Linear Algebra, 4th edition

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Graduate Texts in Mathematics

Linear Algebra

Volume 23 1975

Springer

Graduate Texts in Mathematics 23

Editorial Board: F. W. Gehring P. R. Halmos (Managing Editor) C. C. Moore

Werner Greub

Linear Algebra Fourth Edition

Springer-Verlag

New York Heidelberg Berlin

Werner Greub University of Toronto Department of Mathematics Toronto M5S 1A1 Canada

Managing Editor P. R. Halmos Indiana University Department of Mathematics Swain Hall East Bloomington, Indiana 47401

Editors F. W. Gehring

C. C. Moore

University of Michigan Department of Mathematics Ann Arbor, Michigan 48104

University of California at Berkeley Department of Mathematics Berkeley, California 94720

AMS Subject Classifications 15-01, 15A03, 15A06, 15A18, 15A21, 16-01 Library

Greub,

of Congress Cataloging in Publication Data Werner Hildbert, 1925—

algebra. (Graduate texts in mathematics; v. 23) Bibliography: p. 445 1. Algebras, Linear. 1. Title. II. Series. QA184.G7313 1974 512'.S 74-13868 Linear

All rights

reserved.

No part of this book may be translated or reproduced in any form without written permission from Springer-Verlag. © 1975 by Springer-Verlag New York Inc. Softcover reprint of the hardcover 4th edition 1975 ISBN 978-1-4684-9448-8 DOl 10.1007/978-1-4684-9446-4

ISBN 978-1-4684-9446-4 (eBook)

To Roif Nevanlinna

Preface to the fourth edition This textbook gives a detailed and comprehensive presentation of linear algebra based on an axiomatic treatment of linear spaces. For this fourth edition some new material has been added to the text, for instance, the intrinsic treatment of the classical adjoint of a linear transformation in Chapter IV, as well as the discussion of quaternions and the classification of associative division algebras in Chapter VII. Chapters XII and

XIII have been substantially rewritten for the sake of clarity, but the contents remain basically the same as before. Finally, a number of problems covering new topics — e.g. complex structures, Caylay numbers and symplectic spaces — have been added.

I should like to thank Mr. M. L. Johnson who made many useful suggestions for the problems in the third edition. I am also grateful to my colleague S. Halperin who assisted in the revision of Chapters XII and XIII and to Mr. F. Gomez who helped to prepare the subject index. Finally, I have to express my deep gratitude to my colleague J. R. Vanstone who worked closely with me in the preparation of all the revisions and additions and who generously helped with the proof reading.

Toronto, February 1975

WERNER H. GREUB

Preface to the third edition The major change between the second and third edition is the separation

of linear and multilinear algebra into two different volumes as well as the incorporation of a great deal of new material. However, the essential

character of the book remains the same; in other words, the entire presentation continues to be based on an axiomatic treatment of vector spaces.

In this first volume the restriction to finite dimensional vector spaces has been eliminated except for those results which do not hold in the infinite dimensional case. The restriction of the coefficient field to the real and complex numbers has also been removed and except for chapters VII to XI, § 5 of chapter I and § 8, chapter IV we allow any coefficient

field of characteristic zero. In fact, many of the theorems are valid for modules over a commutative ring. Finally, a large number of problems of different degree of difficulty has been added.

Chapter I deals with the general properties of a vector space. The topology of a real vector space of finite dimension is axiomatically characterized in an additional paragraph. In chapter lIthe sections on exact sequences, direct decompositions and duality have been greatly expanded. Oriented vector spaces have been incorporated into chapter IV and so chapter V of the second edition has disappeared. Chapter V (algebras) and VI (gradations and homology)

are completely new and introduce the reader to the basic concepts associated with these fields. The second volume will depend heavily on some of the material developed in these two chapters.

Chapters X (Inner product spaces) XI (Linear mappings of inner product spaces) XII (Symmetric bilinear functions) XIII (Quadrics) and XIV (Unitary spaces) of the second edition have been renumbered but remain otherwise essentially unchanged. Chapter XII (Polynomial algebra) is again completely new and developes all the standard material about polynomials in one indeterminate. Most of this is applied in chapter XIII (Theory of a linear transformation). This last chapter is a very much expanded version of chapter XV of the

second edition. Of particular importance is the generalization of the

Preface to the third edition

results in the second edition to vector spaces over an arbitrary coefficient field of characteristic zero. This has been accomplished without reversion

to the cumbersome calculations of the first edition. Furthermore the concept of a semisimple transformation is introduced and treated in some depth.

One additional change has been made: some of the paragraphs or sections have been starred. The rest of the book can be read without reference to this material.

Last but certainly not least, I have to express my sincerest thanks to everyone who has helped in the preparation of this edition. First of all I am particularly indebted to Mr. S. HALPERIN who made a great number of valuable suggestions for improvements. Large parts of the book, in particular chapters XII and XIII are his own work. My warm thanks also go to Mr. L. YONKER, Mr. G. PEDERZOLI and Mr. J. SCHERK who did the proofreading. Furthermore I am grateful to Mrs. V. PEDERZOLI

and to Miss M. PETTINGER for their assistance in the preparation of the

manuscript. Finally I would like to express my thanks to professor K. BLEULER for providing an agreeable milieu in which to work and to the publishers for their patience and cooperation.

Toronto, December 1966

WERNER H. GREUB

Preface to the second edition Besides the very obvious change from German to English, the second

edition of this book contains many additions as well as a great many other changes. It might even be called a new book altogether were it not

for the fact that the essential character of the book has remained the same; in other words, the entire presentation continues to be based on an axiomatic treatment of linear spaces. In this second edition, the thorough-going restriction to linear spaces of finite dimension has been removed. Another complete change is the

restriction to linear spaces with real or complex coefficients, thereby removing a number of relatively involved discussions which did not really contribute substantially to the subject. On p. 6 there is a list of those chapters in which the presentation can be transferred directly to spaces over an arbitrary coefficient field.

Chapter I deals with the general properties of a linear space. Those concepts which are only valid for finitely many dimensions are discussed in a special paragraph.

Chapter II now covers only linear transformations while the treatment of matrices has been delegated to a new chapter, chapter III. The discussion of dual spaces has been changed; dual spaces are now introduced abstractly and the connection with the space of linear functions is not established until later. Chapters IV and V, dealing with determinants and orientation respectively, do not contain substantial changes. Brief reference should

be made here to the new paragraph in chapter IV on the trace of an endomorphism — a concept which is used quite consistently throughout the book from that time on. Special emphasis is given to tensors. The original chapter on Multilinear Algebra is now spread over four chapters: Multilinear Mappings (Ch. VI), Tensor Algebra (Ch. VII), Exterior Algebra (Ch. VIII) and

Duality in Exterior Algebra (Ch. IX). The chapter on multilinear mappings consists now primarily of an introduction to the theory of the tensor-product. In chapter VII the notion of vector-valued tensors has

been introduced and used to define the contraction. Furthermore, a

XII

Preface to the second edition

treatment of the transformation of tensors under linear mappings has been added. In Chapter VIII the antisymmetry-operator is studied in greater detail and the concept of the skew-symmetric power is introduced. The dual product (Ch. IX) is generalized to mixed tensors. A special paragraph in this chapter covers the skew-symmetric powers of the unit tensor and shows their significance in the characteristic polynomial. The paragraph "Adjoint Tensors" provides a number of applications of the duality theory to certain tensors arising from an endomorphism of the underlying space.

There are no essential changes in Chapter X (Inner product spaces) except for the addition of a short new paragraph on normed linear spaces.

In the next chapter, on linear mappings of inner product spaces, the orthogonal projections 3) and the skew mappings 4) are discussed in greater detail. Furthermore, a paragraph on differentiable families of automorphisms has been added here. Chapter XII (Symmetric Bilinear Functions) contains a new paragraph dealing with Lorentz-transformations. Whereas the discussion of quadrics in the first edition was limited to quadrics with centers, the second edition covers this topic in full. The chapter on unitary spaces has been changed to include a more thorough-going presentation of unitary transformations of the complex plane and their relation to the algebra of quaternions. The restriction to linear spaces with complex or real coefficients has of course greatly simplified the construction of irreducible subspaces in chapter XV. Another essential simplification of this construction was achieved by the simultaneous consideration of the dual mapping. A final paragraph with applications to Lorentz-transformation has been added to this concluding chapter. Many other minor changes have been incorporated — not least of which are the many additional problems now accompanying each paragraph.

Last, but certainly not least, I have to express my sincerest thanks to everyone who has helped me in the preparation of this second edition. First of all, I am particularly indebted to CORNELIE J. RHEINBOLDT

who assisted in the entire translating and editing work and to Dr. WERNER C. RHEINBOLDT who cooperated in this task and who also made a number of valuable suggestions for improvements, especially in the chapters on linear transformations and matrices. My warm thanks

also go to Dr. H. BOLDER of the Royal Dutch/Shell Laboratory at Amsterdam for his criticism on the chapter on tensor-products and to Dr. H. H. KELLER who read the entire manuscript and offered many

Preface to the second edition

XIII

important suggestions. Furthermore, I am grateful to Mr. GI0RGI0 PEDERZOLI who helped to read the proofs of the entire work and who collected a number of new problems and to Mr. KHADJA NESAMUDDIN KHAN for his assistance in preparing the manuscript.

Finally I would like to express my thanks to the publishers for their patience and cooperation during the preparation of this edition. Toronto, April 1963

WERNER H. GREUB

Contents Chapter 0. Prerequisites

Chapter I. Vector spaces § 1. Vector spaces § 2. Linear mappings § 3. Subspaces and factor spaces § 4. Dimension § 5. The topology of a real finite dimensional vector space

37

Chapter II. Linear mappings § 1. Basic properties § 2. Operations with linear mappings § 3. Linear isomorphisms § 4. Direct sum of vector spaces § 5. Dual vector spaces § 6. Finite dimensional vector spaces

41 41

Chapter III. Matrices § 1. Matrices and systems of linear equations § 2. Multiplication of matrices § 3. Basis transformation § 4. Elementary transformations

83 83 89 92 95

Chapter IV. Determinants § 1. Determinant functions § 2. The determinant of a linear transformation § 3. The determinant of a matrix § 4. Dual determinant functions § 5. The adjoint matrix § 6. The characteristic polynomial §7. The trace § 8. Oriented vector spaces Chapter V. Algebras § 1. Basic properties

51

55 56

63 76

104 109 112 114 120 126 131

Change of coefficient field of a vector space

144 144 158 163

Chapter VI. Gradations and homology § 1. G-graded vector spaces § 2. G-graded algebras § 3. Differential spaces and differential algebras.

167 167 174 178

Chapter VII. Inner product spaces § 1. The inner product § 2. Orthonormal bases

186 186

§2. Ideals § 3.

191

Contents

XVI

Normed determinant functions Duality in an inner product space Normed vector spaces § 6. The algebra of quaternions Chapter VIII. Linear mappings of inner product spaces § 1. The adjoint mapping § 2. Selfadjoint mappings § 3. Orthogonal projections § 4. Skew mappings § 5. Isometric mappings

195

§ 3. § 4. § 5.

§ 6.

Rotations of Euclidean spaces of dimension 2, 3and4

§ 7.

Differentiable families of linear automorphisms.

202 205

208 216

216 221

226 229 232

237 249

Chapter IX. Symmetric bilinear functions § 1. Bilinear and quadratic functions § 2. The decomposition of E § 3. Pairs of symmetric bilinear functions § 4. Pseudo-Euclidean spaces § 5. Linear mappings of Pseudo-Euclidean spaces.

261

Chapter X. Quadrics § 1.

296 296

§

301

261 265

272

.

§

281 288

Affine spaces 2. Quadrics in the affine space 3. Affine equivalence of quadrics

§ 4.

310 316

Quadrics in the Euclidean space

Chapter XI. Unitary spaces § 1. Hermitian functions § 2. Unitary spaces § 3. Linear mappings of unitary spaces § 4. Unitary mappings of the complex plane § 5. Application to Lorentz-transformations

325 325 327 334 340

Chapter XII. Polynomial algebra . § 1. Basic properties § 2. Ideals and divisibility § 3. Factor algebras § 4. The structure of factor algebras.

351 351 357

Chapter XIII. Theory of a linear transformation § 1. Polynomials in a linear transformation § 2. § 3.

345

366 369

.

Generalized eigenspaces Cyclic spaces

.

383 390

397

§ 4. Irreducible spaces

402

Application of cyclic spaces Nilpotent and semisimple transformations 7. Applications to inner product spaces .

§ 5.

§ 6. §

383

.

.

415 425 436

Bibliography

445

Subject Index

447

Interdependence of Chapters Vector spaces F

Linear mappings J

Matrices

Determinants

Inner product spaces

Gradations and homology

Linear mappings of inner product spaces

Symmetric bilinear functions

r

Unitary spaces

Algebras

Polynomial algebra

Quadrics

Theory of a linear transformation

Chapter 0

Prerequisites 0.1. Sets. The reader is expected to be familiar with naive set theory

up to the level of the first half of [11]. In general we shall adopt the notations and definitions of that book; however, we make two exceptions. First, the word function will in this book have a very restricted meaning, and what Halmos calls a function, we shall call a mapping or a set mapping. Second, we follow Bourbaki and call mappings that are one-to-one (onto, one-to-one and onto) injective (surjective, bijective). 0.2. Topology. Except for § 5 chap. I, § 8, Chap. IV and parts of chapters VII to IX we make no use at all of topology. For these parts of the book the reader should be familiar with elementary point set topology as found in the first part of [16]. 0.3. Groups. A group is a set G, together with a binary law of composition x G—+G which satisfies the following axioms (x, y) will be denoted by xy): 1. Associativity: (xy)z = x (yz) 2. Identity: There exists an element e, called the identity such that

xe = ex = x. 3. To each element xeG corresponds a second element

xx' =

such that

= e.

The identity element of a group is uniquely determined and each element has a unique inverse. We also have the relation

(xy)' As an example consider the set of all permutations of the set { 1. n} and define the product of two permutations a, t by . .

(crt)i=a(ti) In this way

i=1...n.

becomes a group, called the group of permutations of n objects. The identity element of is the identity permutation.

I

Greub. linear Algebra

Chapter 0. Prerequisites

Let G and H be two groups. Then a mapping G —*

H

is called a homomorphism if

X,yEG.

=

A homomorphism which is injective (resp. surjective, bijective) is called a monomorphism (resp. epimorphism, isomorphism). The inverse mapping of an isomorphism is clearly again an isomorphism. A subgroup H of a group G is a subset H such that with any two elementsyeHand zeH the productyz is contained in Hand that the inverse of every element of H is again in H. Then the restriction of p to the subset Hx H makes H into a group. A group G is called commutative or abelian if for each x, yeG xy=yx.

In an abelian group one often writes x+y instead of xy and calls x+y the sum of x and y. Then the unit element is denoted by 0. As an example consider the set of integers and define addition in the usual way. 0.4. Factor groups of commutative groups.* Let G be a commutative

group and consider a subgroup H. Then H determines an equivalence relation in G given by x

x'

if and only if

x—

x' e H.

The corresponding equivalence classes are the sets {H+x} and are called the cosets of H in G. Every element xeG is contained in precisely one coset ?. The set G/H of these cosets is called the factor set of G by H and the surjective mapping G

G/H

defined by XEX

called the canonical projection of G onto G/H. The set G/H can be made into a group in precisely one way such that the canonical projection becomes a homomorphism; i.e., is

it(x + y) = irx + 7ry. To define the addition in G/H let

e G/H, 9 e G/H be arbitrary and choose

xeG and yeG such that and

my=9.

*) This concept can be generalized to non-commutative groups.

Chapter 0. Prerequisites

Then the element it (x +y) depends only on

two other elements satisfying whence

3

and 9. In

and icy' =9

we

fact, if x', y' are

have

x'—xeH and y'—yeH (x' + y') — (x

+ y)eH

and so ir(x'+y')—ic(x+y). Hence, it makes sense to define the sum by

is easy to verify that the above sum satisfies the group axioms. Relation (0.1) is an immediate consequence of the definition of the sum in G/H. Finally, since it is a surjective map, the addition in G/H is uniquely deterIt

mined by (0.1). The group G/H is called the factor group of G with respect to the subgroup H. Its unit element is the set H. 0.5. Fields. Afield is a set F on which two binary laws of composition, called respectively addition and multiplication, are defined such that 1. F is a commutative group with respect to the addition. 2. The set F — {0} is a commutative group with respect to the multiplication.

3. Addition and multiplication are connected by the distributive

law,

The rational numbers the real numbers 11 and the complex numbers C are fields with respect to the usual operations, as will be assumed without proof. A homomorphism p:F—÷F' between two fields is a mapping that preserves addition and multiplication. A subset A F of a field which is closed under addition, multiplication and the taking of inverses is called a subfield. If A is a subfield of F, F is called an extension field of A. Given a field F we define for every positive integer k the element ke (c unit element of F) by

The field F is said to have characteristic zero if ke + 0 for every positive integer k. If F has characteristic zero it follows that kc+k'c whenever k+k'. Hence, a field of characteristic zero is an infinite set. Throughout this book it will be assumed without explicit mention that all fields are of characteristic zero.

Chapter 0. Prerequisites

4

For more details on groups and fields the reader is referred to [29]. 0.6. Partial order. Let cl be a set and assume that for some pairs X. Y (XE.ci. YE.cl) a relation, denoted by X Y, is defined which satisfies the following conditions: (i) XX for every Xe.cl (Reflexivity) (ii) if X Y and Y X then X = Y (Antisymmetry) (iii) If X Y and Y Z, then X Z (Transitivity). Then is called a partial order in d.

A homomorphism of partial/v ordered sets is a map go:

such

that goX go Y whenever X Y

Clearly a subset of a partially ordered set is again partially ordered.

Let d be a partially ordered set and suppose A Ed is an element such that the relation A X implies that A = X. Then A is called a maximal element of ci. A partial ordered set need not have a maximal element. A partially ordered set is called linear/v ordered or a chain if for every

pairX, YeitherXYor be a subset of the partially ordered set ci. Then an element if X A for every XEc11. A ed is called an upper hound for In this book we shall assume the following axiom: A partially ordered set in which every chain has an upper bound, contains a maximal element. This axiom is known as Zorn's lemma, and is equivalent to the axiom Let

of choice (cf. [11]).

d be a 0.7. Lattices. Let d be a partially ordered set and let subset. An element Aed is called a least upper hound (l.u.b.) for d1 if 1) A is an upper bound for cl1. 2) If X is any upper bound, then A X. It follows from (ii) that if a l.u.b. for d1 exists, then it is unique. In a similar way, lower bounds and the greatest lower bound (g.1.b.) for a subset of d are defined. A partially ordered set ci is called a lattice, if for any two elements X, Y the subset {X, Y} has a 1.u.b. and a g.l.b. They are denoted by X v Y and X A Y. It is easily checked that any finite subset (X1, ..., Xr) of a lattice

has a 1.u.b. and a g.l.b. They are denoted by V

and A X,.

As an example of a lattice, consider the collection of subsets of a given set, X, ordered by inclusion. If U, V are any two subsets, then

UAV=UflV and UvV=UUV.

Chapter 1

Vector Spaces § 1.

Vector spaces

1.1. Definition. A vector (linear) space, E, over the field F is a set of elements x, y, ... called vectors with the following algebraic structure:

I. E is an additive group; that is, there is a fixed mapping Ex E—+E denoted by (1.1)

and satisfying the following axioms: 1.1. (x +y) +z = x + (y + z) (associative law)

1.2. x+y=y+x (commutative law) 1.3. there exists a zero-vector 0; i.e., a vector such that x+0= O+x=x for every xeE. 1.4. To every vector x there is a vector —x such that x+(—x)=O.

II. There is a fixed mapping F x

denoted by

(2,x)—÷Ax

(1.2)

and satisfying the axioms: 11.1. (associative law) 11.2.

A(x + y) = Ax + Ay (distributive laws) 11.3.

unit element of F)

(The reader should note that in the left hand side of the first distributive law, + denotes the addition in F while in the right hand side, + denotes the addition in E. In the sequel, the name addition and the symbol + will continue to be used for both operations, but it will always be clear from the context which one is meant). F is called the coefficient field of the vector space E, and the elements of F are called scalars. Thus the mapping

Chapter 1. Vector spaces

(1.2) defines a multiplication of vectors by scalars, and so it is called scalar multiplication.

If the coefficient field 1' is the field l1 of real numbers (the field C of complex numbers), then £ is called a real (complex) vector space. For the rest of this paragraph all vector spaces are defined over a fixed, but arbitrarily chosen field F of characteristic 0. If {x1 ,...,x,1} is a finite family of vectors in E, the sum x1 + will often be denoted

Now we shall establish some elementary properties of vector spaces. It follows from an easy induction argument on a that the distributive laws hold for any finite number of terms,

A) x .

Ax

holds if and only if

=

=

0

2=0 or x=0.

Proof Substitution of pc=O in the first distributive law yields Ax = Ax + Ox

whence Ox =0. Similarly, the second distributive law shows that

20 = 0. Conversely, suppose that Ax=O and assume that 2+0. Then the associative law 11.1 gives that

lx = (A1A)x = A1(Ax) = Ab0 = 0 and hence axiom 11.3 implies that x=O. The first distributive law gives for Ax + (— A)x =

(A

—

—A

A)x =

whence (— A)

x = — Ax.

0

§ 1. Vector spaces

In the same way the formula 2(— x) = — 2x is

proved.

1.2. Examples. 1. Consider the set F "=

of n-tuples

and define addition and scalar multiplication by I

+

1

+

=

1,

+

and

Then the associativity and commutativity of addition follows at once from the associativity and commutativity of addition in F. The zero vector is the n-tuple (0, ..., 0) and the inverse of is the n-tuple ..., Consequently, addition as defined above makes the set ..., F" into an additive group. The scalar multiplication satisfies I1.i, 11.2, and 11.3, as is equally easily checked, and so these two operations make F" into a vector space. This vector space is called the n-space over F. In particular, F is a vector space over itself in which scalar multiplication coincides with the field multiplication.

2. Let C be the set of all continuous real-valued functions, f, in the

interval I:Ot 1, R.

1ff, g are two continuous functions, then the function f+g defined by (1 + g) (t) is

= f (t) + g (t)

again continuous. Moreover, for any real number 2, the function 2f

defined by

(1f)(t) = 2.1(t) is

continuous as well. It is clear that the mappings (1' g) —p

f +g

and

(2,1)

2]

the systems of axioms 1. and 11. and so C becomes a real vector space. The zero vector is the function 0 defined by satisfy

0(t) =

0

Chapter 1. Vector spaces

and the vector —f is the function given by

(- f)(t) = - f(t). Instead of the continuous functions we could equally well have considered the set of k-times differentiable functions, or the set of analytic functions.

3. Let X be an arbitrary set and E be a vector space. Consider all and define the sum of two mappings f and g as the xeX (f + g)(x) = f(x) + g(x) and the mapping )f by

mappings J: mapping

(i.[)(x) =

XEX.

