Eigenspace dimension of a compact operator

Question

Let $K$ be a compact operator on a Hilbert space $H$. By basic theory such that $K$ can have infinite dimensional kernel, but all other eigenspaces must have finite dimension. Is there a conceptual way to see why this is true?

Answer 1

Here's an argument I like: the restriction of any compact operator to a subspace should be compact. However, the restriction of $K$ to the eigenspace $V$ associated with $\lambda$ is given by $$ K|_V:V \to V\\ Kx = \lambda x $$ If $\lambda \neq 0$, then the map $x \mapsto \lambda x$ is only compact if $V$ is finite dimensional.