Chapter I Operators on Finite-Dimensional Spaces

Chapter I

Operators on FiniteDimensional Spaces

1. Spectrum of an operator. Let E be a finitedimensional, complex, normed space, and let dim E = d . Let also T be a linear operator from E into itself. First, we observe that T is automatically continuous : indeed, take an algebraic basis ( e l , . . . ,e d ) of E , d and write z = El c r j e j , y = Cf &e;. Then :

with K = IfI I T c j l l . We define the operator norm of T :

Since the unit ball is compact (see for instance B.B. [ I ] ) , and since T is continuous, this supremum is actually attained, and is a maximum. We will usually drop the subscript "opn, and we just write [(TI(. We denote by t ( E )the set of linear continuous operators on E , equipped with this norm. When a basis ( e l , . . . ,e d ) is chosen, T can be represented by a matrix

M

= ( t i , j ) i j = l , . _ _dr,

the j - t h column of which is made with the components of T e , on the basis, that is T e j = El t , , , e , . If a vector z is given by its components X = ( 2 1 , . . . ,zd) on the basis, the vector T x has MX as components on the same basis. Of course, the matrix depends on the choice of the basis, the operator norm does not.

Chapter I

6

We are not going here to deal with the theory of matrices, which is a huge theory in itself ; we will instead restrict ourselves t o the notions related to our future study of operators on infinite dimensional Banach spaces. T h e characteristic polynomial of T is the determinant of T - X I : c=(X)

= det

(T- XI) , for X

E

(c,

and this polynomial is also the determinant of the matrix M - X I , if M r e p resents T is a given basis. It is easily seen t o be independent from the choice of the basis. This determinant is of the form :

It has degree d , and therefore has d complex roots, not necessarily distinct. If X is one of them, det (T - X I ) = 0 , which means that T -- X I is not invertible. Since the space is finite dimensional, this is equivalent to the fact that T - X I is not injective (or not surjective). Consequently, there exists a vector z, z # 0 , such t h a t (T - XI)z = 0 , or Tz = Xz. T h e vector z is called an cigenucctor associated with the eigenualue A . To simplify our notation, we write T - X instead of T

-

XI.

. . ,A,

be an enumeration of the roots ( m 5 d ) . T h e set is called the spectrum of T . It is a finite subset of the complex plane, consisting of a t most d points. It is never empty, b u t may consist in a single point : the spectrum of the identity is (1). T h e spectrum of any projection is { O , 1 } . Let now XI,.

a(T) = {XI,.

. . ,A,}

Conversely, given any d points XI,. . . ,X d in the plane, it is easy t o see that there is an operator T on C d with a(T) = {XI , . . . , A , + } . Indeed, let ( e l , . . . , ed) be the canonical basis of C d , and define T e , = X j e , . Therefore, the spectrum has a simple structure. This does not mean, however, that all questions in finite-dimensional operator theory are necessarily easy. Among the hardest (and, quite often, with no satisfactory solution), let’s mention : precise computations of the operator norm, approximation of a given operator by operators in a given class, and so on.

Finite Dimensional Spaces

7

2. Minimal Polvnomial. Let p(z) =

1," aktk be a polynomial with complex coefficients.

We de-

fine :

We will show th a t there is a "smallest" polynomial m, such t hat m ( T ) = 0. This polynomial will be called the minirnol polynomial of T . For this, let ( e l , . . . , e d ) be a basis of E. T h e vectors e l r T e l , .. . ,Tdel cannot b e independent : there exists a linear combination : d 0

that is, by definition ( I ) , a polynomial s1, d"s1 5 d , with sl(T)el = 0. T h e same way, we find polynomials s2,. . . ,S d , with : sj(T)ej =

0

for all j

,

= 2,.

. . ,d.

P u t s = ~ 1 . ~ 2 . .sd. . Then s ( T ) e , = 0 for all j = 1,. s(T)z= 0 for all x in E.

.,, d ,

and therefore

We have found a polynomial s, such t hat s ( T ) = 0 . But s has t o be reduced : its degree may be d Z , and we will see later t hat such a polynomial exists with dos 5 d . We factor .(A) = Q M (A - A,)". , and we may assume Q = 1 . If a point X i is not in the spectrum, the corresponding term T - A i l is invertible, so can be removed from s. Let

n,=, .(A)

=

(A - A,)Q'.