Under these operations the set of all mappings f:

becomes a vector space, which will be denoted by (X; E). The zero vector of (X; E) is the function f defined by f(x)=0, XEX. 1.3. Linear combinations. Suppose E is a vector space and x1 are vectors in E. Then a vector XEE is called a linear combination of the vectors x1 if it can be written in the form x=

be any family of vectors. Then a vector More generally, let if there is a family xEE is called a linear combination of the vectors only finitely many different from zero, such that of scalars, x= where the summation is extended over those We shall simply write =

for which

cLEA

and it is to be understood that only finitely many are different from zero. In particular, by setting =0 for each we obtain that the 0-vector is a linear combination of every family. It is clear from the definition that if x is a linear combination of the family {xj then x is a linear combination of a finite subfamily. Suppose now that x is a linear combination of vectors cteA

x= rzeA

and assume further that each x2 is a linear combination of vectors

§ 1. Vector spaces

9

JJeB2,

=

e F. p

Then the second distributive law yields X

=

x2

r

=

Yap'

=

hence x is a linear combination of the vectors A subset ScEis called a system of generators for Eif every vector xeE is a linear combination of vectors of S. The whole space E is clearly a system of generators. Now suppose that S is a system of generators for E and that every vector of S is a linear combination of vectors of a subset Tc: S. Then it follows from the above discussion that T is also a system of generators for E. 1.4. Linear dependence. Let be a given family of vectors. Then a non-trivial linear combination of the vectors is a linear combination where at least one scalar is different from zero. The family {xa} and

is called linearly dependent if there exists a non-trivial linear combination of the that is, if there exists a system of scalars such that (1.3)

and at least one + 0. It follows from the above definition that if a subfamily of the family is linearly dependent, then so is the full family. An equation of the form (1.3) is called a non-trivial linear relation. A family consisting of one vector x is linearly dependent if and only if x 0. In fact, the relation

1.0=0 that the zero vector is linearly dependent. Conversely, if the vector x is linearly dependent we have that 2x=O where 2+0. Then Proposition I implies that x=0. It follows from the above remarks that every family containing the zero vector is linearly dependent. shows

is linearly dependent if and Proposition II: A family of vectors is a linear combination of the vectors x2, cx+fl. only if for some fleA, Proof: Suppose that for some fleA, xp

Chapter 1. Vector spaces

10

Then setting

—1

we obtain

=

o

and hence the vectors x2 are linearly dependent.

Conversely, assume that

=

0

and that

for some fleA. Then multiplying by view of II.! and 11.2 0= +

we obtain in

2*/i

()ji)_l)2x

xp=— a*f1

Corollary: Two vectors x, y are linearly dependent if and only ify=Ax (or x==)Ly) for some AcT. 1.5. Linear independence. A family of vectors (x2)26A is called linearly

independent if it is not linearly dependent; i.e., the vectors x2 are linearly independent if and only if the equation

= implies that A2=O for each

0

It is clear that every subfamily of a line-

arly independent family of vectors is again linearly independent. If flEA,

is a linearly independent family, then for any two distinct indices is injective. and so the map

Proposition III: A family (x2)2EA of vectors is linearly independent if and only if every vector x can be written in at most one way as a linear combination of the x2 i.e., if and only if for each linear combination

X>22X2

(1.4)

the scalars A2 are uniquely determined by x. Proof. Suppose first that the scalars A2 in (1.4) are uniquely determined by x. Then in particular for x=0, the only scalars A2 such that =0

are the scalars A2=O. Hence, the vectors x2 are linearly independent. Con-

§ 1. Vector spaces

versely,

suppose that the

11

are linearly independent and consider the

relations Then

=0

—

whence in view of the linear independence of the i.e., 1.6. Basis. A family of vectors

in E is called a basis of E if it is simultaneously a system of generators and linearly independent. In view of Proposition III and the definition of a system of generators, we have that A is a basis if and only if every vector xeE can be written in precisely one way as

The scalars

are called the components of x with respect to the basis

As an example, consider the n-space, ['fl, over F defined in example 1, sec. 1.2. It is easily verified that the vectors

i=1...n

x1=(O,...,O,l,O...O)

form a basis for F". We shall prove that every non-trivial vector space has a basis. For

the sake of simplicity we consider first vector spaces which admit a finite system of generators. Proposition IV: (i) Every finitely generated non-trivial vector space has a finite basis (ii) Suppose that S=(x1 ,...,Xm) is a finite system of generators of E and that the subset R c S given by R = (x1 Xr) (r m) consists of

linearly independent vectors. Then there is a basis,

of E such that

RcTcS. Proof: (i) Let x1 the vectors x1

x,, be a minimal system of generators of E. Then are linearly independent. In fact, assume a relation

=

0.

Chapter I. Vector spaces

12

If

= 0 for some 1, it follows that

;eT and so the vectors

(1.5)

generate E. This contradicts the minimality

of ii. (ii) We proceed by induction on ,i

(nr). If n=r then there is nothing to prove. Assume now that the assertion is correct for ii — I. Consider the vector space, F, generated by the vectors x1 Xr+i Then by induction, F has a basis of the form X1

Xr,

where V1ES

Ys

1

(j = 1

Now consider the vector x,1. If the vectors x1

). 1

x, are

linearly independent, then they form a basis of E which has the desired property. Otherwise there is a non-trivial relation

= 0.

+ Since the vectors x1

are

that y+O. Thus

linearly independent, it follows

r

x,7= Q=l

71 generate E. Since they are linearly

Hence the vectors x1

independent, they form a basis. Now consider the general case. Theorem I: Let E be a non-trivial vector space. Suppose S is a system

of generators and that R is a family of linearly independent vectors in E such that R S. Then there exists a basis, of E such that R S. Proof Consider the collection .W(R, S) of all subsets, X, of E such that

1) RcXcS 2) the vectors of X are linearly independent. The a partial order is defined in d(R, 5) by inclusion (cf. sec. 0.6). We show that every chain, in .&(R, 5) has a maximal element A. In fact, R

A

set A =

We

have to show that A ed(R, S). Clearly,

S. Now assume that x5,eA.

(1.6)

§ 1. Vector spaces

Then, for each assume that

i,

for some

is a chain, we may

Since

(i=1...n).

(1.7)

are linearly independent it follows that Since the vectors of (v = ... n) whence A ed(R, S). Now Zorn's lemma (cf. sec. 0.6) implies that there is a maximal element, L in d(R, S). Then R c: Tc:S and the vectors of Tare linearly 1

independent. To show that T is a system of generators, let xeE be arbitrary. Then the vectors of T U x are linearly dependent because otherwise it would follow that x U Ted(R, S) which contradicts the maximality of T Hence there is a non-trivial relation ,tVET, Since

the vectors of T are linearly independent, it follows that

whence

xv.

This equation shows that T generates E and so it is a basis of E.

Corollary I: Every system of generators of E contains a basis. In particular, every non-trivial vector space has a basis.

Corollary II: Every family of linearly independent vectors of E can be extended to a basis. 1.7. The free vector space over a set. Let X be an arbitrary set and such that f(x)+0 only for finitely many XEX. consider all maps f:

Denote the set of these maps by C(X). Then, if fe C(X), ge C(X) are scalars, 2f+ is again contained in C(X). As in example 3, sec. 1.2, we make C(X) into a vector space. For any aeX denote by the map given by and ).,

x=a

51

Then the vectors (aEX) form a basis of C(X). In fact, let fe C(X) be the (finitely many) distinct points be given and let a1 a, such that [(a) 0. Then we have

f where

= [(a)

i=1

(1 =

1

n)

Chapter I. Vector spaces

14

and so the element a relation

(aEX) generate C(X). On the other hand, assume

= 0. j=1

Then we have for each 1(1= ... ii) 1

0

1

=

... a). This shows that the vectors

(aeX) are

linearly independent and hence they form a basis of C(X). Now consider the inclusion map X—*C(X) given by

ix(a)=Ja

aEX.

This map clearly defines a bijection between X and the basis vectors

of C(X). If we identify each element aeX with the corresponding map then X becomes a basis of C(X). C(X) is called the free vector space over X or the vector space generated by X.

Problems 1. Show that axiom 11.3 can be replaced by the following one: The equation 2x=O holds only if 2=0 or x=0. 2. Given a system of linearly independent vectors (x1, ..., xe,), prove that the system (x1, .. i4j with arbitrary 2 is again linearly independent. 3. Show that the set of all solutions of the homogeneous linear differential equation d2y dt2

+

dy + qy = 0 dt

p and q are fixed functions of t, is a vector space. 4. Which of the following sets of functions are linearly dependent in the vector space of Example 2? where

a)f1=3t; f2=t+5; f3=2t2; b)f1=(t+1)2;f2=t2—1; f3=2t2+2t—3 f2=et; f3=e_t c) f1=1; d)f1=t2; e)

f1=1—t;

f2=t;

f3=1

f2=t(1—t); f3=1—t2.

§ 1. Vector spaces

5. Let E be a real linear space. Consider the set E x E of ordered pairs

(x, y) with xeE and yeE. Show that the set Ex E becomes a complex vector space under the operations: (x1,y1) + (x2,y2) =

(x1

+ x2,y1 + Y2)

and + i fJ)(x, y) =

x—

,8

y, y + J3 x)

/3 real numbers).

6. Which of the following sets of vectors in (a generating set, a basis)?

linearly independent,

a) (1, 1, 1, 1), (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1) b) (1, 0, 0, 0), (2, 0, 0, 0) c) (17, 39, 25, 10), (13, 12, 99, 4), (16, 1,0,0) 0, 0, d) (1, 1, 0, 0), (0, 0, 1, 1), (0, 4, 4, 1),

Extend the linearly independent sets to bases. 7. Are the vectors x1 (1, 0, 1); x2 = (i, 1, 0), x3 = (1, 2, 1 +1) linearly independent in C3? Express x = (1, 2, 3) and y = (1, 1, 1) as linear combibinations of x1, x2, x3. is defined by a map 8. Recall that an n-tuple given by

(i=1...n). Show that the vector spaces C{l...n} and F" are equal. Show further that the basis J defined in sec. 1.7 coincides with the basis x defined in sec. 1 .6.

9. Let S be any set and consider the set of maps

f: S

F"

such thatf(x)=0 for all but finitely many xeS. in a manner similar to that of sec. 1.7, make this set into a vector space (denoted by C(S, F')). Construct a basis for this vector space. be a basis for a vector space E and consider a vector 10. Let

a= that for some fleA. again a basis for E. Suppose

Show that the vectors

form

Chapter I. Vector spaces

11. Prove the following exchange theorem of Steinitz: Let be a a system of linearly independent vectors. basis of E and Then it is possible to exchange certain p of the vectors by the vectors a such that the new system is again a basis of F. Hint: Use problem 10. 12. Consider the set of polynomial functionsf:

f (x) =

xi.

Make this set into a vector space as in Example 3, and construct a natural basis. § 2.

Linear mappings

In this paragraph, all vector spaces are defined over a fixed but arbitrarily chosen field F of characteristic zero.

1.8. Definition. Suppose that E and F are vector spaces, and let (p:

be a set mapping. Then 'p will be called a linear mapping if

'p(x+y)='px+'py x,yeE

(1.8)

and

)LEF,xEE

= 2çox

(1.9)

(Recall that condition (1.8) states that 'p is a homomorphism between abelian groups). If F=F then 'p is called a linear function in E. Conditions (1.8) and (1.9) are clearly equivalent to the condition =

'p

'p

and so a linear mapping is a mapping which preserves linear combinations. From (1.8) we obtain that for every linear mapping, 'p, 'p 0

whence 'p (0)

=

'p (0

+0) =

'p (0)

+ 'p (0)

0. Suppose now that

=0 is a linear relation among the vectors 'p x1

=

'p

)!

Then we have

=

'p0

=0

§ 2. Linear mappings

whence

= 0.

17

(1.11)

Conversely, assume that ço:E—*F is a set map such that (1.11) holds whenever (1.10) holds. Then for any x, yeE and )LEF set

u=x+y and v=2x. Since

u—x—y=0 and v—2x=0

it follows that x

(x + y) —

—

y= 0

and ç (2 x) —2 (p x =

0

and hence p is a linear mapping. This shows that linear mappings are precisely the set mappings which preserve linear relations. In particular, it follows that if x1 . ..v,. are linearly dependent, then so If X1 .• . are linearly independent, it does not, are the vectors çox1 however, follow that the vectors cOXi...cOXr are linearly independent. In fact, the zero mapping defined by (px=0,xEE,is clearly a linear mapping which maps every family of vectors into the linearly dependent set (0). A bijective linear mapping (p: E3F is called a linear isomorphism and will be denoted by p: Given a linear isomorphism (p: the set mapping (p 1: E+—F. It is easy to verify that p again satisfies the .

. . .

conditions (1.8) and (1.9) and so it is a linear mapping. p' is bijective and hence a linear isomorphism. It is called the inverse isomorphism of (p. Two vector spaces E and F are called isomorphic if there exists a linear

isomorphism from E onto F. is called a linear transformation of E. A A linear mapping (p bijective linear transformation will be called a linear automorphism of E. 1.9. Examples: 1. Let E=F" and define (p:E—+E by =

+

satisfies the conditions (1.8) and (1.9) and hence it is a linear transformation of E. 2. Given a set S and a vector space E consider the vector space (S; E) be the mapping given by defined in Example 3, sec. 1.2. Let (S;

Then

(pf=f(a) fe(S;E) where aES is a fixed element. Then (p is 2

Greub. Linear Atgebra

a linear mapping.

Chapter

18

1.

Vector spaces

where )LeT is a be the mapping defined by 3. Let fixed element. Then p is a linear transformation. In particular, the ideaWv map i:E—+E, ix=x, is a linear transformation. 1.10. Composition. Let p: E—*F and i/i: F—*G be two linear mappings. Then the composition of ço and /i0ço:E—*G

is defined by XEE. The relation

x) =

(/9

ço x1)

k/i

= p is a linear mapping of E into G. k/Jo p will often be denoted simply by k/1p. If p is a linear transformation in E, then we denote (/9 o (p by (p2. More generally, the linear transformation p ... p is denoted by

shows that k/Jo

i. A linear transformation, p, satisfying (p2 = i is called an involution in E. 1.11. Generators and basis. S—*F Proposition I: Suppose S is a system of generators for E and is a set map (Fa second vector space). Then Po can be extended in at most one way to a linear mapping We extend the definition to the case k=O by setting (p°=

F.

(p: E

A necessary and sufficient condition for the existence of such an extension is that 0

whenever

= 0.

Proof If (p X1

is

an extension of

we

have for each finite set of vectors

ES

=

=

Since the set S generates E it follows from this relation that is (/9 uniquely determined by (po Moreover, if

XES

§ 2. Linear mappings

it follows that =

=

=

cOO

=

0

and so condition (1.12) is necessary. Conversely, assume that (1.12) is satisfied. Then define

by (1.13)

To prove that

is

a well defined map assume that

Then =o

—

whence in view of (1.12)

0

—

and so

= The linearity of p follows immediately from the definition, and it is clear that p extends cOo.

Proposition II: Let be a basis of E and cOo: be a set can be extended in a unique way to a linear mapping map. Then (p:

F.

Proof. The uniqueness follows from proposition I. To prove the existence of p consider a relation = 0. Since the vectors

are linearly independent it follows that each

whence

= 0.

Now proposition I shows that

can be extended to a linear mapping

Corollary: Let S be a linearly independent subset of E and be a set map. Then cOo can be extended to a linear mapping

Chapter I. Vector spaces

20

Proof. Let T be a basis of F containing S (cf. sec. 1.6). Extend c°o in an T—+F. Then k/I0 may be extended to a linear arbitrary way to a set map mapping t/i:E—+F and it is clear that k/i extends

Now let be a surjective linear map, and suppose that S is a system of generators for E. Then the set = is a system of generators for F. In fact, since çü is surjective, every vector yEF can be written as = x

for some xeE. Since S generates E there are vectors such that x=

and scalars

whence

y=

=

This shows that every vector yEF is a linear combination of vectors in

'p (5) and hence 'p (5) is a system of generators for(5). 'p Next, suppose that p: E—÷ F is injective and that S is a linearly independent subset of E. Then 'p (5) is a linearly independent subset of F. in

fact the relation implies that

Since 'p

is

= 0,

= 0.

injective we obtain

=

0

whence, in view of the linear independence of the vectors 'p (5) is a linearly independent set. In particular, if çø: E—+F is a linear isomorphism and for E, then is a basis for F.

A! =0. Hence

is a basis

Proposition III: Let ço:E—+Fbe a linear mapping and be a basis of E. Then 'p is a linear isomorphism if and only if the vectors = form a basis for F. Proof. If 'p is a linear isomorphism then the vectors form a linearly independent system of generators for F. Hence they are a basis. Converse-

§ 2. Linear mappings

ly, assume that the vectors

21

form a basis of F. Then we have for every

yEF y= and so

is

=

surjective.

Now assume that jf Xci.

Then it follows that o=

—

= Since the vectors each and so

—

are linearly independent, we obtain that

for

1?

It follows that 'p is injective, and hence a linear isomorphism.

Problems 1. Consider the vector space of all real valued continuous functions defined in the interval a t b. Show that the mapping 'p given by (p:X(t)—* is

tX(t)

linear.

2. Which of the following mappings of F4 into itself are linear transformations?

-

a) b)

+

c)

+

+

Let E be a vector space over F, and letf1 . given by E. Show that the mapping p: 3.

.

be linear functions in

= (ft(x), ...,fr(X)) is

linear.

4. Suppose

is a linear map, and write 'p x = (f1 (x), .. , fr (x)).

Show

that the

are linear functions in E.

Chapter I. Vector spaces

22

5. The unirersal property of C(X). Let X be any set and consider the free vector space, C(X). generated by X (cf. sec. 1 .7).

(i) Show that if j: then there is a unique linear map p:

X into a vector space F such that (p0

where X—*C(X) is the inclusion map. (ii) Let Y be a set map. Show that there is a unique linear C(X)—*C(Y) such that the diagram map

tX1.

C(X)f'4 C(Y)

is a second set map prove the composition

commutes. If [3: formula

(fib

=

(iii) Let E be a vector space. Forget the linear structure of E and form

the space C(E). Show that there is a unique linear map mE:

C(E)—÷E

such that mE ° 1E

(iv) Let E and F be vector spaces and let E—*F be a map between the underlying sets. Show that is a linear map if and only if ltFO(p*_ (POThE (v) Denote by N(E) the subspace of C(E) generated by the elements of the form a, beE, /IEF — (cf.

part (iii)). Show that

kermE = N(E).

6.Let v=O

be a fixed polynomial and letf be any linear function in a vector space E. Define a function P(f): E-÷F by

P(f)x =

(x)".

Find necessary and sufficient conditions on P that P(f) function. § 3.

be

again a linear

Subspaces and factor spaces

In this paragraph, all vector spaces are defined over a fixed, but arbitrarily chosen field F of characteristic 0.

§ 3. Subspaces

and factor spaces

23

1.12. Subspaces. Let E be a vector space over the field F. A non-empty subset, E1, of E is called a subspace if for each x, yEE1 and every scalar

x+yEE1

(1.14)

and

Equivalently, a subspace is a subset of E such that Ax + /IyEE1

whenever x, yeE1. In particular, the whole space E and the subset (0) consisting of the zero vector only are subspaces. Every subspace E1 E contains the zero vector. In fact, if x1 e E1 is an arbitrary vector we have that 0= x1 — x1 E E1. A subspace E1 of E inherits the structure of a vector space from E. Now consider the injective map i: E1 —* E defined by

ix=x,

xeE1.

In view of the definition of the linear operations in E1 i is a linear mapping, called the canonical injection of E1 into E. Since i is injective it follows from (sec. 1.11) that a family of vectors in E1 is linearly independent (dependent) if and only if it is linearly independent (dependent) in E. Next let S be any non-empty subset of E and denote by E5 the set of linear combinations of vectors in S. Then any linear combination of vecis a linear combination of vectors in S (cf. sec. 1.3) and hence tors in Thus is a subspace of E, called the subspace generated it belongs to by 5, or the linear closure of S. In particular, if the set S is Clearly, S is a system of generators for linearly independent, then S is a basis of We notice that = S if and only if S is a subspace itself. 1.13. Intersections and sums. Let E1 and E2 be subspaces of E and consider the intersection E1 n E2 of the sets E1 and E2. Then E1 fl E2 is again a subspace of E. In fact, since OeE1 and OeE2 we have OEE1 fl E2 and so E1 fl E2 is not empty. Moreover, it is clear that the set E1 n E2 satisfies again conditions (1.14) and (1.15) and so it is a subspace of E. E1 n E2 is called the intersection of the subspaces E1 and E2. Clearly, E1 fl E2 is a subspace of E1 and a subspace of E2. The sum of two subspaces E1 and E2 is defined as the set of all vectors of the form x1eE1,x2EE2 (1.16) x=x1 +x2,

Chapter 1. Vector spaces

24

and is denoted by E1 +122. Again it is easy to verify that E1 +E2 is a subspace of E. Clearly E1 + E2 contains E1 and E2 as subspaces. A vector x of E1 + E2 can generally be written in several ways in the form (1.16). Given two such decompositions

x=x1+x2 and it follows that x1

—

=

x'2 —

x2.

Hence, the vector z = x1

—

fl E2. Conversely, let x=x1 +x2, x1 eE1, x2eE2 be a decomposition of x and z be an arbitrary vector of E1 fl E2. Then the vectors

is contained in the intersection

=

— zEE1

=

and

x2

+ zEE2

form again a decomposition of x. It follows from this remark that the decomposition (1.16) of a vector xeE1 +E2 is uniquely determined if and only if E1 n E2 0. In this case E1 + E2 is called the (internal) direct sum of E1 and 122 and is denoted by Now let S1 and be systems of generators for E1 and E2. Then clearly is a system of generators for E1 +E2. If T1 and T2 are respectively S1 U

bases for E1 and E2 and the sum is direct, E1 fl 122=0, then T1 U T2 is a To prove that the set T1 U T2 is linearly independent, basis for suppose that = 0, + Then

=

=

fl

—

0

whence and

Now the x1 are linearly independent, and so )J=O. Similarly it follows

that Suppose that

E=E1$E2

(1.17)

is a decomposition of E as a direct sum of subspaces and let F be an arbitrary subspace of E. Then it is not in general true that

F=F

n E1

Fn

E2

§ 3. Subspaces

and factor spaces

the example below will show. However, if E1 In fact, it is clear that

as

25

then (1.18) holds. (1.19)

On the other hand, it'

y=x1 +x2

x1eE1,x2eE2

is the decomposition of any vector yEF, then

x1eE1=Ffl E1, x2=y—x1eFfl E2. It

follows that

FcFfl

E2.

(1.20)

The relations (1.19) and (1.20) imply (1.18). Example 1: Let E be a vector space with a basis e1, e2. Define E1, E2 and F as the subspaces generated by e1, e2 and e1 + e2 respectively. Then

E =E1 while on the other hand

Fn

E1

=F

n E2

= 0.

