Let $X$ be an infinite-dimensional complex [Banach space](/page/Banach%20Space) and let $T \in \mathcal{L}(X)$ be compact. If $\lambda \in \sigma(T)$ and $\lambda \ne 0$, then $\lambda$ is an eigenvalue of $T$. Moreover, the eigenspace $\ker(T-\lambda I)$ is finite-dimensional.