0

A, E u ( T )

Now, we look a t all z ' s such that (T- X1)alz = 0. If for all of them we also have (T - X 1 ) Q l - l z = 0 , we replace a1 by a1 - 1 in u. We start again with a1 - 1, and so on, until we cannot proceed further. We then pass to a2,and so on. More precisely, we define, for i = 1,. . . ,m (where m is the number of points in a ( T ) ,m 5 d ) : vi = inf{k E IN ; ( T

- A , ) ~ Z=

Then we put :

0, for a11 z s.t. (T - ~ , ) ' + ' z =

rI(A m

m(A) =

i= 1

~

A;)".,

01.

Chapter I

8

and we have obtained a polynomial m , still satisfying m(T) = 0 . A0

E

By construction, all roots A i of m(A) belong to a(T). Conversely, let o(T), yo a corresponding eigenvector. Then :

so rn(A0)

= 0 , and the roots of m(A) are exactly the points of a ( T ) .

T h e polynomial m(T) is called the minimal polynomial of T ,for the following reason :

Proposition 2.1. p(T) = o .

-

The polynomial m divides every polynomial p s u c h that

Proof. - Let p be such a polynomial. First, as we already observed, p must have the points in a ( T ) as roots ; we eliminate the others, and write :

We now show that a, 2 u,, for i = I , . . . , m . Assume on the contrary that, for instance, a1 < P I . Then, by the definition of vl , there is a point z1 such that :

(T-

A p + l Z l

= 0,

y1 = ( T - Al)(r’z, # O .

From (3), we deduce

p(A) = (A - A p q ( A 1 , with q ( A 1 )

# 0.

We have Tyl

= Alyl,

so :

which contradicts p ( T ) = 0 and proves our claim.

Corollary 2.2. - If p , q are two polynomials, we have p ( T ) = q ( T ) if and only if every A, E a ( T ) is a root of p - q of order 2 v i . Indeed, this says that p - q is divisible by the minimal polynomial.

Finite

Dimensional Spaces

9

Among all polynomials satisfying p ( T ) = 0 , the most noticeable one is the characteristic polynomial, which we have already defined :

c(X) = det (T - A).

Theorem 2.3 (Cayley-Hamilton). - The characteristic polynomial satisfies c(T) = 0.

Proof. - Elementary linear algebra (see for instance I . N . Herstein 111) allows us t o write in a proper basis, the matrix of T in a triangular form :

[; A{

al,2

... ...

... ...

;;]

al,d

where A:, . . . , A > are the points of a(T), but this time each of them repeated according t o its multiplicity. If ( u l , . . . , u d ) is this basis, we get : Tu1 =

Y p l ,

+ X:WZ,

=

a l p I

A{)V,

=

0,

(T- x;)vz

=

01,2Ulr

(T - A)d)Ud

= al,dV1 f

Tu2

which means :

(T-

... f

ar,dU, $.

.. . f

ad-1,dvd-1

Therefore ( T - A : ) u l = 0 , ( T - X { ) ( T - X 9 u 2 = 0,.. . , (T-X{)."(T-A&)ud = 0. So the product (T - A',) - . . (T - A:) annihilates all the vectors of the basis. Since the X i ' s are the roots of the characteristic polynomial, we get c ( T ) = 0 . From Theorem 2.3 and Proposition 2.2 follows obviously that d"m 5 d . There are obvious examples in which the minimal polynomial h a s degree strictly less than the degree of the characteristic polynomial (which is d , the dimension of the space). For example, a projection always satisfies T 2 - T = 0, that is m(A) = A' - A .

Chapter

10

I

3. T h e analytic functional calculus.

We have already defined p ( T ), when p is a polynomial. We now extend this definition t o a larger class of functions. Let f be a function from C into itself. We say t hat f belongs to the space 3 ( T ) if there exists a neighborhood V of a ( T ) on which f is analytic (for an elementary theory of analytic functions, we refer the reader to H. Cartan 111). We recall th at f is said to be analytic o n a compact set if it is analytic on some neighborhood of this compact set. The neighborhood does not need t o be connected, and depends on the function. Let f E 3 ( T ) . For every A, E a ( T ) ,we consider the derivatives f ( k ) ( X , ) , k < Y , (there are v1 + ... Y, = d such derivatives).