Hence

F+Fn

E2.

1.14. Arbitrary families of subspaces. Next consider an arbitrary family of subspaces E, Then the intersection flEa is again a subspace of E. The sum EE1 is defined as the set of all vectors which can be written as finite sums,

(1.21)

and is a subspace of E as well. If for every

then each vector of the sum

can be uniquely represented in the form

(1.21). In this case the space

is called the (internal) direct sum of the

and is denoted by

subspaces

If

is a system of generators for

then the set

If the sum of the

is direct and

generators for then

is a basis for

is a system of

is a basis of

Chapter 1. Vector spaces

26

Example 2. Let by Then

be a basis of E and E2 be the subspace generated

E= Suppose (1.22)

is a direct sum of subspaces. Then we have the canonical injections We define the canonical projections

determined by

= where

It is clear that the are surjective linear mappings. Moreover, it is easily verified that the following relations hold: c'

XEE.

1.15. Complementary subspaces. An important property of vector spaces is given in the

Proposition I: If E1 is a subspace of E, then there exists a second subspace E2 such that E = E1 E2. E2 is called a complementary subspace for E1 in E. Proof We may assume that E1 E and E1 (0) since the proposition is trivial in these cases. Let (xx) be a basis of E1 and extend it with vectors to form a basis of E (cf. Corollary TI to Theorem I, sec. 1.6). Let E2 Then be the subspace of E generated by the vectors

E=

E1

E2.

In fact, since (xx) U (yp) is a system of generators for E, we have that

E=E1+E2.

(1.23)

On the other hand, if XEE1 fl E2, then we may write

x=

x=

Yfi II

and factor spaces

§ 3. Subspaces

27

whence

= 0.

— 13

Now since the set (x2) U

(y13) is

linearly independent, we obtain and

whence x=0. It follows that E1 fl E2=O and so the decomposition (1.23) is direct. As an immediate consequence of the proposition we have Corollary I. Let E1 be a subspace of E and : E1 —* F a linear mapping (Fa second vector space). Then ço1 may be extended (in several ways) to a linear map (p: E—*F.

Proof Let E2 be a complementary subspace for E1 in E,

E=E1Ej3E2

(1.24)

and define p by çox =

where

x—y+z is the decomposition of x determined by (1.24). Then 'p

=

+

'p

=

+

=

= and so q, is linear. It is trivial that 'p extends 'pr.

As a special example we have:

Corollary II: Let E1 be a subspace of E. Then there exists a surjective linear map —*

E1

such that

çox=x

xEE1.

Chapter I. Vector spaces

28

Proof Simply extend the identity map i : E1 —÷E1 to a linear map (p:E-*E1. 1.16.

Factor spaces. Suppose E1 is a subspace of the vector space E.

Two vectors XE E and x' e E are called equivalent mod E1 if x' — XE E1. it

is easy to verify that this relation is reflexive, symmetric and transitive and hence is indeed an equivalence relation. (The equivalence classes are the cosets of the additive subgroup E1 in E(cf. sec. 0.4)). Let E/E1 denote the set of the equivalence classes so obtained and let be

the set mapping given by

7rXX, where

XEE

is the equivalence class containing x. Clearly m is a surjective

map.

Proposition II: There exists precisely one linear structure in E/E1 such that m is a linear mapping. Proof: Assume that E/E1 is made into a vector space such that m is a linear mapping. Then the equations

m(x + y) =

mx

+ my

and m (2 x) =

2

mx

show that the linear operations in E/E1 are uniquely determined by the linear operations in E. it remains to be shown that a linear structure can be defined in E/E1 such that m becomes a linear mapping. Let i and 9 be two arbitrary elements of E/E1 and choose vectors XEE and yEE such that

my=9. Then the class m (x + y) depends only on

and 9. Assume for instance that

x'EE is another vector such that mx'=i. Then mx' = mx and hence we may write

x'=x+z,

zEE1.

it follows that

x' + y = (x + y) + z whence

m(x' + y)= m(x + y).

§ 3. Subspaces and factor spaces

We now define the sum of the elements where

29

and 9eE/E1 by

and 9=7ry.

(1.25)

It is easy to verify that E/E1 becomes an abelian group under this operation and that the class = E1 is the zero-element. Now let be an arbitrary element and 2eF be a scalar. Choose xeE such that Then a similar argument shows that the class ir(Ax) depends only on (and not on the choice of the vector x). We now define the scalar multiplication in E/E1 by where

(1.26)

Again it is easy to verify that the multiplication satisfies axioms 11.1—11.3

and so E/E1 is made into a vector space. It follows immediately from (1.25) and (1.26) that

ir(x+y)=itx+iry

x,yeE

= 2itx i.e., iv is a linear mapping.

The vector space E/E1 obtained in this way is called the factor space of E with respect to the subspace E1. The linear mapping iv is called the canonical projection of E onto E1. If E1 = E, then the factor space reduces to the vector 0. On the other hand, if E1 =0, two vectors xeE and yeE are equivalent mod E1 if and only if y = x. Thus the elements of E/(0) are the singleton sets {x} where x is any element of E, and iv is the linear isomorphism Consequently we identify E and E/(0). 1.17. Linear dependence mod a subspace.. Let E1 be a subspace of E, and suppose that (x2) is a family of vectors in E. Then the will be called linearly dependent mod E1 if there are scalars not all zero, such that

If the

are not linearly dependent mod E1 they will be called linearly

independent mod E1. Now consider the canonical projection

It follows immediately from the definition that the vectors dependent (independent) mod E1 if and only if the vectors dependent (independent) in E/E1.

are linearly are linearly

Chapter 1. Vector spaces

1.18. Basis of a factor space. Suppose that U (:fl) is a basis of E form a basis of E1. Then the vectors 7r:p form a such that the vectors basis of E/E1. To prove this let E2 be the subspace of E generated by the vectors zp . Then

(1.27)

Now consider the linear mapping

defined by

goz=rrz Then

ZEE2.

is surjective. In fact, let

be an arbitrary vector. Since

it:E—*E/E1 is surjective we can write

x—irx,

xEE.

In view of (1.27) the vector x can be decomposed in the form

x=y+z

yEE1,ZEE2.

(1.28)

Equation (1.28) yields

= irx = icy + Trz = and so çø is surjective. To show that çø is injective assume that

çoz=çoz'

=

z,z'eE2.

Then —

z) = p(z' — z) = 0

and hence z'—zeE1. On the other hand we have that z'—zeE2 and thus

z'—zeE1 fl is a linear isomorphism and now PropoIt follows that sition Ill of sec. 1.11 shows that the vectors irzp form a basis of E/E1.

Problems 1. Let be an arbitrary vector in F3. Which of the following subsets are subspaces? a) all vectors with b) all vectors with =0 c) all vectors with d) all vectors with = F,,, Fd generated by the sets of problem 1, 2. Find the subspaces and construct bases for these subspaces.

§ 3. Subspaces and factor spaces

31

3. Construct bases for the factor spaces determined by the subspaces of problem 2.

4. Find complementary spaces for the subspaces of problem 2, and construct bases for these complementary spaces. Show that there exists more than one complementary space for each given subspace. 5. Show that

a) 13=Fa+Fb b)

13=Fb+FC

c) 13=Fa+Fc Find the intersections Fa fl Fb, Fb fl

Fa n

and decide in which cases

the sums above are direct.

6. Let S be an arbitrary subset of E and let be its linear closure. is the intersection of all subspaces of E containing S. Show that 7. Assume a direct composition Show that in each class of E with respect to E1 (i.e. in each coset there is exactly one vector of E2. 8. Let E be a plane and let E1 be a straight line through the origin. What is the geometrical meaning of the equivalence classes with respect to E1. Give a geometrical interpretation of the fact that x

y

x' + i.

Suppose S is a set of linearly independent vectors in E, and suppose T is a basis of E. Prove that there is a subset of T which, together with S, is again a basis of E. and E_ defined 10. Let w be an involution in E. Show that the sets 9.

by

= {xeE; wx = x}, are

E_ = {xeE; wx = — x}

subspaces of E and that

E=

E..

11. Let E1, E2 be subspaces of E. Show that E1 + E2 is the linear closure of E1 u E2. Prove that

E1 + E2 =

E1

U E2

if and only if

E1DE2 or E2DE1. 12. Find subspaces E1, E2, E3 of r3 such that

i) En E1=0 (i+j) ii) E1 + E2 + E3 = p3 iii) the sum in ii) is not direct.

Chapter I. Vector spaces

32

§ 4.

Dimension

In this paragraph all vector spaces are defined over a fixed, but arbitrarily chosen field F of characteristic 0. 1.19. Finitely generated vector spaces. Suppose E is a finitely generated vector space, and consider a surjective linear mapping (p:E—+F. Then F is a system of generators for is finitely generated as well. In fact, if generate F. In particular, the factor space E, then the vectors çox1, ..., of a finitely generated space with respect to any subspace is finitely generated. Now consider a subspace E1 of E. In view of Cor. II to Proposition I, sec. 1.15 there exists a surjective linear mapping ço:E—÷E1. It follows that E1 is finitely generated.

1.20. Dimension. Recall that every system of generators of a nontrivial vector space contains a basis. It follows that a finitely generated non-trivial vector space has a finite basis. In the following it will be shown that in this case every basis of E consists of the same number of vectors.

This number will be called the dimension of E and will be denoted by dim E. E will be called a finite-dimensional vector space. We extend the definition to the case E=(0) by assigning the dimension 0 to the space (0). If E does not have finite dimension it will be called an infinite-dimensional vector space.

Proposition 1. Suppose a vector space has a basis of 11 vectors. Then every family of (n+ 1) vectors is linearly dependent. Consequently, ii is the maximum number of linearly independent vectors in E and hence every basis of E consists of n vectors. Proof: We proceed by induction on ii. Consider first the case n= 1 and let a be a basis vector of E. Then if x 0 and y 0 are two arbitrary vectors we have that whence

and y=pa, ii+O —

= 0.

Thus the vectors x and y are linearly dependent. Now assume by induction that the proposition holds for every vector space having a basis of 1 vectors. = 1. ii) be a basis of E and a family of ii + I vectors. We may assume that + 0 because

Let F be a vector space, and let alL x1 . .

. .

1

§ 4. Dimension

33

otherwise it would follow immediately that the vectors x1 . 1 were linearly dependent. Consider the factor space E1 =E/(xn+ i) and the canonical projection

it: E —÷ E/(xn+i) Since the system where (xn+ i) denotes the subspace generated by generates E1 it contains a basis of E1 (cf. Cor. Ito Theorem I, a1, ...,

sec. 1.6). On the other hand the equation Xn+i

implies that

vi

AV

=0

and so the vectors (a1, ..., an) are linearly dependent. It follows that E1

has a basis consisting of less than n vectors. Hence, by the induction hypothesis, the vectors are linearly dependent. Consequently, there exists a non-trivial relation

=0 and so

=

Xn+i.

This formula shows that the vectors Xi...Xn+i are linearly dependent and

closes the induction.

Example: Since the space vectors it follows that

n

(cf. Example 1, sec. 1.2) has a basis of n

= n.

Proposition II: Two finite dimensional vector spaces E and F are isomorphic if and only if they have the same dimension. Proof: Let be an isomorphism. Then it follows from Proposition III, sec. 1.11 that p maps a basis of E injectively onto a basis of F and so dim E=dim F. Conversely, assume that dim E=dim F—n and let and = 1.. .n) be bases of E and F respectively. According to Propo-

sition II, sec. 1.11 there exists a linear mapping

such that

1...n). Then p maps the basis onto the basis it is a linear isomorphism by Proposition Ill, sec. 1.11.

and hence

3

(Ireub. Lineur Algebra

Chapter I. Vector spaces

34

1.21. Subspaces and factor spaces. Let E1 be a subspace of the n-dimen-

sional vector space E. Then E1 is finitely generated and so it has finite dimension in. Let Xi...Xm be a basis of E1. Then the vectors xi...xm are linearly independent in E and so Cor. II to Theorem I, sec. 1.6 implies that the vectors may be extended to a basis of E. Hence

dimE1 dimE.

(1.29)

if equality holds, then the vectors xi...xm form a basis of E and it follows that E1 = E. Now it will be shown that

dimE=dimE1 +dimE/E1.

(1.30)

If E1 =(0) or E1 =E (1.30) is trivial and so we may assume that E1 is a proper non-trivial subspace of E,

0< dimE1 , is called the scalar product of and x, The scalar and the bilinear function is called a scalar product between E* and E. 2.22.

Examples.

1

Let E= E* = F and define a mapping

by

2,peF.

=2ji

Clearly is a non-degenerate bilinear function, and hence F can be regarded as a self-dual space.

2. Let E= E* = F" and consider the bilinear mapping defined by = 1=1 Greub. Li near AJgcbra

Chapter II. Linear mappings

where

x=

...,

It is easy to verify that the bilinear mapping is non-degenerate and is dual to itself.

hence

3. Let E be any vector space and E* = L (E) the space of linear functions in E. Define a bilinear mapping by

=f(x),

JEL(E),XEE.

Sincef(x)=O for each XEE if and only iff=O, it follows that NL(E)=O. On the other hand, let aeE be a non-zero vector and E1 be the onedimensional subspace of E generated by a. Then a linear function g is defined in E1 by g(x)=)L where In view of sec. 1.15, g can be extended to a linear function f in E. Then

a> =

f (a) =

g

1 +0.

(a)

It follows that NE = 0 and hence the bilinear function is non-degenerate.

This example is of particular importance because of the following

Proposition I: Let E*, E be a pair of vector spaces which are dual with respect to a scalar product Then an injective linear map & E*_*L(E) is defined by =

E E*, XE E.

x>

Proof? Fix a vector a*EE*. Then a linear function,

ja*k) = Since

is

bilinear,

X>

XE E.

(2.39)

is defined in E by (2.40)

depends linearly on a*. Now define cP by setting (2.41)

To show that 1i is injective, assume that i(ci*)=O for some a*EE*. Then

X>=0 for every XEE. Since

is

non-degenerate. it follows

that a*=0. Note: It will be shown in sec. 2.33 that P is surjective (and hence a linear isomorphism) if E has finite dimension.

§ 5. Dual vector spaces

67

e E * and xe E are 2.23. Orthogonal complements. Two vectors called orthogonal if = 0. Now let E1 be a subspace of E. Then the of E*. vectors of E* which are orthogonal to E1 form a subspace is called the orthogonal complement of E1. In the same way every subspace

determines an orthogonal complement (Er)' E. The fact that the bilinear function is non-degenerate can now be expressed by the relations

E'=O and (E*)±=O. It follows immediately from the definition that for every subspace E1

E

(2.42)

E1

Suppose next that E*, E are a pair of dual spaces and that F is a sub-

space of E. Then a scalar product is induced in the pair E*/F', F by VEF,

where is a representative of the class In fact, let be the restriction of the scalar product to E* x F. Then the nullspaces of cli are given by

and NF=O. Now our result follows immediately from sec. 2.21.

More generally, suppose Fc E and H* E* are any subspaces. Then a scalar product in the pair H*/H* fl F', F/Fn (H*)', is determined by

= as

a similar argument will show.

2.24. Dual mappings. Suppose that E, E* and F, F* are two pairs of dual spaces and p:E—÷F, are called dual if 'p and

are linear mappings. The mappings

= exists at most one dual mapping.

To a given linear mapping

If

and

are dual to 'p we have that =

=

and

whence —

=

0

XEE,Y*EF.

Chapter II. Linear mappings

68

This implies, in view of the duality of E and E*, that

=

whence

an example of dual mappings consider the dual pairs E*, E and is a subspace of E (cf. sec. 2.23) and let it be the canonical projection of E* onto E*/EjL, As

E*/EjL, E1 where E1

E*. Then the canonical injection

i:E1

is dual to it. In fact, if xeE1, and y*EE* are arbitrary, we have

= =

=

and thus it

= i*.

2.25. Operations with dual mappings. Assume that E*, E and F*, F are two pairs of dual vector spaces. Assume further that p: E—*F and i/i:E-+F are linear mappings and that there exist dual mappings p*: and ,/,*:E*÷_F*. Then there are mappings dual to and 2p and these dual mappings are given by = (p* + 1/1*

(p +

(2.43)

and (2.44)

(2.43) follows from the relation + VJ*)y*,x> =

+ +

—

=

+

and (2.44) is proved in the same way. Now let G, G* be a third pair of F* +— G * be a pair of dual mappings. Then dual spaces and let x: F—* G, the dual mapping of x p exists, and is given by :

(xoco)* =

In fact, if z*EG* and xeE are arbitrary vectors we have that = .

§ 5. Dual vector spaces

69

For the identity map we have clearly —1 —

\*

Now assume that p: E—+F has a left inverse 'Pi : F—+E, (2.45) and that the dual mappings

and

exist. Then we

obtain from (2.45) that = (1E)*

(2.46)

In view of sec. 2.11 the relations (2.45) and (2.46) are equivalent to p injective,

surjective

injective,

(p* surjective.

and

In particular, if p and

are inverse linear isomorphisms, then so are (p* and 'pt. 2.26. Kernel and image space. Let p : E—*F and (p*: be a pair of dual mappings. In this section we shall study the relations between the subspaces

kerpcE, ImqcF

and ker p*

F

Im (p* c E*.

First we establish the formulae

kerp* = (lmço)'

(2.47)

kerp =(Imp*)±.

(2.48)

In fact, for any two vectors y*eker p*, pxelm çü we have = =

0

and hence the subspaces ker and Im 'p are orthogonal, ker Now let y*E(Im p)' be any vector. Then for every xeE EJ. j*i

Chapter II. Linear mappings

74

is again non-

Moreover, the restriction of the scalar product to E1, degenerate, and

Proposition V has the following converse: Proposition VI. Let E1 c E be any subspace, and let subspace dual to E1 such that

L (E) be a

= Then (2.56)

and

L(E) =

(2.57)

Proof: We have

(E1 + Er)' =

n

(Er)' =

n

=0

whence

E=O'=(E1

(2.58)

On the other hand, since E1 and E1 n

are dual, it follows that =0

which together with (2.58) proves (2.56). (2.57) follows from Proposition V and (2.56).

Problems 1. Given two pairs of dual spaces E*, E and F*, F prove that the spaces

and EG3F are dual with respect to the bilinear function = + •

2. Consider two subspaces E1 and E2 of E. Establish the relation (E1

n

E

consider the mapping ).:

fined by

Prove that is injective. Show that

de-

aEE, jeL(E). is surjective if and only if dim E
= cp(x*, x)

E

E*, xe E

and so a mapping (p : E—÷E is defined. The linearity of çø follows immediately from the bilinearity of and Suppose now that are two

linear transformations of F such that cli(x*,x) =

x>

and

cli(x*,x) =

(p2x>

Then we have that

= 0 whence Proposition III establishes a canonical linear isomorphism between the

spaces B(E*, E) and L(E; F),

L(E; E). Here B(E*, F) is the space of bilinear functions &E* x dition and scalar multiplication defined by (tIi1 + tli2)(x*,x) =

and

(2 k)(x*, x) =

with ad-

+ cP2(x*,x) çji (x*, x).

2.34. The rank of a linear mapping. Let (p : E—÷F be a linear mapping of finite dimensional vector spaces. Then ker cp c E and Tm (p F have finite dimension as well. We define the rank of (p as the dimension of Tm (p

r((p) = dim Im (p. In view of the induced linear isomorphism Im(p

we have at once r((p) + dim ker(p = dimE.

(2.64)

(p is called regular if it is injective. (2.64) implies that (p is regular if and only if r ((p) = dim F.

Chapter 11. Linear mappings

In the special case dim E= dim F (and hence in particular in the case of a linear transformation) we have that is regular if and only if it is surjective. 2.35. Dual mappings. Let E*, E and F*, F be dual pairs and çø: be a linear mapping. Since E* is canonically isomorphic to the space L (E) there exists a dual mapping Hence we have the relations (cf. sec. 2.281

Im

= (ker

and (ker(p)L.

The first relation states that the equation = has

a solution for a given yEF if and only if y satisfies the condition

=O forevery

x*ekerp*.

The second relation implies that dual mappings have the same rank. In fact, from (2.63) we have that

dimlmp* = dim(kerp)' = dimE — dimkerço = dimlmp whence r(co*)

= r(p).

(2.65)

Problems (In problems 1—10 it will be assumed that all vector spaces have finite dimension).

1. Let E, E* be a pair of dual vector spaces, and let E1, E2 be subspaces of E. Prove that (E1 n

Hint: Use problem 2, § 5. 2. Given subspaces Uc E and V* c E* prove that

dim(U' n V*) + dim U = dim(U

n

+ dim

3. Let E, E* be a pair of non-trivial dual vector spaces and let 'p : be a linear mapping such that 'p = (z*) - 'p for every linear auto1

morphism t of E. Prove that 'p =0. Conclude that there exists no linear mapping which transforms every basis of E into its dual basis.

§ 6. Finite dimensional vector spaces

(v = 1. . .n) of E, E* show that the

4. Given a pair of dual bases

••,

bases (x*' + again dual. v2

...,

and (x1,

are

5. Let E, F, G be three vector spaces. Given two linear mappings prove that

ço:E—*F and

and

6. Let E be a vector space of dimension n and consider a system of n linear transformations such that

(i,j=1...n). a) Show that every

has rank 1 a second system with the same property, prove that there exists a linear automorphism t of E such that

b) If

=

t.

a

— 1

7. Given two linear mappings p:

r(q) -

and

: E—>F prove that

r(ço) +

+

Show that the dimensions of the spaces M' (ço), are given by

8. § 2

dimM'(p) dim

(dim F —

(p) in problem 3,

r(p))dimE

((p) = dim ker çø dim F.

9. Show that the mapping

defines a linear isomorphism,

cP:L(E;F)3 L(F*;E*). 10. Prove that c1'M'(ço) =

and = Ml(ço*) where

the notation is defined in problems 8 and 9. Hence obtain the

formula

11. Let p:

r(ço) = r(p*).

be a linear mapping (E, F possibly of infinite dimension). Prove that Im ço has finite dimension if and only if ker 'p has finite codimension (recall that the codimension of a subspace is the dimension 6

Greub, Lineur Algebra

Chapter II. Linear mappings

of the corresponding factor space), and that in this case

codimkerp = dimlmp. 12. Let E and F be vector spaces of finite dimension and consider a bilinear function P in Ex F. Prove that dim E — dim NE = dim F — dim NF

where NE and NF denote the null spaces of P. Conclude that if dim E=dim F. then NE=O if and only if

Chapter III

Matrices In this chapter all vector spaces will be defined over afixed, but arbitrarily chosen field F of characteristic 0.

§ 1. Matrices and systems of linear equations 3.1. Definition. A rectangular array A—_(

(3.1)

)

is called a matrix of n rows and m columns or, in of nm scalars an n x m-matrix. The scalars are called the entries or the elements of the matrix A. The rows (v—_1...n) and therefore are called the

can be considered as vectors of the space row-vectors of A. Similarly, the columns —

...

considered as vectors of the space

(it = 1 ... m)

are called the column-vectors of A.

