On linear maps preserving spectrum, spectral radius and essential spectral radius

Transcription

1 Islamic University Of Gaza Deanery Of Higher Studies Faculty Of Science Department Of Mathematics On linear maps preserving spectrum, spectral radius and essential spectral radius Presented by Huda A. D. AlSerr Supervised By Prof. As'ad Y. As'ad Submitted in Partial Fulfillment of the Requirement for the Degree of Master of Science The Islamic University of Gaza 2016

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4 Dedicated To My parents, My husband, My lovely princes Hala, My brothers, My friends, And all knowledge seekers... i

5 Acknowledgements On the very outset of this work, I would like to thank Allah for giving me power to complete this work. Also, I am grateful to my lovely parents and brothers for their immolation, subsidization, love and guidance. Without them this work would never have come to existence. I would like to give my special thanks to my husband whose patience, love and care enabled me to complete this work. I am ineffably indebted to Prof. As'ad Y. As'ad, my supervisor for his guidance and his helpful suggestions during my work to complete this thesis. I wish to thank my teachers in the Department of Mathematics in the Islamic University for their encouragement and for supporting me in my studying. I really extend my sincere and heartfelt obligation to wards all the people and especially all my friends who have helped me in this endeavor. Without their active guidance, help, cooperation and encouragement, I would not have made this work come true. ii

6 Contents Dedicated Acknowledgements Abstract i ii iii Introduction 1 1 Preliminaries Banach spaces Hilbert spaces Banach algebra Linear maps preserving spectrum Spectrum and some of its properties Linear maps preserving spectrum Linear maps preserving spectral radius Spectral radius and some of its properties Linear maps preserving spectrum Linear maps preserving essential spectral radius Essential spectral radius and some of its properties Linear maps preserving essential spectral radius Bibliography 63 iii

7 Abstract In this thesis, we focus our study on a part of the linear maps on algebras of operators that preserve certain properties of operators. We study linear maps preserving spectrum. Let X and Y be Banach spaces. We show that a spectrum preserving surjective linear map ϕ from B(X) to B(Y) is either of the form ϕ(t) = ATA -1 for an isomorphism A of X onto Y or the form ϕ(t) = BT B -1 for an isomorphism B of X / onto Y. After this, we study linear maps preserving the spectral radius. Let X be a complex Banach space. We show that if ϕ: B(X) B(X) is a surjective linear map such that T and ϕ(t) have the same spectral radius for every T B(X), then ϕ = cθ where θ is either an algebra-automorphism or an antiautomorphism of B(X) and c is a complex constant such that c =1. Finally, we characterize linear maps from B(H) onto itself that preserve the essential spectral radius. Where B(H) is the algebra of all bounded linear operators on an infinite dimensional complex Hilbert space H. iv

8 Introduction Functional analysis is a branch of mathematics, and specially of analysis, concerned with the study of spaces of functions and operators acting on them. It has its historical roots in the study of transformations, such as the Fourier transform, and in the study of differential and integral equations. This usage of the word functional goes back to the calculus of variations, implying a function whose argument is a function. Its use in general has been attributed to mathematician Stefan Banach. In the modern view, functional analysis is seen as the study of complete normed vector spaces over the real or complex numbers. Such spaces are called Banach spaces. An important example is a Hilbert space, where the norm arise from an inner product. These spaces are of fundamental importance in many areas, including the mathematical formulation of quantum mechanics. [13] Spectral theory is one of the main branches of modern functional analysis and its applications. Roughly speaking, it is concerned with certain inverse operators, their general properties and their relations to the original operators. [12] By a linear preserver we mean a linear map of an algebra A into itself which, roughly speaking, preserves certain properties of some elements in A. Linear preserver problems concern the characterization of such maps. Automorphisms and antiautomorphisms certainly preserve various properties of the elements. Therefore, it is not surprising that these two types of maps often appear in the conclusions of the results. Over the last few decades there has been a considerable interest in the so called linear preserver problems. Over the past few years, there has been a considerable interest in linear mappings on algebras of operators that preserve certain properties of 1

9 operators. In particular, a problem how to characterize linear maps that preserve the spectrum of each operator has attracted the attention of many mathematicians. In 1985, Jafarian and Sourour proved that a surjective linear spectrumpreserving map of B(X) onto B(Y), where X and Y are Banach spaces and B(X) the algebra of all bounded linear operators on X, is either an algebra-isomorphism or an antiisomorphism. A similar result for finite-dimensional spaces was obtained much earlier by Marcus and Moyls in 1959.Recently in 1994, Aupetit and Mouton extended the result of Jafarian and Sourour to primitive Banach algebras with minimal ideals. In 1995, Semrl and Bresar characterize the surjective linear maps preserve the spectral radius. Recently in 2007, Mbekhta characterized linear maps from B(H) onto itself that preserve the set of Fredholm operators in both directions. Where B(H) is the algebra of all bounded linear operators on an infinite dimensional complex Hilbert space H. Further, Mbekhta and Šemrl characterized linear maps from B(H) onto itself preserving semi-fredholm operators in both directions, and improved the recently obtained characterization of linear preservers of generalized invertibility that they obtained jointly with Rodman in In 2007, Bendaoud, Bourhim and Sarih extend the above mentioned results to more general settings where a surjective linear map preserves the essential spectral radius. [20] In this work, we focus our study on a surjective linear map preserving the spectrum and the spectral radius of each operator in B(X). It is our goal to describe the general form of these linear maps, where it is introduced in [1] and [29]. Next, we characterize linear maps from B(H) onto itself that preserve the essential spectral radius, where it is introduced in [21]. This thesis is organized as follows. Chapter 1 consists of three sections. where we will introduce some concepts that are necessary for understanding this thesis and we will study some important definitions in functional analysis, this includes the Banach spaces, bounded linear operators, Hilbert spaces, Banach algebra and some related definitions and results. Chapter 2 consists of two sections. In the first one we will study the Spectrum of bounded linear operators and some of its properties. In the second section we will talk about a surjective linear map preserving spectrum. 2

10 Chapter 3 consists of two sections. In the first one we will study the Spectral radius of bounded linear operators and some of its properties. In the second section we will characterize a surjective linear map preserving spectral radius. Chapter 4 is devided into two sections, section 1 is talking about essential spectral radius and some related definitions, and section 2 is talking about a linear map from B(H) onto itself that preserve the essential spectral radius. Throughout our study, all vector spaces and algebras are assumed to be over C, the complex field. R and N denotes the set of; real numbers and natural numbers, respectively. B(X) will denote the algebra of all bounded linear operators on a complex Banach space X. We write X / for the dual of X and T* for the adjoint of T B(X). B(H) is the algebra of all bounded linear operators on an infinite dimensional complex Hilbert space H. Rλ(T), ρ(t), σ(t), r σ (T) and r e (T) denotes the; resolvent operator, resolvent, spectrum, spectral radius and essential spectral radius of T B(X), respectively. x f will denote the rank one operator, where x X and f X /. 3

