If an operator A on a Hilbert space has compact resolvent, is Ker($\lambda-A$) finite dimensional?

Question

If an operator A on a Hilbert space has compact resolvent, is Ker($\lambda-A$) finite dimensional, for any $\lambda$ in A's spectrum?

P.S: What I know now is that the spectrum of A is discrete.

Answer 1

Assuming that $\lambda_0$ is in the resolvent set of $A$, then $(A-\lambda_0 I)^{-1}$ is a bounded compact linear operator on $X$, and $$ (A-\lambda I)(A-\lambda_0 I)^{-1}=I+(\lambda_0-\lambda)(A-\lambda_0 I)^{-1} $$ The right side has the form $I+K$ where $K$ is a compact operator. Because $\lambda_0$ is in the resolvent set, then $$ \mathcal{R}(A-\lambda I)=\mathcal{R}((A-\lambda I)(A-\lambda_0 I)^{-1})=\mathcal{R}(I+K). $$ And, $$ \mathcal{N}(A-\lambda I) = (A-\lambda_0 I)\mathcal{N}(I+K). $$