Interchanging rows and columns we obtain from A the transposed matrix

A*=(

).

(3.2)

In the following, matrices will rarely be written down explicitly as in This notation has the (3.1) but rather be abbreviated in the form A = disadvantage of not identifying which index counts the rows and which the columns. It has to be mentioned in this connection that it would be very undesirable — as we shall see — to agree once and for all to always let

Chapter III. Matrices

the subscript count the rows, etc. If the above abbreviation is used, it will be stated explicitly which index indicates the rows. 3.2. The matrix of a linear mapping. Consider two linear spaces E and With the aid of F of dimensions ii and rn and a linear mapping p: (v 1. . n) and bases = 1. in) in E and in F respectively, every can be written as a linear combination of the vectors vector (1u:=l...rn), . .

.

(v

1... n).

(3.3)

where v In this way, the mapping p determines an n x rn-matrix counts the rows and ,u counts the columns. This matrix will be denoted by M(p, or simply by M(tp) if no ambiguity is possible. Conversely, every n x in-matrix determines a linear mapping p:E-+F by the equations (3.3). Thus, the operator M('p)

M: 'p

defines a one-to-one correspondence between all linear mappings ço:E—*F and all n x rn-matrices.

3.3. The matrix of the dual mapping. Let E* and F* be dual spaces of E and F, respectively, and p: E-+F, a pair of dual mappings. Consider two pairs of dual bases X*V, (v = 1. n) and (/2 1.. . in) of E*, E and F*, F, respectively. We shall show that the two corresponding matrices and M('p*) (relative to these bases) are transposed, i.e., that M('p*) = M((p)*. (3.4) :

. .

The matrices M('p) and M('p*) are defined by the representations 'p

=

and

=

Note here that the subscript v indicates in the first formula the rows of

the matrix and

and in the second the columns of the matrix in the relation

Substituting

=

(3.5)

=

(3.6)

we obtain Now (p

=

YK> =

(3.7)

§ 1. Matrices and systems of linear equations

and

=

(3.8)

The relations (3.6), (3.7) and (3.8) then yield */1_ — V• JL

V

— as stated before — that the subscript v indicates rows of and columns of (ce) we obtain the desired equation (3.4). 3.4. Rank of a matrix. Consider an n x ni-matrix A. Denote by r1 and by r2 the maximal number of linearly independent row-vectors and co-

Observing

lumn-vectors, respectively. It will be shown that r1 =r2. To prove this let E and F be two linear spaces of dimensions n and iii. Choose a basis x. (v 1.. .n) and y,2 (p = 1.. rn) in E and in F and define the linear mapping p: F by (/1X% = Besides

'p, consider the isomorphism f3:

F

F"'

defined by where

y= Then

j3 o p

is a linear mapping of E into F"'. From the definition of fi it into the v-th row-vector,

follows that f3 a p maps

=

/3

Consequently,

the rank of flap

is

equal to the maximal number r1 of

linearly independent row-vectors. Since /3 is a linear isomorphism, has the same rank as p and hence r1 is equal to the rank r of p. Replacing çø by we see that the maximal number r2 of linearly independent column-vectors is equal to the rank of ço*. But (p* has the same rank as ço and thus r1 = r2 = r. The number r is called the rank of the matrix A. 3.5. Systems of linear equations. Matrices play an important role in the discussion of systems of linear equations in afield. Such a system

=

(p = 1... in)

(3.9)

Chapter III. Matrices

of in equations with ii unknowns is called inhomogeneous if at least one is different from zero. Otherwise it is called homogeneous. From the results of Chapter II it is easy to obtain theorems about the existence and uniqueness of solutions of the system (3.9). Let E and F be two linear spaces of dimensions ii and in. Choose a basis (v = I .n) of .

E as well as a basis

(p =

1

. . .

in)

.

of F and define the linear mapping

F by

(p:

= Consider two vectors (3.10)

and

(3.11)

Then (3.12)

Comparing the representations (3.9) and (3. 12) we see that the system (3.9) is equivalent to the vector-equation X

= j.

Consequently, the system (3.9) has a solution if and only if the vector y

is contained in the image-space Im tp. Moreover, this solution is uniquely determined if and only if the kernel of ço consists only of the zero-vector. 3.6. The homogeneous system. Consider the homogeneous system =

0

(p

1... in).

(3.13)

is a From the foregoing discussion it is immediately clear that solution of this system if and only if the vector x defined by (3.10) is contained in the kernel ker p of the linear mapping p. In sec. 2.34 we have shown that the dimension of ker 'p equals n—r where r denotes the rank of 'p. Since the rank of 'p is equal to the rank of the matrix we therefore obtain the following theorem:

A homogeneous system of m equations with n unknowns whose coefficient-

matrix is of rank r has exactly n — r linearly independent solutions. In the special case that the number in of equations is less than the number n of unknowns we have ii— r n — in 1. Hence the theorem asserts that the system (3. 13) always has non-trivial solutions if ni is less than n.

§ I. Matrices and systems of linear equations

87

3.7. The alternative-theorem. Let us assume that the number of equations is equal to the number of unknowns,

=

= 1... n).

(3.14)

Besides (3.14) consider the so-called "corresponding" homogeneous system =0 (3.15) (ii = 1... n).

The mapping introduced in sec. 3.5 is now a linear mapping of the ndimensional space E into a space of the same dimension. Hence we may apply the result of sec. 2.34 and obtain the following alternative-theorem: If the homogeneous system possesses only the trivial solution (0.. .0), the inhomogeneous system has a solution

every choice of the right-

..

hand side. If the homogeneous system has non-trivial solutions, then the inhoinogeneous one is not so/cable for eterv choice of the ;7v(v = . ii). From the last statement of section 3.5 it follows immediately that in the first case the solution of (3.14) is uniquely determined while in the second case the system (3.14) has — if it is solvable at all — infinitely many solutions. 1

.

3.8. The main-theorem. We now proceed to the general case of an arbitrary system

=

(1u

= 1 ... m)

(3.16)

of m linear equations in n unknowns. As stated before, this system has a solution if and only if the vector

y= is

contained in the image-space Im ço. In sec. 2.35 it has been shown that

the space Im p is the orthogonal complement of the kernel of the dual mapping In other words, the system (3.16) is solvable if and only if the right-hand side (p=l...m) satisfies the conditions (3.17)

for all solutions

=I

.. ni)

of the system

(v=1...n). We formulate this result in the following

(3.18)

Chapter III. Matrices

Main-theorem: An inhoniogeneous system of n equations in in unknowns has a solution if and only if every solution 1. in) of the transposed homogeneous system (3.18) satisfies the orthogonality-relation (3.17). . .

Problems 1. Find the matrices corresponding to the following mappings:

a) px=O. b) çox=x. c) px=2x. d)

x=

where

(v =

1,

..., n) is a given basis and in n is a

given number and

2. Consider a system of two equations in n unknowns

Find the solutions of the corresponding transposed homogeneous system.

3. Prove the following statement: The general solution of the inhomogeneous system is equal to the sum of any particular solution of this system and the general solution of the corresponding homogeneous system. 4. Let and be two bases of E and A be the matrix of the basistransformation Define the automorphism c' of E by Prove that A is the matrix of as well with respect to the basis as with respect to the basis 5. Show that a necessary and sufficient condition for the n x n-matrix A= to have rank 1 is that there exist elements and ..., j31 •.., f3fl such that f32 = (v = 1,2, ...,n; = 1,2, ...,n). if A +0, show that the elements cc and 13P are uniquely determined up to = 1. constant factors ). and p respectively, where as 6. Given a basis a linear space E, define the mapping p : p

=

Find the matrix of the dual mapping relative to the dual basis. 7. Verify that the system of three equations:

+ + = 3, —11—4' = 4, + 3ij + 34' = 1

§ 2. Multiplication of matrices

89

has no solution. Find a solution of the transposed homogeneous system which is not orthogonal to the vector (3, 4, 1). Replace the number 1 on the right-hand side of the third equation in such a way that the resulting system is solvable. 8. Let an inhomogeneous system of linear equations be given,

(ii=1,...,in). augmented matrix of the system is defined as the in x (n+ 1)-matrix ..., çi"). Prove that obtained from the matrix by adding the column the above system has a solution if and only if the augmented matrix has the same rank as the matrix (cd). The

§ 2.

Multiplication of matrices

3.9. The linear space of the n x m matrices. Consider the space L(E; F) of all n x rn-matrices. of all linear mappings ço:E—÷F and the set Once bases have been chosen in E and in F there is a 1-1 correspondence between the mappings p: E—÷ F and the n x rn-matrices defined by (3.19)

This correspondence suggests defining a linear structure in the set MnXrn such that the mapping (3.19) becomes an isomorphism. We define the sum of two n x rn-matrices and

A= as

B

the n x rn-matrix

A+B=

+

and the product of a scalar 2 and a matrix A as the matrix 2A

is immediately apparent that with these operations the set is a linear space. The zero-vector in this linear space is the matrix which has only zero-entries. Furthermore, it follows from the above definitions that It

i.e., that the mapping (3.19) defines an isomorphism between L (E; F) and the space X

Chapter III. Matrices

3.10. Product of matrices. Assume that and

are linear mappings between three linear spaces E, F, G of dimensions ii, in and I, respectively. Then /i is a linear mapping of E into G. Select in) and a basis (v I .. ii), (p = .. .1) in each of the three spaces. Then the mappings and i/i determine two matrices and by the relations = 1

. . .

1

and

These two equations yield = Consequently, the matrix of the mapping

ço

relative to the bases x%, and

is given by

(3.20)

The n x 1-matrix (3.20) is called the product of the ii x rn-matrix A = and the rn x /-matrix B= and is denoted by A B. It follows immediately from this definition that

= M(tp)M(t/i).

(3.21)

Note that the matrix M p) of the product-mapping p is the product of the matrices M(ço) and in reversed order of the factors. It follows immediately from (3.21) and the formulas of sec. 2.16 that the matrix-multiplication has the following properties:

A(.)LB1 +pB2)=).AB1 +pAB2 (AB)C = A(BC) (AB)* = B*A*. 3.11. Automorphisms and regular matrices. An ii x n-matrix A is called regular if it has the maximal rank /1. Let 'p be an automorphism of the n-dimensional linear space E and A = M ('p) the corresponding /1 x nmatrix relative to a basis (v = .. n). By the result of section 3.4 the rank of 'p is equal to the rank of the matrix A. Consequently, the matrix A is regular. Conversely, every linear transformation p: E—* E having a regular matrix is an automorphism. 1

.

§ 2. Multiplication of matrices

To every regular matrix A

A' such that

there

exists an in verse matrix, i.e., a matrix

AA' =

=

.1,

where J denotes the unit matrix whose entries are In fact, let cp be the E automorphism of E such that M((p)=A and let be the inverse automorphism. Then = 1

whence

M(p)M(q,)' =

o(p)

= M(i)

J

and

M(ço')M(p)= These

M(t)= J.

equations show that the matrix

A' = is the inverse of the matrix A.

Problems 1. Verify the following properties: a) b)

(L4)*=2A*.

c) 2. A square-matrix is called upper (lower) triangular if all the elements below (above) the main diagonal are zero. Prove that sum and product of triangular matrices are again triangular.

3. Let p be linear transformation such that 'p2 = 'p. Show that there exists a basis in which 'p is represented by a matrix of the form: O...O

1 1

.

rn

O...O

1

o

0

n—rn 0

o n

Chapter III. Matrices

92

4. Denote by A13 the matrix having the entry I at the place (i,j) and zero elsewhere. Verify the formula =

Prove that the matrices form a basis of the space M"

X

Basis-transformation

§ 3.

3.12. Definition. Consider two bases and E. Then every vector (v = 1. ii) can be written as

l...n) of the space

. .

=

(3.22)

Similarly, (3.23)

The two n x n-matrices defined by (3.22) and (3.23) are inverse to each other. In fact, combining (3.22) and (3.23) we obtain — xi,— A

This is equivalent to

5:A = 0

— A

p

and hence it implies that

In a similar way the relations

are proved. Thus, any two bases of E are connected by a pair of inverse matrices. Conversely, given a basis (v = 1. n) and a regular n x n-matrix another basis can be obtained by = . .

To show that the vectors

are linearly independent, assume that

= 0. Then

=0

§ 3. Basis-transformation

and

93

hence, in view of the linear independence of the vectors

Multiplication with the inverse matrix

yields

(K=1...n). 3.13. Transformation of the dual basis. Let E* be a dual space of E, the dual of the basis (v I.. .n). Then the dual basis of and = where

is

(3.24)

a regular n x n-matrix. Relations (3.23) and (3.24) yield

;>.

x%) =

(3.25)

Now xi,> =

and

=

Substituting this in (3.25) we obtain I-'v

This shows that the matrix of the basis-transformation is the inverse of the matrix of the transformation The two basis-transformations and = (3.26) = are called contragradient to each other. The relations (3.26) permit the derivation of the transformation-law

for the components of a vector xeE under the basis-transformation Decomposing x relative to the bases

and

we obtain

and

From the two above equations we obtain in view of (3.26). =

=

(3.27)

Comparing (3.27) with the second equation (3.26) we see that the components of a vector are transformed exactly in the same way as the vectors of the dual basis.

Chapter III. Matrices

94

3.14. The transformation of the matrix of a linear mapping. In this section it will be investigated how the matrix of a linear mapping (p: E—÷ F is changed under a basis-transformation in E as well as in F. Let M(p; xi,, and M (p; be the ii x in-matrices of çø relative = = to the bases xv,, (v = .. in), respectively. Then and .1?, p = 1

.

1

(v=1...n).

and

Introducing the matrices A=

and

of the basis-transformations we then have the relations

-

(3.28)

B= and their inverse matrices,

and

=

=

=

=

(3.29) YK.

Equations (3.28) and (3.29) yield

and we obtain the following relation between the matrices =

and (n): (3.30)

Using capital letters for the matrices we can write the transformation formula (3.30) in the form = A M (p;

M ((p;

It shows that all possible matrices of the mapping p are obtained from a particular matrix by left-multiplication with a regular 11 x n-matrix and right-multiplication with a regular in x in-matrix.

Problems 1. Let f be a function defined in the set of all ii x n-matrices such that

f(TAT1)=f(A) for every regular matrix T. Define the function F in the space L (E; E) by

F (q) =

f (M (ço;

xv))

§ 4. Elementary transformations

95

where E is an n-dimensional linear space and (v = 1.. . n) is a basis of E. Prove that the function F does not depend on the choice of the basis is a linear transformation E—+E having the same 2. Assume that matrix relative to every basis (v = 1. n). Prove that p = 2i where 2 is a scalar. 3. Given the basis transformation . .

= X2

—

x2

—

x3

— X2

X3 = 2x2 + x3

find all the vectors which have the same components with respect to the bases x1, and

1,

§ 4.

2, 3).

Elementary transformations

3.15. Definition. Consider a linear mapping p: Then there exists a basis ..., n)ofE and a basis ..., n) ofFsuch that the corresponding matrix of 'p has the following normal-form:

r (3.31)

b

.

.

where r is the rank of 'p. In fact, let a basis of E such form a basis of the kernel. Then the vectors that the vectors ar+ bQ 'p aQ = 1, ..., r) are linearly independent and hence this system can be extended to a basis (b1, ..., bm) of F. It follows from the construction and that the matrix of 'p has the form (3.31). of the bases 1, ..., n) and 1, ..., m) be two arbitrary bases of Now let E and F. It will be shown that the corresponding matrix M('p; can be converted into the normal-form (3.31) by a number of elementary basis-transformations. These transformations are:

Chapter III. Matrices

(1.1.) Interchange of two vectors and x1(i+j). (1.2.) Interchange of two vectors Yk and Yi (k 1). (11.1.) Adding to a vector an arbitrary multiple of a vector (11.2.) Adding to a vector Yk an arbitrary multiple of a vector Yi (1+ k). it is easy to see that the four above transformations have the following effect on the matrix M((p): (1.1.) Interchange of the rows I andj. (1.2.) Interchange of the columns k and 1. (11.1.) Replacement of the row-vector by (11.2.) Replacement of the column-vector bk by It remains to be shown that every n x rn-matrix can be converted into the normal form (3.31) by a sequence of these elementary matrix-transformations and the operations be the given ii x rn-matrix. 3.16. Reduction to the normal-form. Let otherwise the It is no restriction to assume that at least one matrix is already in the normal-form. By the operations (1.1.) and (1.2.) this element can be moved to the place (1, 1). Then 0 and it is no restriction to assume that = 1. Now, by adding proper multiples of the first row to the other rows we can obtain a matrix whose first column consists of zeros except for Next, by adding certain multiples of the first column to the other columns this matrix can be converted into the form

o...o

1

0*

*

(3.32)

0* if all the elements

*

(v = 2.. .n, R = 2.. rn) are zero, (3.32) is the normal-

form. Otherwise there is an element This can be moved to the place (2,2) by the operations (1.1. and (1.2.). Hereby the first row and the first column are not changed. Dividing the second row by and applying the operations (11.1.) and (11.2.) we can obtain a matrix of the form 1

0

0

1

0 *

00*

*

In this way the original matrix is ultimately converted into the form (3.31.).

§ 4. Elementary transformations

97

3.17. The Gaussian elimination. The technique described in sec. 3.16 can be used to solve a system of linear equations by successive elimination. Let (3.33)

be a system of in linear equations in n unknowns. Before starting the elimination we perform the following reductions: If all coefficients in a certain row, say in the i-th row, are zero, consider the corresponding number on the right hand-side. If the i-th equation contains a contradiction and the system (3.33) has no solution. If the i-th equation is an identity and can be omitted. Hence, we can assume that at least one coefficient in every equation is different from zero. Rearranging the unknowns we can achieve that 40. Multiplying the first equation by — and adding it to the p-th equation we obtain a system of the form 1

+

= (3.34)

which is equivalent to the system (3.33).

Now apply the above reduction to the (rn—. 1) last equations of the system (3.34). If one of these equations contains a contradiction, the system (3.34) has no solutions. Then the equivalent system (3.33) does not have from the a solution either. Otherwise eliminate the next unknown, say reduced system.

Continue this process until either a contradiction arises at a certain step or until no equations are left after the reduction. In the first case, (3.33) does not have a solution. In the second case we finally obtain a triangular system 132+0 (3.35)

which is equivalent to the original system*).

*) If no equations are left after the reduction, then every n-tupie solution of (3.33). 7

Greub. Linear Algebra

...

is a

ChaI)ter I II. \4atrices

The system (3.35) can be solved in a step by step manner beginning

with (3.36)

1

—

Inserting (3.36) into the first (r— 1) equations we can reduce the system to a triangular one of r— equations. Continuing this way we finally obtain the solution of (3.33) in the form 1

(v=1...r)

JLr+ 1

In) are arbitrary parameters.

where the

Problems 1. Two n x ni-matrices C and C' are called equivalent if there exists a regular n x n-matrix A and a regular in x rn-matrix B such that C' = A CB. Prove that two matrices are equivalent if and only if they have the same rank. 2. Apply the Gauss elimination to the following systems: a) —

= 2.

+

b)

2,71+ 172+4173+ 3,71+4,72+ 2,7' + + 5,73 c)

+

2c'+c2

—

= 1,

= 3.

Chapter IV

Determinants In this chapter, except for the last paragraph, all vector spaces will be defined over afixed but arbitrarily chosen field F of characteristic 0.

§ 1. Determinant functions 4.1. Even and odd permutations. Let X be an arbitrary set and denote Let for each 1 by X" the set of ordered p-tuples (x1 Y

be a map from

to a second set, Y. Then every permutation

a new map

acP: X"-* Y

defined by

= It follows immediately from the definitions that (4.1)

and (4.2)

where i is the identity permutation. Now let X = Y= ZL and define cli by cP(x1

=

fl

—

•

x1 E ZL.

in.

Proposition II: Assume that E is an n-dimensional vector space and let P be a skew symmetric n-linear map from E to a vector space F. Then P is completely determined by its value on a basis of E. In particular, if P vanishes on a basis, then = 0. a, of E and write Proof: In fact, choose a basis a1

x)=

(,.= 1 ...

n).

Then çJ5

(x1

.. .

v,,) =

=

1)(a,

"P

1)

=

1)

.

a,,).

Since the first factor does not depend on P, the proposition follows. 4.3. Determinant functions. A determinant function in an n-dimensional vector space E is a skew symmetric n-linear function from E to F. In every n-dimensional vector space E (n 1) there are determinant functions which are not identically zero. In fact, choose a basis, f of the space L(E) and define the n-linear function P by

=

P(x1. ....

...

E E.

is the dual basis in E,

Then. if a1

(4.3)

Now set 4

Then A is a skew symmetric. Moreover, relation (4.3) yields A(a1

and so A

0.

=

§ 1. Determinant functions

103

Proposition III: Let E be an n-dimensional vector space and fix a nonzero determinant function 4 in E. Then every skew symmetric n-linear map ço from E to a vector space F determines a unique vector b E F such that = zl(x1

(p(x1

b.

(4.4)

of E so that zl (a1 = Then the n-linear map i/i: given by

Proof: Choose a basis a1

b=

p(a1

a,1).

= 4(x1

.

1

and set (4.5)

b

agrees with on this basis.Thus, by Proposition II. ,/i = p, i.e., = 4 Clearly, the vector b is uniquely determined by relation (4.4).

.

b.

Corollary: Let zl be a non-zero determinant function in E. Then every determinant function is a scalar multiple of 4.

Proposition IV: Let 4 be a determinant function in E (dim E=n). Then, the following identity holds: XEE, (4.6)

If the vectors x1 are linearly dependent a simple calculation shows that the left hand side of the above equation is zero. On the other hand, by sec. 4.2, 4(x1 Thus we may assume x,, form a basis of E. Then, writing

that the vectors x1

x we

have x1

=

(—

1

4 (xi,, x1

=4(x1

x1,

(4.7)

xv). x.

Problem Let E*, E be a pair of dual spaces and 4+0 be a determinant function in E. Define the function 4* of n vectors in E* as follows: Ifthevectorsx*v(v = .n)arelinearlydependent,then4* (x* 1•• .x*'l)= 0. 1

. .

Chapter I V. Detcrtiiinants

104

Ifthe vectors

(v 1 where

=

.n)are linearly independent, then zi * (x*

1

1

n) is the dual basis. Prove that /1* is a deter-

I

minant function in E*. § 2.

The determinant of a linear transformation

4.4. Definition. Let p be a linear transformation of the n-dimensional linear space E. To define the determinant of (p choose a non-trivial determinant function z1. Then the function defined by /14,

=

(x1 ...

A

x1

...

obviously is again a determinant function. Hence, by the uniquenesstheorem of section 4.3, /14, =

LI,

is a scalar. This scalar does not depend on the choice of A. In fact, if A' is another non-trivial determinant function, then zI'=2A and where

consequently Thus, the scalar

A ,,

=

2

=

is uniquely determined by the transformation p. It is

called the determinant of (p and it will be denoted by det (p. So we have the

following equation of definition: /14,

where A is an arbitrary non-trivial determinant function. In a less condensed form this equation reads = det (p A (x1 ...