11 Chapter 1 Preliminaries This chapter consists of three sections, where we give some basic definitions and theorems that are related to Banach spaces and Banach algebra. These concepts and results are necessary and important to understand this thesis. In section 1.1 we give the definitions of Banach space and linear bounded operator. Also we give some related results. In section 1.2 we define Hilbert space and some related concepts that are used in the next chapters. In section 1.3 we we give some facts about Banach algebra. 1.1 Banach Space Definition [12] A metric space is a pair ( X, d ), where X is a non-empty set and d is a metric on X; that is a function on XX such that for all x, y, z X, we have: (1) d( x, y ) 0. (2) d( x, y ) = 0 if and only if x = y. (3) d( x, y ) = d( y, x ). (4) d( x, y ) d( x, z ) + d( z, y ). Definition [12] A sequence (x n ) in a metric space ( X, d ) is said to be convergent if there is x X such that lim n d( x n, x ) = 0. In this case x is called the limit of x n or (x n ) converges to x and we write lim n x n = x or simply, x n x. If (x n ) is not convergent, it is said to be divergent. 4

12 Definition [12] A sequence (x n ) in a metric space ( X, d ) is said to be Cauchy if for every > 0 there is k = k() N such that d( x n, x m ) < for all n, m > k. Definition [12] The space (X, d) is said to be complete, if every Cauchy sequence in X converges in X. Definition [12] A normed space X is a vector space with a norm defined on it. A norm on a real or complex vector space X is a real-valued function on X whose value at an xx is defined by x which has the properties (1) x 0. (2) x = 0 if and only if x = 0. (3) x = x. (4) x + y x + y. For all x, yx and scalar. Definition [12] A complete normed space is said to be Banach space. Note. A norm on X defines a metric d on X which is given by d(x, y) = x y where x, yx and called the metric induced by the norm. Hence normed spaces and Banach spaces are metric spaces. Definition [12] Let X and Y be normed spaces both real or both complex. An operator T : X Y is called linear operator if for all x, y D(T) and all scalars T(x + y) = Tx +Ty The operator T is called a bounded linear operator if T is linear and there is a number c such that, Tx c x for all x X. The smallest c such that the above inequality holds for all nonzero xd(t) is defined as T, thus Tx T = sup{ x : x D(T), x 0 }. If D(T) = {0} then we define T = 0. Definition [12] Let ( X, d ) and (Y, ~ d ) be metric spaces. A mapping T : X Y is said to be continuous at x 0 X if for every > 0 there is > 0 such that ( Tx, Tx 0 ) < for all x satisfying d( x, x 0 ) <. T is said to be continuous if it is continuous at every point of X. ~ d 5

13 Definition [9] Let X and Y be normed spaces and let T be a linear operator from X to Y, then the kernel of T( also null space of T) is denoted by KerT and defined by KerT = { x X : Tx = 0}. The cokernel of T is the quotient space Y range(t) of the codomain of T by the image of T. The dimension of the cokernel is called the corank of T. Theorem If X and Y are normed spaces, then B(X,Y) the space of all bounded linear operators from X into Y, is a normed space with norm defined by T = sup{ T( x ) x : x X, x 0 } = sup{ Tx : x X, x = 1 } for each T B(X,Y). Furthermore, if Y is complete, then B( X, Y) is complete. Proof. See [12] Definition [12] A bounded linear functional f is a bounded linear operator with domain in a normed space X and range in the scalar field K of X. Definition [12] Let X / denote the Banach space of all bounded linear functionals on X which is called the dual space of X. Note. [3] If x / X / then the value of x / at the point x is denoted by x / x. It will sometimes be convenient to have another notation. we shall write < x, x / > for the value of x / at x. Definition [3] Let X and Y be normed spaces and T: X Y be a bounded linear operator. The operator T : Y / X / given by T y / =x / or its defined by the formula < x, T y / > = < Tx, y / > for each xx, x / X / and y / Y /. The operator T is called the transpose of T (or the adjoint of T). Theorem ( Hahn-Banach Theorem) If X is a normed space, {x 1, x 2,, x n } is a linearly independent subset of X, and a 1, a 2,.., a n are arbitrary scalars, then there is an f in X / such that f (x k ) = a k for all 1 k n. Proof. See [16] Theorem (Bounded linear functionals). Let X be a normed space and let x o 0 be any element of X. Then there exists a bounded linear functional f on X such that f =1 and f (x o ) = x o. 6

14 Proof. See [12] Definition [12] Let X and Y be normed spaces and T : D(T) Y be a linear operator with D(T) X. Then T is called a closed linear operator if its graph G(T) = { (x, y) : x D(T), y = Tx } is closed in the normed space XY, where the two algebraic operations of the vector space XY are defined as (x 1, y 1 ) + (x 2, y 2 ) = (x 1 + x 2, y 1 + y 2 ); (x, y) = (x, y) ( is scalar) and the norm on XY is defined by (x, y) = x + y. Theorem Closed Graph Theorem. Let X and Y be Banach spaces and let T : D(T) Y be a closed linear operator with D(T) X. If D(T) is closed in X, then T is bounded. Proof. See [12] Theorem Let T: D(T) Y be a linear operator with D(T) X and X and Y be normed spaces. Then, T is closed if and only if it has the following property: if x n x, where x n D(T) for all n and Tx n y, then x D(T) and Tx = y. Proof. See [12] Lemma Let T : D(T) Y be a bounded linear operator with D(T) X, where X and Y are normed spaces. Then, (a) If D(T) is a closed subset of X, then T is closed. (b) If T is closed and Y is complete, then D(T) is a closed subset of X. Proof. See [12] Lemma If the inverse T -1 of a closed linear operator exists, then T -1 is a closed linear operator. Proof. See [12] Theorem Open Mapping Theorem, Bounded Inverse Theorem. A bounded linear operator from a Banach space X onto a Banach space Y is an open mapping. Hence if T is bijective, then T -1 is continuous and thus is bounded. Proof. See [12] 7

