Let $\{e_n\}$ be the orthonormal basis for $l^2$ and $\{\alpha_n\} \in \ell^{\infty}$.
Define $Ae_n=\alpha_n e_n$.
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Find $\sigma(A)$ and show that if $K \subset \mathbb{C}$, then there exists some $T \in B(\ell^2)$ such that $\sigma(T)=K$.
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Find the eigenvalues of $A$,
Attempt:
- Let $E$ be the closure of $\{ \alpha_n\}$.
From Spectrum of Diagonal Operator in $\ell^2$, I am guessing that $E= \sigma(A)$.
But I'm not sure if this is the case. I don't see why we have $(\alpha_n) \subset E$.
I know that for any $T \in B(\ell^2)$, $\sigma(T)$ is compact. But I'm not sure how to construct a $T \in B(\ell^2)$ such that $\sigma(T)=K$.
- I know that the matrix of $A$ relative to the orthonormal basis $\{e_n \}$ is the diagonal matrix $diag(\alpha_n)$. So I think that the eigenvalues will be $\{\alpha_n \}$. But I'm not sure about this.
Any help is appreciated!
Thank you in advance.
Best Answer
Hints:
Using only the definition, show that each $\alpha_n$ is an eigenvalue of $A$. Since $\sigma(A)$ is closed, $\overline{\{\alpha_n\}}\subseteq \sigma(A).$ Now show by direct calculation, that if $\lambda\notin \overline{\{\alpha_n\}}$, then $A-\lambda$ is invertible. Finally, if $K$ is given, choose a countable dense subset of $K$, which will be in $\ell^\infty$ since $K$ is compact (being a spectrum).