Whay does an operator commuting with a finite rank operator have eigenvalues?

Question

Let $A,B\in B(H)$ be such that $AB=BA$ where $B\neq 0$ is a finite rank operator. Does it follow that $A$ has eigenvalues? If yes, why please?

Thanks a lot.

Answer 1

Hint: $A$ maps the range of $B$ (which is a finite-dimensional space) to itself.