Let $X$ be a Banach space and $A:D(A)\subset X\to X$ be a linear and closed operator with $\rho(A)\neq \emptyset$. Suppose that the map $$j:\left(D(A),\|\bullet\|_A\right)\hookrightarrow \left(X,\|\bullet\|\right)$$ is compact.
Claim: $\sigma(A)$ consists of a sequence $(\lambda_n)_{n\in \Bbb N}$ of eigenvalues of $A$ with finite-dimensional eigenspaces. Furthermore, $\sigma(A)$ is either finite or $\lim_{n\to \infty} |\lambda_n|=\infty$.
I am stumped by this exercise and don't have much of a good attempt to show for. Considering first the case that $|\sigma(A)|<\infty$, why does every $\lambda \in \sigma(A)$ need to be an eigenvalue? I.e. how do we know that $\ker (\lambda - A)\neq 0$?
Knowing that the resolvent is non-empty, consider $(A-\lambda I)^{-1}$ for some $\lambda\in\rho(A)$. The resolvent operator is compact.