15 Definition [12] Two vector spaces are said to be isomorphic if there is a bijective linear map between the two vector spaces. Two normed spaces are said to be isomorphic if there is a bijective linear map that preserves the norm between the two normed spaces. If the vector space X is isomorphic with a subspace of a vector space Y, then we say that X is embeddable in Y. Lemma Let X be a normed space. Then the canonical mapping C : XX // defined by C(x) = g x, where g x (f) = f(x) ( for all fx / ) is an isomorphism from X onto R(C), the range of C. Proof. See [12] Definition [12] A normed space X is said to be reflexive if R(C) = X //, where C as in Lemma Theorem If a normed space X is reflexive, then it is complete. Proof. See [12] Definition [12] A vector space X is said to be the direct sum of two subspaces Y and Z of X, written X = Y Z, if each xx has a unique representation x = y + z, y Y and z Z. Then Z is called the algebraic complement of Y in X and vice versa, and Y, Z is called a complementary pair of subspaces in X. Proposition If Y is a subspace of a vector space X, then there exists a subspace Z of X such that X = Y Z. Proof. See [12] Definition [12] (Invariant subspace) A subspace Y of a normed space X is said to be invariant under a linear operator T:X X if T(Y) Y. Definition [12] A metric space X is said to be compact if every sequence in X has a convergent subsequence. A subset M of X is said to be compact if M is compact considered as a subspace of X; that is every sequence in M has a convergent subsequence whose limit belongs to M. Definition [12] Let X and Y be normed spaces. An operator T: X Y is called a compact linear operator (or completely continuous linear operator) if T is linear and if for every bounded subset M of X, the image T(M) is relatively compact, that is, the closure T( M ) is compact. Equivalently, an operator T : X Y is said to be compact if for any bounded sequence {x n } in X, {T(x n )} has a convergent subsequence. 8

16 Lemma Let X and Y be normed spaces. Then every compact linear operator T: X Y is bounded, hence continuous. Proof. See [12] Notes. (1) The compact linear operators from a normed space X to a normed space Y form a vector space. Proof. See [12] (2) The set of compact operators from X to Y, denoted K(X,Y), is a closed subspace of B(X,Y). Proof. See [6] Theorem Let T : X X be a compact linear operator and S: X X a bounded linear operator on a normed space X. Then TS and ST are compact. Proof. See [12] Theorem Let T: X Y be a linear operator. If T is compact, so is its adjoint operator T : Y / X / ; here X and Y are normed spaces. Proof. See [12] Definition [32] A polynomial given by p(x) =a n x n +a n 1 x n a 1 x+a 0, is called monic if a n = 1. Definition [32] Let T B(X) and X is a normed space. Then the minimal polynomial of T is the unique monic polynomial p of smallest degree such that p(t) = Hilbert Space Definition [12] An inner product space is a vector space X with an inner product defined on X. An inner product on X is a mapping of X X into the scalar field k of X such that for all x, y and z in X and any scalar we have, (Ip1) < x + y, z > = < x, z > + < y, z > (Ip2) <x, y > = < x, y > 9

17 (Ip3) < x, y > = y, x, the complex conjugate of < y, x > (Ip4) < x, x > 0, and < x, x > = 0 if and only if x = 0. A complete inner product space is said to be a Hilbert space. Notes. (1) An inner product on X defines a norm on X given by x = x, x > and a metric on X given by d( x, y ) = x y = x y, x y >. (2) Inner product spaces are normed spaces and Hilbert spaces are Banach spaces. Example [12] Hilbert sequence space sequences of numbers ( i ) such that 2 i is finite. i 1 This space is a Hilbert spaces with inner product defined by This inner product induces the norm < x, y > =. 1 2 x = 2 ( i ), where x = ( i ) and y = ( i ). which is the set of all Theorem (Riesz-Frechet) Suppose that H is a Hilbert space over a filed (real or complex) F. Then for every continues linear functional f : H F, there exists a unique x 0 H such that Proof : see [12] f = x 0 and f(x) = <x, x 0 > for every x H. Note. By Riesz-Frechet Theorem, for every Hilbert space H, H is isomorphic with H /. Definition Let H 1 and H 2 be two Hilbert spaces and T: H 1 H 2 a bounded linear operator. The Hilbert adjoint operator T* of T is the operator T*:H 2 H 1 such that for all xh 1 and all yh 2, < Tx, y > = < x, T*y >. An operator T is called self adjoint if T* = T. Theorem The Hilbert adjoint operator T* of T exists, is unique, is linear and is bounded with norm T* = T. Proof. See [12] i 1 i 1 i i 2 10

18 Theorem Let H 1 and H 2 be Hilbert spaces, S, T: H 1 H 2 bounded linear operators and any scalar. Then we have, a) < T*y, x > = < y, Tx > for all xh 1 and yh 2. b) (S +T)* = S* + T* and (T )* = T*. c) ( T* )* = T, and in the case H 1 = H 2, (ST)* = T*S*. d) T*T = TT* = T 2. e) T*T = 0 if and only if T = 0. Proof. See [12] Definition [12] An element x in an inner product space X is said to be orthogonal to yx if < x, y > = 0 an we write x y. If A, B are subsets of X, then x A if x a for all aa, and A B if a b for all aa and all bb. Remark Let T B(H 1,H 2 ) where H 1 and H 2 are Hilbert spaces, then the orthogonal complement of range(t), R(T) = Ker(T ), where T* is the Hilbert adjoint of T. Proof. x kert* if and only if T*x = 0 if and only if < T*x,y > = 0 for all y H if and only if < x,t y > = 0 if and only if x (ranget). Definition [12] An orthogonal set M in an inner product space X is a subset M of X whose elements are pairwise orthogonal. An orthonormal subset M of X is an orthogonal set in X whose elements have norm 1. That is, for all x, ym, 0 if x y < x, y > = 1 if x = y A set of basis vectors B = {u 1, u 2, u 3,..,u n } for X is called an orthonormal basis if they are (i) Orthogonal, that is < u i, u j > = 0, for i j. (ii) Normalized, that is u j = 1, for j = 1, 2, 3,., n. Theorem ( Projection Theorem ). Let Y be any closed subspace of a Hilbert space H. Then H = Y Z, where Z = Y = { zh : z Y}. Proof. See [12] 11

19 1.3 Banach Algebra Definition [12] An algebra A over a field k is a vector space A over k such that for all x,ya, a unique product xya is defined with the properties: (1) (xy)z = x(yz) (2) (x+y)z = xz + yz (3) x(y+z) = xy + xz (4) (xy) = (x)y = x(y) for all x,y,z A and scalar k. Notes. (1) A is called an algebra with unity if there exists ea such that ex = xe = x for all xa. A is called unital if e = 1. (2) A is called a commutative algebra if ab = ba for all elements a and b in A. Definition [12] A normed algebra A is a normed space which is an algebra such that xy x y for all x,ya If A has a unit e, then e = 1. The Banach algebra A is a complete normed algebra which is complete considered as a normed space. Example Space B(X). The Banach space B(X) of all bounded linear operators on a complex Banach space X {0} is a Banach algebra with identity I, the multiplication being composition of operators. B(X) is not commutative, unless dim X = 1. Proof. See [12] Definition [30] Let A be an algebra and B A. Then B is said to be a subalgebra if B itself is an algebra with respect to the operations of A. Definition [30] If A is an algebra, a left ideal of A is a subalgebra I of A such that ax I whenever a A, x I. A right ideal of A is a subalgebra I of A such that xa I whenever a A, x I. A (two sided) ideal is a subalgebra of A that is both a left ideal and a right ideal. That is I A is called an ideal (two sided) if (i) I is a subspace i.e., if a, b I and α C, then αa + b I, (ii) I is an ideal in the algebra i.e., a I and x A implies that ax, xa I. An ideal I is said to be maximal if I {0}, I A and if J is any ideal of A such that I J, then either J = I or J = A. 12

