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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
SpringerVerlag
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 1501, 15A03, 15A06, 15A18, 15A21, 1601 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 7413868 Linear
All rights
reserved.
No part of this book may be translated or reproduced in any form without written permission from SpringerVerlag. © 1975 by SpringerVerlag New York Inc. Softcover reprint of the hardcover 4th edition 1975 ISBN 9781468494488 DOl 10.1007/9781468494464
ISBN 9781468494464 (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 thoroughgoing 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 tensorproduct. In chapter VII the notion of vectorvalued 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 antisymmetryoperator is studied in greater detail and the concept of the skewsymmetric power is introduced. The dual product (Ch. IX) is generalized to mixed tensors. A special paragraph in this chapter covers the skewsymmetric 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 Lorentztransformations. 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 thoroughgoing 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 Lorentztransformation 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 tensorproducts 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. Ggraded vector spaces § 2. Ggraded 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. PseudoEuclidean spaces § 5. Linear mappings of PseudoEuclidean 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 Lorentztransformations
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 onetoone (onto, onetoone 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 noncommutative 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 zerovector 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 ntuples
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 ntuple (0, ..., 0) and the inverse of is the ntuple ..., 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 nspace 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 realvalued 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 ktimes 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 0vector 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 nontrivial 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 nontrivial 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 nontrivial 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 nspace, ['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 nontrivial 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 nontrivial 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 nontrivial 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 nontrivial 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 nontrivial 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 nontrivial 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 ntuple 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 nonempty 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 nonempty 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 zeroelement. 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 nontrivial 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 finitedimensional 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 infinitedimensional 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 nontrivial 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 ndimen
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 nontrivial 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 nondegenerate bilinear function, and hence F can be regarded as a selfdual 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 nondegenerate 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 nonzero 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 nondegenerate.
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
nondegenerate. 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 nondegenerate 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 nontrivial 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 mmatrix. 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 rowvectors of A. Similarly, the columns —
...
considered as vectors of the space
(it = 1 ... m)
are called the columnvectors 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 rnmatrix 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 inmatrix determines a linear mapping p:E+F by the equations (3.3). Thus, the operator M('p)
M: 'p
defines a onetoone correspondence between all linear mappings ço:E—*F and all n x rnmatrices.
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 nimatrix A. Denote by r1 and by r2 the maximal number of linearly independent rowvectors and co
Observing
lumnvectors, 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 vth rowvector,
follows that f3 a p maps
=
/3
Consequently,
the rank of flap
is
equal to the maximal number r1 of
linearly independent rowvectors. 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 columnvectors 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 vectorequation X
= j.
Consequently, the system (3.9) has a solution if and only if the vector y
is contained in the imagespace Im tp. Moreover, this solution is uniquely determined if and only if the kernel of ço consists only of the zerovector. 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 nontrivial solutions if ni is less than n.
§ I. Matrices and systems of linear equations
87
3.7. The alternativetheorem. Let us assume that the number of equations is equal to the number of unknowns,
=
= 1... n).
(3.14)
Besides (3.14) consider the socalled "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 alternativetheorem: 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 nontrivial 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 maintheorem. 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 imagespace 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 righthand 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
Maintheorem: 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 orthogonalityrelation (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 nmatrix 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 righthand 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 rnmatrices. of all linear mappings ço:E—÷F and the set Once bases have been chosen in E and in F there is a 11 correspondence between the mappings p: E—÷ F and the n x rnmatrices 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 rnmatrices and
A= as
B
the n x rnmatrix
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 zerovector in this linear space is the matrix which has only zeroentries. 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 1matrix (3.20) is called the product of the ii x rnmatrix 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 productmapping 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 matrixmultiplication 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 nmatrix A is called regular if it has the maximal rank /1. Let 'p be an automorphism of the ndimensional 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 squarematrix 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
Basistransformation
§ 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 nmatrices 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 nmatrix another basis can be obtained by = . .
To show that the vectors
are linearly independent, assume that
= 0. Then
=0
§ 3. Basistransformation
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 nmatrix. 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 basistransformation is the inverse of the matrix of the transformation The two basistransformations and = (3.26) = are called contragradient to each other. The relations (3.26) permit the derivation of the transformationlaw
for the components of a vector xeE under the basistransformation 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 basistransformation in E as well as in F. Let M(p; xi,, and M (p; be the ii x inmatrices 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 basistransformations 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 leftmultiplication with a regular 11 x nmatrix and rightmultiplication with a regular in x inmatrix.
