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
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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) .