A (p X1 ... p

(4.8)

In particular, if (p = 2i, then /14,

=

and hence

det(2i) = It follows from the above equation that the determinant of the identitymap is 1 and the determinant of the zero-map is zero. 4.5. Properties of the determinant. A linear transformation is regular if and only if its determinant is different from zero. To prove this, select a basis (v = 1.. .n) of E. Then A (p e1 ...

p

= det (p A (e1 ... en).

(4.9)

§ 2. The determinant of a linear transformation

If p is regular, the vectors p (v =

1.

. .

105

are linearly independent; hence

n)

(4.10)

Relations (4.9) and (4.10) imply that

+ 0.

det

+ 0. Then it follows from (4.9) that

Conversely, assume that det

Hence the vectors p (v = 1. . n) are linearly independent and Consider two linear transformations and i,Ii of E. Then

is regular.

.

det

det

In particular, if is a linear automorphism and p1 is the inverse automorphism, we obtain

detq1 det(p = deti =

1.

4.6. The classical adjoint. Let E be an n-dimensional vector space and let zi +0 be a determinant function in E. Let çoeL(E; E). Then an n-linear map E)

is given by

v,)x =

px1

j=

E.

.

1

It is easy to check that Ji is skew symmetric. Thus, by Proposition 11,

there is a unique linear transformation, ad(p), of E such that cP(x1

= z1(x1

ad(p)

E;

E

E.

This equation shows that the element ad(p)EL(E; E) is independent of the choice of A and hence it is uniquely determined by 'p. It is called the classical adjoint of'p.

Chapter IV. 1)eterminaiits

106

Proposition V: The classical adjoint satisfies the relations

ad(ço)o(p=1 •detq and i

. det(p.

Proof: Replacing x by pX in (4.14) we obtain x,) x1 = 4(x1

x,

IP

,...,x,) ad

x).

j=1

Now observe that, in view of the definition of the determinant and

Proposition IV.

P1

=det(p j= 1

=detq)

x1EE, XEE.

x

.

Composing these two relations yields

4(x1 ,...,x,) ad (p((px) = det

.

zl(x1 ,...,x,) x .

and so the first relation of the proposition follows. The second relation is established by applying p to (4.12) and using the identity in Proposition IV, with x replaced by (i= ... a). 1

Corollary: If go is a linear transformation with det an inverse and the inverse is given by go1

then go has

det go

Thus a linear transformation of E is a linear isomorphism if and only if its determinant is non-zero. 4.7. Stable subspaces. Let go: E—*E be a linear transformation and assume that E is the direct sum of two stable subspaces,

E=

E1

$E2.

Then linear transformations and

go2:E2—*E2

§ 2. The determinant of a linear transformation

107

are induced by (p. It will be shown that

det(p = detp1detq2.

Define the transformations

and

E—÷E

by

mE1 i

Then (P

and so det(p =

Hence it is sufficient to prove that det

=

det(p1

and

detk//2 = dettp2.

0 be a determinant function in E and h1 ... the function /11, defined by

Let zl

...

be a

basis of E2. Then

p=dimE,

... hqh

is a non-trivial determinant function in E1. Hence

= detp1A1(x1

A1 (Pi x1 ...

On the other hand we obtain from (4.14) ... xi,, b1 ... A1(x1 ... xi,). A (x1

=

bq)

These relations yield det(p1

The second formula (4.13) is proved in the same way.

Problems 1. Consider the linear transformation p:

(v=1...n) where

(v =

1.

.

.n)

is a basis of E. Show that

(4.13)

defined by

(4.14)

I \'.

Deicriiii

2. Let be a linear transformation and assume that E1 is a and stable subspace. Consider the induced transformations Prove that

detp = 3. Let x:E-+F be a linear isomorphism and p be a linear transformation of E. Prove that = detp. 4. Let E be a vector space of dimension ii and consider the space L(E; E) of linear transformations. a) Assume that F is a function in L (E; F) satisfying

F can be written in the form

F(p)= f(detq) wheref:F—+F is a mapping such that

=f(2)f(p). b) Suppose that F satisfies the additional condition that Then, if E is a real vector space,

= detp or F(p) = and if E is a complex vector space F(ço) = detçc.

Hint for part a): Let (i== 1.. .n) be a basis for E and define the transformations and by

v+i v=t and

v=i

.

Show first that =I and that F(ço1)

is

independent of i.

i,j=1...n

§ 3. The determinant of a matrix

4. Let E be a vector space with a countable basis and assume that a function F is given in L (E; E) which satisfies the conditions of problem 3a). Prove that

peL(E;E).

Hint: Construct an injective mapping that

such

and a surjective mapping i/i

=0.

5. Let w be a linear transformation of E such that w2 = i. Show that where r is the rank of the map i — w. — 6. Let j be a linear transformation of E such that = — i. Show that then the dimension of E must be even. Prove that det j = 1. 7. Prove the following properties of the classical adjoint: det w = (

1

(i)

(ii) det (iii) If p has rank n—i, then Im ad((p)=ker cm (iv) If p has rank n—2, then ad(p)=0. (v) 8.

ad(ad p)=(det If n=2 show that

ad(p)=i . trp—p. § 3.

The determinant of a matrix

4.8. Definition. Let ço be a linear transformation of E and sponding matrix relative to a basis (v = 1. . n). Then .

=

'p

Substituting

in (4.8) we obtain A ('p

= det 'p A (e1 ...

... 'p

The left-hand side of this equation can be written as A ('p e1, ... 'p

= =

We thus obtain

A

ep,

r(1)

... c(n)

.A(e ...

en).

the corre-

Chapter IV. Determinants

1 1()

This formula shows how the determinant of is expressed in terms of the corresponding matrix. We now define the determinant of an ii x n-matrix A by

f(n)

detA =

(4.16)

Then equation (4.15) can be written as det(p =

(4.17)

Now let A and B be two n x n-matrices. Then det (A B) = det A det B.

(4.18)

In fact, let E be an n-dimensional vector space and define the linear transformations 'p and of E such that (with respect to a given basis)

M(q)=A and M ('p) M (i/i) = det M p) = det (i/i p) = det M ('p) det M (i/i) = det A det B.

= det 'p det

Formula (4.18) yields for two inverse matrices

detAdet(A1) = detJ =

1

(J unit-matrix)

showing that

det(A1) = (detA)'. Finally note that if an (n x ;i)-matrix A is of the form

A = det

det A2

(4.19)

as follows from sec. 4.7. 4.9. The determinant considered as a function of the rows. If the rows

of the matrix A are considered as vectors of the space = . .. the determinant det A appears as a function of the n vectors (v = n). To investigate this function define a linear transformation 'p of by 1

(v=1...n)

. . .

§ 3. The determinant of a matrix where the vectors

111

are the n-tuples

(v=1...n). Now let A be the deterThen A is the matrix of relative to the basis minant function in r which assumes the value one at the basis (v = 1.. .n),

A

A

A

det

A (a1 ... an).

A

(4.20)

This formula shows that the determinant of A considered as a function of the row-vectors has the following properties: 1. The determinant is linear with respect to every row-vector. 2. If two row-vectors are interchanged the determinant changes the sign.

3. The determinant does not change if to a row-vector a multiple of another row-vector is added. 4. The determinant is different from zero if and only if the row-vectors are linearly independent. An argument similar to the one above shows that

where the bV are the column-vectors of A. It follows that the properties 1—4 remain true if the determinant of A is considered as a function of the column-vectors.

Problems 1. Let A = (cd) be a matrix such that

= 0 if v 0. However,

if det p

0

if p(n — p) is even if p(n—p)is odd.

(4.71)

is positive with respect to the original orienSince the basis of tation, relation (4.71) shows that the induced orientation coincides with the original orientation if and only if p(n—p) is even. 4.30. Example. Consider a 2-dimensional linear space 12. Given a basis (e1, e2) we choose the orientation of E in which the basis e1, e2 is positive. Then the determinant function A, defined by

zl(e1,e2) =

1

1,2) generrepresents this orientation. Now consider the subspace 1,2) with the orientation defined by Then E1 induces in ated by 122 the given orientation, but E2 induces in E1 the inverse orientation.

§ 8. Oriented vector spaces

135

In fact, defining the determinant-functions A1 and A1 (x)

A (e2,

x)

xe E1,

A2 (x) = A

and

A2 in E1 and in E2 by (e1,

x)

x e E2

we find that

4.31.

A2(e2) = A(e1, e2) =

1

Intersections. Let E1

and E2 be

and

A1 (e1) =

A(e2,

e1) = —

two subspaces of E

E = E1 + E2

1.

such

that (4.72)

and assume that orientations are given in E1, E2

and E. It

will be shown

that then an orientation is induced in the intersection E12=E1 n

E2.

Setting

dimE1 = p,

dimE2 = q,

dimE12 = r

we obtain from (4.72) and (1.32) that

r= p + q — n. Now consider the isomorphisms ço:E/E1 and Vi:E/E2

E1/E12.

Since orientations are induced in E/E1 and E/E2 these isomorphisms determine orientations in E2/E12 and in E1/E12 respectively. Now choose two positive bases +1• a,, and hr . bq in E1/E12 and E2/E12 respecand bJEE2 be vectors such that tively and let .

and

where it1 and

it2

denote the canonical projections and

ir2:E2—+E2/E12.

Now define the function A12 by A12 (z1 ... Zr) = A (Z1 ... Zr, ar+ 1

a,,, br+ 1 .. bq).

(4.73)

In a similar way as in sec. 4.30 it is shown that the orientation defined in depends only on the orientations of E1, and E (and not on E12 by the choice of the vectors and ba). Hence an orientation is induced in E12.

Chapter IV. Determinants

Interchanging E1 and E2 in (4.73) we obtain

(:i

•..

(:i

=A

r,

br+ 1

bq,

1

ar).

(4.74)

Hence it follows that =

(q-r)4

1)(P_r)

(n-q)4

= (_

(475)

Now consider the special case of a direct decomposition. Then p+q=n and E12=(0). The function /112 reduces to the scalar (4.76)

Moreover the number—2 depends

It follows from (4.76) that

only on the orientations of E1, E2 and E. It is called the intersection number

of the oriented subspaces E1 and E2. From (4.75) we obtain the relation =(

= 1.. ii) be two bases of E. Basis deformation. Let and is called deformable into the basis if there exist n Then the basis continuous mappings 4.32.

t t1

t

satisfying the conditions 1. (t0) = and (t1) = (t) (v = 2. The vectors 1

The

.

.11) are

linearly independent for every fixed t.

deformability of two bases is obviously an equivalence relation.

Hence, the set of all bases of E is decomposed into classes of deformable bases. We shall now prove that there are precisely two such classes. This is a consequence of the following Theorem: Two bases and (v = 1.. ./1) are deformable into each other if and only if the linear transformation p: E defined by 'p = has positive determinant.

Proof: Let 4+0 be an arbitrary determinant function. Then formula lii) are 4.17 together with the observation that the components continuous functions of shows that the mapping Ex x E—*R defined by A is continuous. into the is a deformation of the basis Now assume that basis Consider the real valued function (t) ...

§ 8. Oriented vector spaces

137

The continuity of the function 4 and the mappings

the function Ji

is

implies that

continuous. Furthermore,

ck(t)+0 the vectors (t) (v = I .n) are linearly independent. Thus the function cP assumes the same sign at t=t0 and at t=t1. But because

. .

= 4(pa1 ...

= A(b1

= det(pA(a1 ...

=

whence

detço >0 and so the first part of the theorem is proved. 4.33. Conversely, assume that the linear transformation a deformation assume first that the vector n-tuple (a1 ...

...

(4.77)

1). Then consider the de-

is linearly independent for every 1(1 composition = By the above assumption the vectors (a1..

are linearly independ-

ent, whence /3" + 0. Define the number c,, by

1+1 if if

L—1

It will be shown that the n mappings

Ix,(t)=a.(v=1...n—1) = define

(1 —

+

(0O.

(4.79)

and by the result of sec. 4.32 Relations (4.78), and (4.79) imply that > 0.

det But whence

+1. Thus, the number of equal to — is even. Rearranging the vectors (v = 1.. n) we can achieve that 1

.

5—i

vl+i

(v=1...2p)

(v=2p+1...n).

§ 8. Oriented vector spaces

139

Then a deformation b1

...

e,,

-÷ (b1 ...

is defined by the mappings

+

=— =—

(v — —

1

—

(v=2p+1...n) 4.34. The case remains to be considered that not all the vector n-tuples (4.77) are linearly independent. Let A + 0 be a determinant function. The linear independence of the vectors (v = 1 . n) implies that . .

Since zl is a continuous function, there exists a spherical neighbourhood that Ua

(v= 1...n).

if XvEUa

Choose a vector a'1 e Uaj which is not contained in the (n — 1)-dimensional subspace generated by the vectors (b2. . Then the vectors (a'1, b2.. are linearly independent. Next, choose a vector a'2 E Ua2 which is not con-

tained in the (n— 1)-dimensional subspace generated by the vectors (a'1, b2. Then the vectors (a'1, a'2, b3. are linearly independent. Going on this way we finally obtain a system of n vectors (v = 1.. n) .

.

.

such that every n-tuple ...

(a'1

is

linearly independent. Since

Hence the vectors

(1= 1 ... ;i)

it follows that

(v = 1. . n) form a basis of E. The n mappings .

define a deformation (4.80)

In fact,

(t)(0 t

1) is contained in Uav whence A

(t) ...

(t)) + 0

(0

t

This implies the linear independence of the vectors

1).

(t)(v = 1.. .n).

140

Chapter IV. Determinants

By the result of sec. 4.33 there exists a deformation ...

... ba).

(4.81)

The two deformations (4.80) and (4.81) yield a deformation (a1 ...

... ba).

This completes the proof of the theorem in sec. 4.32. 4.35. Basis deformation in an oriented linear space. If an orientation is given in the linear space E, the theorem of sec. 4.32 can be formulated as and (v = I n) can be deformed into each other follows: Two bases if and only if they are both positive or both negative with respect to the given orientation. In fact, the linear transformation . .

p:

.

(v = I ... a)

—*

has positive determinant if and only if the bases and b%, (v = I .. are both positive or both negative. Thus the two classes of deformable bases consist of all positive bases and all negative bases. 4.36. Complex vector spaces. The existence of two orientations in a

real linear space is based upon the fact that every real number 2t0 is either positive or negative. Therefore it is not possible to distinguish two orientations of a complex linear space. In this context the question arises whether any two bases of a complex linear space can be deformed into each other. It will be shown that this is indeed always possible. Consider two bases and (v = a) of the complex linear space E. As in sec. 4.33 we can assume that the vector n-tuples 1

(a1...

. .

...

are linearly independent for every 1(1 1). It follows from the above assumption that the coefficient in the decomposition = is

different from zero. The complex number

Now choose a positive continuous function

r(0) =

1,

r(I) = r

can

be written as

1) such that (4.82)

§ 8. Oriented vector spaces

and a continuous function

1) such that

(t)(O t

(4.83)

Define mappings

(t),

(0

t 1)

(v =

=

by 1 ...

n — 1) )

and (t)

=t

Then the vectors (t) (v = fact, assume a relation

1 ..

(4.84)

1.

t

flV

+ r(t) .

n)

are linearly independent for every t. In

0. Then

+

r (t)

+

=

0

whence

(v==1...n—1) and = 0.

1, the last equation implies that )J' = 0. r (t) + 0 for 0 t (n— 1) equations reduce to in— 1). It follows from (4.84), (4.82) and (4.83) that

Since

Hence the

first

=

and

=

Thus the mappings (4.84) define a deformation (a1 ...

—÷

Continuing this way we obtain after 1...n). into the basis

(a1 n

...

ba).

steps a deformation of the basis

Problems 1. Let E be an oriented n-dimensional linear space and

(v = 1.. .11) be

the subspace generated by the vectors a positive basis; denote by is positive with respect Prove that the basis (x1 to

the orientation induced in

by the vector

l)i —

1

xL.

Chapter IV. Determinants

142

2. Let E be an oriented vector space of dimension 2 and let a1, a2 be two linearly independent vectors. Consider the 1-dimensional subspaces E1 and E2 generated by a1 and a2 and define orientations in E. such that the bases a are positive (i= 1,2). Show that the intersection number of E1 and E2 is + 1 if and only if the basis a1, a2 of E is positive.

3. Let E be a vector space of dimension 4 and assume that (v

= ... 1

4)

is a basis of E. Consider the following quadruples of vectors:

I. e1+e2,e1+e2+e3,e1+e2+e3+e4,e1—e2+e4 II. e1+2e3, e2+e4, e2—e1+e4, e2

III.

e2+e4, e3+e2, e2—e1

1V.

e1+e2—e3, e2—e4, e3+e2, e2—e1

V. e1—3e3, e2+e4, e2—e1—e4, e2. a) Verify that each quadruple is a basis of E and decide for each pair of bases if they determine the same orientation of E. b) If for any pair of bases, the two bases determine the same orientation, construct an explicit deformation. c) Consider E as a subspace of a 5-dimensional vector space E and assume that (v = 1, . .5) is a basis of E. Extend each of the bases above to a basis of which determines the same orientation as the basis (v= 1, ..., 5). Construct the corresponding deformations explicitly. 4. Let E be an oriented vector space and let E1, E2 be two oriented subspaces such that E== E1 + E2. Consider the intersection E1 fl E2 together with the induced orientation. Given a positive basis (c1, ..., Cr) of extend it to a positive basis (c1, ..., Cr, ar+ E1 fl a positive basis (C1, ..., Cr, br+ i, ..., bq) of E2. Prove that then (C1, ..., Cr, ar+ 1' ..., ap, br+ i, ..., bq) is a positive basis of E. .

5. Linear isotopices. Let E be an n-dimensional real vector space. will be called linearly Two linear automorphisms p: and i/i: (I the closed unit isotopic if there is a continuous map 1: I x and such that for each interval) such that x)=p(x), x) is a linear automorphism. tEl the map given by (i) Show that two automorphisms of E are linearly isotopic if and only if their determinants have the same sign. (ii) Let j: E—+E be a linear map such that = — Show that j is linearly isotopic to the identity map and conclude that detj= 1. 6. Complex structures. A complex structure in real vector space E is a linear transformation j: E—+E satisfying j2 = — 1.

§8. Oriented vector spaces

143

(i) Show that a complex structure exists in an n-dimensional vector space if and only if ii is even. (ii) Let j be a complex structure in E where dimE=2n. Let 4 be a determinant function. Show that for (v= 1 ... 11) either 4(x1

or 4(x1 ,...,x,1,jx1 The natural orientation of (E, J) is

defined to be the orientation re-

presented by a non-zero determinant function satisfying the first of these.

Conclude that the underlying real vector space of a complex space carries a natural orientation. (iii) Let be a vector space with complex structure and consider the complex structure (E, —j). Show that the natural orientations of and (E, —j) coincide if and only if n is even (dimE=2n).

Chapter V

Algebras In paragraphs one and two all vector spaces are defined over a fixed, but arbitrarily chosen field F of characteristic 0.

§ 1. Basic properties 5.1. Definition: An algebra, A, is a vector space together with a mapping A x such that the conditions (M1) and (M2) below both hold. The image of two vectors xeA, yeA, under this mapping is called the product of x and y and will be denoted by xy. The mapping A x A-+A is required to satisfy: (M1) (M2)

(2x1 + jix2)y = 2(x1 y) + ,u(x2y) + 11y2)

2(xy1) + ji(xy2).

As an immediate consequence of the definition we have that

0x = x0 0. Suppose B is a second algebra. Then a linear mapping p : a homomorphism (of algebras) if çø preserves products; i.e.,

p(xy)=pxpy.

is called (5.1)

A homomorphism that is injective (resp. surjective, bijective) is called a monomorphism (resp. epimorphism, isomorphism). If B=A, p is called an endomorphism. Note: To distinguish between mappings of vector spaces and mappings of algebras, we reserve the word linear mapping for a mapping between vector spaces satisfying (1.8), (1.9) and homomorphism for a linear mapping between algebras which satisfies (5.1). Let A be a given algebra and let U, V be two subsets of A. We denote by U V, the set

§ 1. Basic properties

145

Every vector aeA induces a linear mapping p(a):A—*A defined by

1u(a)x=ax

(5.2)

4u (a) is called the multiplication operator determined by a.

An algebra A is called associative if

x(yz)=(xy)z and commutative if

xy=yx

x,y,zEA

x,yeA.

From every algebra A we can obtain a second algebra (x y)OPP

by defining

=yx

AON' is called the algebra opposite to A. It is clear that if A is associative

then so is A°". If A is commutative we have A°"=A. If A is an associative algebra, a subset Sc: A is called a system of generators of A if each vector xeA is a linear combination of products of elements in 5, 1... x= e S, 1... Vp ... (v)

A unit element (or identity) in an algebra is an element e such that for every x

xe=ex=x.

(5.3)

if A has a unit element, then it is unique. In fact, if e and e' are unit elements, we obtain from (5.3) e = e e' = e'. Let A be an algebra with unit element eA and ço be an epimorphism of A onto a second algebra B. Then eB = eA is the unit element of B. In fact,

if yEB is arbitrary, there exists an element xeA such that y= çox. This gives

yeB =

=

=

= y.

In the same way it is shown that An algebra with unit element is called a division algebra, if to every element a 0 there is an element a' such that a a — = a — = c. 1

5.2. Examples: 1. Consider the space L(E; E) of all linear transformations of a vector space E. Define the product of two transformations by 10

= i/i o Greub Linear Algebra

Chapter V. Algebras

146

The relations (2.17) imply that the mapping (p, i)—*çlip satisfies (M1) and (M2) and hence L(E; E) is made into an algebra. L(E; E) together with this multiplication is called the algebra of linear transformations of E and is denoted by A (E; E). The identity transformation i acts as unit element in A (E; E). It follows from (2.14) that the algebra A (E; E) is associative.

However, it is not commutative if dim E 2. In fact, write E= (x1) and (x2) are the one-dimensional subspaces generated by two linearly independent vectors x1 and x2, and F is a complementary subspace. Define linear transformations 'p and by

'px1=O,

(px2=x1,

py=O,yeF

ç/ix2=O;

çliy=O,yeF.

and Then

= 'p0 = 0 while (pX2

=

çlix1

= x2

whence

Suppose now that A is an associative algebra and consider the linear mapping defined by

p(a)x=ax.

(5.4)

Then we have that

p(a b)x = a bx = p(a)p(b)x whence

p(ab) = p(a)p(b). This

relation shows that p is a homomorphism of A

into A (A; A).

M X be the vector space of (n x mi)-matrices for a given integer ii and define the product of two (mi x n)-matrices by formula Example 2: Let

(3.20). Then it follows from the results of sec. 3.10 that the space is made into an associative algebra under this multiplication with the unit matrix J as unit element. Now consider a vector space E of dimension n with a distinguished basis (v = 1. n). Then every linear transforE determines a matrix M ('p). The correspondence 'p —+ M(co) mation 'p . .

