or

In 1873 the

German mathematician

point

Georg

Xo , Fro-

for treating such equations. We shall method (1849-1917) developed a useful describe the theorem of Frobenius but we shall not present its proof. t In the next section we give the details of the proof for an important special case, the Bessel equation. Frobenius' theorem splits into two parts, on the nature of the roots of the depending

benius

quadratic

equation)

t(t

( 6.46))

- 1) +

P(xo)t

+

Q(x o)

= o.)

This quadratic

is called the indicial equation of the differential equation given equation The in coefficients and are the constant terms the ex(6.45). P(xo ) Q(x o) power-series of and Let and roots of P denote the the indicial Theseroots (Xl (X2 Q. pansions equation. be real or complex, equal or distinct. The type of solution obtained may by the Frobenius

method dependson whether t For a proof see E. Hille, Introduction

to

Ordinary

or

Analysis,

Differential

not

these

roots

Vol. II, Blaisdell Equations,

differ

by

Publishing 1961.)))

Prentice-Hall,

an integer.)

Co.,

1966, or

E. A.

Coddington,

An

Linear

182)

FIRST CASE OF

6.14.

THEOREM

indicial

and assume

equation

(6.45) has

two

-

(Xl

U2

00

(6.47))

=

uI(x)

-

Ix

xolCll.2 an(x

of

-

and

(Xl

Then

an integer. the form)

not

is

CX2

UI and

solutions

independent

Let

THEOREM.

FROBENIUS'

that

equations)

differential

(X2 the

roots

the

be

differential

xo)n,

with

ao =

1,)

- xo)n,

with

bo =

1

of

the

equation

n=O)

and) 00

-

= Ix

u 2(x)

(6.48))

bn(x

xolCl2.2

.)

n=O)

series

Both

in

converge

o < Ix -

r


0, so that

valid

keep

y'

=

tx

t-

l

x

with

t =

Ixl

the

xt

00

.2 anx

n=O

.

possible exception of Differentiation

00 n

+

xt

.2

n=O

nanx

n- l

of x

= O.

(6.50) gives us)

00

= xt-l

cylindrical

researches

of

= 1 , and Q(x) = x2 - cx 2 , analytic on the entire real line,

0, P(x) Q are

vibrations in

Bessel functions also arise is named after the German

earlier

it appeared

(1732)and Euler (1764). = Bessel has the form (6.45) with Xo equation is a P point Xo regular singular point. Since and to find

Bessel equation)

This equation is usedin problems concerning and propagation of electric currents cylinders,

in

Daniel Bernoulli so the

solve the

to

Frobenius

constant.

nonnegative

of membranes, heat flow conductors. Some of its in

C mayor lnay and the solutions

r.)


0, the constant - xol < r, Ix

.2 (n

n=O)))

+

t)anx

n

.

Bessel

The

183)

equation)

Similarly, we obtain) =

y\"

x

00

t- 2

+

! (n

- 1)an x n .

t

+

t)(n

n=O)

If

= x 2y\" + xy' +

L(y)

=

L(y)

xt

-

(x 2

find)

we

(X2)y,

00

!

t)(n +

+

(n

-

t

n

l)anx

x

+

n=O

x

!

nx

a

power of x will

t

first the choice become)

coefficients

[(1 +

(6.52))

Since

2.

n >

be written

-

(X)2

(X

t

=

(X2]a

1

a

a5 = a2 =

=

(X2

-

=

nx

n

a n x n +2 .

t

+ x !

n=O)

of each Since we seek a

the coefficient

that

so

=

(X2)ao

o.

o.)

(X

the

are

these

of

first

t the

only possible

values of

remaining equations for

[(n +

and)

0)

-

(X)2

(X2]a

that a 1 =

implies

a n -2

=-

n

a7

that

t

can

the

determining

a n -2 =

n +

0)

o. The secondformula

+

o.

-aO

2(2 +

(X)2

-ao

=

2

2 (1

2(X)

n(n +

(X2)

+

a 4

,)

can

4(4 +

-

( -1)

we have)

subscripts

2ao

(-1)

-

-a2

, 2(X))

even

with

--

(X)

-a4 6(6 + 2(X))

a n -2

=_

For the coefficients

\302\267 \302\267 \302\267 =

=

a 6 -and,

-

this

For

(X.

(n

as =

(X2]a

as)

(6.53)) so

-

t)2

type.

0, the

>

n=O

Its roots (X and

desired

+

that)

t2

solution of the

00

x t ! [(n

determine the an constant term we need (t2 -

this requires

indicial equation.

n =

nx

try to

and

))

give us a Consider

for

the

For

vanish.

a o =/:. 0,

with

This is the

n

t)anx 00

- x ! (X2a n=O

n +2

0, cancelxt ,

L(y) =

we put

(6.51

+

(n

00

00

t

n=O

solution

!

n=O)

+

Now

00

t

4

(1 +

2 2!

2(X)

, (X)(2

+

(X)

3ao ,)

6 2 3!

(1 +

(X)(2

+

(X)(3

+

(X)

in general,)

( -1)n

=

a2n

2

Therefore

the

t =

choice

(X

gives

2n

n! (1

us the

+

(X)(2

+

ao

\302\267 \302\267 \302\267 \302\267

(n

(X)

+

(X))

solution)

00 y

The

ratio

test shows

=

aox\302\2531

+

(

that

the

power

( \037 2

2n

n! (1

+

1X)(2

series appearing

1 ) nx2n +

\302\267 \302\267 \302\267 IX)

(n

+

in this formula

. IX)) )

convergesfor

all

real

x.)))

Linear

184)

In t

=

by (-x)t. then obtain

we

rx

assumed that x > o. If x < 0 we can repeat the discussion with We again find that t must satisfy the equation t 2 - (X2 = o. Taking the same solution, exceptthat the outside factor xex is replaced by

we

discussion

this

x t replaced

Therefore the function/ ex given

(-x)ex.

equations)

differential

by the

equation)

00

(6.54))

is a I\037(O)

2n

2

\037

I:

[(1 -

-

(X)2

-

t =

root

=

l

(X2]a

r/..

and)

0)

(1

-

l

2(X)a

=

and)

0)

an

ex)) )

For those values We obtain, in

equation.

-

(X)2

Since this

2.

n >

are led to

the

(6.55))

recursion

of

(X

for

which

-

(X2]a

an-

n +

place

2

=

of (6.52),

the

0,)

= ao

1 +

lxi-a

( for

all real

solution

an-

2

=

O.)

and)

a n -2

is the

-

2r/..))

same as

(6.53),with

(X

replaced

by

-(X,

we

L

'>

n=l

x

(_1)n

\037

2..nn!

(1

-

(X)(2

-

2n

-

\302\267 \302\267 \302\267

(n

(X)

(X)) )

x=/:.O . obtained under for I-ex is meaningful

the hypothesis

was

I-ex

the

n +

solution)

f-a(x)

The

2(X)a

= 0

al

us

=-

formula

-

n(n

n(n

However,

+

(n

ex)

x=/:.o. O.

[en

these equations give

not an integer

2(X is

valid

\302\267 \302\267 \302\267

+

become)

which

for

2n

x =

for

indicial

1)n x

ex)(2

all real

for

of the

+

n! (1

(0) exist the solution is also valid

Now consider the equations)

If

(

-

(

1 +

of the Bessel equation valid

solution and

a o Ixl a

=

fa(x)

series

integer. It can

even

if

is

2(X

be verified

that

2rx

a positive

a positive integer. integer, so long as (X is

is not

such (X. that I-ex satisfies the Bessel equation for and series solution have the Therefore, lex, given by Equation (6.54); if (X is not a nonnegative integer we have found another solution/_ ex given by Equation (6.55). The two solutionslex and I-ex are independent, since one of them \037oo as x \037 0, and the other does not. Next we shall simplify the form of the solutions. To do this we need some to recall these properties.) and we digress briefly function, properties of Euler's gamma

not

a positive

for

each

For each real s

(X

>0

>

we

0 we

define

res)

by the

improper

integral)

OO

res)

=

i 0+))) tS-le-t

dt .

Bessel

The

This integral converges if functional

parts

by

Integration

leads

to the

equation)

res + 1) = s res).)

(6.56)) This

< o.

diverges if s

0 and

s >

185)

equation)

that)

implies

I)r(s + I) = 3) = (s + 2)r(s+ = (s

2) = (s +

res +

res +

(s

2)

I)s res),

+

(s + I)s r(s),)

+ 2)

in general,)

and,

res +

(6.57))

for every positive integer

=

n)

Since

n.

(s +

=

reI)

-

n

1)

S\037

\302\267 \302\267 \302\267

(s

e- t dt

I)s res))

+

we put s

= I, when

= 1 in

(6.57)

we

find)

the

Thus,

of the factorial

function is an extension

gamma

= n! .)

+ I)

r(n

integers

to positive

of res)

to negative

from

function

real numbers.

The functional values of

can be

(6.56)

equation

are not

s that

integers.

used to extend the

write

We

+ 1).

r(s r(s) =

(6.58))

definition

in the form)

(6.56)

s)

The

is meaningful if s + 1 > 0 and s=/:.o. we can use this Therefore, -1 < s < o. The right-hand member of (6.58) is now meanings \037 0, and we can use this equation to define for - 2 < -1, res) in this manner, we can extend the definition of res) by induction to -n < s < -n + 1, where of the form n is a positive integer. The in and its extension are now valid for all real s for which (6.56) (6.57)

member

right-hand

equation to define res) if s + 2 > 0, s =/:. ful

s
0,

+

X

(6.59))

rJ..)

the solution

J,ix)=

( \"2)

\302\253

\037

in

( -I)n

\037n!r(n+1+Ot)\"2

series for fa. in Equation express this product

This gives

r( n + 1

Equation for x >

can

We

(6.57).

r(1 Therefore,

The

6quation. + rJ..).

\302\267 \302\267 \302\267 rJ..)

+

+

.

rJ..))

(6.54)

0 can x

rJ..)

in

us)

and denote the resulting

be written

2n \302\267)))

( )

as)

Linear

186)

The

the first

J

is given

p

J

function

of

rx

by this

defined

n=O

have

of the Besselequation The graphs of the

a solution constructed.

is also

This

been

called the Besselfunction = P , the Bessel function

r/..

2n+ p

-x

=

(p

+ p)! ( 2) )

n! (n

0 is

>

r/..

integer, say

nonnegative

( _l)n

\037 P J\037)=\037

is a

r/..

> 0 and

for x

equation

kind of order rJ... When by the power series)

equations)

differential

2, . . .).)

o. Extensive tables of Besselfunctions J o and J 1 are shown in Figure

x


+ I 0,

r/..

a new rJ..)

=/:. 1,

Graphs of the J-

function

. . . , we

1

-

rJ..

in

that the series

we see

and

-

2 Ot/r(I

is not

equation for

x

>

0

>

r/..)

=

2)

(I

for J_rx(x)

\037 n=O

-

rJ..)

is the

if

o.

stant

rJ..

is a

multiples

nonnegative valid

r/..

in

(6.59),

Equation

if

r/..

integer. Therefore, if

a positive

is such x

>

0

x 2n

n!