20 Proposition If I is a closed ideal in a Banach algebra, then A I is a Banach algebra with the quotient norm a a I inf a b for any a A. Proof. See [30] bi Proposition The set of all compact operators on a Hilbert space H, denoted K(H), is a closed two-sided ideal in B(H). Proof. See [30] Theorem (N. Jacobson) Let A be an algebra with unit 1 and let x, y A, α C, with α 0. Then α xy is invertible in A if and only if α yx is invertible in A. Proof. See [30] Theorem Let A be a Banach algebra with unit 1. Then the following sets are identical: 1) The intersection of all maximal left ideals of A. 2) The intersection of all maximal right ideals of A. 3) {x A : 1 zx is invertible for all z A}. 4) {x A : 1 xz is invertible for all z A}. Proof. See [30] Definition [30] If A is a unital Banach algebra, then the set having properties (1) - (4) is called the radical (or more exactly, the Jacobson radical) rad(a) of A. If rad(a) = {0}, we say that A is semi-simple. Definition [20] If A and B are algebras over a field K, then we will call a map φ : A B a homomorphism if for all α in K and x,y in A. φ(αx) = αφ(x) φ(x + y) = φ(x) + φ(y) φ(xy) = φ(x) φ(y) If φ is bijective then φ is said to be an isomorphism between A and B. Definition [20] A bijective linear map φ between two algebras, A and B is called an anti-isomorphism if φ(xy) = φ(y) φ(x) for all x and y in A. 13

21 Definition [20] An automorphism is an isomorphism from an algebra to itself. An inner-automorphism of an algebra X is a function φ : X X defined for all x in X by φ(x) = a 1 xa, where a is a given fixed element of X. Definition [20] Let A be a Banach algebra. An element x A is called nilpotent if x m = 0 for some natural number m 1. Definition [6] Let A be a Banach algebra. An involution on A is a map : A A satisfying the following properties: (i) (a ) = a for all a A, (ii) (αa + βb) = a + b for all a, b A and all α, β C, (iii) (ab) = b a for all a, b A. The pair (A, ) is called an involutive Banach algebra. Notes. [6] (1) If A is an involutive Banach algebra and L A has the property that x L whenever x L, then we say that L is self-adjoint. (2) If A and B are involutive Banach algebras and ϕ : A B is a homomorphism satisfying ϕ(x ) = ϕ(x), for all x A, we say that ϕ is a - homomorphism. (3) If A is an involutive Banach algebra and for all x A we have x* = x, we say that the involution is isometric. Definition [6] By a C -algebra we shall mean an involutive Banach algebra A which satisfies the C -equation for every x A. x*x = x x* = x 2 Remarks.[6] (1) In a C*-algebra the involution is isometric. Proof. Using the C* equation, we have that for any x A, x*x = x x* = x 2, so x* = x. (2) If a C*-algebra A has a unit e, then e = e*. Proof. In any unital algebra, the unit is unique. Now, for all a A, ae* = (ea*)* = (a*)* = a and similarly a = e*a so e* is the identity, showing that e = e*. Examples [6] 14

22 (1) From Theorem an example of a C*-algebra is the algebra B(H) of bounded linear operators defined on a complex Hilbert space H; here x* denotes the adjoint operator of the operator x : H H. Furthermore, any subalgebra A of B(H) that is closed under adjoints (that is, T* A whenever T A) and is closed in the norm sense (hence complete) is an example of a C*-algebra. (2) Let H be a Hilbert space and T B(H) be a compact operator. Denote by K(H) the set of all compact operators in B(H). K(H) is a closed ideal of B(H) by proposition Moreover if T K(H), then T* K(H) by Theorem Therefore K(H) is a C -subalgebra of B(H) by (1) above. (3) The algebraic quotient of a C*-algebra by a closed proper two-sided ideal, with the natural norm, is a C*-algebra. Definition [6] A C -algebra A is simple if it has no nontrivial closed ideals. Definition [6] A C -algebra A is prime if, whenever J and K are ideals of A with J K = {0}, either J or K is {0}. (i.e {0} is a prime ideal in A ). 15

23 Chapter 2 Linear Maps Preserving Spectrum This chapter consists of two sections, in the first one we study spectrum and some of its properties. In the second section we study linear maps that preserve spectrum. 2.1 Spectrum and some of its properties Definition [12] Resolvent. Let X {0} be a complex normed space and T: D(T) X a linear operator with domain D(T) X. With T we associate the operator T λ =T λi where λ is a complex number and I is the identity operator on D(T). If T λ has an inverse, we denote it by R λ (T), that is R λ (T) = T λ -1 = ( T λi ) -1, and call it the resolvent operator of T or, simply, the resolvent of T. Definition [12] (Regular value, resolvent set, spectrum). Let X {0} be a complex normed space and T: D(T) X a linear operator with domain D(T) X. A regular value λ of T is a complex number such that (Rl) R λ (T) exists, (R2) R λ (T) is bounded, (R3) R λ (T) is defined on a set which is dense in X. The resolvent set ρ(t) of T is the set of all regular values λ of T. Its complement σ(t) = C ρ(t) in the complex plane C is called the spectrum of T, and a λ σ(t) is called a spectral value of T. Remark [12] The spectrum σ(t) is partitioned into three disjoint sets: 16

24 (i) The point spectrum or discrete spectrum σ p (T) is the set such that R λ (T) does not exist. A λ σ p (T) is called an eigenvalue of T. A scalar λ C is an eigenvalue of T if there exists a nonzero vector x in X such that Tx = λx. (ii) The continuous spectrum σ c (T) is the set such that R λ (T) exists and satisfies (R3) but not (R2), that is, R λ (T) is unbounded. (iii)the residual spectrum σ r (T) is the set such that R λ (T)exists (and may be bounded or not) but does not satisfy (R3), that is, the domain of R λ (T) is not dense in X. Notes. [12] (1) Each pair of ρ(t), σ p (T), σ c (T) and σ r (T) are disjoint and their union is the whole complex plane: C = ρ(t) σ(t) = ρ(t) σ p (T) σ c (T) σ r (T). (2) In the case of a finite dimensional normed space X we have σ c (T) = σ r (T) = and so σ p (T) = σ(t). (3) If X is infinite dimensional, then T can have spectral values which are not eigen values. This can be seen from the following example. Example. [12] The right-shift operator. Let T : 2 2 be defined by T((ξ1, ξ2, ξ3, )) = (0, ξ1, ξ2, ξ3,.. ) where x = (ξ 1, ξ 2, ξ 3,. ) is any element in 2. Then 0 σr (T). Proof. First note that T is bounded, because for any x Tx 2 2 j x 2 and so Tx x j 1. Hence, T = 1. 2, Let T 0 x = 0. Then ( 0, ξ 1, ξ 2, ξ 3,.. ) = (0, 0, 0, 0,.. ) and so ξ 1 = ξ 2 = ξ 3. = 0; that is x = 0. Therefore ker(t 0 ) = 0, so that R 0 (T) = T 0-1 = T -1 exists and R 0 (T) : T(X) X is given by T -1 ((0, ξ 1, ξ 2, )) = (ξ 1, ξ 2, ξ 3,.. ). Hence 0 σ p (T). Since (1, 0, 0, 0,.. ) 2 but (1, 0, 0, 0,.. ) T X, then R(T 0 ) = T(X) = { (η i ) 2 : η 1 = 0 } is not dense in 2. 17