§ 1. Basic properties

determines a linear isomorphism of A (E; E) onto

147 X

In view of sec.

3.10 we have that

M is an isomorphism of the algebra A (E; E) onto the opposite algebra (M" X n)0PP. Example 3: Suppose F1 cF is a subfield. Then F is an algebra over F1. We show first that F is a vector space over F1. In fact, consider the mapping F1 x F—+F defined by

(2,x)-÷2x, It satisfies the relations

2eF1,xEF.

+ x =2x+ x 2(x + y) = Ax + 2y (24u)x = 2(1ux)

lx = x where 2,

x, yeF. Thus F is a vector space over F by (x, y) x y (field multiplication).

Then M1 and M2 follow from the distribution laws for field multiplication.

Hence F is an associative commutative algebra over F1 with 1 as unit element.

Example 4: Let cr be the vector space of functions of a real variable t which have derivatives up to order r. Defining the product by

(1 g) (t) = f (t) g (t) we obtain an associative and commutative algebra in which the function 1(t) = 1 acts as unit element. 5.3. Subalgebras and ideals. A subalgebra, A1, of an algebra A is a linear subspace which is closed under the multiplication in A; that is, if x and y are arbitrary elements of A1, then xyeA1. Thus A inherits the structure of an algebra from A. It is clear that a subalgebra of an associative (commutative) algebra is itself associative (commutative). Let S be a subset of A, and suppose that A is associative. Then the subspace B A generated (linearly) by elements of the form

is

SEES Si...Sr, clearly a subalgebra of A, called the subalgebra generated by S. It is

easily verified that

where the 10*

B= A containing S.

Chapter V. Algebras

A right (left) ideal in an algebra A is a subspace I such that for every xEI, and every yEA, xyEI(yxEI). A subspace that is both a right and left ideal is called a tivt'o-sided ideal, or simply an ideal in A. Clearly, every

right (left) ideal is a subalgebra. As an example of an ideal, consider the subspace A2 (linearly generated by the products xy). A2 is clearly an ideal and is called the derived algebra. The ideal I generated by a set S is the intersection of all ideals containing S. If A is associative, I is the subspace of A generated (linearly) by elements of the form

s,as,sa

sES,aEA.

In particular every single clement a generates an ideal principal ideal generated by a.

4

is

called the

Example 5: Suppose A is an algebra with unit element e, and let (p:F-÷A be the linear mapping defined by ço2 = Ae.

Considering F as an algebra over itself we have that =

e

=

e)(R e) =

then ).e=0 whence p is a homomorphism. Moreover, if )L=0. It follows that p is a monomorphism. Consequently we may identify F with its image under p. Then F becomes a subalgebra of A and scalar multiplication coincides with algebra multiplication. In fact, if). is any scalar, then = (2e).a = = Example

6: Given an element a of an associative algebra consider

the set, Na, of all elements xEA such that ax =0. If xENa then we have for every VEA

and so xyENa. This shows that Na is a right ideal in A. It is called the right annihilator of a. Similarly the left annihilator of a is defined. 5.4. Factor algebras. Let A be an algebra and B be an arbitrary subspace of A. Consider the canonical projection

It will be shown that A/B admits a multiplication such that ir is a homomorphism if and only if B is an ideal in A. Assume first that there exists such a multiplication in A/B. Then for

§ 1. Basic properties

149

every xeA, yeB, we have

ir(xy) =

irxiry =

it follows that and so B must be an ideal. Conversely, assume B is an ideal. Then define the multiplication in

A/B by

= ir(xy)

(5.5)

where x and y are any representatives of

and 7 respectively.

It has to be shown that the above product does not depend on the choice of x and y. Let x' and y' be two other elements such that it x' = and iry' =7. Then

x'—xeB and y'—yeB. Hence we can write

x'==x+b, It

beB and y'—y+c,

ceB.

follows that

x'y' — xy = by + xc + bceB and so

rr(x'y') = it(xy). The multiplication in A/B clearly satisfies (M1) and (M2) as follows from the linearity of it. Finally, rewriting (5.5) in the form

it (x y) = it x iry

we see that it is a homomorphism and that the multiplication in A/B is uniquely determined by the requirement that it be a homomorphism. The vector space A/B together with the multiplication (5.5) is called the

factor algebra of A with respect to the ideal B. It is clear that if A is associative (commutative) then so is A/B. If A has a unit element e then ë= ire is the unit element of the algebra A/B. 5.5. Homomorphisms. Suppose A and B are algebras and is

a homomorphism. Then the kernel of p is an ideal in A. In fact, if xeker ço and yEA

are

arbitrary we have that

p(xy) =

=

=

0

whence xyEker qi. In the same way it follows that yxEker tp. Next consider the subspace Im pcB. Since for every two elements x, yeA = qi(xy)elmq

it follows that Im 'p is a subalgebra of B.

Chapter V. Algebras

Now let

be the induced injective linear mapping. Then we have the commutative d iagrani

A/kerço

and since

is

a homomorphism, it follows that = = = =

'p

(x y) (x). 'p (y)

is a homomorphism and hence a monomorphism. In particular, the induced mapping This relation shows that

4 Imço is an isomorphism. Finally, assume that C is a third algebra, and let be a homomorphism. Then the composition ti0 p: A —+ C is again a homomorphism. In fact, we have 'ps. = = =

Let 'p:A—*B be any homomorphism of associative algebras and S be by a system of generators for A. Then 'p determines a set map (po:

'p0x=çox,

xeS. In fact, if

The homomorphism 'p is completely determined by

x=

...

E S,

(v)

is

an arbitrary element we have that

... 'poxv

= (v)

I

Vp

e I'

§ 1. Basic properties

151

be an arbitrary set map. Then Po can be Proposition I: Let can be extended to a homomorphism ço:A—+B if and only if ...

whenever

0

po

...

= 0. (5.6)

(v)

(v)

Proof: It is clear that the above condition is necessary. Conversely, assume that (5.6) is satisfied. Then define a mapping ço:A—÷B by

= (v)

...

(5.7)

(v)

To show that çü is, in fact, well defined we notice that if = ... •.. Ypq

(v)

(p)

then —

Ypq

= 0.

(ji)

(v)

In view of (5.6) ...

...

—

(v)

0

(p)

and so

...

It follows from (5.7) that

px=ip0x xeS p(2x + jiy) = 2px + and

'p (x y) = 'p

'p y

and hence 'p is a homomorphism. Now suppose that

be a linear map such

is a basis for A and let

= for each x, j3. Then 'p is a homomorphism, as follows from the relation 'p (x y) =

'p

ep)}

e2) (> p

2,fl

=

'p (ep)) =

'p p

'p

(x) 'p (y).

Chapter V. Algebras

152

5.6. Derivations. A linear mapping 0: A called a derivation if

A of an algebra into itself is

X,yEA.

(5.8)

and As an example let A be the algebra of Cr-functions f: define the mapping 0 by 0:f—+f' wheref' denotes the derivative off. Then the elementary rules of calculus imply that 0 is a derivation. If A has a unit element e it follows from (5.8) that Oe =

Oe

+ 0e

whence Oe=—O. A derivation is completely determined by its action on a

system of generators of A, as follows from an argument similar to that used to prove the same result for homomorphisms. Moreover, if O:A—*A

is a linear map such that

=

+

is a basis for A, then 0 is a derivation in A. For every derivation 0 we have the Leibniz formula

where

on(xy)

(5.9)

= In fact, for n = 1, (5.9) coincides with (5.8). Suppose now by induction that (5.9) holds for some n. Then +

1(x y) =

o on

(x y) or+

-r y +

= x.On+ly +

+

or x

-r+

=

= x.On+ly + n+ 1

and so the induction is closed.

Orx.On+l_ry +

(n+1) Orx.On_r+ly +

§ 1. Basic properties

153

The image of a derivation 0 in A is of course a subspace of A, but it is in general not a subalgebra. Similarly, the kernel is a subalgebra, but it is not, in general, an ideal. To see that ker 0 is a subalgebra, we notice that for any two elements x, yeker 0

0(xy) =

0

whence xy E ker 0.

it follows immediately from (5.8) that a linear combination of derivations 01:A—*A is again a derivation in A. But the product of two derivations 01, 02 satisfies

(0102)(xy) = 01(02x.y + x.02y) (5.10)

and so is, in general, not a derivation. However, the commutator [01,02] = 0102—0201 is again a derivation, as follows at once from (5.10). 5.7. ip-derivations. Let A and B be algebras and 'p : A B be a fixed homomorphism. Then a linear mapping 0: A B is called a 'p-derivation if

0(xy)=

x,yeA.

+

denotes In particular, all derivations in A are i-derivations where i : A the identity map. As an example of a 'p-derivation, let A be the algebra of Cm-functions f: R—* and let B= 11. Define the homomorphism cp to be the evaluation homomorphism : f -÷ (0) and the mapping 0 by

f

0:f Then it follows that

0(fg) = (1 g)'(O) = = 0

f' (0) g (0) + f (0) g' (0)

a 'p-derivation.

More generally, if °A is any derivation in A, then derivation. In fact, 0(xy) = çoOA(xy) + xOAv) = = 'p

+

'p x

0 y.

Similarly, if °B is a derivation in B, then °B° 'p

is

a 'p-derivation.

is

a ço-

Chapter V. Algebras

154

5.8. Antiderivations.

Recall that an involution in a linear space is a

linear transformation whose square is the identity. Similarly we define an involution w in an algebra A to be an endomorphisin of A whose square is the identity map. Clearly the identity map of A is an involution. If A has a unit element e it follows from sec. 5.1 that we=e. Now let w be a fixed involution in A. A linear transformation Q:A—*A will be called an antiderivation with respect to w if it satisfies the relation

Q(xj) =

+

wxQy.

(5.11)

In particular, a derivation is an antiderivation with respect to the involution i. As in the case of a derivation it is easy to show that an antiderivation is determined by its action on a system of generators for A and that ker Q is a subalgebra of A. Moreover, if A has a unit element e, then Qe=O. It also follows easily that any linear combination of antiderivations with respect to a fixed involution w is again an antiderivation with respect to w. Suppose next that Q1 and Q2 are antiderivations in A with respect to 0w1. Then w1 oW2 the involutions w1 and w2 and assume that w1 0w2 is again an involution. The relations

(Q122)(xy)= Q1(Q2xy + w2xQ2y)=

+

and

(Q2Q1)(xy)=

+ w1xQ1y)= Q2Q1xy + + w2Q1xQ2y + Q2w1xQ1y + w2w1xQ2Q1y

yield

(Q1Q2 ± Q2 ± Q2w1)xQ1 y + +(Q1w2 ± w2Q1)xQ2y + W1W2X.(Q1Q2 ± Q2Q1)y. Q2

(5.12)

Now consider the following special cases: 1. w1 Q2 = Q2 w1 and w2 Q1 = Q1 (this is trivially true if w1 = ± i and Q2 —Q2 is an antiderivation = ± i). Then the relation shows that with respect to the involution W1 w2. In particular, if Q is an antiderivation

with respect to w and 0 is a derivation such that 0)0=0W, then OQ—Q0 is again an antiderivation with respect to oi. 2. w1Q2=—Q2W1andw2Q1=—Q1W2.ThenQ1Q2+Q2Q1isanantiderivation with respect to the involution W1 0)2.

§ 1. Basic properties

155

Now let Q1 and Q2 be two antiderivations with respect to the same involution w such that

(i=1,2). Then it follows that Q1 Q2 + Q2 Q1 is a derivation. In particular, if Q is any antiderivation such that wQ = - Qw then Q2 is a derivation. Finally, let B be a second algebra, and let cp: A —÷ B be a homomorphism. Assume that WA is an involution in A. Then a p-antiderivation with respect to WA is a linear mapping Q:A-+B satisfying (5.13) If WB is

an involution in B such that WA

WB 'P

then equation (5.13) can be rewritten in the form

Q(xy) =

+

Problems 1. Let A be an arbitrary algebra and consider the set C(A) of elements aEA that commute with every element in A. Show that C(A) is a subspace

of A. If A is associative, prove that C(A) is a subalgebra of A. C(A) is called the centre of A. 2. If A is any algebra and 0 is a derivation in A, prove that C(A) and the derived algebra are stable under 0. 3. Construct an explicit example to prove that the sum of two endomorphisms is in general not an endomorphism. 4. Suppose 'p A —÷ B is a homomorphism of algebras and let ). + 0, 1 be

an arbitrarily chosen scalar. Prove that 2p is a homomorphism if and only if the derived algebra is contained in ker 'p. 5. Let C1 and C denote respectively the algebras of continuously differentiable and continuous functionsf: (cf. Example 4). Consider the linear mapping d: C' C

given by df=f' wheref' is the derivative off.

Chapter V. Algebras

a) Prove that this is an i-derivation where i: C1—*C denotes the canonical injection. b) Show that d is surjective and construct a right inverse for d.

c) Prove that d cannot be extended to a derivation in the algebra C. 6. Suppose A is an associative commutative algebra and 0 is a derivation in A. Prove that Ox" = px"' 0(x). 7. Suppose that 0 is a derivation in an associative commutative algebra A with identity e and assume that xeA is invertible; i.e.; there exists an element x ' such that

Prove that x" (p 1) is invertible and that

(x")' —(x')" Denoting the inverse of x" by

show that for every derivation 0

0(x") = — px"1 0(x). 8. Let L be an algebra in which the product of two elements x, y is denoted by [x, y]. Assume that [x, y] + [y, x] = z], x] + y], z] +

0

(skew symmetry)

x], y] = 0 (Jacobi identity)

Then L is called a Lie algebra. Let Ad (a) be the multiplication operator in the Lie algebra L. Prove that Ad (a) is a derivation.

9. Let A be an associative algebra with product xy. Show that the multiplication (x, y)—÷[x, y] where

[x,y]

xy — yx

makes A into a Lie algebra. 10. Let A be any algebra and consider the space D (A) of derivations in A. Define a multiplication in D(A) by setting [01,02] = a)

0102 —0201.

Prove that D (A) is a Lie algebra.

b) Assume that A is a Lie algebra itself and consider the mapping given by Show that 'p is a homomorphism of Lie algebras. Determine the kernel of p.

§ 1. Basic properties

11. If L is a Lie algebra and I is an ideal in A, prove that the algebra L/I is again a Lie algebra. 12. Let E be a finite dimensional vector space. Show that the mapping 1: A (E; E) —÷ A (E*; E*)OPP

is an isomorphism of algebras. given by 13. Let A be any algebra with identity and consider the multiplication

operator

ji: A

A (A; A).

Show that p is a monomorphism. If A = L (E; E) show that by a suitable restriction of p a monomorphism

(i= 1

E be an n-dimensional vector space. Show that each basis n) of L(E; E) such that n) of E determines a basis 1

(i) QijOlk = (ii)

i.

Conversely, given n2 linear transformations of E satisfying i) and ii), prove that they form a basis of L(E; E) and are induced by a basis of E. Show that two bases e1 and of E determine the same basis of L (E; E) if and only if e = A 2 eF. 15. Define an equivalence relation in the set of all linearly independent E), in the following way: n2-tuples (p1...pn2), ...

('P1 ...

if and only if there exists an element =

such that (v

= ... 1

,12)

Prove that 2EV

only if 2=1. 16. Prove that the bases of L(E; E) defined in problem 14 form an equivalence class under the equivalence relation of problem 15. Use this to show that every non-zero endomorph ism cP: A (E; E)—*A (E; E) is an inner automorphism; i.e., there exists a fixed linear automorphism of E such that 1

17. Let A be an associative algebra, and let L denote the corresponding

Chapter V. Algebras

158

is Lie algebra (cf. problem 9). Show that a linear mapping derivation in A only if it is a derivation in L. 18. Let E be a finite dimensional vector space and consider the mapping E)—*A(E; E) defined by

= — Prove that is a derivation. Conversely, prove that every derivation in A (E; E) is of this form. Hint. Use problem 14. § 2.

Ideals

5.9. The lattice of ideals. Let A be an algebra, and consider the set 1 of ideals in A. We order this set by inclusion; i.e., if and '2 are ideals if and only if in A, then we write '1 The relation is clearly a partial order in 1 (cf. sec. 0.6). Now let and '2 be ideals in A. Then it is easily cheeked that + and '1 n '2 are again ideals, and are in

fact the least upper bound and the greatest lower bound of '1 and '2 Hence, the relation induces in 5 the structure of a lattice. 5.10. Nilpotent ideals. Let A be an associative algebra. Then an element aEA will be called ni/potent if for some k, (5.14)

The least k for which (5.14) holds is called the degree of ni/potency of a. An ideal I will be called nilpotent if for some k,

,k0

(5.15)

The least k for which (5.15) holds is called the degree of ni/potency of I and will be denoted by deg I. 5.11.* Radicals. Let A be an associative commutative algebra. Then the nilpotent elements of A form an ideal. In fact, if x and y are nilpotent of degree p and q respectively we have that p+q

+

p+q

p

= i=O

and

(xy)" = x"y" = 0.

§ 2. Ideals

The ideal consisting of the nilpotent elements is called the radical of A and will be denoted by rad A. (The definition of radical can be generalized to the non-commutative case; the theory is then much more difficult and belongs to the theory of rings and algebras. The reader is referred to [14]).

It is clear that rad (rad A) = rad A.

The factor algebra A/rad A contains no non-zero nilpotent elements. To prove this assume that A is an element such that for some k. Then x"erad A and hence the definition of rad A yields / such that = (x")' = 0. It follows that xErad A whence by the formula

The above result can be expressed

rad(A/radA) = 0.

Now assume that the algebra A has dimension n. Then rad A is a nilpotent ideal, and (5.16) deg(radA) dim(radA) + 1 n + 1. For the proof, we choose a basis e1, ..., er of rad A. Then each is nilpotent. Let k = max (deg e1), and consider the ideal (rad An arbitrary element in this ideal is a sum of elements of the form where

k

so

...

=0. This shows that

=0 and so rad A is nilpotent. Now let s be the degree of nilpotency of rad A, and suppose that for some m dim (radA)m+l,

m 0 is a constant. 4. Prove that every normal non-selfadjoint transformation of a plane is homothetic. 5. Let be a mapping of the inner product space E into itself such that (pO=0 and çox — (P11 = lx — x. VEE.

Prove that

is then linear.

there exists a continuous 6. Prove that to every proper rotation family (p(0t I) of rotations such that and (P1 = 7. Let be a linear automorphism of an n-dimensional real linear space E. Show that an inner product can be defined in E such that p becomes a rotation if and only if the following conditions are fulfilled:

(i) The space E

can

be decomposed into stable planes and stable

straight lines. (ii) Every stable straight line remains pointwise fixed or is reflected at the point 0.

(iii) In every irreducible invariant plane a linear automorphism induced such that and tn/il 2. Consider a proper rotation t which commutes with all proper rotations. Prove that = i where c = if ii is odd and = ± if n is even. 1

§ 6.

1

Rotations of Euclidean spaces of dimension 2,3 and 4

8.21. Proper rotations of the plane. Let E be an oriented Euclidean plane and let zl denote the normed positive determinant function in E. is determined by the equation Then a linear transformation j: zi (x, i') = (j

i)

x, v E.

Chapter VIII. Linear mappings of inner product spaces

238

The transformation / has the following properties:

I) (jx,v)+(x,jv)=O (jx,jv)=(x,v)

2)

3)

4)

detj=1.

In fact, 1) follows directly from the definition. To verify 2) observe that the identity (7.24), sec. 7.13, implies that (x, y)2 + (/x,v)2 y

= IxV

x we obtain, in view of 1),

(jx, jx)2 = j is injective (as follows from the definition) we obtain

= imply that

Now the relations .1= —.1 and

=—

Finally, we

obtain from 1), 2), 3) and from the definition of /

4(jx,iv) = (/2 x, iv) = — (x,jy) = (jx, v) = zl(x,

y)

whence

det j =

1.

The transformation / is called the canonical complex structure of the oriented Euclidean plane E. Next, let p be any proper rotation of E. Fix a non-zero vector x and denote by 0 the oriented angle between x and p x (cf. sec. 7.1 3). It is determined mod 2 it by the equation

px = x cos 0 +j(x) sin 0.

(8.39)

We shall show that cos 0 = tr 'p

and

sin 0 = 4 tr(jo p).

(8.40)

In fact, by definition of the trace we have

zl(p x, y) + 4(x, py) = 4(x, y). tr 'p. Inserting (8.39) into (8.41) and using properties 2) and 3) of j

(8.41)

we

obtain

2cos0 .A(x, v)=zl(x,y) . trqi and so the first relation (8.40) follows. The second relation is proved in a similar way.

§ 6. Rotations of Euclidean spaces of dimension 2, 3 and 4

239

Relations (8.40) show that the angle (9 is independent of x. It is called the rotation angle of and is denoted by e((p). Now (8.39) reads

cosO((p)+j(x) sinê((p)

XEE.

In particular, we have

O(i)=0,

/ix=x

&(—i)=m, cos

sin

a second proper rotation with rotation angle shows that is

a

simple calculation

= x cos(&(p)+ (9(h)) +j(x) sin((9(p)+ Thus any two proper rotations commute and the rotation angle for their

product is given by (p) = (9((p) +

(mod 2it).

(8.42)

Finally observe that if e1, e2 is a positive orthonormal basis of E then we have the relations (p e1 e1 cos O(p) + e2 sin O((p) (p e2 = — e1 sin O(p) + e2 cos O(p).

Remark: If E is a non-oriented plane we can still assign a rotation angle to a proper rotation (p. It is defined by the equation

cos 9(p) = tr p

(0

it).

(8.43)

Observe that in this case (9(p) is always between 0 and it whereas in the oriented case 9(p) can be normalized between —it and it. 8.22. Proper rotations of 3-space. Consider a proper rotation p of a 3-dimensional inner product space E. As has been shown in sec. 8.19,

there exists a 1-dimensional subspace E1 of E whose vectors remain fixed. If p is different from the identity-map there are no other invariant vectors

(an invariant vector is a vector xeE such that (px=x). In fact, assume that a and b are two linearly independent invariant vectors. Let c(c+0) be a vector which is orthogonal to a and to b. Then (pc—2c where 2= ± 1. Now the equation det (p= implies that 2= + 1 showing that (p is the identity. In the following discussion it is assumed that +1. Then the invariant vectors generate a i-dimensional subspace E1 called the axis of (p. 1

240

Chapter VIII. Linear mappings of inner product spaces

To determine the axis of a given rotation (p consider the skew mapping (8.44)

and introduce an orientation in E. Then

can

t/ix=u xx

be written in the form

UEE

(8.45)

(cf. problem 2, § 4). The vector u which is uniquely determined by (p iS

called the rotation-vector. The rotation vector is contained in the axis of (p. In fact, let a+O be a vector on the axis. Then equations (8.45) and (8.44) yield (8.46)

showing that u is a multiple of a. Hence (8.45) can be used to find the rotation axis provided that the rotation vector is different from zero. This exceptional case occurs if and only if = i.e. if and only if Then (p has the eigenvalues 1, —1 and — 1. In other words, is (p = (p a reflection at the rotation axis. 8.23. The rotation angle. Consider the plane F which is orthogonal to E1. Then transforms F into itself and the induced rotation is again proper. Denote by 0 the rotation angle of (p1 (cf. remark at the end of see 8.21). Then, in view of (8.43),

cosO =

Observing that trço =

+1

trço1

we obtain the formula cosO = 3-(trp

1).

