# Operator whose spectrum is given compact set

Let $A\subset \mathbb{C}$ be a compact subset.

Since $A$ is compact and metric space, it is separable, say $\overline{\lbrace a_n\rbrace_{n=1}^\infty}=A$.

Let $\mathcal{l}^2(\mathbb{Z})$ be the Hilbert space consisting of $L^2$-summable sequences and $\lbrace e_n\rbrace_{n=1}^\infty$ be the canonical basis of $\mathcal{l}^2(\mathbb{Z})$.

Define an operator $T\colon\mathcal{l}^2(\mathbb{Z})\to\mathcal{l}^2(\mathbb{Z})$ by sending $e_n$ to $a_ne_n$. I want to prove that $A=\sigma(T)$, where $\sigma(T)$ is the spectrum of $T$.

What I can prove is that $A=\overline{\lbrace a_n\rbrace_{n=1}^\infty}\subset \sigma(T)$ because each $a_n$ is an eigenvalue of $T$ and $\sigma(T)$ is closed. How can I prove the other inclusion, namely $\sigma(T)\subset A$?

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I'm interested in where is the question from and its background. –  Jack Apr 26 '11 at 22:13

## 1 Answer

Hint: If $\lambda$ in not is $A$, then there is $\epsilon>0$ with $\lvert\lambda-a_n\rvert>\epsilon$ for all $n$. Use this to write down a bounded inverse for $T-\lambda\cdot\mathrm{Id}$.

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I think that now I can solve it. Thank you. –  user8484 Mar 23 '11 at 16:34
@user8484: Please don't forget to accept the answer. –  anonymous Mar 23 '11 at 16:35
@user8484: You are welcome. There is no hurry for accepting answers. In fact, there are drawbacks in accepting answers too quickly. This might be more important for less localized questions that this one. –  Rasmus Mar 23 '11 at 16:44