Suppose $H$ is a hilbert space. Let $T:H\rightarrow H$ be a compact (in the bolzano weistrass sense), self adjoint operator. Then show that if $\lambda$ is an eigenvalue then then eigenspace of $\lambda$ has finite dimension.
Is this a true statement? All counterexamples i've been able to cook up suggest that it is true. However, i'm not sure how to prove it.
It's true if $\lambda\neq 0.$ Indeed, consider $T$ restricted to the $\lambda$-eigenspace (on this space, $Tx=\lambda x$). This is still compact, as the restriction of a compact operator to any subspace is compact. However, this is a scaling of the identity operator, which is only compact for finite dimensional spaces by the Banach-Alaoglu theorem. Thus, it can only be compact if the eigenspace is finite dimensional.
However, this argument clearly breaks down if $\lambda=0$. In fact, the kernel of a compact operator can have infinite dimension.