To find a formula for sin 0 consider the orientation of Fwhicl' is induced by E and by the vector u (cf. sec. 4.29)*). This orientation is represented by the normed determinant function

A1(y,z)= lul

where A is the normed determinant function representing the orientation

of E. Then formula (7.25) yields

sinO=A1(y,(py)='A(u,y,(py) lul

*) It is assumed that u

0.

(8.47)

§ 6. Rotations of Euclidean spaces of dimension 2. 3 and 4

241

where y is an arbitrary unit vector of F. Now

A(u,y,çoy)= = 4(u,co' y,y)= — A(u,y,ço1y) and hence equation (8.47) can be written as (8.48) lul

By adding formulae (8.47) and (8.48) we obtain (8.49)

lxi

21u1

Inserting the expression (8.45) in (8.49), we thus obtain

x y)=

sin0 = lul Since

x lul

(8.50)

y is a unit-vector orcnogonal to u, it follows that iu

xyl=iuilyl=lui

and hence (8.50) yields the formula

sin0=iui. This equation shows that sin 0 is positive and hence that 0 1 and define polynomials by —

g

are relatively prime, and hence there

exist polynomials h1 such that (13.19)

On the other hand, it follows from Corollary II, Proposition II, sec. 13.2 that

K(g) =

j*i

and so, in particular, =

0

xe

j*i

Now let xeE be an arbitrary vector, and let x1EE1

(13.20)

392

Chapter XIII. Theory of a linear transformation

be the decomposition of x determined by (13.18). Then (13.19) and (13.20) yield the relation x=

=

x =

h (go) g1 (go)

h. (go) g. ((p) x1

whence x1

=

I = 1,

...,r.

(13.21)

It follows at once from (13.20) and (13.21) that

=

I

=

1,

...,r

which completes the proof. 13.6. Arbitrary stable subspaces. Let Fc: E be any stable subspace. Then

F=

(13.22)

n

where the are the generalized eigenspaces of go. In fact, since the projection operators are polynomials in go, it follows that F is stable under each 7C1.

Now we have for each xeF that X

=1X=

X

and fl

It follows that

E1,

whence E1.

inclusion in the other direction is obvious, (13.22) is established. 13.7. The Fitting decomposition. Suppose F0 is the generalized eigenspace of go corresponding to the irreducible polynomial t (if t does not divide then of course F0=0). Let F1 be the direct sum of the remaining generalized eigenspaces. The decomposition Since

E=

F0

F1

is called the Fitting decomposition of E with respect to go. F0 and F1 are called respectively the Fitting-null component and the Fitting-one component of E.

§ 2. Generalized eigenspaces

393

Clearly F0 and F1 are stable subspaces. Moreover it follows from the and F1, then definitions thatif 'p is ni/potent; i.e.,

forsomel>O

while Pi is a linear isomorphism. Finally, we remark that the corresponding projection operators are polynomials in (p, since they are sums of the projection operators defined in sec. 13.5. 13.8. Dual mappings. Let E* be a space dual to E and let E*

çp* E*

be the linear transformation dual to 'p. Then if f is any polynomial, it follows from sec. 2.25 that This

=

implies that f(p*) = 0 if and only if f('p) 0. In particular, the

minimum polynomials of 'p and

coincide.

Now suppose that F is any stable subspace of E. Then F' is stable under co*. In fact, if yEF and y*EF' are arbitrarily chosen, we have =

=

'p

0

whence 'p*y*EFI. This proves that F' is stable. Thus 'p* induces a linear transformation ((p*)I.

E*/FI

E*/FI.

On the other hand, let and ('p*)1 are be the restriction of 'p to F. It will now be shown that dual with respect to the induced scalar product between F and E*/FI is a representative of (cf. sec. 2.24). In fact, if yEF is any vector and an arbitrary vektor e E*/FI, then EJ)l

j*i

defined by

i=1,...,r.

Then (13.24)

§ 2. Generalized eigenspaces

are stable under

Then, as shown above, the

395

and (13.25)

It will now be shown that (13.26)

Let y*EF,* be arbitrarily chosen. Then for each

=

=

we have

= 0.

In view of the duality between EL and F1*, this implies thatf/'1(cp*)y* =0;

y*EE* This establishes (13.26). Now a comparison of the decompositions (13.23) and (13.25) yields (13.24).

Problems 1. Show that the minimum polynomial of p is completely reducible (i.e. all prime factors are of degree 1) if and only if every stable subspace contains an eigenvector.

2. Suppose that the minimum polynomial p of ço is completely reducible. Construct a basis of E with respect to which the matrix of ço is lower triangular; i.e., the matrix has the form 0

*

Hint: Use problem 1.

3. Let E be an n-dimensional real vector space and p: E—*E be a regular linear transformation. Show that çø can be written p=c01p2 where every eigenvalue of p1 is positive and every eigenvalue of .p2 is negative.

4. Use problem 3 to derive a simple proof of the basis deformation theorem of sec. 4.32. 5. Let ço:E-*E be a linear transformation and consider the subspaces F0 and F1 defined by F0 =

and

F1 =

fl

Chapter XIII. Theory of a linear transformation

396

a) Show that F0 = U ker

ço1.

1

b)

Show that

c) Prove that F0 and F1 are stable under

and that the restrictions are respectively nilpotent and regular. and cond) Prove that c) characterizes the decomposition

(p0:F0-3F0 and ço1

clude that F0 and F1 are respectively the Fitting null and the Fitting 1-component of E.

6. Consider the linear transformations of problem 7, § 1. For each transformation a) Construct the decomposition of into the generalized eigenspaces. b) Determine the eigenspaces. c) Calculate explicitly polynomials g, such that the g.(p) are the pro-

jection operators in E corresponding to the generalized eigenspaces. Verify by explicit consideration of the vectors that the (ço) are in fact the projection operators. d) Determine the Fitting decomposition of E. 7. Let E= be the decomposition of E into generalized eigenspaces

of p, and let

be the corresponding projection operators. Show that

there exist unique polynomials

(p) = Conclude that the polynomials

such that and

deg

deg

depend only on p. E*

8. Let E* be dual to E and

E* into generalized eigenspaces of co* prove that

=

are the corresponding projection operators and the g, are defined in problem 7. Use this result to show that and are dual and to obtain formula

where the

(13.24).

9. Let FcE be stable under 'p and consider the induced mappings E/F—+ E/F. Let E= be the decomposition of F into

c°F :F—÷F and

be the canonical injection and generalized eigenspaces of 'p. Let Q:E—*E/Fbe the canonical projection. a) Show that the decomposition of F into generalized eigenspaces is given by

F=

where

F

fl

§ 3. Cyclic spaces

397

Show that the decomposition of E/F into generalized eigenspaces of is given by E/F = (ElF)1 where (E/F), = (E1). b)

Conclude that

determines a linear isomorphism (E/F),.

E1/F,

c) If denote the projection operators in E, F and E/F associated with the decompositions, prove that the diagrams

and F

are commutative, and that

are the unique linear mappings for

which this is the case. Conclude that if g are the polynomials of problem 7,then F g1((p) and = —

10. Suppose that the minimum polynomial ji of çü is completely reducible. a) By considering first the case p = (t —

dimE.

degp

b) With the aid of a) prove that of p. § 3.

prove that

x the characteristic polynomial

Cyclic spaces

13.9. The map 0a• Fix a vector aEE and consider the linear map ar,:

given by a

fer[t].

Let K1, denote the kernel of ar,. It follows from the definition that feK,, Ka, where p is if and only if aEK(J). K,, is an ideal in T[t]. Clearly the minimum polynomial of p. Proposition 1: There exists a rector a E E such that K,, = I,, Proof: Consider first the case that p is of the form

p=

k 1.

/ irreducible.

Chapter XIII. Theory of a linear transformation

398

Then there is a vector tIE E such that (13.27)

0.

Suppose now that hEKa and let g be the greatest common divisor of h and p. Since aeK(fl, Corollary Ito Proposition I, sec. 13.2 yields

ueK(g).

(13.28)

Since g, p it follows that g =1' where / k. Hence 1 (p) a = () and relation (13.27) implies that /=k. Thus K(g)=K0. Now it follows from (13.28) that

In the general case decompose p in the form =

J irreducible

...

and let (i= 1 i) be be the corresponding decomposition of E. Let is the induced transformation. Then the minimum polynomial of given by (cf. sec. 13.4) = r) (1= 1

Thus, by the first part of the proof, there are vectors K,, =

I,,

(i =

I

such that

. . .

Now set

(1(l1 Assume that jEK11. Then f(p) a=0 whence

= 0.

Since f(p)

it follows that

f(p) whence fe Ka (i =

1

=0

(1 =

I

... r). Thus •f C

and so tel0. This shows that

•'' fl

=

III

whence K(,=I,L.

13.10. Cyclic spaces. The vector space E is called crc/ic (with respect

to the linear transformation p) if there exists a vector acE such that is surjective. Every such vector is called a geiiclaloI of E. the map In fact, let I cK and let XEE. Then, Ifa is a generator of E. then

§ 3. Cyclic spaces

399

for some geT[t], x=g(p)a. It follows that

=f(p)g(p)a = whence

f(p)a = g(p)O = 0

and so tel1.

Proposition II: If E is cyclic, then

degp = dimE.

Proof Let a be a generator of E. Then, since Ka =

0a induces an

isomorphism E.

It follows that (cf. Proposition I, sec. 12.11)

= degp.

dim

Proposition III: Let a be a generator of E and let deg p = m. Then the vectors

a,p(a)

form a basis of E.

Proof. Let / he the largest integer such that the vectors a, p(a)

(13.29)

are linearly independent. Then these vectors generate E. In fact, every

vector xeE can be written in the form x=f(p)a where I =

is a

polynomial. It follows that i—i

k

X

=

(a) = V=

p' (a). 1)

1)

Thus the vectors (13.29) form a basis of E. Now Proposition II implies that I=ni. Proposition IV: The space E is cyclic if and only if there exists a basis (v=O... n—i) of E such that =

(v = 0 ... ii —2).

Proof: If E is cyclic with generator a set a=a and apply Proposiu—I) is a basis satisfying tion III. Conversely. assume that

the conditions above. Let v= e I'[t] by

be an arbitrary vector and define V

Chapter XIII. Theory of a linear transformation

400

Then

=x as is easily checked and so E is cyclic. (v = 0 ii — I be a basis as in the Proposition CoioI/urr: Let above. Then the minimum polynomial of is given by )

= where the

are

—

determined by

Proof: It is easily checked that p(p)=O and so p is a multiple of the minimum polynomial of p. On the other hand. by Proposition II. the minimum polynomial of has degree ii and thus it must coincide with p. 13.11. Cyclic subspaces. A stable subspace is called a crc/ic siihspacc if it is cyclic with respect to the induced transformation. Every vector UEE determines a cyclic subspace. namely the subspace = Im Proposition V: There exists a cyclic subspace whose dimension is equal to degp. Proof: In view of Proposition I there is a vector u e F such that ker = Then induces an isomorphism E11.

It follows that

dimE, = diml'[t]

'R

= degp.

Throic;,, I: The degree of the minimum polynomial satisfies

degp dimE. Equality holds if and only if E is cyclic. Moreover, if F is any cyclic subspace of E. then

dimF degp.

Proof: Proposition V implies that deg p dimE. If E is cyclic. equality

holds (cf. Proposition III). Conversely. assume that degp=dimE. By Proposition V there exists a cyclic subspace with dim F=degp. It follows that F=E and so E is cyclic.

§ 3. Cyclic spaces and irreducible spaces

401

Finally, let Fc:E be any cyclic subspace and let v denote the minimum polynomial of the induced transformation. Then, as we have seen above, dim F = deg v. But v/p (cf. sec. 13.2) and so we obtain dim F deg p.

Corollary: Let F E be any cyclic subspace, and let v denote the minimum polynomial of the linear transformation induced in F by (p. Then

v=u if and only if

dimF = degp. Proof: From the theorem we have

dimF =

degv

while according to sec. 13.2 v divides p. Hence v=p if and only if deg v=deg ,i; i.e., if and only if

dimF = degp. 13.12. Decomposition of E into cyclic subspaces. Theorem II: There exists a decomposition of E into a direct sum of cyclic subspaces. Proof: The theorem is an immediate consequence (with the aid of an induction argument on the dimension of E) of the following lemma.

E such that

Lemma I. Let dimEa = degp.

Then there is a complementary stable subspace, Fc E, E = Ea

F.

Proof: Let Ea

denote the restriction of (p to Ea, and let (p*:E*

E*

the linear transformation in E* dual to (p. Then (cf. sec. 13.8) stable under (p*, and the induced linear transformation be

*

*

J_

*

(pa:E lEa *—E lEa 26

Greub. Linear Algebra

is

Chapter XIII. Theory of a linear transformations

402

is dual to

with respect to the induced scalar product between Ea and

The corollary to Theorem I, sec. 13.11 implies that the minimum polynomial Of 'Pa is again p. Hence (cf. sec. 13.8) the minimum polynomial

are dual, so that

of p is p. But Ea and

= dimEa = degp.

dim

is cyclic with respect to çü.

Thus Theorem I, sec. 13.11 implies that Now let

be the projection and choose u*eE* so that the element generates E*/E±. Then, by Proposition III, the vectors (p = 0... in — 1) form a basis of Hence the vectors

(p* V

(p = I ... in — I) are linearly independent.

Now consider the cyclic subspace Since I) it follows that On the other hand. Theorem I. sec. 13.11. implies that dim Hence dim

= (p = 0... is injective. Thus 0

Finally, since 7r* TE to

dim

+ dim

=

in —

= m. I

)

it follows that the restriction of But

in + (ii — in)

=

ii

(n =dim E)

and thus we have the direct decomposition E* =

of E* into stable subspaces. Taking orthogonal complements we obtain the direct decomposition

E=

E into stable subspaces which completes the proof.

§ 4.

Irreducible spaces

13.13. Definition. A vector space E is called

!I1(Iecoin/)os(Ib/e or

ii'i'cthicible with respect to a linear transformation 'P. if it can not be expressed as a direct sum of two proper stable subspaces. A stable

§4. Irreducible spaces

403

subspace F E is called irreducible if it is irreducible with respect to the linear transformation induced by (p. Proposition I: E is the direct sum of iireducihle subspaces. Proof: Let

E into stable subspaces such that s is maximized

(this is clearly possible. since for all decompositions we have sdim E). Then the spaces are irreducible. In fact, assume that for some i, =

is a decomposition of

dim

Ft"

> 0, dim

>0

into stable subspaces. Then

E=

F Fe" j*i is a decomposition of E into (s+ I) stable subspaces, which contradicts the maximality of s. An irreducible space E is always cyclic. In fact, by Theorem II, sec. 13.12, E is the direct sum of cyclic subspaces.

E=

If E is indecomposable. it follows that 1=and so E is cyclic. On the other hand, a cyclic space is not necessarily indecomposable. In fact, let E be a 2-dimensional vector space with basis a. h and define p: E—+E by setting (pa=h and (ph =0. Then (p is cyclic (cf. Proposition IV, sec. 13.10). On the other hand, E is the direct sum of the stable subspaces generated by a + b and a — b. 1

Theorem I: E is irreducible if and only if i)

p=fk,

f irreducible;

ii) E is cyclic.

Proof. Suppose E is irreducible. Then, by the remark above. E is E into the cyclic. Moreover, if generalized eigenspaces (cf. sec. 13.4), it follows that r= 1 and so

(f irreducible). Conversely, suppose that (i) and (ii) hold. Let E=

E1

E2

Chapter XIII. Theory of a linear transformations

404

be any decomposition of E into stable subspaces. Denote by and c02 the linear transformations induced in F1 and E, by ço, and let and 112 be the minimum polynomials of and q2. Then sec. 13.2 implies that Hence, we obtain from (i) that p and 111 =

k2 k.

=

(13.33)

Without loss of generality we may assume that k1 k2. Then xEE1

or

XEE7

and so

0.

It follows that

whence k1 k. In view of (13.33) we obtain k1 =k

i.e., It

= Iti•

Now Theorem I, sec. 13.10 yields that

dimE1 degp.

(13.34)

On the other hand, since E is cyclic, the same Theorem implies that

dimE = degjt.

(13.35)

Relations (13.34) and (13.35) give that

dimE = dimE1. Thus E= F1 and F2 = 0. It follows that E is irreducible.

Corollary I. Any decomposition of F into a direct sum of irreducible subspaces is simultaneously a decomposition into cyclic subspaces.

Then a stable subspace Corollary II. Suppose that of F is cyclic if and only if it is irreducible. 13.14. The Jordan canonical matrix. Suppose that F irreducible with respect to p. Then it follows from Theorem I sec. 13.13 that F is cyclic and that the minimum polynomial of ço has the form

p_fk

kl

(13.36)

where f is an irreducible polynomial. Let e be a generator of E and

§4. Irreducible spaces

405

consider the vectors (13.37)

it will be shown that these form a basis of E. Since

dimE =

deg,u

= pk

it is sufficient to show that the vectors (13.37) generate E. Let Fc:E be the subspace generated by the vectors (13.37). Writing

we obtain that

i==1,...,k j=1,...,p—1 e

= f(çp)1_i ço"e =

p—i —

i1,...,k1 These

equations show that the subspace F is stable under 'p. Moreover,

e = a11 e F. On the other hand, since E is cyclic, E is the smallest stable

subspace containing e. It follows that F—E. Now consider the basis ...

a11,

...;akl ...

of E. The matrix of 'p relative to this basis has the form 0 1

(13.38)

0

406

Chapter XIII. Theory of a linear transformation

are all equal, and given by

where the submatrices

01 0

0 1

0•.

j=1,...,k. 0

1

The matrix (13.38) is called a Jordan canonical matrix of the irreducible transformation p. Next let ço be an arbitrary linear transformation. In view of sec. 13.12 there exists a decomposition of E into irreducible subspaces. Choose a basis in every subspace relative to which the induced transformation has the Jordan canonical form. Combining these bases we obtain a basis of E. In this basis the matrix of p consists of submatrices of the form (13.38)

following each other along the main diagonal. This matrix is called a Jordan canonical matrix of çø. 13.15. Completely reducible minimum polynomials. Suppose now that E is an irreducible space and that the minimum polynomial is completely

reducible (p= 1); i.e. that

p = (t

—

2)k.

It follows that the

are (1 x 1)-matrices given by Jordan canonical matrix of ço is given by

21

Hence the

0

2 1. (13.39)

..1 0

In particular, if E is a complex vector space (or more generally a vector space over an algebraically closed field) which is irreducible with respect to 'p, then the Jordan canonical matrix of çü has the form (13.39). 13.16. Real vector spaces. Next, let E be a real vector space which is irreducible with respect to ço. Then the polynomialf in (13.36) has one of the two forms f=t—2 2ER or

§4. Irreducible spaces

407

In the first case the Jordan canonical matrix of p has the form (13.39). In the second case the are 2 x 2-matrices given by 0

Hence

1

the Jordan canonical matrix of 0

has the form

1

-fl

1

- -

1

H 13.17. The number of irreducible subspaces. It is clear from the construction in sec. 13.12 that a vector space can be decomposed in several ways into irreducible subspaces. However, the number of irreducible subspaces of any given dimension is uniquely determined by as will be shown in this section. Consider first the case that the minimum polynomial of 'p has the form

p=f

degf=p

wheref is irreducible. Assume that a decomposition of E into irreducible subspaces is given. The dimension of every such subspace is of the form pK(l as follows from sec. 13.12. Denote by FK the direct sum of the irreducible subspaces of dimension pic and denote by NK the number of the subspaces Then we have

E=

FK.

Comparing the dimensions in (13.40) we obtain the equation = dim E=n.

(13.40)

Chapter XIII. Theory of a linear transformation

405

Now consider the transformation

it follows from (13.40) that

Since the subspaces FK are stable under

(13.41)

By the definition of

we have FK

dim

=

whence

=

Since the dimension of each follows that

decreases by p under k/I (cf. sec. 1 3. 14) it

dimk//Fh =

—

(13.42)

1)NK.

Equations (13.41) and (13.42) yield (r(k/I) = rank t/i)

K=2

Repeating the above argument we obtain the equations j=1,...,k.

Kj+

1

Replacing] byj+ I and]—! respectively we find that

(K—j—l)NK=p (13.43)

and —j + 1)NK =

=

—j)NK +

pINK.

(13.44)

Adding (13.43) and (13.44) we obtain

+

= 2p

(K

=2p Kj+

1

—j)NK +

+

+

§4. Irreducible spaces

409

whence

=

+

j

—

= 1,...,k.

This equation shows that the numbers N1 are uniquely determined by the

(j= l,...,k).

ranks of the transformations In particular.

1.(i/1k_l)>

Nk=

I

if]> k. Thus the degree of is pk where k is the largest integer k and such that Nk>O. In the general case consider the decomposition of E into the generalized eigenspaces E= a a direct sum of irreducible subspaces. Then E every irreducible subspace F, is contained in some E (cf. Theorem I, sec. 13.13). Hence the decomposition determines a decomposition of

each E, into irreducible subspaces. Moreover, it is clear that these subspaces are irreducible with respect to the induced transformation E. Hence the number of irreducible subspaces in E1 of a given dimension is determined by and thus by p. It follows that the number of spaces F, of a given dimension depends only on (p. 13.18. Conjugate linear transformations. Let E and F be u-dimensional vector spaces. Two linear transformations p: E—*E and i/i: F—÷F are called conjugate if there is a linear isomorphism such that (pt: E1

= Proposition II: Two linear transformations (p and u/i

are

conjugate if

and only if they satisfy the following conditions: (i)

The minimum polynomials of (p and

factors /1

u/i

have

the same prime

Ir

(ii)

Proof: If (p and u/i

1=1 ... are

I.

conjugate the conditions above are clearly

satisfied.

To prove the converse we distinguish two cases.

Chapter XIII. Theory of a linear transformation

410

Casc I: The minimum polynomials of p and /i are of the form Pq)

.1

=

and

f irreducible.

.11.

Decompose E and F into the irreducible subspaces I

i=-1 j=1

The numbers =

.\ (/1)

F=>i=1 j=1

and N1(l/J) are given by

-

+

p=degf

and

= (cf.

-

+

sec. 13.17). Thus (ii) implies that

il. Since

i>k and

I>!

it follows that k=l. In view of Theorem I, sec. 13.13, the spaces El and F/ are cyclic. Thus Lemma I below yields an isomorphism such that El

= These

isomorphisms determine an isomorphism 1/I

Case II: and ized eigenspaces

=

0

such that

0

are arbitrary. Decompose E and F into the general-

and set

(i= I ... i). Then

restricts to a linear isomorphism in the subspace

§4. Irreducible spaces

Moreover, for

411

is zero in E,. Thus dimE —

=

= 1 ...

dimF —

=

=

Similarly.