25 Hence by definition of σ r (T), we have 0 σ r (T); that is 0 σ(t) but 0 σ p (T). Lemma [12] Let X be a complex Banach space, T: X X a linear operator, and λ ρ(t). Assume that (a) T is closed or (b) T is bounded. Then Rλ(T) is defined on the whole space X and is bounded. Proof : (a) Suppose that T is closed. Then D(Tλ) = D(T) with x n x and T λ x n y. Then Tx n = (Tx n λxn ) + λx n = T λ x n + λx n. Therefore, lim Tx n = lim T λ x n + lim λx n = y + λx. That is Tx n y + λx and x n x, hence by Theorem , Tx = y + λx, so that T λ x = (T λi)x = y and so x D(T λ ) = D(T). Therefore T λ is closed. By Lemma R λ is closed. But λ ρ(t), then R λ is bounded, so by Lemma (b) D(R λ ) is closed. But D R = X for any λ ρ(t), hence D(R λ ) = D R = X. (b) Since D(T) = X is closed and T is bounded, then by Lemma (a) T is closed and so the statement follows from part (a) of this proof. Theorem [12] Let T B(X), where X is a Banach space. If T < 1, then (I - T) -l exists as a bounded linear operator on the whole space X and (I - T) -l = j T = I + T + T j 0 [where the series on the right is convergent in the norm on B(X) ]. Proof : Since T < 1, then Then j 0 But T j T T j-1 T T T j T j. j T is convergent; that is j 0 T j is convergent. complete, then B(X) is complete by Theorem , therefore say j T = S. j 0 Note that (I - T) (I + T T n ) = (I + T+ + T n )(I- T) = I T n+l. 18 j 0 T j is absolutely convergent. Since X is We now let n, then T n+l 0 because T < 1. Hence we obtain j 0 T j is convergent

26 (I T)S = S(I T) = I. Therefore, (I T) -1 = S = j T. j 0 Theorem [12] (Spectrum closed). The resolvent set ρ(t) of a bounded linear operator T on a complex Banach space X is open; hence the spectrum σ(t) is closed. Proof. If ρ(t) =, then it is open. Let ρ (T). For a fixed λ o ρ(t) and any λ C we have T λi = T λ o I (λ λ o )I = (T λ o I) [I (λ λ o )(T λ o I) -l ]. Let V = I (λ λ o )(T λ o I) -l, then Tλ = Tλo V.. (1) Since λ o ρ(t) and T is bounded, then by Lemma (b) Rλo = Tλo -l B(X). Now we show that U(λ o ) = { λ C : λ - λ o < 1 R } is contained in ρ(t), and so ρ(t) will be open, because λ o was arbitrary and fixed in ρ(t). o Let λ U(λ o ). Then (λ λ o ) Rλo < 1 and so by Theorem 2.1.5, V = I (λ λ o ) Rλo has an inverse, V -l = [( ) ] j o R o = j 0 ( ) j j o R o and V -l B(X). j 0 But Tλo -l = Rλo B(X), then by (1) Rλ = Tλ -l = ( Tλo V) l = V -l Tλo -l = V -l Rλo exists. That means λ ρ(t). Therefore U(λ o ) ρ(t). Hence ρ(t) is open, so that its complement σ(t) = C ρ(t) is closed. Theorem [12] The spectrum σ(t) of a bounded linear operator T: X X on a complex Banach space X is compact and lies in the disk given by λ T Hence the resolvent set ρ(t) of T is not empty. [ Recall that a subset of the complex plane is compact if and only if it is closed and bounded ] 19

27 Proof. Let λ { λ C : λ > T }. Then Theorem Rλ = (T λi) -l 1 T T < 1, and so by = = 1 1 ( I T ) l 1 1 ( ) j T and Rλ B(X). j 0 Hence λ ρ(t). Therefore { λ C : λ > T } ρ(t) which implies that { λ C : λ T } σ(t), so that σ(t) is bounded and by Theorem σ(t) is closed. Hence σ(t) is compact. Definition [12] A metric space is said to be connected if it is not the union of two disjoint nonempty open subsets. A subset of a metric space is said to be connected if it is connected regarded as a subspace. Definition [12] A domain G in the complex plane C is an open connected subset G of C. Definition [12] A complex valued function h of a complex variable λ is said to be holomorphic (or analytic) on a domain G of the complex λ -plane if h is defined and differentiable on G, that is, the derivative h' of h, defined by h' (λ) = ( ) ( ) lim h h 0 exists for every λ G. The function h is said to be holomorphic at a point λ o C if h is holomorphic on some ε-neighborhood of λ o. Notes. [12] (1) h is holomorphic on G if and only if at every λ o G it has a power series representation h (λ) = j 0 c j ( ) j o with a nonzero radius of convergence. (2) Here, holomorphicity is defined over an open set, however, differentiability could only at one point. If h is holomorphic over the entire complex plane, we say that h is entire. 20

28 Definition [12] A vector valued function or operator function is a mapping S : Λ B(X) where Λ is any subset of the complex λ-plane and X is anormed space. Definition [12] Let Λ be an open subset of C and X a complex Banach space. Then S: Λ B(X) given by S(λ) = S λ is said to be locally holomorphic on Λ if x X and f X / the function h : Λ C defined by h(λ) = f(s λ x) is holomorphic at every λ o Λ. S is said to be holomorphic on Λ if S is locally holomorphic on Λ and Λ is a domain. S is said to be holomorphic at a point λ o C if S is holomorphic on some ε-neighborhood of λ o. Theorem For any Banach space X, T B(X) and every λ o ρ(t), the resolvent R λ (T) is represented by j 1 R λ (T) = j 0 o j R o The series being absolutely convergent for every λ in the open disk given by λ - λ o < Proof. See [12] 1 R o in the complex plane. This disk is a subset of ρ(t). Theorem [12] The resolvent R λ (T) of a bounded linear operator T: X X on a complex Banach space X is holomorphic at every λ o ρ(t). Hence it is locally holomorphic on ρ(t). Proof. By Theorem , if λ o ρ(t), then j o o.... (1) j 0 j 1 R λ (T) = R ( T ) Converges absolutely for all λ satisfying λ - λ o < 1 R. o Now, x X and f X /, define h : ρ(t) C by h(λ) = f(r λ (T)(x)). Then by (1) ( ( ) j o o x) ( since f is continuous ) j 1 h(λ) = f(r λ (T)(x)) = f R T j 0 = j 0 o j f 1 ( R ( T j o ) x) 21