-

+ 1

r(n

r/..)

(2 )

.)

(2

-

as

same r/..

not

is

-

\302\267 \302\267 \302\267

(n

r/..)

that

for

rJ..)

f-rx(x)

a positive J -rx(x)

and

the

r(1

-

rJ..))

in Equation

integer, J- Ot is

(6.55) a solution

with

of

are linearly independent on the of the Bessel solution general

is)

y

If

-

(_l)n

Therefore, r/..), the Bessel equation for x > o. J rx(x) and If r/.. is not an integer, the two solutions is not axis their ratio real constant) (since positive x

by

we obtain)

(6.57)

r(n + 1 -

ao =

r/..

-x -a\037 (

s =

r/..

replacing

J 1.)

define)

J -a (\037=

Taking

by

that is, if

is meaningful;

2, 3,

rx

J o and

functions

Bessel

for

=

c 1J Ot(x)

+

c 2J_ rx(x).)

J p and its coninteger, say r/.. = P , we have found only the solution x > o. Another solution, independent of this one, can be found)))

Bessel

The

+ PlY' + y\" of Ul is given

P 2y

=

0 that

the

by

4 of Section6.16.This states that if U l is a solution of never vanishes on an interval I, a second solution U 2 independent

in Exercise

described

method

the

by

integral)

x =

u 2 (x)

where

=

Q(x)

and a

187)

equation)

second solution

U 2 is

f

C

Q(t)

we have PI(x) =

equation given by the formula) x

(6.60))

=

u 2 (x)

J p(x)

2 dt,

[ul(t)])

Bessel

the

For

e-JP1(X)dX.

ul(x)

Ijx, so

=

Q(x)

Ijx

1 dt ,

2

Jc

t[J

pet)])

I in which J p does not vanish. and x lie in an interval This second solution can be put in other forms. For example, from

if c

may

Equation (6.59)we

write)

1

1 gv(t),) 2p t

=

[J v(t)]2

where gp(O)

-:;f

In the

O.

interval I the

g p has

function

a power-series

expansion)

00

be determined we assume the existence of could

which

by

=

t[J p (t)]2 from this formula term by term the power t- l) plus a seriesof the

Integrating

takes the

in the

the

expansion,

1

(from

Antn

coefficients

equating

such an

! n=O)

=

gp(t)

2p

in (6.60)

integrand

=

[J p(t)]2

takes the

t 2p

.

If

log

x

form)

00

-1 t

g pet)

identity

\037

+l

Ln. A

t

n

n=O)

c to

a x we obtain n 2P ! Bnx .

logarithmic Therefore

x-

form

term

A 2p

(6.60)

Equation

form) 00

u 2(x)

= A 2p J p (x) log

X

+

Jp(x)x-2p!Bnxn. n=O)

be shown

can

It

solution

the

that

is denoted

by

coefficient

Kp(x)

and

A 2p -:;f o. has the form)

If we multiply

u 2 (x)

by IjA

2p

the

resulting

00

Kp(x)

=

Jp(x) log x

n + x- P ! Cnx

.

n=O)

This

is the

Having

form of the arrived

at this

solution promisedby the formula, we can verify

by substituting the right-hand member in C n so as to satisfy the equation. efficients

the

second that

Bessel

case of

Frobenius' theorem.

of this form actually and determining equation

a solution

The details of this

calculation

are

exists

the co-

lengthy

and)))

Linear

188)

will

Ki

omitted.

x)

= Jix)logx 0 and

ho =

where

for all real

of the

_l

2 2) -VI

(

h

n

=

hn +

\037

(

2 2 ) V!(_1)n

n ! (n

n=O

hn+v

\037

2n,

+ p)!

( 2) )

11n for n > 1. The series on the right converges for x > 0 by this formula is called the Besselfunction is not a constant multiple of J p , the general solution

x > 0 is)

case for =

+ c2 K

c1Jp(x)

Besselfunctions

of the

properties

!

_

( 2)

+

Y

Further

2n \037

\302\267 \302\267 \302\267

p. Since Kp this

in

1)!

n !

Kp defined

of order

kind

of the Besselequation

-

n

n=O

!+

1 +

-

(p

\037

x. The function

second

can be expressedas)

The final result

be

equations)

differential

p (x).)

set of exercises.)

in the next

discussed

are

6.24 Exercises

1.

(a)

Show

Let f that

be

of the

solution

any

g satisfies the

Bessel equation

differential

(b)

When

=

4(X2

differential

1 the

(c) Deduce Use the

the

series

Use this

2

=

J\037(x)

2.

x.

1 +

sin

in part for

representation

1

-

4(X2

y

4x2

(

> O.

=

O.)

)

equation in (a) becomes y\" + information and the equationt

and)

x)

(b)

from

directly

Bessel

J_Y2,(x)

functions

=

show

= 0;

TTX

its

general

= v;

r(l)

to

solution show

that,

1A

cos

(

the series

to

Y

2

\037

( TTX )

formulas

= xY1(x) for x

let g(x)

and

(X

equation)

y\" +

is y = A cos x + B sin for x > 0,)

of order

x.)

)

for J \037-2(x) and

J_\0372(X).

that)

d

(a)

dx

(xaJa(x\302\273)

=

d

(b) dx (x-aJa(x))

3.

xaJa_1(x),)

= -x-aJa+1(x).)

= Let xaJa(x) and Ga(x) = x-aJa(x)for x > O. Note Fa(x) zero of Fa and is also a zero of Ga. Use Rolle's theorem and That is, there is a zero of tive of J a and J a + 1 interlace. zeros zerosof J a + 1 , and a zero of Ja + 1 between each pair of positive

t The change

of variable

t =

u

2

gives

r(!) = (See

Exercise

16 of

Section

11.28

each

that

J a between zeros of

us oo

f 0+

t-\037fe-l

for a proof

dt = 2

f 0) that

2

S\037

-

oo

e-

e- u2 du

u2

du =

positive

Exercise2 to

V 7T .

= V;.))))

prove

zero of J a is a that the posi-

each pair of positive J a . (See Figure 6.2.))

Exercises) 4.

From the relations

(a)

the recurrence relations)

2 deduce

Exercise

in

189)

ex

ex

-

+

x) Ja.(x)

relations

Use the

(b)

Ja.-1 (x)

5.

in

+

1(b) and

Use Exercise

=

J\037(x)

a

- Ja.(x) x)

2

6. Prove

to

\037how

=

(x)

Ja.+1

2J\037(x).)

that)

-cosx

)

.)

( \037

formula for J_%,(x). Note: Ja.(x)

a similar

)

is an

elementary

odd integer.

half an

Ja.+1 (x).

Sin x

\037

TTX

(

-

Ja.-1 (x)

formula

J%(x) = Find

and)

recurrence

suitable

=

J\037(x)

formulas)

the

2ex

=

(x)

Ja.+1

-

Ja.(x)

x)

deduce

(a) to

part

-

and)

Ja.-1 (x)

function for every

ex

which

is

that)

Id - -2

(x)) =

2 -ex Ja.(x) -

( xJ a.(X)Ja.+1(X)) =

x(Ja.(x) -

2

dx

2

+ Ja.+1

(Ja.(x)

ex+1

x

x)

2 Ja.+1

(x)

and)

d dx

7.

Use the

(a)

6 to

in Exercise

identities

show

2

that)

00 J\037(x)

From

(b)

part

+

2

! n=l)

00

=

J\037(x)

8. Letg\037(x)

=

b

) for

x >

0,

terms of

Besselfunctions

(a) y\" (b) y\"

+

xy

+

x2y

10.Generalize

Exercise

Then find

the

for

x >

O.

(a)

xy\"

+

xy\"

+

(b)

Bessel

= o. = O. general

for

+ (a

2b2 x 2b

the Bessel

of

solution

general

x >

+

!

-

!

V2

for

2 ex b

2

n =

1, 2,

3, . . . , and

all

g a. satisfies

= 0)

)y

equation of order cx. of each of the following

(c) y\" + (d) x2y\"

differential

equations

in

= O.

xmy + (x 4

+ l)y = O. and g a are related by the equation ga(x) = xCfa(ax b ) for x > O. in terms of Besselfunctions solution of each of the following equations

8 when fa

+ Y = o. = O. 6y' + xy function identity exists

(c)

6y'

a and

= !x.

o.

+

xy\"

(d) x2y\" of the J 2 (x)

where

n + 1(x)

nonzero constants. Show that

b are

a and

where

x 2y\"

A

IJn(x)1


9.

2 Ja.+1(X\302\273.)

c are constants. Determine

-

6y'

xy' + (x

form)

-

Jo(x) =

a and

c.)))

+ x 4y = O.

aJ: (x),)

+ l)y = O.

Linear

190)

12. Find

-

13.

a

of the differential + 00. Show that for x > 0 it can be linear second-order differential equation series solution

power < x


should

infinite series defining k tkA, t A2, . . . , tkAn-l. in A of the form)

the

in

n is

equations)

tkI,

n-I (7.27))

=

etA

!

etA,

term

each

tkAk/k! expect that

we can

Hence

with

etA

k ,

qk(t)A

k=O)

coefficients qk(t)

scalar

the

where

expressing

a polynomial

as

etA

depend on t.

Putzer

useful

methods

for

simpler

of the

two

two

developed

The next theorem describesthe

in A.

methods.) AI, .

Let

7.9.

THEOREM

sequenceof polynomials (7.28))

. . , An

in A

Pk(A) =

=1,)

Po(A)

the

be

of an

eigenvalues

X n

n

A, and define

matrix

a

as follows:) k

-

TT (A m=l)

AmI),

k =

for

...,n

1, 2,

.)

we have)

Then

n-l (7.29))

=

etA

!

rk+l(t)Pk(A),

k=O)

coefficients r l (t),

scalar

where

the

linear

differential r\037(t)

(7.30)) r\037+l(t)

Note: (7.27),

. . . , rnet)

are

determined

from

recursively

the

system

rl(O) = 1,)

=

Alrl(t),)

=

Ak+lrk+l(t)

rk+l(O)=

+ rk(t),)

(k

0,)

=

1, 2,

. ..,n

-

1).)

in powers of A as indicated does not express etA directly in (7.29) linear combination of the polynomials Po(A), PI (A), . . . , Pn-1 (A). These are easily calculated once the eigenvAlues of A are determined. Also the

Equation

but

polynomials

as a

this requires 'l(t), . . . , 'n(t) in (7.30) are easily calculated. Although multipliers differential this particular system has a triangular a system of linear equations, in succession.) and the solutions can be determined

Let rl(t),

Proof

matrix

function

of

equations)

F

. . . , rnet)

by the

be

the

functions

scalar

determined

by

(7.30)

solving matrix

and

define

a

that

F

equation)

F(t) =

(7.31))

n-l

!

rk+l(t)Pk(A).

k=O)

that F(O) = rl(O)Po(A) = I. satisfies the same differential equation

Note

Differentiating

(7.31)

and

using

We

will

=

!

k=O

F(t) =

etA

by

namely, F'(t) = AF(t). the recursion formulas (7.30) we obtain) n-l

n-l

F'(t)

that

prove

as etA,

r\037+I(t)Pk(A)

=

!