... r.

i

Now (ii) implies that

1=1 ...r. Let and denote the respectively. The minimum polynomials of

restrictions of 'p and are of the form

and

i=l ... r and Since

+ dimE —

and

= r(f

j

+ dim F — dim

1

—

it follows from (ii) that l

(j(y) = Thus the pair

satisfies

...

the hypotheses of the theorem and so we

are reduced to case I.

Lemma I: Let p: E E and /i: F —* F (dim E = dim F = ii) be linear transformations having the same minimum polynomial. Then, if E and F are cyclic, 'p and are conjugate. Proof: Choose generators a and b of E and F. Then the vectors 01='pi(a) and

ii— I) forma basis ofEand F respectively.

Now set a,='p(a) and h= t/i'(h). Then, for some and

a,, = 1=

h=

()

h1. 1=

()

Moreover, the minimum polynomials of 'p and =

t'

and

—

1=)

=

t'

are given by —

1=0

Chapter XIII. Theory of' a linear transformations

412

(cf. the corollary of Proposition IV, sec. 1 3.10). Since 1 = 0 ...

=

it follows that

I).

n

by setting

define an isomorphism =

=

/ = 0 ...

h1

n —

Then we have

= (a,) = These

= h,. j=()

j=()

equations imply that

j=0...ii— I. This shows that Anticipating the result of sec. 13.20 we also have Corollary I: Two linear transformations are conjugate if and only if they have the same minimum polynomial and satisfy condition (ii) of Proposition II.

Corollary II: Two linear transformations are conjugate if and only if they have the same characteristic polynomial (cf. sec. 4.19) and satisfy (ii).

Proof: Let J be the common characteristic polynomial of p and /i. Write

=

. .

=

...

.

(f irreducible).

and = .1111

Set

the generalized eigenspaces for direct decompositions

and

fr

13(1=1 ... r) denote 1= 1 ... r; let and t/i respectively. Then we have the

§4. Irreducible spaces

it follows that

Since

413

(1=1 ... r). Similarly.

whence

(i=l...r). Thus we may assume that (g irreducible). Then (kni). ,=g'

I

is

of the form f=g" and

the corollary

follows immediately from the proposition.

Corollaiv III: Every linear transformation is conjugate to its dual.

Problems 1.

(i)

Let P: A(E; E) be a non-zero algebra endomorphism. Show that for any projection operator it.

r(P(m))=i(rr), (ii) Use (i) to prove that 2. Show that every non-zero endomorphism of the algebra A (E; E) is an inner automorphism. Hint: Use Problem I and Proposition II. sec. 13.18. 3. Construct a decomposition of into irreducible subspaces with respect to the linear transformations of problem 7, § 1. Hence obtain the Jordan canonical matrices. 4. Which of the linear transformations of problem 7, § make into a cyclic space? For each such transformation find a generator. For which of these transformations is irreducible? 5. Let E1 cE be a stable subspace under p and consider the induced mappings p1: E1 E1 and E/E1 E/E1. Let p, 1u1, ji be the corresponding minimum polynomials. a) Prove that E is cyclic if and only if 1

i) E1 is cyclic ii) E/E1 is cyclic

iii) p=u1ü. In particular conclude that every subspace of a cyclic space is again cyclic.

b) Construct examples in which conditions i), ii), iii) respectively fail, while the remaining two continue to hold. Hint: Use problem 5, § 1. 6. Let j=1 are

E into subspaces and suppose linear transformations with minimum polynomials

Chapter XIII. Theory of a linear transformations

414

Define a linear transformation

of E by

a) Prove that E is cyclic if and only if each relatively prime.

is cyclic and the

are

b) Conclude that if E is cyclic, then each F3 is a sum of generalized eigenspaces for çø.

c) Prove that if E is cyclic and

a generates Eifand only if generates 1. ..s). 7. Suppose Fc: E is stable under p and let (pF:F-+F (minimum polynomial jib.) and (minimum polynomial ji) be the induced transformations. Show that E is irreducible if and only if i) E/F is irreducible. ii) F is irreducible. iii) wheref is an irreducible polynomial. 8. Suppose that Eis irreducible with respect to çø. Letf (f irreducible)

be the minimum polynomial of ço. a) Prove that the k subspaces K(f) stable subspaces of E b) Conclude that lmf(co)K =

K(fk_l) are the only non-trivial

K

k.

9. Find necessary and sufficient conditions that E have no nontrivial stable subspaces. 10. Let be a differential operator in E.

a) Show that in any decomposition of E into irreducible subspaces, each subspace has dimension I or 2. b) Let N1 be the number of/-dimensional irreducible subspaces in the above decomposition (j= 1, 2). Show that

N1-f-2N,=dimE and N1=dimH(E). c) Using part b) prove that two differential operators in E are conjugate if and only if the corresponding homology spaces coincide. Hint: Use Proposition II, sec. 13.18.

11. Show that two linear transformations of a 3-dimensional vector space are conjugate if and only if they have the same minimum polynomial.

§ 5. Applications of cyclic spaces

12. Let be a linear transformation. Show that there exists a (not necessarily unique) multiplication in E such that i) E is an associative commutative algebra ii) E contains a subalgebra A isomorphic to F((p)

iii) If &F(p)4A is the isomorphism, then

13. Let F be irreducible (and hence cyclic) with respect to p. Show that the set S of generators of the cyclic space E is not a subspace. Construct a subspace F such that S is in I — correspondence with the non-zero elements of E/F. 1

14. Let

be

a linear transformation of a real vector space having can not be written in the

distinct eigenvalues, all negative. Show that form § 5. Applications

of cyclic spaces

In this paragraph we shall apply the theory developed in the preceding paragraph to obtain three important, independent theorems. 13.19. Generalized eigenspaces. Direct sums of the generalized eigenspaces of 'p are characterized by the following

Theorem I: Let (13.45)

be any decomposition of E into a direct sum of stable subspaces. Then the following three conditions are equivalent: (i) Each is a direct sum of some of the generalized eigenspaces E, of 'p. (ii) The projection operators

in E associated with the decomposition

(13.45) are polynomials in 'p. (iii) Every stable subspace UcE satisfies U=

U n

Proof: Suppose that (i) holds. Then the projection operators are sums of the projection operators associated with the decomposition of F into generalized eigenspaces, and so it follows from sec. 13.5 that they are polynomials in 'p. Thus (i) implies (ii).

416

Chapter XIII. Theory of a linear transformation

Now suppose that (ii) holds, and let Uc E be any stable subspace. Then i, we have

since

Uc>o1U. Since Uis stable underQ1 it follows that Q1Uc: Un U c:

U

fl

whence

F1.

The inclusion in the other direction is obvious. Thus (ii) implies (iii). Finally, suppose that (iii) holds. To show that (i) must also hold we first prove

Lemma I: Suppose that (iii) holds, and let

E into generalized eigenspaces. Then to every i, such that (1= 1, ...,r) there corresponds precisely one integer], (1 E1 n

Proof of lemma I: Suppose first that s). Then from (iii) we obtain every ./(1

E. =

E, n

=0 for a fixed I and for

fl

=

0

j= 1

which is clearly false (cf. sec. 13.4). Hence there is at least one] such that

To prove that there is at most one] (for any fixed 1) such that E1 n we shall assume that for some f,]1,]2, fl

+0

and E1 n

0,

+0

and derive a contradiction. Without loss of generality we may assume that

E1flF1+0 and E1flF2+0. Choose two non-zero vectors

y1eE1flF1 and y2EE1flF2. Then since Yi' y2eE1 we have that =

§ 5. Applications of cyclic spaces

417

J irreducible, is the decomposition of p. Let and '2 the least integers such that =0 and =0. We may assume that = '2 we simply replace Yt by the vector where ji

. .

be

Then

f

= 0 and

11

+ 0 for k
k

whence k=l== 1. Hencef(p)=O. is simultaneously nilpotent and semisimple, then Corollary I: If (p = 0.

The major result on semisimple transformations obtained in this section is the following criterion:

Theorem I: A linear transformation is semisimple if and only if its minimum polynomial is the product of relatively prime irreducible polynomials (or equivalently, if the polynomials p and ji' are relatively prime).

Remark: Theorem I shows that a linear transformation p is semisimple if and only if it is a semisimple element of the algebra F(p) (cf. sec. 12.17).

Proof: Suppose (p

=

j1ki

is

...

semisimple. Consider the decompositions

1 irreducible and relatively prime

of the minimum polynomial, and set

ffifr

Then

f(p) is nilpotent, and hence by Proposition I of this section,

f(p)=O. It follows that

by definition, we have

This proves the only if part of the theorem.

To prove the second part of the theorem we consider first the special case that the minimum polynomial, ji, of ip is irreducible. To show that (p is semisimple consider the subalgebra, T(p), of A (E; E) generated by (p and i. Since p is irreducible, F(p) is a field (cf. sec. 12.14). E(p) contains F and hence it is an extension field of F, and E may be considered as a vector space over F(p) (cf. § 3, Chapt. V). Since a subspace of the F-vector space E is stable under p if and only if it is stable under every transforma-

tion of F(p), it follows that the stable subspaces of E are precisely the F(p)-subspaces of the space E. Since every subspace of a vector space has a complementary subspace it follows that 'p is semisimple.

Chapter XIII. Theory of a linear transformation

Now consider the general case

P=

fi

•

.

. fr

ft irreducible and relatively prime.

Then we have the decomposition

E into generalized eigenspaces. Since the minimum polynomial of the is preciselyf (cf. sec. 13.4) it follows induced transformation from the above result that is semisim pIe. Now let E be a stable subspace. Then we have, in view of sec. 13.6,

F F is a stable subspace of complementary subspace

=

and hence there exists a stable

(F n E1)

These equations yield =

H=

H is a stable subspace of E it follows that p is semisimple. Corollary I. Let p be any linear transformation and assume that

E=(F E into stable subspaces such that the induced are semisimple. Then p is semisimple.

transformations be the minimum polynomial of the induced transforProof: Let Since is semisimple each p is a product of relatively mation prime irreducible polynomials. Hence, the least common multiple, f, of the is again a product of such polynomials. Butf(p) annihilates E and hence the minimum polynomial, p, of 'p dividesf It follows that p is a product of relatively prime irreducible polynomials. Now Theorem I implies that 'p is semisimple. Proposition H: Let A c F be a subfield, and assume that E, considered as a A-vector space has finite dimension. Then every (F-linear) transfor-

mation, çü, of E which is semisimple as a A-linear transformation is semisimple considered as F-linear transformation.

§ 6. Nilpotent

and semisimple transformations

429

Proof Let p4 be the minimum polynomial of p considered as a zi-linear are relatively

transformation. It follows from Theorem I that and prime. Hence there are polynomials g,rE4[t] such that

(13.55)

On the other hand, every polynomial over A[t] may be considered as a polynomial in F[t]. Since ((p) 0 we have JLr

denotes the minimum polynomial of the (F-linear) transformation p. Hence we may write = p1 h some /1 E F [t] and so = (13.56) + Pr/I'. where

Combining (13.55) and (13.56) we obtain

q/zp1+h'rp1+hrp-= 1 whence

p- = I

(q Ii + h 'r) Pr + (Ii

are relatively prime. This relation shows that the polynomials and Now Theorem I implies that the r-linear transformation ço is semisimple.

Theorem II: Every linear transformation p can be written in the form

where c°s is semisimple and p and

are given by

and

k=max(k1 Moreover, if N

any decomposition of p into a semisimple and nilpotent transformation such that then is

and Proof': For the existence apply Theorem I and Theorem IV. sec. 12.16

with A=F(p).

430

Chapter XIII. Theory of a linear transformation

To prove the uniqueness let

be

any decomposition of (p

into a semisimple and a nilpotent transformation such that with Then the subalgebra of A(E; E) generated by i.

commutes is and

commutative and contains (p. Now apply the uniqueness part of Theorem IV. sec. 12.16.

13.25. The Jordan normal form of a semisimple transformation. Suppose that E is irreducible with respect to a semisirn pie transformation Then it follows from sec. 13.13 and Theorem I sec. 13.24 that the minimum polynomial of (p

has

the form IL =

where

f is irreducible. Hence the Jordan canonical matrix of (p

has

the

form (cf. see. 13.15) o i.••

(13.57) o

p

•i

p=

deg p.

It follows that the Jordan canonical matrix of an arbitrary semisimple transformation consists of submatrices of the form (13.57) following each other along the main diagonal. Now consider the special case that E is irreducible with respect to a semisimple transformation whose minimum polynomial is completely reducible. Then we have that p= 1 and hence E has dimension 1. It follows that if is a semisimple transformation with completely reducible minimum polynomial, then E is the direct sum of stable subspaces of dimension 1; i.e., E has a basis of eigenvectors. The matrix of with respect to this basis is of the form

(13.58)

§ 6. Nilpotent

and semisimple transformations

431

are the (not necessarily distinct) eigenvalues of p. A linear transformation with a matrix of the form (13.58) is called diagonalizable. Thus semisimple linear transformations with completely reducible minimum polynomial are diagonalizable. Finally let p be a semisimple transformation of a real vector space E. where the

Then a similar argument shows that E is the direct sum of irreducible subspaces of dimension 1 or 2. 13.26. * The commutant of a semisimple transformation.

Theorem III: The commutant C(p) of a semisimple transformation is a direct sum of ideals (in the algebra C((p)) each of which is isomorphic to the full algebra of transformations of a vector space over an extension

field ofF. Proof: Let

EE1EI3"EBEr

(13.59)

be the decomposition of E into the generalized eigenspaces. It follows from sec. 13.21 that the eigenspaces E are stable under every transformation I/JEC((p). Now let be the subspace consisting of all transformations t/i such that

is stable under each element of C(p) it follows that is an ideal in the algebra C((p). As an immediate consequence of the definition, we Since

have

fin

j=1,...,r.

k*j

(13.60)

be arbitrary and consider the projection operators associated with the decomposition (13.59).

Now let

Then (13.61)

where (13.62)

It follows from (13.62) that

Hence formulae (13.60) and (13.61)

imply that C ('p) =

It is clear that C

is the restriction of 'p to

Chapter XIII. Theory of a linear transformation

432

where the isomorphism is obtained by restricting a transformation to

induced by (p. Since the Now consider the transformations minimum polynomial of is irreducible it follows that E we obtain from chap. V, § 3 a vector space over that = of linear 13.27. Semisimple sets of linear transformations. A set { (p transformations of E will be called seniisimple if to every subspace F1 c: E which is stable under each there exists a complementary subspace F2 which is stable under each }

Suppose now that {co2} is any set of linear transformations and let be the subalgebra generated by the Then clearly, a subspace Fc:E is stable under each if and only if it is stable under is semisimple if and only if the each L'eA. In particular, the set algebra A is semisimple. be a set of commuting semisimple transforTheorem IV: Let mations. Then is a semisimple set.

Proof: We first consider the case of a finite set of transformations and proceed by induction on s. If s= I the theoreni is trivial. Suppose now it holds for s--I and assume for the moment that the minimum polynomial of is irreducible. Then E may be considered as a they may F((p1)-vector space. Since the ...,s) commute with be considered as F(ço1)-linear transformations (cf. Chap. V, § 3). Moreover, Proposition II, sec. 13.24 implies that the considered as linear transformations, are again semisimple. Now let F1 E be any subspace stable under the I, ...,s). Then since F1 is stable under (pt, it is a F(p1 )-subspace of F. Hence. by the induction hypothesis, there exists a F(p1 )-subspace of E, F2, which is stable under and such that

E=

F1

F2.

Since F2 is a F((p1)-subspace, it is also stable under stable subspace complementary to F1. of Let the minimum polynomial simple. we have P1 =

fi

. .

. Jr

and so it is a p

f1 irreducible and relatively prime.

is semi-

§ 6. Nilpotent and semisimple transformations

433

Let

E into the generalized eigenspaces of (pt.

Now assume that F1 cE is a subspace stable under each (1=1, ...,s). According to sec. 13.21 each E1 is stable under each It follows that the subspaces F1 n are also stable under every Moreover the restrictions of the to each are again semisimple (ci. sec. 13.24) and in particular, the restriction of ço1 to has as minimum polynomial the

irreducible polynomial f1. Thus it follows that the restrictions of the form a semisimple set, and hence there exist subspaces which are stable under each co and which satisfy to

= (F1

P

fl

Setting F2

I, ...,

=P

we have that F2 is stable under each

E=

j=

F1

and that F2.

This closes the induction, and completes the proof for the case that the are a finite set. If the set is infinite consider the subalgebra A c: A (E; E) generated by the 'pa. Then A is a commutative algebra and hence every subset of A consists of commuting transformations. In view of the discussion in the beginning of this section it is sufficient to construct a semisimple system of generators for A. But A is finite dimensional and so has a finite system of generators. Hence the theorem is reduced to the case of a finite set. Theorem IV has the following converse: Theorem

V: Suppose Ac:A(E; E) is a commutative semisimple set.

Then for each çoEA, p is a semisimple transformation. Proof Let coEA be arbitrary and consider the decomposition

of its minimum polynomial p. Define a polynomial, g, by

Since the set A is commutative, K(g) is stable under every t/ieA. Hence 28

Greub. Linear Algebra

Chapter XIII. Theory of a linear transformation

434

there exists a subspace E1 c:Ewhich is stable under every /JEA such that

E= Now let

Ii =

(13.63)

1k1—1

Then we have

h(ço)E c: K(g). On the other hand, since E1 is stable under h(p), h((p)E1

E1

whence

K(g)

h((p)E1

n E1

= 0.

It follows that E1 c: K(h). Now consider the polynomial

1,1).

where

Since

k, —

1

it follows that

p(p)x=O

xeE1

(13.64)

xEK(g).

(13.65)

and from 1, 1 we obtain

p(p)x=O

In view of (13.63), (13.64) and (13.65) imply that p(p)=O and so Now it follows that

max(k1—1,l)k,

i=1,...,r

whence

i = 1, ...,r. k, = I Hence ço is semisimple. Corollary: If çü and are two commuting semisimple transformations, then p+i/i and IJco are again semisimple.

Proof. Consider the subalgebra A cA(E;E) generated by p and Then Theorem IV implies that A is a semisimple set. Hence it follows

from Theorem V that ço +

and

p

are semisimple.

Problems I. Let p be nilpotent, and let be the number of subspaces of dimension 2 in a decomposition of E into irreducible subspaces. Prove that dim kerp = EN).

§ 6. Nilpotent

and semisimple transformations

435

2. Let p be nilpotent of degree k in a 6-dimensional vector space E. For each k (1 k 6) determine the possible ranks of p and show that k and r(co) determine the numbers N) (cf. problem 1) explicitly. Conclude that two nilpotent transformations ço and are conjugate if and only if

and 3. Suppose ço is nilpotent and let p*:E*÷_E* be the dual mapping. Assume that E is cyclic with respect to p and that p is of degree k. Let a be a generator of E. Prove that E* is cyclic with respect to and that

is of degree k. Let a*EE* be any vector. Show that

if and only if a* is a generator of E*. 4. Prove that a linear transformation p with minimum polynomial 1u is diagonalizable if and only if I) p is completely reducible ii) p is semisimple Show that i) and ii) are equivalent to

where the

are

distinct scalars.

5. a) Prove that two commuting diagonalizable transformations are simultaneously diagonalizable; i.e., there exists a basis of E with respect to which both matrices are diagonal. b) Use a) to prove that if p and t/i are commuting semisimple transformations of a complex space. then a complex space E. Let a p be the decomposition of £ into generalized eigenspaces, and let

be the corresponding projection operators. Assume that the minimum is Prove polynomial of the induced transformation that the semisimple part of 'p is given by cos

Let £ be a complex vector space and 'p be a linear transformation a), not necessarily distinct. Given an arbitrary polynomialf prove directly that the linear transformationf('p) has the eigenvalues (v= 1 n) (n=dimE). 7.

with eigenvalues

28*

I

Chapter XIII. Theory of a linear transformation

8. Give an example of a semisimple set of linear transformations which contains transformations that are not semisimple.

9. Let A be an algebra of commuting linear transformations in a complex space E. such that for any pEA, a) Construct a decomposition E1 is stable under 'p and the minimum polynomial of the induced transformation is of the form (t

(p) —

b) Show that the mapping A—÷C given by

'peA is a linear function in A. Prove that preserves products and so it is a homomorphism. c) Show that the nilpotent transformations in A form an ideal which is precisely rad A (cf. chap. V, § 2). Consider the subspace T of L(A) Prove that generated by the radA = T1 d) Prove that the semisimple transformations in A form a subalgebra, A5. Consider the linear functions in A5 obtained by restricting to A5. Show that they generate (linearly) the dual space L (A5). Prove that the is a linear isomorphism. T4 L(A5). mapping

10. Assume that E is a complex vector space. Prove that every commutative algebra of semisimple transformations is contained in an ndimensional commutative algebra of semisimple transformations. 11. Calculate the semisimple and nilpotent parts of the linear transformations of problem 7, § 1. § 7. Applications

to inner product spaces

In this concluding paragraph we shall apply our general decomposition theorems to inner product spaces. Decompositions of an inner product space into irreducible subspaces with respect to selfadjoint mappings,

skew mappings and isometries have already been constructed

in

chap. VIII.

Generalizing these results we shall now construct a decomposition for a iiorinal transformation. Since a complex linear space is fully reducible with respect to a normal endomorphism (cf. sec. 11.10) we can to real inner product spaces. restrict

§ 7. Applications

to inner product spaces

437

13.28. Normal transformations. Let E be an inner product space and (p: E—*E be a normal transformation (cf. sec. 8.5). It is clear that every

polynomial in ço is again normal. Moreover since the rank of (p1' (k =2, 3 ...)

equal to the rank of p, it follows that p is nilpotent only if = 0. Now consider the decomposition of the minimum polynomial into its prime factors, (13.66) = ... is

and the corresponding decomposition of E into the generalized eigenspaces,

(13.67)

Since the projection operators m1 associated with the decomposition (13.67) are polynomials in they are normal. On the other hand and so it follows from sec. 8.11 that the and X1EE1 be arbitrary. Then

arc selfadjoint. Now let

=0 i + I; = i.e., the decomposition (13.67) is orthogonal. It follows from Now consider the induced transformations sec. 8.5 that the are again normal and hence so are the transformations On the other hand, is nilpotent. It follows thatf(pj=0 and (xi,

= (xi,

hence all the exponents in (13.66) are equal to I. Now Theorem I of sec. 13.24 implies that a normal transformation is semisimple. Theorem I: Let E be an inner product space. Then a linear transformation ço is normal if and only if i) the generalized eigenspaces are mutually orthogonal. are homothetic (cf. sec. 8.19). ii) The restrictions

Proof: Let p be a normal transformation. It has been shown already that the spaces E1 are mutually orthogonal. Now consider the minimum polynomial. of the induced transformation p.. Since j is irreducible over it follows that either (13.68)

or

ft =

+

;t +

— 4,8k

In the first case we have that consider the case (13.69). Then

+