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Good evening, I have a question concerning non-invertible operators. Let $H$ be a Hilbert space and $T$ a non-invertible bounded operator on $H.$ Is it true that $T$ is the limit of some sequence of invertible bounded operators on $H$? We can see that is true when $H$ is finitely dimensional. So the interesting case is infinitely dimensional. Does anyone know any information about this? Thanks in advance, Duc Anh |
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This is true if $0$ belongs to the boundary of the spectrum of $T$ (more or less from the definition of the spectrum) but fails in general. The unilateral left shift $S: \ell^2({\mathbb N})\to\ell^2({\mathbb N})$ satisfies $S^\ast S=I\neq SS^\ast$. If $(T_n)$ were a sequence of invertible operators with $T_n\to S$ in norm, then $S^*T_n\to I$ in norm, which means that for all sufficiently large $n$ the operator $S^*T_n$ is invertible, which means $S^*$ is invertible, a contradiction. |
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