{rk(t) k=O)))

+

Ak+Irk+l(t)}Pk(A),

showing

method

Putzer's

be o.

defined to

ro{t) is

where

We

calculating

for

in the form)

this

rewrite

n-l

n-2

F'{t) :z::\037 rk+l{t)Pk+l{A)

+

\037 k=O)

k=O

AnF(t) =

subtract

then

(7.32))

Therefore

F'{t)

The

-

=

AnF{t)

shows

which that

-

'

we find F (t) F{t) = etA.) 1.

EXAMPLE

its eigenvalues Solution.

An)Pk{A)

-

(A

F'(t) from

-

equal to

=

first-order

A2

(7.33))

Po{A)

= I

-

=

(A

-

+

Ak+lI)Pk{A)

-

(Ak+l

An)Pk{A)

AnI)Pk{A).)

rk+l{t)Pk{A)

-

(A

=

(A

-

=

(A

-

=

AnI)F(t)

=

F{O)

-

AnI)F{t)

of I

last

-

AF{t)

I, the

-

AnI){F{t)

0, so the

=

P n{A)

that

=

Arl{t),

=

Ar 2{t)

equations

and Pl{A) = etA =

eAt]

are to

A, we

=

in

rl{t) = Since

(A

\302\267

An)Pk{A)}

so)

Ak+lI)Pk(A),

a linear combination

r\037{t)

these

=

-

r n{t)P n-l{A)} r n{t)P n{A).) becomes)

equation

AnF{t),)

uniqueness theorem (Theorem 7.7)

and A if

is a

A

2 x

2 matrix

with

both

A.)

r\037(t)

Solving

-

= AF{t).Since

Al =

Writing

\037 k=O)

implies

AnF{t)

etA as

Express

(A

n-2 AnI)

theorem

Cayley-Hamilton

=

+ (Ak+l

Pk+l{A)

rk+l{t){

becomes)

(7.32)

Equation

\037 k=O)

Pk+l{A)

+ (Ak+l

Pk+l(A)

the relation)

n-2 =

AnF(t)

see that

(7.28) we

from

But

Ak+lrk+l{t)Pk{A),

to obtain

Anrk+l{t)Pk(A) \037\037:\037

F'(t) -

207)

etA)

A

+

-

the

solve

=

rl(O)

+

r 2 (0)

rl{t),)

=

r 2{t)

=

-

AI)

O.)

teAt.)

AI, the required formula teAt{A

1,

we find)

succession

eAt,)

of differential

system

=

eAt(l

-

At)I

for etA +

is)

teAtA.)))

equations)

208)

of differential

Systems

case the system of differential

In this

Solution.

=

r\037(t)

are given

/lr

=

r 2 (0)

rl(t),)

A =;C

/l.)

is)

rl(O)

2(t) +

/l, where

A and

are

equations

Arl(t),

=

r\037(t)

Its solutions

of A

1 if the eigenvalues

Solve Example

2.

EXAMPLE

equations)

1,

=0.)

by)

_-

r l ( t)

eAt ,)

r 2 (t)

eAt _

=

ellt

.

A-/l)

Since Po(A)

(7.34

= I and etA =

))

eAt] +

-

=

A

eAt

- eIlt

PI(A)

the

AI

'] Ilt Ae

AI) =

_

(A

the

I+

eAt -

Then

A.

-

/l =

2if3

so

oc

=

/l

ifJ,)

-

oc

A.

ellt

multiplying

will

be

also

I and

A

in

fJ\037O.)

ifJ,)

(7.34) becomes)

Equation

etA =

+

Ilt

eAt and

complex numbers. But if (7.34) will be real. For example,suppose) A =

e

A-/l)

are complex numbers, the exponentials A and /l are complex conjugates, the scalars

A, /l

eigenvalues

- /le At

is)

etA

A-/l

A-/l

If

for

formula

required

e(Hi(J)tI +

e(a+ip)t

e (a-ip)t

-

[A

+

(ex

if3)I]

2\037f3

.

= ellt

(A

+

{

= ellt

e- ipt

_

eipt

e'(JtI

i sin f3t)I

f3t +

(cos

i cancel

involving

etA =

(7.35))

-

exI

if3I) }

+

{

The terms

-

2if3 (A

-

exI

-

if3I)}.)

Si\037f3t

and we

-eat

get)

cos

{({3

f3t

-

oc

sin

f3t)1 +

sin

{3t

A}

.

{3)

7.14

Alternate

methods

for calculating

etA

in

special

cases

completely general becauseit simplest method to use in certain special cases. In this section we give simpler methods for computing etA in all the ei gen val ues 0f A are equal , (b) when all the eigenval ues three special cases: ( a) When of A are distinct, and (c) when A has two distinct eigenvalues, exactly one of which has Putzer's method

is valid

multiplicity

for expressingetA

for all square

1.)))

matrices A.

A

a polynomial method general as

in A is

is not always the

methods

Alternate

If A

7.10.

THEOREM

is an

X n

n

calculating

for matrix

equal to

its eigenvalues

all

with

209)

cases)

in special

etA

A,

}ve have)

then

n-l (7.36))

\037

=

etA

e).t2

Since the

Proof.

matrices

and

AtI

-

teA

- A/)k.

(A

k!)

k=O

00

k

(e).t/)2 .!..-(A k!)

=

etA = eWetCA-H)

we have)

commute

AI)

A/)k.

k=O

theorem impliesthat

The Cayley-Hamilton

-

(A

AI)k

0 for

=

every k >

n, so the

theorem

is proved.)

If A

7.11.

THEOREM

we

is an

X n

n

matrix

n distinct

with

eigenvalues

AI'

A2

, . . . , An'

then

have)

n

etA =

where

is a

Lk(A)

polynomial

A

in

11 j=1

The

a matrix

k =

for

the formula)

1, 2,

. ..,n

.)

coefficients.)

interpolation

the equation)

F by

function

by

A,,j./)

Lk(A) are calledLagrange

polynomials

We define

Proof.

Ail -

Ak

j*k Note:

-

- 1 given

n

of degree A

LiA} =

\037 et).kLk(A), k=1)

n

(7.37)) and

=

F(t)

that

verify

equation F'(t) =

F satisfies the differential

F(O) = I. From (7.37)we

see

\037 et).kLk(A) k=1)

and

AF(t)

the

initial

condition

that) n

- F'(t) = \037 et).k(A - AkI)Lk(A).

AF(t)

k=1)

the

By

theorem

Cayley-Hamilton

the differential equation To complete the proof which

AkI)Lk(A) =

-

we have (A

= AF(t). F'(t) we need to show

satisfies

F

that

0

for

the initial

each

k,

so F

satisfies

condition F(O) =

I,

becomes)

n (7.38))

\037Lk(A)

= J

.

k=1)

A

of (7.38) is

proof

The next which

has

theorem

multiplicity

outlined treats 1.)))

in

Exercise

16 of

the case when

A

Section 7.15.) has

two

distinct

eigenvalues,

exactly one of

210)

of differential

Systems

has

A

be an

I.Jet A n multiplicity

7.12.

THEOREM

where

n-2

etA = eA.t

k

AI )

ellt

k

+

proof of Theorem

As in the

00

=

etA

eAt

2:

=

eA.t

2:

\037k! k=O

+

AI)k

series over r

t

- III =

find)

we

(A

-

AI)n-l(A

The left memberis 0

the

by

relation

-

-

AI)n

-

(Il

2:

r=O (n

- 1

+ r)!

(Il

-

A)r(A

The explicitformula explicit

easy

3

-

A)(A

-

-

A)r(A

AI)n-l.)

1

AI)n-l =

(Il -

n-l

-tk

2: k=n-l k !

A)

A)k(A -

-

n-2

1

1

-

(Il

_

et(Il-A.)

At- {

6

AI)n-l

k

\037 \037

x

7.12 can

Theorem

in

are more

also be deducedby

(Il

-

k!)

(A

A )k

_ AI)n-l .

}

1. If

formulas

in

3 case often

a

3

x

applying

Putzer's

method,

complicated. Theorems

arises

7.11 and 7.12 cover all in this the formulas

7.10, in

practice,

reference.)

CASE

AI)n-l.)

the proof.)

completes

but the details The

Cayley-Hamilton theorem.

r becomes)

(ft

Since the

AI)n-l+r.

find)

= This

k AI)

AI)n-l.)

00

-

-

(A k!)

so)

-

00

tn-l+r

2:

A)l

- (Il -

A)(A

- AI)n-l+r = (Il

(A

- (Il -

AI

theorem

=

AI)n

we

repeatedly

Therefore the series over

(A

_

(A

by using the

-

Cayley-Hamilton

(A

U sing this

- Ill) =

A

n-l. AI)

n - 1+r

form

Since)

A

-

(A

\037

e).t

+

AI)k

L(n-l+r)! r=O)

closed

in

k A)

}

k=n-l

\037

eA.t

Il,

by writing)

-

(A

k!

00

-

(A

-

k=O

\037

eAt

k=O

n-2 k \037 -t

we evaluate the

Now

(A

k!

k=O

AI)k =

-

and

kOOk

n-2

k

-.!....

tk

At- 12: ,(ft k.)

(_ Il

A

eigenvalues

we have) n-2

eA.t

we begin

7.10

distinct

two

Then

1.

_

_ At- 1

{ ( Il

k=O

Proof.

3) with

- 1and Il has multiplicity

A 2: k! ( -t

(n >

matrix

X n

n

equations)

3 matrix

etA =

A has eA.t{I

+

eigenvaluesA, teA

-

AI)

+

A,

2 !t (A

A, then)

-

AI)2}.)))

matrices

of order

n < 3.

case are listed below for

211)

Exercises)

2. If

CASE

a3

- /l1)(A

;'t (A

etA =e

-

(A

3.

CASE

EXAMPLE.

/l)(A

-

vI)

+

eAt{I

has

A

+

AI)}

(/l

=

A

etA =

(7.39)

(7.40)

=

etA

stage we or (7.40) to write this

At

et{1 +

powers of

By collecting

=

2t e )1

+

-

o)

1

t

2e 2t

-2(t

+ l)e t

+

-2(t

+ 2)et

+ 4e

2t)

of the

For each

A

=

-

2t

(e

- 1)2-

et)(A

A)

3 gives -

tet(A

us)

1)2.)

as follows,)

this

2)e t

and

2e 2t }A - {(t

+

2)e

(3t

+

5)e

(3t

+

8)e

-

+ l)et

the indicated

perform

(3t

1 through

in Exercises

matrices

A

e2t }A

2

.)

operations

in

-

2e 2t

-(t

+ l)e

t

-

4e 2t

-(t

+ 2)et

+

2e 2t .

t

-

8e 2t)

-(t

+ 4)et

+

4e 2t)

t

(7.39)

e 2t

t

6, expressetA

=

+

0

0

0

1 .

