Let $y\in c_0$ and define the operator from $l^2 \rightarrow l^2$ as the following $$T\bigg(\sum x_n e_n\bigg) \mapsto \sum y_n x_n e_n.$$ I have shown that the operator is continuous, compact and selfadjoint, but I am not so sure about its spectrum.
My attempt:
From definition we look for $\lambda$ such that $\lambda I - T$ is not inevitable. And I know a theorem says that for a compact operator, the spectrum is just its eigenvalues and zero.
To find its eigenvalues, let $h$ be a non zero element of $l^2$, then set $$\lambda h - \sum y_n h_n e_n = 0$$ $$\sum (\lambda - y_n)h_n e_n = 0$$ Then I can see that each $y_n$ is an eigenvalue with $e_n$ being its eigenvector.
Questions:
Is this a good way of finding the spectrum of an operator in general? We pick a non-zero $x$ in the space and solve for the equation $$\lambda x - T(x) = 0$$
Are there any other propositions about spectrum that would be useful? Like the result about the spectrum of a compact operator.
Thank you very much!
Your operator is already diagonalized, and it is clear that $Te_{n}=y_{n}e_{n}$ for $n \ge 1$. So $\{ y_{n}\}$ are eigenvalues. The eigenvectors of a selfadjoint operator are orthogonal for different eigenvalues. Nothing except the $0$ vector is orthogonal to $e_{n}$ for all $n$. So there can't be any other eigenvalues. Now you know that the spectrum of $T$ is $\{ y_{n} \}\cup \{0\}$ because $T$ is compact.