Prove that every eigenvector of exp(A) is also an eigenvector of A

Question

In Gerald Teschl's book on ODE's, he states the following:

Lemma 3.3 A vector u is an eigenvector of A corresponding to the eigenvalue $\alpha$ if and only if u is an eigenvector of exp(A) corresponding to the eigenvalue $e^{\alpha}$.

I can clearly see one way of this argument. However, given an arbitrary eigenvector u of exp(A), I am unable to prove that u is necessarily an eigenvector of A. Any help would be appreciated. Thanks!

Answer 1

This is not true in general. Let $A = \begin{bmatrix} 4\pi i & 0\\ 0 & 2\pi i\end{bmatrix}$. Then $exp(A) = Id$ and $v = \begin{bmatrix}1\\1\end{bmatrix}$ is an eigenvector of $exp(A)$ but not of $A$.