# 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

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 $T-\lambda\cdot\mathrm{Id}$.