+

Let p be a polynomial such t hat all k < v,. We then put :

/(T)

f(k)(X,) =

= P ( ~ ) ( X , ) , for all A, E a ( T ) ,

P(T)

This definition does not depend on the choice of the polynomial p : if q is another one with the same properties, then p ( T ) = q ( T ) ,by Corollary 2.2. We now list some elementary properties of this definition :

Theorem 3.1. - If f,g E 3(T), a , B E Q:,

~f + P g E 3 ( T ) ,and (4 + P g ) ( T ) = a / ( T )+ P g ( T ) , f-4 f - 9 E 3 ( T ) ?and f.g(T) = f(T)S(T) c) if f(x) = 1 ;:akXk, then /(T)= a k T k ,

a)

c," 7

d) / ( T ) = 0 if and only if f ( k ) ( X , ) = 0 , for all X i E a ( T ) ,all k

< v;.

T h e proof is left to the reader. T h e first quality of the class 7(T)is t hat it contains functions with values 0 and 1 only, thus allowing us t o build projections which commute with T : For every A, E a ( T ) , let e,(A) defined by : .,(A) = 1 on some neighborhood of A,, = 0 on some neighborhood of all other A] 's, J # i. We put E, = e , ( T ) ; this is a n operator, with the following properties :

Proposition 3.2. - For a)

E: I=

j = 1,.

. . ,m , we have :

= E,,

b) E,E, = 0 , for C)

1,

1 7E , .

t

#

J ,

These properties follow immediately from Theorem 3.1.

Finite Dimensional Spaces

11

For i = 1,. . . ,m, we call X , the image of the operator Ei . Then we get the formula :

E

=

X1@..*$Xm.

xr

Indeed, by Proposition 3.2, c ) every ~ z can be written as z = E i z , and this decomposition is unique by b). Since E; is in fact defined as a polynomial in T , it clearly commutes with it. Therefore :

i = 1,...,m ,

TX,c X i , which means that XI is invariant by

T.

Also, we have :

(T-

=

0.

Indeed, El= p,(T), where pI is a polynomial satisfying : PI(A1)

= 1

, P,( k )(A,) Vj

p,(k)(X,) = 0 ,

Therefore, pl factors as :

P*(A) =

= 0,

4x1

15 k

< Vl

# i, V k , 0 5 k < uJ .

n(A ~

A,)”,

7

J f l

and (A - X,)”~p,(X) is a multiple of the minimal polynomial m, which proves (1)-

So we get :

( T -A,)”~X,= 0.

(2)

For : = 1,. . . , m , we put N, = Ker(T - A , ) ” ~ , N: = Ker I7,+,(T

Lemma 3.3. Proof.

~

For i = I , . ..,rn,N,n N,’

=

(0).

~

X,)”J.

n,,,(A

The polynomials p,(X) = (A - A,)”,, and q l ( A ) = - A,)”, have no roots in common. By Bezout Identity, there exist polynomials rl(X), r2 ( A ) such that : -

tlP1

+rzq, = 1

which implies :

from which Lemma 3.3 follows obviously.

Chapter I

12

We may now prove :

Proposition 3.4. - For i = I , . . . ,m, X, = N,.

Proof. - 1) We have seen that (T - A,)"# X, = 0 , thus Xi c K e t ( T - X,)"i

.

2) By Lemma 3.3, the sum of the N , is a direct sum ; the sum of the X i is also direct. Since the latter is E (prop. 3.2,c ) , so is the former. The proposition follows.

The projections E, , i = 1,. . . ,m , will also allow u s to give an expression of any function f ( T ) , f E f ( T ) :

Proposition 3.5.

-

If f E 3 ( T ) ,we may write .-

Proof. - We consider the function :

One checks immediately that, for i = I , ,

dk)(X,)=

. . , m , and k < v , ,

f'k'(X,) ,

and therefore f ( T ) = g ( T ) ,by Theorem 3.1, d). We now study the convergence of a sequence of operators fn(T) :

Proposition 3.6. - Let ( f n ) n > ~ be a sequence of functions in 3 ( T ) . Then f n ( T ) converges in operator norm if and only if the d sequences , for i = 1,. . . ,m, k < v , , converge in a .

(fik'

Proof. - 1) Assume that the d complex sequences converge. Then Proposition 3.5 indicates that fn(T) is Cauchy, and therefore converges. 2) Assume that the sequence (fn(T))nEmconverges in L ( E ) . We know that (T - X 1 ) " l - ' E 1 # 0, and therefore we may find z such that (T -

Finite Dimemional Spaces Since y = E l ( z ) , all terms with i

#

13

1 are 0 , by Proposition 3.2, b). So :

All terms with k 2 1 are 0 by the definition of r = v1 - 1, and formula (2). Since moreover E l y = y , we get : In(T)yr = fn(Al)(T

AI)'Y .