A is

Find

a corresponding

A.)

in

polynomial

2

0

0 1 3 .

0 0 1 -1

3

5.

A

=

1

0

2

1 -1

-11 -6 matrix

as a

1 1 0 0

1

A 3 x 3

=

-\037J.

[\037

etA = (b)

3. A

=

0 -6 (a)

2. A

-1 -2J.

[:

7.

- AI)2.

(A

-

of Case

formula

1

4.

;'t

Exercises

7.15

1.

te

.

then)

A. =;c p\"

-

Ill) Il))

.)

2, so the

write

2t + e

with

/l,

-

4)

-5

1, 1,

-

evt (A - AI)(A (v A)(v

+

/l

0)

1)2 or A2 x 3 matrix,)

as a 3

result

the

-

v, then)

/l,

- U)2

(A

o

l)} + also

;'t \0372 Ii)

1

+ {(3t +

can calculate (A

-2te etA

-

teA

we can

A

t

(-2te

A are

of

-

o)

2

Solution. The eigenvalues

A, A,

eigenvalues e Jlt

-

t(A

eigenvaluesA,

- Al)(A - vI) - A)(/l - v)

(A

Jlt

(/l

etA when

Compute

+e

v)

x 3 matrix

If a 3

etA =

A has distinct

3 matrix

x

known to have

!e At { (A 2t 2 formula

-

all

2At +

if A

is a

its

6.

1 .

2 equal to

eigenvalues

2)1 + ( -

2At

4 x 4 matrix

2

+

2t)A

with

0

0

-2

1

0

0

3 0

=

A

0 0 0

4

Prove

that)

A.

+ t 2A2}.)

all its

eigenvalues equal to

)..)))

212

15, solve the

8 through

Exercises

of

each

In

equations

of differential

Systems

=

y'

system

A

to the

Y subject

initial

given

condition.

-5

8. A =

-:].

[: 10. A

=

12.

A =

0

0

0

1

14. A

16.

=

be

o

-6

0

0

3 0

0

0

0

=

Y(O)

A

n

=

Lk(A)

A1

, . . . , An are

(a) Prove

n distinct

-

A

TT j=1

8

-6

=

Y(O)

0

0

0 1 0 =

Y(O)

2 1)

of Theorem

7.11. Let Lk(A)

,) Aj

scalars.

that)

G) Let Yl,

. . . , Yn

be

n arbitrary

if

Ai \037 A k ,)

if

Ai =

=

Lk(Ai)

(b)

2 -3 1

.

A.3

-

Ak

j*k) where

in the proof the by equation)

- 1 defined

n

C2

C3

0 0 0 4 0 0 1 0

used

(7.38)

Equation

of degree

2

=

Y(O)

2

1

in

C1

1

-2 0 0 2 0 1 0 0

=

15. A

1

outlines a proofof

1

-1

0 0

4

the polynomial

13. A =

.

0

1 1 0 0 0 2 1 0

This exercise

=

0

0 1 0

-2

1 Y(O)

0

0

2

1

-11

11. A =

.

\302\267

C] 2

-1

2

0

-6

=

Y(O)

=

\037].

1

1

-1

1

Y(Q)

-15

[

1

0

2

A =

9.

\302\267

[::J

-1

3

=

Y(Q)

scalars, and

Ak.)

let)

n

peA)

=

!

YkLk(A).

k=1)

Prove

that p( A)

is the

only polynomial peAk)

(c)

Prove

that

!\037=l

Lk(A)

= 1 for

=

of degree < n

Yk)

every

A,

for

k =

and

deduce

n

!

k=l)

where

I is

the

identity

matrix.)))

Lk(A)

=

I,

-

1 which

1, 2,

satisfies

the

n equations)

. . . , n.)

that for every

square matrix

A we have)

.

linear

Nonhomogeneous

Nonhomogeneous linear systems with

7.16

J. Here

as an n

(regarded

obtain

an

n

is an

A

1 column

X

formula

explicit

First we multiply both differential equation in

problem)

yea) =

yet) + Q(t),)

B,)

Q is an n-dimensional vector function on J, and a is a given in J. We can point by the same process used to problem

matrix,

matrix) continuous for the solution of this of (7.41)

members the

-

e-tA{Y'(t)

of (7.42)is the derivative integrate both membersof (7.42)from a to e- xA e xA

obtain

we

e- iA

matrix

exponential

=

AY(t)}

The left member

by

the

by

and

the

rewrite

form)

(7.42))

Multiplying

coefficients

constant

n constant

X

213)

coefficients)

case.

scalar

the

treat

= A

Y'(t)

(7.41))

an interval

constant

\037t'ith

next the nonhomogeneous initial-value

We consider

on

systems

product e- yet). where x E J, we obtain)

the explicit

L1

of the x,

- e-aAyea)

Y(x)

e-iAQ(t).)

=

t

dt

e-tAQ(t)

if we

Therefore,

.)

formula (7.43) which

in the

appears

following

theorem.) THEOREM

function

=

Y'(t) a unique

has

solution on J

(7.43))

As

in the

homogeneous

that the

A

yea)

Y(t),

= B.

term,

EXAMPLE.

Solve

the

Q be an

vector

n-dimensional

problen1)

yea) =

Q(t),)

B ,)

explicit formula)

e(x-a)AB +

exA

in

difficulty

e(x-a)AB,

The second term

Theorem

illustrate

+

initial-value

r

e-tAQ(t) dt

.)

this formula

applying

in

practice

lies in

the

Y'(t)

=

matrices.

Y'(t)

We

=

yet)

the

case, the

exponential first

A

by

given

Y(x)

calculation of the Note

matrix and let

7.13. Let A be an n X n constant continuous on an interval J. Then the

=

is the

solution of the

is the

solution

A yet)

7.13 with an initial-value

Y'(t) =

+ Q(t),)

homogeneous

yea) = o.)

example.)

problem) A

yet)

+

Q(t),)

problem

of the nonhomogeneous

Y(O) =

B,)))

problem)

214)

(- 00 , +

interval

the

on

Solution.

2

1 = exA

of

Section

are

A

0)

7.13, the solution is given dt =

e-tAQ(t)

fox

4. To

t

by

dt

eeX-tJAQ(t)

e

calculate

xA we

.)

of Case 3,

use the formula

to obtain)

7.14,

exA

- 21)}+ I(e 4x - e2X )(A - 21)2 - !xe2X (A - 21)2 - 21) + !(e2X - 2x - 1)(A - 21)2} .

e 2X { 1 + x(A

=

= e2X { / can

B= 0 .

2t

te

2, and

2,

0

0

3

to Theorem

Y(x)

eigenvalues

Q(t) =

-1

3

According

(7.44))

The

e 2t

1

-1

0

A=

equations)

(0), where

2

We

of differential

Systems

+

x by x

replace

x(A

-

this

in

t

formula

to obtain

-

l[e

the

Therefore

e(x-t)A.

in

integrand

(7.44) is 2 e(x-OAQ(t)= e (x-t){

I +

-

(x

t)(A

21) +

1 = e

2X

0

- 21)e

+ (A

we

Y(x)

IX

21)

2 2X

e

\037(A

e2x

2

2x

te-

!

!e2X _

(A

A

2/ =

-1

_ te2X

1

0

1 -1

2

1

(A

- 2/)2

=

=

e 2x

+ e2x)

0

ie

2x

-

= e 2x -8 3e 2x + Je

of this

-6 x

tx2)

rows

3)

2X

-

2

1

+

X)

3

Be 3

lx

! + !x 3

3x

+

+ 3)

3

-\"8-

t]

- ie2X

+

2x

-

Q 8e

3 2 4\"X

+

J

- Ix + !X2

matrix are the required functions

2

i + Q 8

-

\302\267)

Yl , Y2 , Ya .)))

.

0

2)

-

!x + !X2+ Q

4x

!X2

2

3 3 2 - :Ix -:Ix

!X2)

\"8

4\"

e 2x

2x

X

0 -2 ,

2 lx

-

X

0

-2

1

x)

-

! _ !x _

2 and

1 2\"

find)

Y(x)

t)

-

0 [

0

-

2X

21)2e

we have

-

.

- 2t(x -

2t

6X3 ] [ \037\0372

4

The

e

_ 21)e2x

+ (A

\037

+

we

0

[

t) ]

[ ix ]

Since

- 21)2}Q(t) 2(x - t) - 1

find

e(x-tJAQ(t) dt =

=

1](A

e2Xe-2t

-

+

-

[t(x

-

t)

t

0

[t ] Integrating,

-

X 2X

-

- 2(x

2 (x-t)

-

Q

4x

2

1

6 X)

3

lx

3)

3 - .l 6)x

_

3)

lx

]

215)

Exercises)

Exercises)

7.17

1. Let Z be

of the

a solution

nonhomogeneous system) ==

Z'(t)

J

on an interval

homogeneous

value

initial

with

J with

value

initial

yea)

are often

methods

Special

one

is only

there

that

non-

the

of

solution

system)

and that ==

yet)

the given

Prove

Z(a).

y' (t) on

1r (2(t),)

\037Z(t)

== \037 yet)

1r e(t-a)A{

Z(t)

available for ==

e7.tC, Q(t) == tmC, and (2(t) If the particular solution Z(t) Z(t) as indicated in Exercise

1

- Z(a)}.) solution Z(t) which such methods for (2(t) == C and D are constant where

a particular 7 indicate

rxt)C 1r

(sin rxt)D, obtained does not have the required to obtain another solution yet)

(cos

so

formula)

yea)

determining

Exercises2, 3, 5, and

function Q(t).

(2(t))

by the

is given

it

1r

initial

resembles C,

we modify

value,

the required

with

=

(2(t) vectors.

initial

value.

n x

2. (a) Let \037 be a constant solution of the system)

n

== \037 yet)

y'(t)

on ( -

1r (0)

00,

is given

by

the

If

\037 is

show that when) explicitly

nonsingular,

(c) Compute

Y(x)

3.

Let

x n scalar. given

\037 be

be a

an n

(a) Prove that

uA

dU)C.)

(a) has

{e(x-a)A

- I}A-l.

O.)

eJ

Gl let Band

value

the

a=

B=

=

\037l

matrix,

(f:-a e

in part

integral

C

constant

the

yea) ==B,)

1r C,)

eCx-a)AB+

the

A = [=\037

vectors. Provethat

formula)

Y(x) =

(b)

C constant n-dimensional

Band

matrix,

C be n-dimensional

constant

vectors,

and let

ex

== of the form \037Z(t) 1r erxtC has a solution system Z' (t) nonhomogeneous == C. and \037)B Z(t) only if, (rxI that the vector B can always be chosen so that the (b) If rx is not an eigenvalue of \037, prove == e7.tB. form in (a) has a solution of the Z(t) system 2t Y'(t) == \037yet) 1r e C (c) If rx is not an eigenvalue of \037, prove that every solution of the system tA == == B e where has the form yet) ( Y(O) B) + e(J.tB, (rxI \037)-lC. ==

4. Use the y' (t)

the

e(J.tB if,

method

== \037 yet)

suggested

2t 1r e C, with)

A

by Exercise

=

3 to

C [\037

\037l)

find

of the

a solution

=

Y(O)

[

=:l)

nonhomogeneous system

=

.))) [\037]

216)

5.

of differential

Systems Let A be an n x n constant matrix, m be a positive integer. (a) Prove that the nonhomogeneous of the form) solution

Y(t) = Bo, B 1 ,

where

. . . , Bm

are

constant

vectors,

(b) If A

the coefficients

in (a)

is nonsingular,

has a

6. Considerthe

of

solution

and

if

+ tmC,

Y(t)

. . . + tm Bm

B2 +

-, m.)