29 = j 0 j c o j. Where c j = f( R λ0 (T) j+1 x ) with radius of convergence 1 R o 0. Hence R λ is holomorphic at λ 0 ρ(t) and so it is holomorphic at every λ ρ(t). Hence R λ is locally holomorphic on ρ(t). Theorem [12] If X {0} is a complex Banach space and T B(X), then σ(t). Proof : If T = 0, then Tλx = 0 for x 0 only when λ = 0, that is σ(t) = {0}. Let T 0, then T 0. Now if λ 2 T, then T and hence by Theorem T Rλ = But 1 1 ( T) j is convergent. Moreover, Rλ j 0 T T T ( T )( I... ) 2 n 1 1 T..... (1) j 0 2 n 2 n1 T T T T = T... T... n1 n j = T n 1 n = n 1 T (1 ). Then as n, we have 1 ( T ) T because j 0 j T 1. 2 Hence T..... (2) T j 0 j By (1) and (2), R 1 1 T T..... (3) because 2 T. 22

30 Hence the series is absolutely convergent. Now, suppose that σ(t) =. Then ρ(t) = C. Hence Rλ is holomorphic for all λ C by Theorem Consequently, for a fixed x X and a fixed f X /, the function h : C C defined by h(λ) = f(r λ x) is holomorphic on C, that is, h is an entire function. Since holomorphy implies continuity, h is continuous and thus bounded on the compact disk {λ C : λ 2 T }. But by (3), for λ 2 T, we have, h( ) = f ( Rx) f Rx f R x f x T Hence h is bounded on { λ C : λ 2 T }. Therefore h is bounded on C which is entire, then by Liouville's theorem, which states that an entire function which is bounded on the whole complex plane is a constant. Since x X and f X /, in h were arbitrary, h = constant implies that Rλ is independent of λ, and so is R λ -1 = T λi is independent of λ which is impossible. Therefore σ(t). Theorem [12] Spectral Mapping Theorem for Polynomials. Let X be a complex Banach space, T B(X) and p(λ) = a n λ n + a n-1 λ n-l a 0 (a n 0), then σ(p(t)) = p(σ(t)) That is, the spectrum σ(p(t)) of the operator p(t) = a n T n + a n-1 T n-l a 0 I consists precisely of all those values which the polynomial p assumes on the spectrum σ(t) of T. Proof : By Theorem σ(t). If n = 0, then p(λ) = α 0, so that p(σ(t)) = { p(λ) C : λ σ(t) } = {α 0 } and σ(p(t)) = σ(α 0 I) = { λ C : (α 0 I λi)x = 0, x 0 } = { λ C : (α 0 λ)ix = 0, x 0 } 23

31 = { λ C : λ = α 0 } = {α 0 } Therefore, p(σ(t)) = σ(p(t)) = {α 0 } in the case n = 0. If n > 0, we show that (a) σ(p(t)) p(σ(t)) and (b) p(σ(t)) σ(p(t)). (a) Let μ C be fixed and arbitrary, S = p(t) and let Sμ = P(T) μi. Suppose that S μ -1 exists. Then S μ -1 is the resolvent operator of p(t). Since X is complex, then the polynomial given by S μ (λ) = p(λ) μ must be factored completely into linear terms say, S μ (λ) = p(λ) μ = α n ( λ γ 1 ) ( λ γ 2 ) ( λ γ n ) (1) where γ 1, γ 2,., γ n are the zeros of S μ.. Corresponding to (1) we have S μ = p(t) μi = α n ( T γ 1 I)( T γ 2 I) ( T γ n I). If each γ i ρ(t), then each T γ i I has a bounded inverse which, by Lemma is defined on all of X, because T B(X) and X is complex Banach space. -1 So that S μ is bounded and is defined on all of X and -1 1 S μ = ( T γ ni) -1 ( T γ 1 I) -1 n Hence, μ ρ(p(t)). Therefore, we conclude that if μ σ(p(t)), then μ ρ(p(t)) and so γ i ρ(t) for some i ; that is γ i σ(t) for some i. Now by (1) s μ (γ i ) = p(γ i ) μ = 0, hence μ = p(γ i ) p(σ(t)). Therefore, σ(p(t)) p(σ(t)). (b) We show that p(σ(t)) σ(p(t)). Let κ p(σ(t)). Then κ = p(β) for some β σ(t). There are two cases: (A) T βi has no inverse or (B) T βi has an inverse. (A) If T βi has no inverse, since κ = p(β), then p(β) κ = 0 and so β is a zero of the polynomial given by s κ (λ) = p(λ) κ. It follows that we can write 24

32 s κ (λ) = p(λ) κ = (λ β )g(λ), where g(λ) denotes the product of the other n 1 linear factors and α n. Corresponding to this representation we have S κ = p(t) κi = (T βi)g(t) (2) Since the factors of g(t) all commute with (T βi), we also have S κ = g(t)(t βi). Now we show that S κ -1 does not exist ( by contradiction ). Suppose that S κ -1 exists, then I = (T βi)g(t)s κ -1 = S κ -1 g(t)(t βi). Hence (T βi) has an inverse, which is a contradiction. Therefore S κ -1 does not exist, so κ σ(p(t)). In this case we have p(σ(t)) σ(p(t)). (B) If (T βi) -1 exists, then for the range of T βi we must have R(T βi) X (3) because if R(T βi) = X, then by Theorem applied to T βi, (T βi) -1 is bounded and so β ρ(t) which is a contradiction. By (2) and (3) we obtain R(S κ ) X. Then κ σ(p(t)), because if κ σ(p(t)), then κ ρ(p(t)) and so Lemma 2.1.4(b) applied to p(t) implies that R(S κ ) = X. Therefore, in this case also p(σ(t)) σ(p(t)). From all of the above we have, p(σ(t)) = σ(p(t)). Remark: (1) In particular, σ(t n ) = σ(t) n for every n 0, that is, ν σ(t ) n = { λ n C : λ σ(t ) } if and only if ν σ(t n ), and σ(αt) = ασ(t ) for every α C, 25

