Separable Hilbert space implies countable number of eigenvalues

Question

Given a separable Hilbert space H, and a linear operator $T:{\cal D}(T) \to $T. If $T$ is symmetric I can show that $T$ can have at most a countable number of eigenvalues. Is this still true in the non-symmetric case?

Answer 1

If $T$ is symmetric and $Tv=\lambda v$, $Tw=\mu w$ with $\lambda\ne\mu$, it is easy to check that $\lambda,\mu\in\mathbb R$. Then $$ \lambda \langle v,w\rangle=\langle Tv,w\rangle=\langle v,Tw\rangle=\overline\mu\,\langle v,w\rangle=\mu\langle v,w\rangle. $$ As $\lambda\ne\mu$, it follows that $v\perp w$. As $H$ is separable, there can only be countably many pairwise orthogonal vectors, and so $T$ has at most countably many eigenvalues.

If you don't require $T$ to be symmetric, then the answer is no. Take $\{v_t\}_{t\in [0,1]}$ be an (uncountable!) Hamel basis for $H$ and define $Tv_t=tv_t$ and extend by linearity. Then each $t\in [0,1]$ is an eigenvalue of $T$.