Bo, B1, . . . , Bm

=

Y(O)

B, has

,)

Am+lB.

vector

a solution. B can always

so that

be chosen

=

3Yl +

the

system

=

2Yl + 2Y2

t3

+

Y2

+

t

3

.)

2 of the form Y(t) = Bo + tB l + t B2 + t3B3. a particular solution = = . 1 with the of (b) Y2(O) Yl(O) system constant let B, C, D be n-dimensional Let A be an n x n constant matrix, Prove that the ex be a number. nonzero real nonhomogeneous system) given

a solution

y' (t)

has a

= A Y(t)

E and

where

F are constant

E and

the form.

vector

initial

Find

F in

terms

of A,

B can always

a particular

+

solution

2 ex [)B

of

the

Y\037

a solution

of the

system

+

=

-

(sin

Y(O) =

and let

B,)

C

(A

+

=

Yl + Yl

-

with Yl (0)

3Y2

exD).)

Note

system system)

nonhomogeneous =

ext)F,)

if)

B, C for such a solution. be chosen so that the

Y\037

Find

(sin ext)D,)

(cosext)E

if and only

vectors,

(A2

Determine

+

vectors,

of the form)

solution

particular

+ (cosext)C

Y(t) =

(b)

particular

Find

Find

8. (a)

a

let

system)

nonhomogeneous

Y\037

(a)

and

vectors,

if)

only

such

for

prove that the initial the specified form.

y\037

7.

2

t

A

constant

1

C =

Determine

+

tB l

+

=

y' (t)

system

Bo

C be n-dimensional

Band

let

equations)

+

4 sin 2t)

Y2.)

=

Y2(O)

=

1 .)))

that

has a

if A2 solution

+

2 ex [

of

is nonsingular, the specified

The

of Exercises9 through

In each

subject to

the

given

[ -2

-3 -I}

-5 -3 -I}

0 0 -1 -1 , 0 0 -1

=

The

7.18

general

[ -3e

t

=

Q(t)

X n

+

--4-42

=

Y(O)

[ 221 1007]

2

2t

6

Y(O) =

,

.

-2 1)

Y'(t) =

P(t)Y(t)

formula

for the

= AY(t)

+ Q(t),)

+

and Q(t),

matrix

.

707

12 27]

Q(t)

solution of the

yea) =

yet) are n

X

linear

system)

B,)

matrices.

1 column

We

turn

case) Y'(t)

(7.45))

,

t

explicit

a constant n more general

A is to the

+ Q(t)

.

-

t

Y'(t)

now

tet

linear system

Theorem 7.13 gives an

where

A Y(t)

[::]

-1

-1 12. A

=

=

=

Y(O}

2t [e }

Q(t) = 2

[

system y' (t)

[\037l

=

Q(t}

=

A

217)

Q(t))

e

1

[

Y(O}

[ -2e t}

:}

-5

10. A =

11.

0

Q(t} =

=

A

+

the nonhomogeneous

solve

12,

= pet)yet)

system Y'(t)

condition.)

initial

4

9.

linear

general

= pet)

yet) +

yea) =

Q(t),)

B,)

necessarily constant. theorem J, a general open interval existence-uniqueness which we shall a in that a in J later sectiontells us for each and each initial vector B prove there is exactly one solution to the initial-value In this section we use this problem (7.45). result to obtain a formula for the solution, generalizingTheorem 7.13. In the scalar case (n = 1) the differential equation (7.45)can be solved as follows. We let A(x) = S\037pet) dt, then multiply both members of (7.45)by e-A(t) to rewrite the differential equation in the form) where

the

If P and

n

X

pet) is not on an continuous

n matrix

Q are

e-A(t) {Y'(t)

(7.46))

-

pet)

Y(t)}

=

e-A(t)Q(t).)

Now the left member is the derivative of the product e-A(t) yet). Therefore, from a to x, where both members a and x are points in J, to obtain) e-AC'dy(X) Multiplying

(7.47))))

by eA(x)

we obtain the Y(x)

=

-

e-ACa>Y(a)

=

t

e-A(t)Q(t)dt.)

explicit formula

eAC\"')e-ACa>Y(a) +

eA a in




a in

and

for

all x

Y m + 1(x)

in

-
k-III.)

that)

- l)

Mrx k


1

which is the same

(x)

I T(qJk)




J and

in

-

ICPk+I(X)

for

-

19'>k+1(X)

))

which

for T.

constant

a contraction

is

rx

inequality)

=

II qJoll

1 we

have)

M,)

for k

holds

(7.75)

=

k

+ 1 if

it holds

for k

we note

that

IqJk+1(X)

Since

-

qJk(X)

is valid for

this

I

=

each x

I T(qJk)

in

J we

(x)

If we

(7.75) by induction. qJ(x) denote its sum we

proves let

-

I

T(qJk-1)(X)

must also

II9'>k+1

This

-

rx

IlqJk

-

qJk-111

qJ(x)

= lim

Therefore the series in

(7.73)

converges

have)

n\03700

qJn(x) =

CfJo(x)

+

Mrx k .)

Mrx k .)

00

(7.76))




I.

in Volume

extensively 1 and

1.

=

1 the

a scalar

field.

m

When

or, more briefly,

function is called a When m > 1 it is

a vector field. variable, or simply continuity, and derivative to 10 and 11 extend the concept of the integral.)

a vector-valued function This chapter extends the

called

vector

>

vector variable

of a

function

real-valued

studied n

be

both used or f(XI'

of

a vector

concepts

of

limit,

type, and

denoted at

by light-faced a point x = (Xl'

to denote the . , xn ) for

..

...,

value

off

x

n)

vectors

bold-faced

by

in Rn, the

type.

notations f(x) and

at that

particular point. value at x. We shall

the function

scalar and

Iff

a vector

is

the inner

use

product) n

= X \302\267 Y

!

XkYk

k=l)

and

(YI,

the

corresponding

\302\267 . , \302\267

Yn).

Points

in

norm Ilxll R2 are usually

=

(x

denoted

\302\267

where

x)\037,

by (x, y)

(x, y, z) instead of (Xl' X 2 , x 3). on subsets Scalar and vector fields of R2 defined cations of mathematics to scienceand engineering.

x = (Xl' . . . , xn ) and instead of (Xl' X 2 ); points

y = in R3

by

atmospherewe assigna real numberf(x)which

and

R3 occur

frequently

For example,if

represents

at

the temperature

each

in

the

point

at x, the

appli-

x of the function)

243)))

244)

defined is a

f so that

scalar field.

we

If

an example of problems dealing with

we obtain

point,

of scalar

calculus

Differential

a vector

assign

a vector

and vector which

fields)

velocity at

wind

the

represents

field.

or vector fields it is important scalar either to know how the field changes as we move from one point to another. In the one-dimensionalcase is the mathematical the tool used to study such derivative Derivative changes. theory in one-dimensional case deals with functions on open intervals. To extend the the defined to Rn we consider generalizationsof open called open sets.) intervals theory In physical

8.2 Open balls and open in Let a be a given point

Rn and let r

be a given

The set

number.

positive

of all points x

that)

such

Rn

in

sets

-

Ilx




The exterior Ilxll = I

with

of S

is the

set

of all

x

with

.)

Exercises)

8.3

field defined on a set S and let c be a given real number. The set of all points a level set off that f(x) = c is called (Geometric and physical problems dealing scalar fields, with in this chapter.) level sets will be discussed later For each of the following S is the whole space RH. Make a sketch to describe the level sets corresponding to the given values of c. c = 0, 1,4, 9 . (a) .(x, y) = x 2 + y2, - -2 e-1 1 XY = e C - e , , , , e, e2 , e3 . (b) .(x, y)

1. Let f

x

in

be a

scalar

S such

(c) f(x, (d)

.(x,

(e)

.(x,

(f) .(x,

y) = cos (x + y), = x + y + z, y, z) = x2 + 2y2 + 3Z2, z) y, 2 y, z) = sin (x + y2 + z2),)

c = -1, 0,t, !J2,

c = c =

1

.

-1, 0, 1 . 0, 6, 12.

c= -

1,-!,

O,!

,/2,

1 .)))

246)

each of

2. In

the

cases, let a sketch

following

Make

inequalities.

given

whether or not S is open. Indicate 2 (a) x + y2 < 1 . 2 (b) 3x + 2y2 < 6. (c) Ixl < 1 and Iyl < 1 . x > 0 and y > 0 . (d) (e) Ixl :::; 1 and Iyl < 1 . xy

(g)

each

inequalities

(a)

Ixl

(c)

x + y


o.

boundary

(h)

(i)

S and

(j) x > y. > y.

x

(k)

2 y > x

1, and

1 , Iy I
o. the given of the fo1lowing, let S be the set of a1l points (x, y, z) in 3-space satisfying and determine whether or not S is open. - x2 - 1 > O. y2

(f) x > 0 and

3. In

and vector

of scalar

calculus

Differential

real

A,

B is

and

line

set A

the

that

show

- {x}, obtained

a closed subinterval

by

that

A, show

of

open.t

(c) If

B are open intervals closed interval on

A and

(d) If A

is a

real line) is open.

5. Prove (a)

the

(b) Rn

set 0 is open.

(d) The (e) Give

of open sets in

example

Closed sets. A exercises discuss

of

the

Rn is

let S

cases,

following

a sketch

showing

open

and

the

be the set S

closed, or

+ y2

:::;

2. :::;2.

x 2 + y2 :::; 2. (f) (a) If A is a closed set in 1


< 4. < 4.

:::; y < y

x2 .

(k)

2 Y > x

and

Ixl

< 2.

(I)

2 Y > x

and

Ixl

:::;

2.

A u {x} is also closed. prove is a closedset. that A u B and A \" B are closed.) real line, show

x is a

interval [a, b] on

B are sets, the difference of A which are not in B.)))

and

co1lection

Rn

its complement

(g) 1 :::;x (h) 1 :::;x

+ y2 :::; 1 . 1 < x 2 + y2 < 1 x 2

(b) Prove

are open.

open setsis open.

(c) x2

7.

\" B

Rn:

called closedif of closed sets.) properties

S is open, closed,both (a) x 2 + y2 > o. (b) x 2 + y2 < O.

(e)

A

complement (relative

collection of open setsis open. to show that the intersection of an infinite

Sin

set

Make

conditions.

(d)

and

open.

necessarily

In each

B

A u

that

show

of a finite

intersection an

co1lection of

of any

union

line,

line, show that its

real

is open.

