Linear Maps Preserving Operators of Local Spectral Radius Zero |
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Authors: | Constantin Costara |
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Institution: | 1. Faculty of Mathematics and Informatics, Ovidius University of Constan?a, Mamaia Boul. 124, 900527, Constan?a, Romania
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Abstract: | Let X be a complex Banach space and let B(X){\mathcal{B}(X)} be the space of all bounded linear operators on X. For x ? X{x \in X} and T ? B(X){T \in \mathcal{B}(X)}, let rT(x) = limsupn ? ¥ || Tnx|| 1/n{r_{T}(x) =\limsup_{n \rightarrow \infty} \| T^{n}x\| ^{1/n}} denote the local spectral radius of T at x. We prove that if j: B(X) ? B(X){\varphi : \mathcal{B}(X) \rightarrow \mathcal{B}(X)} is linear and surjective such that for every x ? X{x \in X} we have r
T
(x) = 0 if and only if rj(T)(x) = 0{r_{\varphi(T)}(x) = 0}, there exists then a nonzero complex number c such that j(T) = cT{\varphi(T) = cT} for all T ? B(X){T \in \mathcal{B}(X) }. We also prove that if Y is a complex Banach space and j:B(X) ? B(Y){\varphi :\mathcal{B}(X) \rightarrow \mathcal{B}(Y)} is linear and invertible for which there exists B ? B(Y, X){B \in \mathcal{B}(Y, X)} such that for y ? Y{y \in Y} we have r
T
(By) = 0 if and only if rj( T) (y)=0{ r_{\varphi ( T) }(y)=0}, then B is invertible and there exists a nonzero complex number c such that j(T) = cB-1TB{\varphi(T) =cB^{-1}TB} for all T ? B(X){T \in \mathcal{B}(X)}. |
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