• Bulletin of the Korean Mathematical Society
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• v.48 no.6
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• pp.1129-1135
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• 2011

In this note, we prove that every subscalar operator with finite spectrum is algebraic. In particular, a quasi-nilpotent subscala operator is nilpotent. We also prove that every subscalar operator with property (${\delta}$) on a Banach space of dimension greater than 1 has a nontrivial invariant closed linear subspace.

• Journal of the Chungcheong Mathematical Society
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• v.24 no.1
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• pp.19-34
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• 2011

A bounded linear operator T on a complex Banach space X is said to have property (I) provided that T has Bishop's property (${\beta}$) and there exists an integer p > 0 such that for a closed subset F of ${\mathbb{C}}$ ${X_T}(F)={E_T}(F)=\bigcap_{{\lambda}{\in}{\mathbb{C}}{\backslash}F}(T-{\lambda})^PX$ for all closed sets $F{\subseteq}{\mathbb{C}}$, where $X_T$(F) denote the analytic spectral subspace and $E_T$(F) denote the algebraic spectral subspace of T. Easy examples are provided by normal operators and hyponormal operators in Hilbert spaces, and more generally, generalized scalar operators and subscalar operators in Banach spaces. In this paper, we prove that if T has property (I), then the quasi-nilpotent part $H_0$(T) of T is given by $$KerT^P=\{x{\in}X:r_T(x)=0\}={\bigcap_{{\lambda}{\neq}0}(T-{\lambda})^PX$$ for all sufficiently large integers p, where ${r_T(x)}=lim\;sup_{n{\rightarrow}{\infty}}{\parallel}T^nx{\parallel}^{\frac{1}{n}}$. We also prove that if T has property (I) and the spectrum ${\sigma}$(T) is finite, then T is algebraic. Finally, we prove that if $T{\in}L$(X) has property (I) and has decomposition property (${\delta}$) then T has a non-trivial invariant closed linear subspace.