(c) The

6.

properties

following

The empty

the real

on the

the

point

not

in A,

that

real line

- B (calledthe

complement

of B

relative

to A)

is the set of

all

and

Limits

8. Provethe following (a) The empty set (b) Rn is closed. (c) The

union

(e) Give

an

necessarily

of a

may use

You

Rn.

results

the

of Exercise

5.

of closed setsis closed.

of any collection number of

closed sets is closed. the union of an infinite

finite

to show

example closed.

S be a subset

9. Let

sets in

is closed.

0

intersection

The

(d)

of closed

properties

247)

continuity)

that

of closed

collection

sets is

not

of Rn.

sets. (a) Prove that both int S and ext S are open of disjoint sets, and use this to deduce (b) Prove that Rn = (int S) u (ext S) u as, a union as is always a closed set. that the boundary 10.Given a set Sin Rn and a point x with the property that every ball B(x) contains both interior of S. Is the converse x is a boundary point to S. Prove that exterior points of S and points of S necessarily have this property? true? statement That is, does every boundary point 11. Let S be a subset Prove that ext S = int(Rn of Rn. S). 12. Prove that a set S in Rn is closed if and only if S = (int S) u as.)

Limits and

8.4 The

fields. extended to scalar and vector are easily of limit and continuity the definitions for vector fields; they apply also to scalar fields. If a ERn and bERm S is a subset of Rn. a function I: S \037 Rm, where

concepts

shall formulate consider

We

continuity

We

we

te)

wri

= b

Jim I(x)

(8.1))

\037

(or,/(x)

x

b as

\037

a))

x-+a)

to mean

that)

111(x)-

Jim

(8.2))

=

bll

O.

Ilx-all-+O)

limit

The

symbol

not required

it is

If we write

h =

in equation (8.2) is the usual limit that I be defined at the point a x

- a,

Jim Ilh

For

points

in R2 we

write

(x,

y) for

11-0)

11/(a +

x and

h)

-

(a, b) for

=

bll

a and

Jim

in R3 we

x

put

=

(x,y,

I(x,

y) =

b.

a = (a, b, c)

z) and lin1

(X,1I,Z) -+

A

function

I is

said to be continuous

say lis

continuous

on

a set

write)

= b.

I(x, y, z) at

S if lis

and

(a ,b,c))

a if

lim I(x) x-+a)

We

definition

the limit relation

express

(x,y)-+(a,b))

points

this

o.

follows:)

For

In

becomes)

(8.2)

Equation

calculus.

of elementary

itself.

I is defined = I(a)

continuous at

at

a and

if)

\302\267

each

point

of

S.)))

(8.1) as

248)

is not

are

definitions

these

Since

case, it

that

and vector

familiar

many

fields)

in the one-dimensional of limits and continuity can and limits of concerning continuity fields. For vector fields, quotients are of

extensions

straightforward

to learn

surprising

of scalar

calculus

Differential

those

properties

For example, the usual theorems and sums, quotients also hold for scalar products, but we have the following theorem concerning sums,multiplication not defined inner products, and norms.) be extended.

also

THEOREM

lim

(a)

x\037a

x\037a

lim

(c)

x\037a

Iim

(d)

lim If x\037a

[f(x)

+ g(x)]

Af(x) =

Jim

(b)

8.1.

x\037a)

\302\267

II f(x)

II

lor

= b

g( x)

=

Ilbll

lim x\037a

=

g(x)

scalar

every

e, then we

also have:

A.

\302\267

e .

\302\267

(c) and (d);

only parts

prove

band

scalars,

= b + e.

Ab

f( x)

We

Proof.

f(x) =

by

of

proofs

(a)

and

(b) are left

as exercisesfor

reader.

the

(c) we write)

To prove

g(x)

- b.

Now we use the

triangle

f(x).

o
o.

its value

compute

a 2f

ayax)

=1

of x and y. of t, say

a function x,

g(u,

v).

and

Y.

Compute

the

282) 7. The

(a)

assume equality II Vf(x,

-

x =

bles

varia

og

a (

8.

F and G

functions

Two

fields)

uv, y = \037(U2 v ) transfornls [(x, y) to g(u, l') . of f derivatives and ogj ou, ogj ov, 0'1;/( ou ov) in terms of partial of nlixed partials.) 2 = 2 for all x and y, determine constants a and b such that) y)11

change of Calculate

(b) If

scalar and vector

calculus of

Differential

2

2

2

- b

ou )

og ov

(

of one variable

(You may

u 2 + v2

=

z of

a function

and

.)

)

related

are

variables

two

by

the

equation)

[F(x) + whenever

A field

bounded and continuous on R as follows:)

g is defined

= g(u, v) (a)

=

A(y)

(b)

the open

on

u in

(y))

a rectangle

dx

]

R =

y) is

D 2 ,lZ(X,

derivative

never zero.

encountered.)

[a, b]

X

A new

[c, d].

scalar

dy.)

A defined on [c, d] by the [a, b] the function [c, d]. Use this fact to prove that og/ov exists (a, b) X (c, d) (the interior of R).

on

rectangle S

=

that)

Assume

dy =

]:[J:f(x,y)dx] for

2F' (x)G'

all derivatives

f(x, y)

ilJ:

f\037 [(x,

and is continuous

of on

that for each fixed y) dx is continuous

be shown

can

It

equation

=

mixed partial

+ G(y) rf o. Show that the the existence and continuity

F(x)

may assume field f is scalar

(You

9.

G(y)]2eZ(x,y)

all (u, v)

dx)

differentiable on S and that the mixed partial derivatives and are equal to feu, v) at each point of s. u and v are expressed parametrically as follows: u = A \302\253() , B(t)]. g[A(t), in terms of [, A', and B'. in terms of t when [(x, y) = eX + Y and A(t) = B(t) = (2. (Assume R lies

that g is

Prove

R.

in

J:[tf(x,y)dy]

D 1,2g(U, v) and D2 ,lg(U, v) exist 10. Refer to Exercise 9. Suppose v = B(t); and let gJ(t) =

(a) Determine gJ' (t)

(b) Compute in the

12. Let

(a)

quadrant.)

z) = (r

11. Iff(x,y, show

gJ'(t)

first

that

r = xi A .

(b) B.

+ yj + zk

and

1

.

,

=

-

( ) A.

V

(

.

A)

where r = xi + yj + zk and A) + A x (r X B). = IIrli. If A and B are constant

(r X B),

z) = B x (r

Vf(x,y,

-

V

X

let,

A and

show

vectors,

vectors,

that:)

- . A

r

,3)

1 V ( )

,

= )

3A

.

r B .

,5

r

A .B

--.

,3)

the set of all points for which the two spheres (a, b, c) in 3-space 2 (z - C)2 = 1 and x + y2 + Z2 = 1 intersect (Their orthogonally. perpendicular at each point of intersection.) 14. A cylinder whose to the surface equation is y = [(x) is tangent common to two the Find surfaces. points [(x).))) 13.

B are constant

X

Find

(x

- a)2 +

Z2

+

2xz

+

-

(y

b)2

should

planes

tangent

Y

=

+

be

0 at all

9)

9.1

differential

Partial

This chapter illustrates their use in and extremum functions, implicit

applications. tial

equations, remarks

tary An tial

concerning involving Two simple

equation equation.

order

(9.1

equations

of differential calculus developedin

theorems

The

CALCULUS)

OF DIFFERENTIAL

APPLICATIONS

some

examples

problems.

partial differential equations. a scalar fieldf and its partial derivatives

examples

in

a function

f is

which

8 have

Chapter

We

related begin

a wide variety of to partial differenwith some elemen-

is called a partial differenof two variables are the first-

equation) y) =

af(x, ))

second-order

the

and

0

ax equation)

a 2f(x, ox2

(9.2)) Each

of these

form

L(f)

is a

= 0,

')

homogeneous linear

where L is a

y)

+

2 a f(x,

ol differential

partial differential

linear

y) -_ 0 \302\267)

operator

equation. involving

That is, one or

each has the more partial

called the two-dimensionalLaplaceequation. of linear ordinary can be extended to partial differential equations For it is for differential to that each of Equations (9.1) equations. example, easy verify and the set of solutions is a linear is an important difference there However, (9.2) space. between and linear that should differential be realized at the equations ordinary partial outset. We illustrate this difference the differential equation (9.1) by comparing partial derivatives.

Equation of the theory

Some

the

with

ordinary

(9.2) is

differential

equation)

f'(x) =

(9.3)) The

most

In other

function

general

function

words, the satisfying

(9.1)

satisfying

(9.3)

O.)

is f(x)

= C,

solution-spaceof (9.3)is one-dimensional.

where C is an But

constant.

arbitrary

the

most

general

is)

f(x,y)

=

g(y),) 283)))

284)

of differential calculus)

Applications

where g is any

we can easily obtain an infinite set of of y. Since g is arbitrary CY = take e and c all real we can let over For g(y) vary independent example, of (9.1) is infinite-dimensional. numbers. Thus, the solution-space in general. I n some respects this example is typical of what happens Somewhere in the an is required to of a differential first-order equation, integration process solving partial the an function is in each this introduced At remove derivative. arbitrary step partial function

solutions.

solution space. This results in an infinite-dimensional In many problems involving it is necessary to select from the differential partial equations of solutions a particular solution As or conditions. wealth one more satisfying auxiliary or on existence the of has effect the be nature these conditions a profound might expected, will in not this of solutions. A of such be attempted systematic study problems uniqueness solution.

book. Instead, we

will

9.2

partial

first-order

A

treat

some special

differential

the first-order partial

Consider

3

(9.4))

the

All

of

solutions

left member

this

equation

differential

as a dot product,and

+

(3;

This tells us that point (x, y). But these level curves of f are the lines)

must

be

f(x, y) is constant when

2x

-

- 3Y

some

function

does,

indeed,

form)

= o.)

curves

words,

This suggests

-

= g(2x

2j at each

Hence curves

of j:

the level

that)

3y))

g. for

that,

each differentiable

function

g,

the

satisfy (9.4). Using the chain rule to compute

scalar the

fieldf defined

partial

find)

-

af,

ax

=

(2x

2g

3 -of + 2 -af,= ax

Therefore,

3; +

vector level

c.)

is constant.

3y

f(x,y)

Now we verify

=

We express the

considerations.

in the

Vf(x, y) is orthogonal to the Vf(x, y) is orthogonal to the straight lines parallel to 3; + 2j. In other

(9.5))

for

o.

geometric

by

equation

Vf(x,y)

2j).

=

oy)

the

2x

Therefore

2 af(x, y)

vector gradient also know that

the we

in Chapter8.)

introduced

ideas

coefficients

constant

with

be found write

the

equation)

af(x, y) + dx can

equation

cases to illustrate

f satisfies

(9.4).)))

ay)

- 3y),

6g

(2x

af, - = -3g

(2x

-

ay

- 3y)

' - 6g(2x

3y )

3y), =

o.

derivatives

by

(9.5)

off we

A

show

we can

Conversely,

the form

have

that

To do this,

x =

(9.6))

This transforms

a

y) into

j'(x,

Au +

the constants

choose

shall

=

)'

u and

of

v,

C, D so that

A, B,

r)

ah(u,

must

necessarily

change of variables,)

Dv.)

