If $e^\lambda$ is an eigenvalue of $e^A$, then $\lambda$ is an eigenvalue of $A$

Question

Let $A$ be an $n\times n$ matrix. I want to show that if $e^\lambda$ is an eigenvalue of $e^A$, then $\lambda$ is an eigenvalue of $A$.

I know the converse is true, but I'm not sure how to go the other way. Our assumption seems to be equating power series to each other, which seem more difficult to work with.

Thanks!

Answer 1

As Omnomnomnom says, you can only prove that $\lambda+2i\pi k$ is an eigenvalue for some $k\in\mathbb{Z}$. This follows immediately from the following useful fact:

Let $\lambda_1,\ldots,\lambda_k$ denote the (complex!) eigenvalues of a matrix $A$. Then the (complex!) eigenvalues of $e^A$ are $e^{\lambda_1},\ldots,e^{\lambda_k}$. (This can be proved using Jordan's theorem).