If $v$ is an eigenvalue of $e^{t A}$ for all $t \geq 0$, is it also an eigenvalue of $A$? I'm currently writtng a proof and I wont use that, if $v$ is an eigenvector of $e^{tA}$ for all $t \geq 0$ where $A$ is some generating matrix, then $v$ is an eigenvector of $A$ itself. However, I found neither the proof nor a contradiction in available sources.
The case of diagonalisable matrices is self-evident as  a popular computation approach utilises this fact. However, it is still unclear for me if  the same is true for non-diagonalisable matrices at least for the case of basic (non generalised) eigenvectors. Recalling Jordan structure  I think that all kinds of eigenvectors are preserved. 
But are there any better rigorous proof for this conjecture to make me totally confident? Can this fact be assumed as obvious? 

If $Av = \lambda v$, it's easy to see $\exp(tA)v = \exp(t\lambda) v$. 
Conversely, if $v$ is an eigenvector of $\exp(tA)$ for all non-zero $t$, then $v$ is an eigenvector of $\frac{1}{t}(\exp(tA) - I) = A + O(t)$ for all $t \neq 0$, and by continuity $v$ is an eigenvector of $A$.
Erratum: Though the question stipulates "...for all $t \geq 0$" (so the preceding argument suits the OP's needs), my initial wording, "if $v$ is an eigenvector of $\exp(tA)$ for some (hence all) non-zero $t$...", is generally incorrect. For example, if
$$
A = \left[\begin{array}{@{}rr@{}}0 & -1 \\ 1 & 0 \\ \end{array}\right],
$$
then $\exp(2\pi A) = I$, while $A$ has no real eigenvectors. (As noted in the question, the "some (hence all)" assertion does hold under additional hypotheses, such as if all eigenvalues of $A$ are real.)