Cu +

say)

Cu + Dr).)

+ Br,

r) =/{Au

h(u,

We

Bv,)

function

a linear

introduce

we

285)

coefficients

satisfies (9.4)

which

differentiable!

every

(9.5) for some g.

constant

M'ith

equation

diJjerential

partial

first-order

the simpler

h satisfies

eq uation)

= o.

(9.7)) au)

we shall

Then

Usingthe

solve

this

find)

oh

of ox

=

satisfies

x =

(9.8))

For this

choice

of A

and C, the

of

ax

of

A +

= aj

au

ax

+

_

A

\037

\037C.

h satisfies

function

A =

Taking

r = 2u

Br,)

becomes)

ah/au

)

2)

=

for

equation

.

C

(

c.

ay

- (3/2)(aj/ax),sothe

ah

3u

=

ayau

al/ay =

have

oy

satisfy (9.7) if we choose A

h will

Therefore,

we

(9.4)

of

+

all

ax

au

Since I

form.

the required

show thatf'has

and

equation

we

rule

chain

3 and

C = 2 we

find)

+ Dr.) so h(u, r) is a

(9.7),

of

function

r alone,

say)

h(u,

for some

2x

function

- 3)' =

D =

g.

-

(2B

1. For

this

r in terms Now we choose

To express

3D)r. choice

the

.l(x,y)

(9.9)

9.1.

Let g

and.y we eliminate

= g(l') = g(2x-

solution,l of (9.4)has proves

be differentiable

on

the

Rl,

following

and

let j'

equation)

!(x,

make

2B

u

from

- 3D

(9.8)

= I,

(9.6) is nonsingular; we have r

= h(u, r)

shows

THEOREM the

of x

transformation

that every differentiable the same type of argument Exactly t coefficien ts.) with constan

This

g(v))

Band D to

hence)

and

l') =

.;r) =

g(bx -

ay),)))

the

and obtain say

the

=

2, 3y,

3)').)

form

(9.5).

theorem for first-order

be

B

= 2x -

scalar

field

defined

equations

011 R2

b)'

286)

of differential calculus)

Applications b are

a and

where

constants,

Then f

zero.

both

not

satisfies

the

partial

first-order

differential

equation)

a a/ex, y) ax

(9.10)) in R2.

everywhere

al(x,

b

+

y) =

0

ay)

Conversely, every differentiable

solution

has

necessarily

(9.10)

of

the

form

some g.)

(9.9) for

9.3 Exercises)

In this I.

of exercises

set

of the

solution

that

Determine

you may assume

differential

partial

of (x,

4

y)

satisfies the condition that solution

which

2. Determine

0) =

f(x,

of the

of(x,y)

sin x for

satisfies

which

the conditions

= f(xy), prove

of (x, y)

equation) of (x,

-2

D1[(x,

the

u satisfies ou

such

a solution

(b) If v(x,

y)

=

that

for y

f(xfy)

x 4 e x2

x) =

u(x, rf

0,

such satisfies the

a solution

that

v(l,

partial

1) = 2 differential

v

g(u,

v) =

qJl

(u)

+

qJ2(V),

the partial-differential

equation)

oy)

and D1v(x, I/x)

= I/xfor

all

x

O.

rf

equation) v)

= 0,)

ou ov

of

equation)

ov

02g (U,

prove that

all x.

x.

satisfies

v

ov

Find

for

eX

differential

partial

for all

that

prove

0) =

oy)

x-+y-=O. ox

4. If g(u, v)

=0

ou

x--y-=O. ox Find

y)

oy)

[(0, 0) = 0 and that

0)

x.

all

ox

3. (a) If u(x, y)

=

oy

differential

partial

5

equation)

+ 3

ox

under consideration.

functions

of all

differentiability

where

qJl (u)

of u

is a function

alone

and

qJ2(V)

is a

function

alone.

5. Assume

f satisfies

the

partial

differential

o\037f

or Introduce

are constant,

the

linear

and

change

let g(u, v)

equation)

- 2

of variables,

= f(Au

- 3 o'i

02[ ox

+ Ev,

oy

x = Cu

=

o.)

oy2

+ Bv, + Dv).

Au

Y

= Cu

+

Dv,

where

Compute nonzero

integer

A, B, values

C, D of)))

A first-order

mine f

6.

A

such that g (Assume equality u is defined

C, D

B,

A,

function

partial

differential

satisfies

a 2g/(

au av) =

o. Solve this

by

an

partial

differential

au

G(x,

y).

to

known

satisfy

show

that g

satisfies

the

-

a\037 + 2 ax

=

.{(x, y)

ax

ay2

of

a 2g 0/

0 and

every

x in

S for

is known

expressed

the

S

, e

t

).

If f is

Y

-if =

0,

ay)

2

= o.)

We

say that

fis

homogeneous

tx E

S.

For a homogeneous scalarfield

of degree

p

= p f(x))

x

each

for

in

If x

functions.

homogeneous

s.)

= (Xl'

. . . , xn )

it can

be

fixed x,

_aj' aXl

define

+... get)

+

=

Xn

_af = aX n) p f(XI

f(tx)

, . . . , X n) \302\267

and compute g' (1).])

= of Euler's theorem. That is, iff satisfies x . Vf(x) p f(x) for all x in an must be of over S. [Hint: For fixed x, define f homogeneous degree p - t Jl'(x) and compute g' (t).] the following for homogeneous extension of Euler's theorem functions of degree p in 2-dimensional case. (Assume equality of the mixed partials.)) the

converse

open set S, = f(tx) g(t)

10. Prove

f(e

= t Pf(x))

which

as Euler'stheorenl.{or

For

[Hint:

Prove

t) =

as)

Xl

9.

g(s,

have)

x . Vf(x) This

+

on an open set Sin Rn.

is differentiable

f(tx)

for every / > show that we

where

equation)

partial-differential

that

into g(s, t),

-a\037 + x -if

y2

+ as 2

a scalar field degree p over S if)

G(x,y)u,)

equation)

a 2g

8. Let.{be

of the form)

y2

= e S,

x2

)

oy

t y = e converts the partial differential

x

substitution

deter-

thereby

au

-

ox find

and

\302\267)

equation

x2

and 7. The

for g

Y

xy

(

u satisfies a

equation

mixed partials.) of the form) equation x +

that

287)

coefficients

of the

u(x,y)=xyf

Show

constant

with

equation

then

x2

a\037 2

ox

a2 +

f

2xy

ox oy

+

a 2f y2

oy2

=

p(p

-

l)f.)))

288)

9.4 The

wave

one-dimensional

calculus)

of differential

Applications equation

to vibrate in the length stretched along the x-axis and allowed = of at the the the vertical f(x, t) string xy-plane. by y point x displacement at time t. We assume that, at time t = 0, the string is displaced a along prescribed curve, is shown in Figure 9.1(a). Figures 9.1(b)and (c) show y = F(x). An example possible displacement curves for later values of t. We regard the displacementf(x, t) as an unknown A mathematical model for this problem function of x and t to be determined. (suggested we shall not discuss here) is the partial differential by physical considerations which equation)

a

I magine

infinite

of

string

denote

We

o2f

2

-=c2

-

o2f ox

ot

2 ')

a positive constant dependingon the physical characteristics is will solve called the wave We one-din1ensional equation equation. certain auxiliary conditions.)

of the string.

c is

where

this

This

subject to

equation

y)

x)

-

0

1

(a) t

= 0)

FIGURE

9.1

(c) t

\037)

= j'(x,

is the prescribed

,{(x, 0) = also

We

assume

that

oy/ot, the velocity

0

-2-1

0

The displacement curve y

displacement satisfying the condition)

x)

x) 1

(b) t =

initial

the

Since

-

1

of the

t)

shown

2)

= 1)

for various

values of

curve y = F(x), we

seek

t.)

a solution

F(x).) vertical

is prescribed

displacement,

at time

t=O,say)

DJ(x, 0) =

G(x),)

should G is a given function. It seems reasonable to expectthat this information suffice to determinethe subsequent motion of the string. We will show that, indeed, this is true by determining in a form is expressed the functionj'in terms of F and G. The solution and philosopher.) mathematician given by Jean d'Alembert (1717-1783),a French where

THEOREM

functions

given

(9.1t)

9.2.

such

D'ALEMBERT'S

that G

SOLUTION

OF

WAVE EQUATION.

THE

is differentiable and F is twice

Let F and Then the

on RI.

differentiable

by the forn1ula

f(x,

t)

=

F(x +

ct) +

2

F(x -

ct)

+

1.

x+ctC(S)ds

2c 1 x-ct)))

G be given functionf

satisfies

the

one-dimensionallvave

equation)

a 2f -=c2

(9.12))

-

f(x,O) = F(x),)

(9.13))

reader One linear

t) into

f(x,

a

the constants

+

Au

Bv,)

u and

of

function

v) =

au this

Solving u =

for g

equation

and ((J2(V) x + ct , v

u alone

=

x

that

- ct , from

=

v) =

g(u,

({Jl(U) +

({Jl(U) is

where

({J2(V),

B,

C, D can

a

function

be chosen so

we obtain)

which

ct) +

+

({Jl(X

({J2(X

-

ct).)

determine the functions

({Jl

and

({J2

in

terms

of

and G.

the

linear

first-order L1 =

-a -

c-a

solution of (9.15)and

equation

differential

')

ax

at

use of

makes

t) =

the

Theorem 9.1 and avoids form)

given

operators

L2 =

-a + c -a at

let)

u(x,

in

= 0,)

L 1(L 2 f)

(9.15))

be a

the simpler equation)

The constants A,

obtain (9.14) by another method which of variables. First we rewrite the wave

are

Dv),)

O.

will

where Ll and L2

Cu +

av)

(9.13) to

conditions

initial

the given functions F

Let f

the

say)

satisfies

g

of valone.

f(x, t) =

Then we use the

change

we find

is a function

(9.14))

We

(9.11) to

introduce a

wave equation,

the

by left

Cu + Dv,)

v,

f(Au + Bv,

C, D so that

A, B,

=

f

a2 g

that

function j' given This verification is

the

neces-

of variables,)

transforms

choose

and (9.13)

(9.12)

satisfies

that exercise to verify and the initial conditions. equation given . We shall prove the converse. of way to proceed is to assume that,fis a solution

g(u,

of

M'hich

It is a straightforward

x =

and

= G(x).)

2

the wave

change

which

D f(x,0)

equal mixed partials

j' \",ith

Conversely, any function sarily has the fOr/11 (9.11). Proof

2)

ax

conditions)

the initial

satisfies

2

2 a f

at

and

289)

u'ave equation)

one-dimensional

The

L 2 f(x,

t).)))

ax)

.

by)

the

290)

Equation (9.15) states

the

satisfies

u

that

u(x, t) for some

function

Let

cpo

be

=

-1

L2(V) =

that

show

-ov = -1

L 2 (f). '

o. Hence,

by

2c)

=

-ov + c -ov

of

ct))

+

(x

2c)

(y)

say

cp,

=

ds,

S\037cp(s)

and let)

ct).

