So, I've been working on this question for a while now and cannot get it right.
I have a compact, self adjoint, positive definite operator $K$ on infinite dimensional Hilbert space. I know that $0\in spec(K)$ and I also know that $spec(K)$ is at most countably infinite and every $0\neq\lambda\in spec(K)$ is an eigenvalue of K. But the proof I am reading says that it follows that $spec(K)$ is countably infinite. How do I justify this conclusion?
Since $K$ is positive definite, $0$ is not an eigenvalue, and since $K$ is compact, every nonzero eigenvalue has finite multiplicity. If the spectrum were finite, the sum of all the eigenspaces of $K$ would be finite dimensional, and thus its orthogonal complement $V$ would be nontrivial. Since $K$ is self-adjoint, it restricts to an operator on $V$ which is still compact, self-adjoint, and positive definite. But by definition of $V$, the restriction of $K$ to $V$ has no eigenvalues, which is a contradiction.