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CHAPTER I

SPECTRUM PRESERVING LINEAR MAPS

In this chapter the structure of spectrum

preserving linear maps between 43(X) and (3(Y) is

studied, where X and Y are locally convex topological

vector spaces over the field e of complex numbers.

This is a generalisation of the work of Ali A Jafarian

and A.R. Sourour [14], where they considered spectrum

preserving linear maps on P(X) where X is a complex

Banach space. Section 1.1 deals with this. In

Section 1.2, spectrum preserving linear maps on P(X)

which preserves eigen values of operators in P(X) are

studied.

1.1. SPECTRUM PRESERVING LINEAR MAPS ON P(X)

THEOREM (A LI A JA FARIAN and A.R. SOUROUR)

Let X and Y be Banach spaces and I:P (X) -p p(Y)

be a spectrum preserving surjective linear mapping. Then

either

(i) there is a bounded invertible operator A:X --'^ Y

such that T(T) = ATA-1 for all T in 43(X) , or

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(ii) there is a bounded invertible operator

B from X* (the dual of X) to Y such that

(T) = BT*B-1 for every T E. P(X).

As specified we establish the same result,

when X and Y are locally convex topological vector

spaces over 0. Since the method adopted is the same

as in [14], we start with generalising various

technical results proved in [14].

Even though the proofs are exactly similar,

we supply the details. Through out this section X

and Y denote locally convex topological vector spaces

over C and p(X) the set of all continuous linear

mappings on X.

LEMMA 1.1.1

Let A be in 3(X). Then a(T+A) G a(T) for every

T in P (X) if and only if A= 0.

PROOF

A= 0 a(T+A) = cs(T) for all T.

Now assume that a(T+A) c a(T) for all T.

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To show that A = 0.

Let if possible A $ 0. Then there exists

x E. X, x A 0 such that A (x) = y A 0. By Hahn

Banach theorem in locally convex topological vector

spaces, there exists f E X* (X*- the dual of X)

such that

f(x) = 1 and f(y) A 0

Let a be a nonzero complex number and let

T = (ax-y)(& f, where

T(z) = ((ax-y)(&f)(z) = f(z) (ax-y), z E X

Continuity of T follows from the continuity of f.

Now

(T+A)(x) = T(x) + A(x) = ax-y+y = ax

Hence a is an eigen value of T+A.

Now one can easily show that, for P E ( , T-PI is

not invertible in P(X) if and only if either P = 0

or P = f(ax -y). Therefore a(T) = tO, f(ax -y) I

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Since f ( ax-y) = a-f ( y) A a, we have

6(T+A) a(T)

This proves our assertion.

LEMMA 1.1.2.

Let ^: p(X) -'^ P (Y) be a spectrum

preserving linear mapping . Then j is injective.

PROOF

c(T) = a((^(T)) for all T in 3(X) .

Suppose J(A) = T(B), A,B E P(X)-

To show that A = B

v(T+A -B) = a (^(T+A-B))

= cs ((I(T) )

a (T)

for all T in 13(X). Hence by lemma 1.1.1, A-B = 0.

LEMMA 1.1.3.

If T: 13(X) to 13(Y) is a spectrum

preserving surjective mapping, then j(IX) = IY,

where IX (or IY) denote the identity mapping on X

(or Y respectively).

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PROOF

IY

For T

6(I(T-IX+S))

^(IX) + ^(S) )

a(^(T))

Q(T)

Hence by lemma 1.1.1, S = IX.

LEMMA 1.1.4.

Let X be a locally convex topological vector

space and K(X) denote the set of all compact operators

on X. Let A be in S(X) and C be in K(X). If

A E cr(A) is not an eigen value of A, then F cr(A+C) .

PROO F

Let if possible, A+C- 2,.IX is invertible in p(X).

Therefore,

A - 2.IX = (A+C- a•IX) [IX-(A+C- a IX)-lC]

Since (A+C- )IX)-1 is continuous and C is compact,

(A+C- 2, IX)-1.C is compact [ 9 ]

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Case 1

IX- (A+C- A-IX)-1 C is invertible.

In this case A-P•IX is invertible.

Case 2

IX-(A+C- 2•IX)-l is not invertible.