Since the sequence (jn(T)y,)nEmconverges, this implies that converges. Then we write :

fn(T)yr-l =

(fn(X1)),,m

+

fn(x~)~r-l fA(Al)~r,

and therefore (J,I,(Al))converges. The same holds for the other points in the spectrum.

To end this chapter, we give a Cauchy formula for operators : P r o p o s i t i o n 3.7. - Let U be an open set, containing a ( T ) . We assume that the boundary 'I of U consists in a finite number of simple closed curves, oriented in the direct sense. Then, if f E 7(T)is analytic on U , we have :

P r o o f . - For A

4

a ( T ) ,we set R ( z ) = 1/(A R ( T ) = (A

-

~

T)-l

and so :

by the usual Cauchy formula, and finally, = f(T) 9

and Proposition 3.7 is proved.

2).

Then :

Chapter I

14

4. Computing the operator norm on a Hilbert sDace.

The formula 1 ( l ) , given as definition of the operator norm, is not suitable for practical purposes. Instead, practical computations are made the following way : Let M be the matrix representing T in some basis. Let M' be its adjoint, that is M' = t & f . Then the matrix M ' M is self-adjoint, and, by the results of Chapter IV, Section 3, or Chapter VII, it has real, positive eigenvalues, and one of them, X , satisfies X = llM'Mll = IIMII'.

So

fi is the required value of

IlTll.

Programs d o exist to find the eigenvalues of a matrix ; however the computations become obviously longer and less precise when the dimension increases.

15

Finite Dimensional Spaces Exercises on Chapter I.

Exercise 1. - Show t h a t the following are equivalent : a)

t I,"-' ~k converges,

b) gT" converges, a ( T ) is contained in D and .(A)

c)

= 1 , for all X E a ( T ) ,with

(Hint : consider t h e sequences of functions /,,(A) Xn/n).

=

x:-'Xk,

1x1 = 1. gn(X) =

Exercise 2. Let T be a n operator on a Banach space E , and assume that T has a cyclic vector 2 0 , t h a t is : ~

E = s p a n ( z 0 , T Z O ,T'zo,

...I

Assume moreover t h a t the minimal polynomial of T can be written (with d=dim E) : 70

+ 7lx 4-

' - '

Find a basis in which the matrix of

0

..

I

+ 7 d - ] X d - l + Ad .

T

is of the form :

.. . . ..

... ... Prove t h a t in this case, t h e characteristic polynomial of minimal polynomial ( u p to a change of sign).

T is equal

to the

Exercise 3. - Let E , d i m E = d , and T an operator such t h a t T d= 0. Show of E and k 1 integers 1 n o < nl < that there exists a basis ( 2 1 , . . . , q ) < n k = d , such t h a t Tr; = r ; + l ,except if i E {no,nl,...,nk},in which

+

case

Ti;= 0.

Exercise 4. - In

a! n , what

is the operator norm of the matrix whose entries

are all 1's ?

Exercise 5. - Let

M = ( a Find A such t h a t e A = M .

[ 2:)

16

Chapter I

Exerciee 6. - Let

M = ( i

i i)

Compute the distance between M and the set of diagonal matrices.

Finite Dimensional Spaces

17

Notes and Comments. T h e results of this chapter are mostly reproduced from Dunford-Schwartz [I],vol.1, with a few modifications in the presentation. The reader may consult this book, as well as the book by P. Halmos [ I ] .

Complements on Chapter I. Exercise 4 involves the computation of the operator norm of the matrix, whose entries are all 1's. In a " , the result is n . But if, instead of 1, one puts ~ ; , ,j ( i , i = 1,.. . ,n), independent random variables with values f1 , then the operator norm is, with great probability, of the order of fi (see Benett Goodman - Newman Ill), thus dropping considerably. An open problem (communicated by David Larson) : We consider 3 x 3 matrices. Let P be a "diagonal projection", that is a matrix of the form :

P

=

(H !)

where a , b and c take the values 0 or 1.

For any matrix M , let : a ( M ) = sup{II(Z - P)MPII ; P is a diagonal projection} and

d(M)

=

inf{J\M- DJJ;

D is a diagonal matrix}

Compute :

and then :

K 3 = s u p { K ( M ) ; M is a non - diagonal 3 x 3 matrix}