33 that is, ν ασ(t) = {αλ C : λ σ(t )} if and only if ν σ(αt). (2) If T is a nilpotent operator (i.e., if T n = 0 for some n 1), then σ(t) = σ p (T) = {0}, and so rσ(t ) = 0. Proof. Suppose T is nilpotent, with T n = 0 for some n 1. If λ σ(t), then λ n σ(t n ) = σ(0) = {0} by (1) above, so that λ = 0. Definition [31] Let X and Y be two normed spaces then T B(X, Y ) is said to be an operator of finite rank if dim R(T) is finite. Definition [18] We say that an operator T: X X is a rank one operator if there exist x X and f X / so that Ty = <y, f>x. We use the notation T= x f. Note [1] The duality between a Banach space and its dual is denoted by <x,f> = f(x), for any x X and any f X /. Lemma If K : H B is a finite rank operator where H and B are Hilbert spaces, then K is compact. Proof. See [31] Theorem Let X be a normed space, and let T B(X) be compact and let λ be a nonzero complex number. Then, exactly one of the followings hold: )1) T λi is invertible (2) λ is an eigenvalue of T. This result, known has Fredholm Alternative, Proof. See [12]. Lemma [1] Let X be Banach space and A B(X). Then σ(t+ A) σ(t) for every T B(X) if and only if A = 0. Proof. Let A= 0, then σ(t+ A) = σ(t+0) = σ(t). Conversely assume that A 0, and let x be a vector in X such that Ax= y 0. Since y 0, then there exists an f X / such that <x, f> = 1 and <y, f> 0. Let T= (x y) f, then (T+A)x = Tx + Ax = (x y) f x + Ax = <x, f>(x y) + y 26

34 = (x y) + y (since <x, f> = 1 ) = x so 1 σ(t+a). But σ(t) = {0, <x y, f> } and <x y, f> = <x, f> <y, f> = 1 <y, f> l, so 1 σ(t). Therefore, σ(t+ A) σ(t). 2.2 Linear Maps Preserving Spectrum Definition [1] Let X and Y be Banach spaces, we say that ϕ :B(X) B(Y) is a spectrum-preserving linear map from B(X) to B(Y) if σ(ϕ(t)) = σ(t) for every T B(X). Lemma [1] Let X and Y be Banach spaces. If ϕ is a spectrumpreserving linear map from B(X) to B(Y), then ϕ is injective. Proof. Let ϕ be a linear map from B(X) to B(Y) which preserves the spectrum. Let A B(X) such that ϕ(a) = 0, then for every T B(X), σ(t+ A) = σ(ϕ(t+a)) = σ(ϕ(t)+ ϕ(a)) = σ(ϕ(t)+ 0) = σ(ϕ(t)) = σ(t) By Lemma , A = 0. Therefore ϕ is injective. 27

35 In what follows, we will assume that ϕ is a surjective linear map preserving the spectrum from B(X) to B(Y). Lemma [1] Let X, Y and ϕ be as in Lemma If ϕ is surjective, then ϕ(ι) = Ι. Proof. Since I B(Y) and ϕ is surjective, there exists an S B(X) such that ϕ(s) = Ι. For any T B(X), We have σ(t+(s Ι)) = σ(ϕ(t+s Ι)) (by Definition ) = σ(ϕ(t Ι)+ ϕ(s)) = σ(ϕ(t Ι)+ I ) = σ(ϕ(t Ι)) + 1 (by Theorem where a 0 =1 and ϕ(t Ι) is the operator ) = σ(t Ι) + 1 (by Definition ) = 1+ σ(t) 1 ( by Theorem ) = σ(t) for every T B(X). Then S Ι = 0 by Lemma Therefore S = I and so ϕ(ι)= Ι. Lemma [1] Let X be Banach space. Then for T B(X), x X, f X / and λ σ(t), we have λ σ(t+x f) if and only if <(λ T) -1 x, f> = l. Proof. First of all since λ σ(t), then (λ T) -1 B(X). Let <(λ T) -1 x, f)>= l.... (1), then (T+ x f)( λ T) -1 x = T(λ T) -1 x + x f (( λ T) -1 x) = T(λ T) -1 x + <(λ T) -1 x, f)> x = T(λ T) -1 x + x ( by (1) ) = (λ T) -1 (Tx +( λ T)x ) = λ (λ T) -1 x, 28

36 Therefore, λ is an eigenvalue of T + x f, thus λ σ(t+x f). Conversely, Let λ σ( T + x f ), then by the invariant of the fredholm alternative λ is an eigenvalue of T + x f and so there exists a nonzero vector u X such that ( T+ x f )u = λ u. Then, λ u = Tu +( x f)u = Tu + <u, f>x. Then <u, f>x = λ u Tu = (λ T)u Therefore u = <u, f> ( λ T) -1 x = < (λ T) -1 x, f> u It follows that < (λ T) -1 x, f> = l. Definition [10] An operator is Fredholm if its range is closed and its kernel and cokernel are finite-dimensional. The essential spectrum of T, usually denoted σ e (T), is the set of all complex numbers λ such that T λi is not a Fredholm operator. Remark. The essential spectrum is a subset of the spectrum σ, and its complement is the discrete spectrum, so σ e (T) = σ(t) σ p (T). The following theorem, which may be of independent interest, gives a spectral characterization of rank one operators. Theorem [1] Let Let X be Banach space and A B(X), A 0. The following conditions are equivalent. (i) A has rank 1. (ii) σ(t+a) σ(t+ca) σ(t) for every T B(X) and for every scalar c 1. Proof. (i) (ii) Suppose A has rank 1, then by Definition , A = x f for any x X and any f X /. Let T B(X) and c 1, suppose that λ σ(t), we show that λ σ(t+a) σ(t+ca) for all c 1. Since λ σ(t), then (λ T) -1 B(X). Now, if λ σ(t+ca), then by Lemma 2.2.4, c <(λ T) -1 x, f> = 1. But c 1, then < (λ T) -1 x, f> 1 and so by Lemma 2.2.4, λ σ(t+a). 29

37 Therefore, λ σ(t+a) σ(t+ca). Hence, σ(t+a) σ(t+ca) σ(t). (ii) (i) Suppose for contraposition that rank A 2. We will show that condition (ii) is not satisfied. Case I : A is a scalar αι, α 0. Let T be an operator with σ(t) ={0, α}. But σ(a) ={α} so σ(t+a) ={ α, 2α} and σ(t+2a) ={ 2α, 3α}. Thus, σ(t+a) σ(t+2a) = {2a} σ(t). Case II : A is not a scalar. We will construct a nilpotent operator N with N 3 = 0 and a scalar c 1 such that σ(n + A ) σ( N + ca) contains a nonzero scalar and so we conclude σ(n + A ) σ( N + ca) σ(n) = {0} Subcase II.1: Suppose that there exists a vector u in X such that u, Au, A 2 u are linearly independent. Let U be the linear span of {u, Au, A 2 u} and let V be a (closed) complement of U in X (by Proposition ). Define alinear perator N on X by Nu = u Au, NAu = Au 2A 2 u, NA 2 u = -u/2 + 3Au/2 2A 2 u, Nν = 0 for ν V. We have, Nu = u Au u Au u c u ( since A B(X), so a scalar c such that Au c u ) = k u where k = 1 c NAu = Au 2A 2 u Au 2 Au 2 = Au (1 2 Au ) Au (1 2c u ) 30

