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Let $T\in \text{Aut}(\ell^2(\mathbb{C}))$ and $T(x)=(a_1 x_1, a_2 x_2,\ldots)$ where $a=(a_i)_i \in \ell^\infty(\mathbb{C})$. How can I easily see what is $\sigma(T)$ and $\sigma_p(T)$ (that are the eigenvalues of $T$)?


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To begin, not that the $a_i$ are eigenvalues of $T$. – Davide Giraudo Nov 14 '12 at 16:19

$T$ defined in this way is a diagonal operator, i.e., you just put many scalars together, so the analysis is relatively easy.

Obviously the $a_j$ are eigenvalues. Also, if $\lambda\neq a_j$ for all $j$ and $(T-\lambda)x=((a_j-\lambda)x_j)=0$ then it is easy to see $x_j=0$ for all $j$ so $x=0$.

Therefore $\sigma_p(T)=\{a_j\}$.

Also for diagonal operators, $\sigma(T)=\overline{(\sigma_p(T))}$. It is obvious that $\sigma(T)\supset \overline{(\sigma_p(T))}$ because the spectrum is closed and it contains all eigenvalues. On the other hand, if $\lambda$ is not in the closure of point spectrum, then we can separate $\lambda$ from the point spectrum by a positive distance. Using this we can show \begin{equation} S(x)=((a_j-\lambda)^{-1}x_j) \end{equation} is a bounded linear operator and it is the inverse of $T-\lambda$. So $\lambda$ is not in $\sigma(T)$.

So in conclusion $\sigma_p(T)=\{a_j\}$ and $\sigma(T)=\overline{\{a_j\}}$.

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If $Tx=\lambda x$, you get the equations $$ (a_1-\lambda )x_1=0,\ (a_2-\lambda) x_2=0,\ \ldots $$ If $x_j\ne0$ for some $j$, then $\lambda=a_j$. By taking $x$ such that $x_j=1$, $x_k=0$ for $k\ne j$, we can get $a_j$ as an eigenvalue. So $\sigma_p(T)=\{a_1,a_2,\ldots\}$.

Now suppose that $a$ is an accumulation point of $\{a_1,a_2,\ldots\}$. Then $a\in\sigma(T)$, since the spectrum is closed.

Finally, if $\lambda$ is not an accumulation point of $\{a_1,a_2,\ldots\}$, then there exists $\delta>0$ with $|\lambda-a_j|>\delta$ for all $j$. Then the map $$ x\mapsto (\frac1{\lambda-a_1}\,x_1,\frac1{\lambda-a_2}\,x_2,\ldots) $$ is a bounded inverse of $T-\lambda I$, so $\lambda\not\in\sigma(T)$.

In conclusion, $\sigma_p(T)=\{a_1,a_2,\ldots\}$, and $\sigma(T)=\overline{\sigma_p(T)}$.

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