We have)

ct )

(x +

ox

+

cp(x

primitive

any

v(x, t) =

will

equation Ll(u) =

first-order

9.1 we have)

Theorem

We

calculus)

of differential

Applications

-1 ' ( x +

-ov =

and)

ct

2

ot

)

')

so)

L2v

In

other

words,

ot

=

-

v

,

ct) =

+

(x

the difference /

9.1 we

must

t)

have/ex,

-

first-order

the

satisfies

L 2 (fBy Theorem

ct) =

+

cp(x

u(x,

t)

=

L 2f.

ox)

v) =

vex,

o.)

=

t)

equation)

1p(x

- ct) for

function

some

1p.

There-

fore)

f(x,

t) =

t) +

v(x,

1p(x

-1

- ct) =

(x

ct) + 1p(x

+

2c)

This

(9.14) with

proves

({Jl

=

\037

Now we use the initial conditions the given functions F and G. The

({J2

(9.13) relation/ex,

(9.16)) The other

and

2c

+

CPl(X)

initial

D 2 /(x,

condition,

(9.17))

Differentiating

ccp\037(x)

(9.16) we

cp\037(x)

and (9.18)for ((J{(X)

=

'IjJ.

to determine the functions 0) = F(x) implies)

G(x), -

=

CP2(X)

CPl

and

F(x).)

implies)

=

ccp\037(x)

G(x).)

obtain)

(9.18))

Solving (9.17)

0) =

=

ct).

1.

(x) cP\037

2 F'(x)

+

and

+

cp\037(x)

cp\037(x)

.1 G(x), 2c)

we

=

F'(x).)

find)

((J\037(X)

=

F'(x) \037

-

;c

G(x).)))

CP2

in

terms

of

Integrating these relations

we

the

In

first

- ct.

F(O) to

we

add

0 = 1(0,0), whereas points = 0 x have and second and fourth < us/(x, 1(0,0). y) opposite signs, giving y quadrants is function the are points at which of the origin there in every Therefore, neighborhood is a saddle the function exceeds 1(0, 0), so the origin less than 1(0, 0) and points at which

Saddle

3.

EXAMPLE

point. surface

Near the origin the derivatives ai/ax and

z

y)

y)

x)

z =

(a)

9.4

FIGURE

The presence of some of the level curves

point.

Level curves:

(b)

xy)

Example

saddle

the

near

3. Saddle point

= c)

the origin.)

at

revealed

is also

point

xy

Figure

by

(0, 0). These are hyperbolashaving

9.4(b), x-

the

which

shows

and y-axes

as

asymptotes.) Saddle

4.

EXAMPLE

of a mountain

the appearance as

to

referred

illustrated

are

surrounded

minimum

at

the

9.6(b)]

minimum. z = four mountains, by since I(x, y) origin,

are hyperbolas

to those in on all its values nonnegative are similar

6.

EXAMPLE cylinder

cut

by

with planes

generators parallel

as suggestedby

> 1(0, 0) for

having

all

x- andy-axes 3.

In this

surface

This

9.6(a).

Figure

has the appearance of There is an absolute

The level curves [shown as asymptotes. Note that these

in

(x, y).

case, however, the

function

assumes

level

only

curves.)

level

z = f(x, y) to the y-axis,

parallel to the x-axis

attheoriginbecause/(x,y) family of parallel straight

has 3xy 2. Near the origin, this surface of three peaks. This surface, sometimes in Figure Some of the level curves 9.5(a). is a saddle point at the origin.) there

f(x,y) = x2y2.

the

Example

maximum.

Relative

-

in the vicinity

pass

a \"monkey saddle,\" is shown in Figure 9.5(b). It is clear that

a valley curves

3 f(x, y) = x

5. Relative

EXAMPLE

Figure

z =

point.

= 1as

shown

x

2

In

. in

this

Figure

case the surface is a Cross sections 9.7(a).

an absolute maximum are parabolas. Thereis obviously = 1 - x2 < 1 =/(0, 0) for all (x,y). The level curves form

lines

as shown

in Figure

9.7(b).)))

a

z)

y)

(a)z

= x3-

3 xy 9.5

FIGURE

2.)

Level curves:

(b)

4. Saddle

Example

at

point

-

x3

3 xy2 = c.)

the origin. y)

z)

x)

y)

x

(a) z

=

X

2 2)

(b)

y

9.6

FIGURE

Example

S. Relative

Level curves: at

minimum

X

2

y

2 = C)

the origin.) y)

z)

j

Tangent

at (0,0, I))

plane

-

-..)

x)

y)

x)

(a)

z = 1

-

FIGURE

X2)

9.7

(b)

Example

Level curves:

6. Relative maximum at

the

1

-

X2

= C)

origin.) 307)))

308)

Applications

9.10 Second-orderTaylor

formula

If a point

x = a

have

we

I(a + y) a stationary

At

field I has a stationary point the by algebraic sign of the difference the first-order Taylor formula)

determined

+ y,

- I(a) =

\302\267

V/(a) =

point,

+

y

V/(a)

0 and the

- I(a) =

Ily

derivatives

partial

L

\037

i=l

a term

order

of smaller

than

partial

second-order

derivatives

j=l)

Di(Djf),

calledthe

H(x) = whenever

notation

Hessian

!

where y = (Y1, . . . , Y rJ matrix.

1 column

X

Djifand the matrix Taylor's formula, following

field

Rn

(9.34))

be written

can

more simply

in

matrix

!

j

Dijf(a)YiY

= yH(a)yt,

is considered as a 1 x n row matrix, When the partial derivatives Dijf are

H(a)

is symmetric.

giving

a quadratic

and yt is its

+ y) -

to I(a

approximation

transpose, an have

we

continuous

I(a),

now

Diif takes

=

the

form.)

9.4.

THEOREM with

[Dijf(x)]\037j=l)

n

i=lj=l)

in

evaluated

The quadratic form

exist.

the derivatives as follows:)

n

y

of the quadratic form are the at a. The n x n matrix of and is denoted by H(x). matrixt

coefficients

The

Ily112.

Dof(x) is

the

about

second-order

we have)

Thus,

n

y).)

Did(a)YiY;

L

Dijf =

derivatives

second-order

If

a.

n

n

plus

stationary near

E(a,y)\037O asy\037O.)

the algebraic sign of I(a + y) - I(a) we need more information if I has continuous The next theorem shows that II E(a, y). at a, the error term is equal to a quadratic form,)

To determine

term

x

for

becomes)

E(a,

Ilyll

- I(a)

j{x)

formula

Taylor

the nature of the

at a,

where

y),)

Ilyll E(a,

f(a + y)

error

fields

scalar

for

scalar

differentiable

is

calculus)

of differential

such

second-order

that a

f(a + y)

TAYLOR

SECOND-ORDER

continuous

-

+ y f(a)

E

R(a)

=

FOR

FORMULA

derivatives

partial

SCALAR

Dijf

in

FIELDS.

an

n-ball

Let f

be a scalar all

Then for

R(a).

we have)

Vf(a)

.

y +

-1 2!)

t Named for Ludwig Otto Hesse (1811-1874), a the theory of surfaces.)))

yH(a

German

+

cy)yt,

mathematician

where)

who made

O

Al and A 2 Then for each let

th

u

0 as y --+0, there is a < r. For such y we have)

(9.39) shows

formula

>

IlyI12IE2(a, y)1
0 as above so that IE2(a, y)1 < !h whenever0 < Ily II < r. as above, for such y the algebraic sign of f(a + y) - f(a) is the we see that Then, arguing same as that of Q(y). Since both positive and negative as y -+ 0, f has a occur values saddlepoint at a. This completes the proof.) Note: If concerning such

treat

the

all

the

point.

stationary

but we

examples,

9.12 Second-derivativetest In

the case

sign of the

algebraic

THEOREM

in

=

A

and

and

DI.lf(a)

of a

point

a 2-ba/1

gives

no information can be used to

also be determined

the determinant

scalar fieldf(xl'

x2 )

by

the

of the Hessian matrix.) with

continuous

second-

Let)

B(a).

B =

D1.1f(a),)

9.6

derivatives

variables)

two

of the stationary point can

be a stationary

derivatives

of

functions

of

extrema

for

second derivative

Let a

9.7.

order partial

shall

2 the nature

=

n

of H(a) are zero, Theorem Tests involving higher order them here.) not discuss

eigenvalues

C = D2 . 2 f(a),)

DI.2f(a),)

let)

:]=AC-B2.)

/)\",=detH(a)=det[;

Then we have:

(a) If 6..< (b) If L\\ >

0,f

0 and

A

(c) If (d) If

0 and

A

Proof.

L\\ Ll

> =

In

0, the this

a saddle

has

0, f < 0, f

>

at a.

point

has a has a

relative

minimum

relative

maximum

at a. at a.

test is inconclusive.)

case the characteristic

equation det

[AI

-

H(a)] =

0

IS

a

quadratic

equation,) A

The eigenvalues

AI'

A2

are

related

Al +

2

-

(A +

to the

A2

=

C)A

+

L\\

=

coefficients by

A +

C,)

AIA2

o.)

the

=

equations)

6...)

0 the eigenvalues have oppositesigns,sofhas a saddle at a, which proves (a). point If 6..> 0, the eigenvalues have the same sign. In this case AC > B2 > 0, so A and C have the same sign. This sign must be that of Al and A 2 since A + C = Al + A2 . This proves and (b) (c). To prove (d) we refer to Examples 4 and 5 of Section 9.9. In both these exampleswe have \037 = 0 at the 5 it is a origin. In Example 4 the origin is a saddle point, and in Example relative minimum.))) If L\\
O. = O. [Hint:

are relative

DJ(x, y)

minima?

Which

answers. or

minimum

maximum on

an absolute

the

whole

plane?

Give

answers.

all the relative and absolute extreme values 2 = xy(1 - x - y2) on the square 0 \037 x \037 1 , f(x, y) Determine constants a and b such that the integral)

18. Determine 19.

y).

- y)(x + y- 3).

of the

(c)

\037

-

.

all points (x, y) in the a as factor.] y)

are

2

4x 2y + y2. Show that on every line y = mx the function has a minimum is no relative minimum in any two-dimensional neighborhood of the a sketch indicating the set of points (x, y) at which f(x, y) > 0 and the set at

3x4

y)
the following paths:

z =

(X2(t),)

dx

as Sell

, or

equations,)

two-dimensional vector field

be a

Letl

da as Sell

y, z)

fl(X,

fe

12dy

parametric

y =

X=(XI(t),)

(X2(t),)

dx +

as Sell

written

is

y =

(XI(t),)

t3

)j.

0

aCt) =

take

17

3

(.jt+t

+t)dt=-. 12)

t

2

; +

t 3j.

Then a'(t)

= 2t; +

3t2j

and

Therefore)

f[a(t)] . a'(t) =

2t

5A

3t 8 + 3t 5 ,)

+

so)

(0,1)

'I