Here, since (A+C- ?.IX )-1C is compact, 1 is

an eigen value 'of (A+C- T IX)-1C [9 ] .

Hence, there exists a non zero vector x such

tha t

IX - (A+C-A.IX)-lC (x) = 0.

Therefore (A_A-IX) (x) = 0.

In either case the conclusions are contradictions

to the assumption that a E a(A) but not an eigen value

of A.

This completes the proof.

LEMMA 1.1.5.

For T in P(X), x E. X, f E X* and A not in o(T)

we have A E a(T+ x Of) if and only if f(( 2^IX_T)-1(x))=1.

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PROO F

Assume that f(;N - 1 x-T )- '(x) = 1

Then (x®f) ((a-IX-T)-1(x)) = (f( A.IX-T)- 1(x)).x

= x

Hence

(T+x®f)(a•IX-T)-1(x)

= T((;^-IX-T)-1(x)) + (x®f)(A IX-T)-1(x)

= T(T-IX-T)-1(x) + x

_ (T( A-IX-T)-1 + IX)(x)

(T+;A. IX-T) ( ;k. IX-T) -1 (x)

=A (A.IX-T)-1(x)

Therefore A is an eigen value of T+x ®f.

Conversely assume that A £ a(T+x ®f) . Then by

lemma 1.1.4, A is an eigen value of T+x ®f. Hence

there exists a non zero vector u in X such that

(T+ x®f)(u) = A u

i.e., T(u)+f(u).x -A u

since A ^ a(T), f(u) A 0

. ' . (\ IX-T ) -'(X) = f u

i.e., f((2^IX-T)-1(x))= 1

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THEOREM 1.1.6.

Let A €13(X) , A A 0. Then the following

conditions are equivalent.

(1) A has rank 1

(2) a(T+A) f a(T+cA) c a(T) for every T in 13(X)

and every c A 1.

PROOF

Assume that A is of rank 1. Hence there

exists x E X and f . X* such that

A = x ®f.

Now let T be in 13(X) and A not in a(T). Then by

lemma 1.1.5 a is in a(T+cA) if and only if

f((A IX-T)-1(x))=1. Hence T does not belong toc-(T + cA) foX

two distinct values of c. Hence (1) implies (2).

Now to show that (2) implies (1).

Assume that rank A > 2.

Case 1

A = a.IX for some nonzero scalar a.

Let T in p(X) be such that a(T) = {O,aa

It is enough to take T = y ® g for suitable y E X and

g in X.

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Then

a(T+A) = [a ,2aj and

a(T+2A ) = 12a,3a' .

Therefore a(T+A) n a (T+2A) = L2a) which is not

contained in a(T). This completes the proof of

case 1.

Case 2.

A A aIX for any a in C and rank A Z 2.

Case 2' .

There exists a vector u v- X such that

tu,A u,A2uf is linearly independent . Let U be the

linear span of ^u ,A u,A2uj and V be a closed complement

of U in X . It is enough to take

V = ker fAu [1 ker fA2u f ker fu

where fu , fAu' fA2u are bounded linear functionals

on X such that

Amu' = Sm n' m,n = 0 ,1,2,...

where,

m,n= 0 if m n

= 1 if m = n

Put Nu = u - Au

N(Au) = Au - 2Au

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and

Nv = 0 for all v in V and extend it linearly.

Clearly N E(3(X) and N3 = 0, (N+A)(u)=u and

(N+2A) (Au) = Au.

Therefore, 1 E6(N+A) no(N+2A)

But a(N) =301.

Thus 6(N+A)( (N+2A) is not contained in c(N).

This establishes case 2'.

Case 2"

ju,Au,A2u)are linearly dependent for every u in X.

Let A2u = au + PAu for some scalars a and P.

First we assume that a ^ 0.

Let N(x) = x for every x E V, and

N(u) -Au

N(Au)= A2u = au + 13Au

Clearly N is invertible.

Also (N+A)(u) = 0 and (N-A)(Au) = 0

Thus 0 E a(N+A) f\ a(NA) whereas 0 is not in a(N) .