38 = k Au where k =1 2c u Similarly, NA 2 u k A 2 u and Nv = 0 = 0. Therefore N B(X) We have N 3 v = 0, v V and N 3 u = N 2 (u Au ) = N ( Nu NAu ) = N (u Au Au + 2A 2 u) = Nu 2NAu + 2 NA 2 u = u Au 2 (Au 2A 2 u) +2(-u/2 + 3Au/2 2A 2 u) = u 3Au + 4A 2 u u + 3Au A 2 u = 0 Similarly, N 3 Au = 0 and N 3 A 2 u = 0. Therefore, N 3 x = N 3 (α 1 u + α 2 Au + α 3 A 2 u +βv ) for some scalars α 1,α 2,α 3,β = α 1 N 3 u + α 2 N 3 Au + α 3 N 3 A 2 u +β N 3 v = 0, x X. Now, (N + A) u = Nu + Au = u Au + Au = u and (N + 2A) Au = NAu +2AAu = Au 2A 2 u +2A 2 u = Au. But u 0 and Au 0, therefore, 1 σ( N + A) and 1 σ(n + 2A), that is; 1 σ( N + A) σ(n + 2A) σ(n) = {0} Subcase II.2: For every x X, the vectors x, Ax, A 2 x are linearly dependent. First we show that A satisfies a quadratic polynomial equation p(a) = 0. Since A is not a scalar, there exists a vector u 1 in X such that u 1 and Au 1, are linearly independent. Therefore the minimal polynomial p of u 1 is quadratic. Now let x X and consider the restriction of A to the invariant subspace U 1 = span (u l, Au 1, x, Ax}. Let q be the minimal polynomial of A U 1. By a standard result in linear algebra, there exists a vector u U 1 such that q is also the minimal polynomial of u and so by our assumption, deg q 2. On the other hand, p divides q, so q = p and p(a U 1 ) = 0; in particular, p(a) x = 0. Since x is arbitrary, we have p(a) = 0. We now consider four subcases according as p(t) = (t α)(t β) or (t-α) 2 or t(t α)or t 2, where α 0 β α. 31

39 By the standard decomposition of algebraic operators and since rank A 2 and A is not a scalar, we see that A has a finite-dimensional invariant subspace W such that A W has a matrix representation respectively ,, 0 0 or We consider a complement Z of W in X and an operator N such that N Z= 0 and N W has matrix representation respectively ,, or Let c be αβ -1 or 4 or 2 or 2 or 2, respectively. Then N 2 = 0 and σ(n+ A) σ(n+ ca) includes a nonzero scalar, namely α or 2α or 2α or 2, respectively. Corollary [1] Let X be a Banach space and let A B(X). The following conditions are equivalent. (i) A has rank 1 or is a scalar. (ii) σ(n + A ) σ( N + ca) {0} for every nilpotent operator N satisfying N 3 = 0 and every scalar c 1. Proof. By Theorem and its proof. In infinite-dimensional spaces, this result can be refined as follows: Corollary [1]. If X is infinite-dimensional, then rank A 1 if and only if σ(n + A ) σ( N + ca) = {0} for every nilpotent operator N with N 3 = 0 and every scalar c 1. Furthermore, A is a nonzero scalar if and only if the above intersection is empty. Proof. This follows from the fact that for a nilpotent N and a compact K, the essential spectrum of N+ K is {0} and so the spectrum includes 0. 32

40 We now return to our study of the mapping ϕ. Lemma [1]. Let X be a Banach space and ϕ be a spectrum preserving linear map from B(X) to B(Y). If R B(X) has rank one, then ϕ(r) has rank 1. Proof. Since ϕ is injective by Lemma 2.2.2, then ϕ -1 exists and so if T B(X), then S B(X) such that T = ϕ -1 (S). But ϕ preserves the spectrum, so σ(t) = σ(ϕ(t)) which implies σ(ϕ -1 (S)) = σ(ϕ(ϕ -1 (S))) = σ(s); that is ϕ -1 preserves the spectrum. Now, if R B(X) has rank one, then by Theorem σ(t+r) σ(t+cr) σ(t) for every T B(X) and every c 1. Hence, σ(t+ ϕ(r)) σ(t+ cϕ(r)) = σ(ϕ -1 (T+ ϕ(r))) σ(ϕ -1 (T+ cϕ(r))) = σ(ϕ -1 (T)+ ϕ -1 (ϕ(r))) σ(ϕ -1 (T)+ cϕ -1 (ϕ(r))) = σ(s + R) σ(s + cr) σ(s) (by Theorem ) = σ(ϕ(t)) Therefore by Theorem ϕ(r) has rank 1. = σ(t) (since ϕ preserves the spectrum) Theorem [1] Let X and Y be Banach spaces. If ϕ: B(X) B(Y) is a spectrum-preserving surjective linear mapping, then either (i) there is a bounded invertible operator A: X Y such that ϕ(t) = ATA -1 for every T B(X) or (ii) there is a bounded invertible operator B: X / Y such that ϕ(t) = BT B -1 for every T B(X). Proof. For every nonzero x X and every nonzero f X / consider the sets L x ={x h : h X / } and R f ={u f :u X). Each of L x and R f is a linear subspace of B(X) consisting of rank one operators and is maximal among such spaces. 33

41 It follows that for every x X, ϕ(l x ) ={ ϕ(x h) : h X / } some y Y or an R g for some g Y /. is either an L y for Furthermore, we cannot have ϕ(l u ) = L y and ϕ(l ν )= R g simultaneously for some u and ν in X since L y R g = y g is a one-dimensional space while L u L ν has dimension 0 if u ν or dim X / if u = ν. So we have two cases: Case I. For every x X, there exists a y Y such that ϕ(l x ) = L y, so ϕ(x f) = y g. The mapping f g is linear, so g = C x f for a linear transformation C x : X / Y /. Claim : the space {C x : x X} has dimension 1. If this is not the case, then there exist x 1, x 2 X, y 1, y 2 Y and two linearly independent transformations C 1, C 2 such that ϕ(x 1 f) = y 1 C 1 f and ϕ(x 2 f) = y 2 C 2 f for every f. It follows that y 1 C 1 f + y 2 C 2 f = ϕ(x 1 f) + ϕ(x 2 f) and so has rank 1 for every f. = ϕ(x 1 f +x 2 f) (since ϕ is linear) = ϕ((x 1 +x 2 ) f) Since C 1 and C 2 are linearly independent we must have that y 1 and y 2 are linearly dependent. and so there exists a scalar k such that y 2 = ky 1. Now, L y1 ={ y 1 h : h Y / } = {ky 1 h : h Y / } = {y 2 h : h Y / } = L y2, implying that ϕ -1 (L y1 )=L x1 = ϕ -1 (L y2 ) = L x2 dependent. and so x 1 and x 2 are linearly However, in this case, we get that C 1 and C 2 are linearly dependent which is acontradiction. This establishes the fact that dim{c 1 : x X} = 1 and so by absorbing a constant in the first term of the tensor product, we have one linear transformation C: X / Y / such that ϕ(x f) = y Cf. 34