Let $X$ be an infinite-dimensional complex [Banach space](/page/Banach%20Space) and let $T \in \mathcal{K}(X)$. Then $0 \in \sigma(T)$. Every nonzero $\lambda \in \sigma(T)$ is an eigenvalue of $T$, and the eigenspace $\ker(T-\lambda I_X)$ is finite-dimensional. Moreover, $\sigma(T)\setminus\{0\}$ is at most countable, and its only possible accumulation point is $0$.