If A2(u) = 13(u)Au for every u, we get,

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A2(u-v) = A2u - A2v = P(u)Au - P(v)Av

= 13(u-v)Au - p(u-v)Av

3(u) _ p(v ) = p(u-v) since rank A > 2.

Thus A2 = PA for a fixed scalar P.

i.e., P(A) = 0 where P(t) = t2-P.t

Now rank (A) > 2. Also 0 and P are eigen values of A.

Hence there exists three linearly independent vectors

x,y,Az such that

A(x) = 0

A(y) = Py, and

A(z) A 0

Let W be the three dimensional space generated by x,y

and A(z). Then A(W)5;W andP 0 00 P 0 is the matrix0 0 0

of A/W with respect to jA(z), y,x3. Now we define a

nilpotent operator N as follows. Let Z be a complement

of W in X and let N(Z) = [0} . Let N/W has the matrix

representation

0 0 00 2p 2P with respect to {A(z)y,x) .0 -2P -2P

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Then N is nilpotent. One can easily see that 2P

is an eigen value of (N+A) and (N+2A). Since PLO

we get

6(N+A) n o(N+2A) u(N).

Now we prove the main theorem of this section.

THEOREM 1.1.7.

Let f: P(X) --^i p(Y) be a spectrum preserving

surjective linear mapping. Then either

(i) there is an invertible linear operator A:X--> Y

such that ((T) = ATA-1 for every T in P(X) for

which there is an unbounded sequence in C-o(T)

or

(ii) there is an invertible linear operator B:X*---Y

such that J(T) = BT*B-1 for every T in (3(X),

for which there is an unbounded sequence in

ci -o(T).

PROOF

Let x and f be nonzero elements in X and X*

respectively. Let Lx and Rf be linear subspaces

of P(X) defined by

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Lx = t x ®h : h E X*l and

Rf = [u ®f : u E XI

First we prove the following. Corresponding to

each x in X there is a y E Y such that

J(Lx) = Ly

or corresponding to each x in X, there is a g F- X*

such that

T(Lx) = Rg

Also if T(Lx) = Ly for some x E. X, then

T(LU) Rg for any u e X

This follows from the following observations.

(i) By lemma 1.1.2 and theorem 1.1.6, if R is of

rank one, J(R) is of rank one.

(ii) Lyl R9 is one dimensional where as Lufl Lv

has dimension 0 or dimension X*.

(iii) If (J(Lu) = Ly for some u E X, then

J(Lv) Rg for any v in X.

For,

J(LUn Lv) = J(Lu) n J(Lv) = Lyn Rg

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Since - is one-one and onto, dimension LufLv

should equal dimension J(LufLv) = dim Ly()R9

which is not possible. This leads to two cases.

Case 1.

J(Lx) = Ly(x) for every x E X. Put y(x)=y

for brevity.

Therefore,

I(x ®f) = y (D g for some g e X*

Now let,

C x : X* ) Y* be defined by

Cx(f) = g. Clearly Cx is linear.

Claim

The set , Cx: x E X3 is one dimensional. Let

if possible , there exists two linearly independent

transformations C x and C x , wherel

(I(xl(@ f) = Yl ® Cx1(f) and

(x2 ®f) = Y2 ® Cx2(f)

Now,

T((xl+x2)(D f) = (xl ®f) + V x2 Of)

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Since xl+x2 O f is of rank 1,

J(x1+x2)®f = Y®9

for some y EX and g £ X*. Hence we get

Cx1(f)(u).yl + Cx2(f)(u)'y2 = g(u)•Y

for every u in X. Since Cxl and Cx2 are linearly

independent, yl and y2 should be linearly dependent.

Hence L = LY1 y2

.'. I(Lx1) = I(Lx2)

Since T is one-one we have Lx1=Lx 2. Therefore xl

and x2 are linearly independent. Then Cxl and Cx2

are linearly dependent. This is a contradiction.

dimension tcx: x E X7 = 1

Hence there is a linear operator C such that

^Cx: x E Xj =[A C: E J

Therefore,

(J(x of ) = y ® Cf where y depends on x.

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Put Ax=y. Hence (I(x ®f) = Ax ® Cf. Since T is

bijective both A and C are bijective linear mappings.