Problems 1. Let f be a function defined in the set of all ii x nmatrices 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 ndimensional 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 normalform:
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 normalform (3.31) by a number of elementary basistransformations. 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 rowvector by (11.2.) Replacement of the columnvector bk by It remains to be shown that every n x rnmatrix can be converted into the normal form (3.31) by a sequence of these elementary matrixtransformations and the operations be the given ii x rnmatrix. 3.16. Reduction to the normalform. Let otherwise the It is no restriction to assume that at least one matrix is already in the normalform. 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 ith row, are zero, consider the corresponding number on the right handside. If the ith equation contains a contradiction and the system (3.33) has no solution. If the ith 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 pth 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 ntupie 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 nimatrices C and C' are called equivalent if there exists a regular n x nmatrix A and a regular in x rnmatrix 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 ptuples (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 ndimensional vector space and let P be a skew symmetric nlinear 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 ndimensional vector space E is a skew symmetric nlinear function from E to F. In every ndimensional 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 nlinear 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 ndimensional vector space and fix a nonzero determinant function 4 in E. Then every skew symmetric nlinear 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 nlinear 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 nonzero 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 ndimensional linear space E. To define the determinant of (p choose a nontrivial 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 nontrivial 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 nontrivial 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 zeromap 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 ndimensional vector space and let zi +0 be a determinant function in E. Let çoeL(E; E). Then an nlinear 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 nonzero. 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 nontrivial 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 lefthand 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 nmatrix 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 nmatrices. Then det (A B) = det A det B.
(4.18)
In fact, let E be an ndimensional 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 unitmatrix)
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 ntuples
(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 rowvectors has the following properties: 1. The determinant is linear with respect to every rowvector. 2. If two rowvectors are interchanged the determinant changes the sign.
3. The determinant does not change if to a rowvector a multiple of another rowvector is added. 4. The determinant is different from zero if and only if the rowvectors are linearly independent. An argument similar to the one above shows that
where the bV are the columnvectors of A. It follows that the properties 1—4 remain true if the determinant of A is considered as a function of the columnvectors.
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 2dimensional 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 determinantfunctions 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 =
(qr)4
1)(P_r)
(nq)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 ntuple (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 ntuples (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 ntuple ...
(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 ntuples 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 ndimensional 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 1dimensional 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 5dimensional 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 ndimensional 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 ndimensional 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 nonzero 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 onedimensional 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'osided 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 Crfunctions 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. ipderivations. Let A and B be algebras and 'p : A B be a fixed homomorphism. Then a linear mapping 0: A B is called a 'pderivation if
0(xy)=
x,yeA.
+
denotes In particular, all derivations in A are iderivations where i : A the identity map. As an example of a 'pderivation, let A be the algebra of Cmfunctions 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 'pderivation.
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 'pderivation.
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 pantiderivation 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 iderivation 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 ndimensional 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: n2tuples (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 nonzero 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 noncommutative 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 nonzero 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 nonselfadjoint 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 ndimensional 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 nonzero 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 nonoriented 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 3space. Consider a proper rotation p of a 3dimensional inner product space E. As has been shown in sec. 8.19,
there exists a 1dimensional subspace E1 of E whose vectors remain fixed. If p is different from the identitymap 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 idimensional 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 rotationvector. 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 unitvector 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 Fittingnull component and the Fittingone 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 ndimensional 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:F03F0 and ço1
clude that F0 and F1 are respectively the Fitting null and the Fitting 1component 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 2dimensional 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 2matrices 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 udimensional 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 nonzero algebra endomorphism. Show that for any projection operator it.
r(P(m))=i(rr), (ii) Use (i) to prove that 2. Show that every nonzero 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 nontrivial
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
N1f2N,=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 3dimensional 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 nonzero 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 nonzero 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 Fvector 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 Avector space has finite dimension. Then every (Flinear) transfor
mation, çü, of E which is semisimple as a Alinear transformation is semisimple considered as Flinear transformation.
§ 6. Nilpotent
and semisimple transformations
429
Proof Let p4 be the minimum polynomial of p considered as a zilinear 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 (Flinear) 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 rlinear 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 sI 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 6dimensional 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
+