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Let $\{e_n\}$ be the orthonormal basis for $l^2$ and $\{\alpha_n\} \in \ell^{\infty}$.

Define $Ae_n=\alpha_n e_n$.

  1. Find $\sigma(A)$ and show that if $K \subset \mathbb{C}$, then there exists some $T \in B(\ell^2)$ such that $\sigma(T)=K$.

  2. Find the eigenvalues of $A$,

Attempt:

  1. 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 $\sigma(T)=K$.

  1. 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.

Catcher
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  • $Ae_n = \alpha_ne_n$ means by definition that $\alpha_n$ is an eigenvalue of $A$. So, each $\alpha_n$ is in the spectrum of $A$. Since the spectrum is always closed, you have that $E\subset\sigma(A)$ and it remains to show the converse. – amsmath Feb 28 '21 at 03:30
  • I know that ${ \alpha_n} $ are Eigenvalues. But I'm not sure how to show that they're the only eigenvalues. Any help is appreciated! – Catcher Feb 28 '21 at 19:40

1 Answers1

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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).

Matematleta
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  • If $K$ is given to be compact in $\mathbb{C}$, why does that imply that $K$ is separable? Thank you! – Catcher Feb 28 '21 at 22:36
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    $K$ inherits the metric on $\mathbb C$ and every compact metric space is separable, right? – Matematleta Feb 28 '21 at 23:00
  • Thank you for the hint! I've figured out this part. Now I just have to show that $ { \alpha_n }$ is the set of eigenvalues of $A$. I know that ${\alpha_n}$ is contained in the set of eigenvalues of $A$. But not sure how to show the other way, namely, the set of eigenvalues of $A$ is contained in ${\alpha_n}$. – Catcher Feb 28 '21 at 23:35
  • Hi! do you know what happens if the diagonal operator is instead a map between $\ell^{\infty}$? It seems to me that the spectrum and the point spectrums are the same, but the continuous and residual spectrums are different in comparison to those of $\ell^p$, $p\in [1,\infty)$. –  Dec 28 '21 at 19:28