Now let T E 3(X) be such that there is an unbounded

sequence in (2 -a(T).

((T+x®f) _ T(T) + Ax®Cf

Let A be not in a(T). Hence by lemma 1.1.5 we have

f ( ( 2, -T) x)= 1 if and only if A £ a(I(T)+Ax ®Cf)

and

a(^(T)+ Ax ®Cf) if and only if

Cf(A Iy- J(T))-1 A(x) = 1

Thus for A not in a(T), we get

f ( ( A IX-T)-1(x)) = Cf((A Iy-T(T))-1 Ax)

Replacing A with and using similar argument as in [14

we get,

f((IX-zT)-1(x)) = Cf(IY -zz(T))-1(y), where A(x)=y

That is

f((IX-zT)-1 A-1(Y)) = Cf(Iy-zT (T))-1 (Y)

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Since L`t - a(T) contains an unbounded sequence,

by taking the limit as z ---> 0 we get,

f(A-1(y)) = Cf(y)

Againf[(IX-zT)-1 A-1(y) - A-1 (y)]

z

= Cf[(IY - z^ (T))-1(y)-y I

z

Now letting z tend to 0 we get,

f(TA-1(y)) = Cf(j(T)y) for all f . X*

But we already have,

f(A-1(y)) = Cf(y)

Combining these two we get,

Cf(I(T)(y)) = f(A-1 ^(T)(y)),

Cf(^(T)(y)) = f(TA-1(y))

and Cf(^(T)(y)) = A-1((I(T)(y))

Thus we get,

f(TA-1(y)) = f(A-1(^(T)y) for every f in X*.

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That is, TA-1(y) = A-1 J(T) (y) for all y E Y

That is, ATA-1 = ^(T)

Case 2

Let x X and ^(Lx) = Rg for some g E Y*.

As in case 1, we can show that for each x E X and

f E X* ,

(T(x ®f) = Bf ® Ax

where B: X* --p Y is linear.

As before for T E P (X) , x E X, f EX* and A 4, j(T),

A F a(T+x®f) if and only if f((\IX-T)-1(x)) = 1

and finally for every x E X, f E X* and A 4 a(T)

f ( ( A IX-T)-1(x)) = A(x) ((T Iy _(T))-1 (Bf))

Now for T E p(X) such that @ -6(T) contains an unbounded

sequence , identical arguments leads to the conclusion

f(T(x)) = A(x) (I(T) B(f))

f(x) = A(x) (B(f))

Therefore,

A(x) ^(T) B(f) = f(T (x)) = AT(x ) ( B(f)) = A(x )(BT*f)

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Hence we get

g ((T) (B( f)) = g(BT*(f)) for all g in Y*

Therefore,

UT) B(f) = BT*f for all f in X*

Thus ^(T) = BT*B-1

REMARK 1.1.8.

One does not know whether the operators A or B

obtained in Theorem 1.1.7 is continuous or not. But

when X and Y are Frechet spaces, using closed graph

theorem, continuity of A and B can be established [9].

REMARK 1.1.9.

It is to be observed that Theorem 1.1.6 is a

generalisation of the corresponding theorem of

Jafarian and Sourour [14]. Though the proof goes along

the same line as in [14], our proof is simpler in the

following sense. By considering one more simple case,

we are able to arrive at the quadratic polynomial

P(t) = t(t-P) such that P (A) = 0 directly without using

any existence theorems. Also the other forms of minimal

quadratic polynomials are not needed at all.

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1.2. EIGEN VALUE PRESERVING LINEAR MAPS

In section 1.1 we analysed spectrum preserving

surjective linear maps of p(X) to P(Y), where X and Y

are locally convex topological vector spaces over 0_.

In this section we characterise spectrum preserving

linear maps which preserves eigen values when X and Y

are complex Banach spaces with Schauder basis.

THEOREM 1.2.1.

Let X and Y be Complex Banach spaces with

Schauder basis and ^: P(X)-> P(Y) be a spectrum

preserving , surjective linear mapping. Then

preserves eigen values if and only if it is of the

form ^(T) = ATA -1 for every T in P(X) where A:X -- Y

is a bounded invertible linear operator.

PROOF

Let ^(T) = ATA-1, for all T in P(X), A:X ---Y

an invertible bounded linear map . Let A Ea(T). Then

T(x) = A x for some nonzero x in X.

Let y = A(x). Thus TA-l(y) = a A-1(y)

i.e., ATA -1(y) = I y

Therefore A is an eigen value of T.

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From theorem 1.1.7, either

(i) ( (T) = ATA-1, A:X --'^ Y a bounded invertible

linear map, or

(ii) I(T ) = BT*B-1, where B : X* Y is a bounded

invertible linear map.

We show that if T takes the form (ii), ^ will not

preserve eigen values for all T.

Let if possible a is an eigen value of T

implies a is an eigen value of BT* B-1. Then there

exists a nonzero x in X such that T (x) = a•x.

i.e., (T- ).IX) is not one-one.

Since A is an eigen value of BT*B-1,

BT*B-1 (y) = A•y for some non zero y in X.

Therefore T*B-1(y) =B(y)

i.e., is an eigen value of T*.

Let f = B_ 1(y). Then we get,

(T*f)(z) = a f (z) for all z in X.

Hence,

f((T-AI)( z)) = 0 for all z EX.

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Since f is a non zero continuous linear functional

on X, its null space is a proper closed subspace of

X. Therefore Range (T- ,AI) is not dense in X.

Now we show that there is a bounded linear

operator S on X which is not one-one but onto. In

this case, 0 is an eigen value of S but it is not

an eigen value of ^(S).

Let Jxl,x2,...,xnj be a Schauder basis in X.

For z in X,

00z = E an ( z)xn, where an(z) n=1,2,...

n=1

a2(z) a3(z) an+1(z)...Put 5(z) = 22 x1 + 32 x2 + ... + (n+1) 2 x n +

Then S E (X), S(xl) = 0.

Now we show that Range (S) is dense in X.

COLet y = E an(y)xn1

Let xn = 0.x1 + 22.a1(y)x2 + 32a2(y)x3+ ...+

(n+l) 2 an(y) xn+l

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Then

S(xn) = a1(y)xl + a2(y)x2+ ... + an(y)xn

Therefore,

S(xn)^P y as n

i.e., Range (S) is dense in Y.

REMARK 1.2.3.

Let X and Y be complex Banach spaces and

let 1: ^(X) -3 P(Y) be a linear mapping. If

J(T) = ATA- 1 + K1TK2,

where A : X --) Y a bounded invertible linear map,

K1:X -- Y compact linear mapping and K2 : Y ---^ X

a compact linear mapping , one can easily see that

I preserves the essential spectrum of T, for every T

in P(X).

Now let J(T) = BT*B-1 + K1TK2, where B:X* > Y

a bounded, invertible linear map, and K1,K2 as above.

One can easily prove that ae[I(T)] cae(T) for all T

in P(X), where ae(.) denote the essential spectrum.

The inclusion may be proper as every compact operator

K on X* need not be dual of some compact operator on X.

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The following is an example for that

EXAMPLE 1.2.4.

Let X = Q.l, the Banach space of all summable

sequences of complex numbers with 1 norm. Then

^l C and the closure 11 of 11 under the L norm

is properly contained in I. . Hence there is a

non zero bounded linear functional F on 9 . such that

F(x) = 0 for all x in k l o Let f in - -'1 be such

that F(f)=l.

Now let,

= f®F, where f o F(g) = F(g).f, g £ L .

Then ^c. is a compact linear operator on Q . We show

that T* for any T in P( Q1) .

Let if possible = T* for some T in

Therefore,

(g) (u) = T* (g) (u) for all g in 100 and

for all u in

i.e., F(g ).f(u) = g(Tu) for all u in

Hence 0 = g(T(u)) for all g in k 11

34

Now let h E 11 be arbitrary, and let

T(h) _ (ul,u21 ...Iun ...) Ji. Let g = (ul,u2,...un..) e il•

Then 0 = g(T(h)) = E lun12 un = 0 for all n.

Hence T(h) = 0. But h is arbitrary. Hence T = 0.

Therefore T* = 0 which is not true. Hence -C? T*

for any T in 3 ( .Q.1) .