TITLE: Eigenvector and eigenvalue for exponential matrix
QUESTION [19 upvotes]: $X$ is a matrix. Let $v$ be an eigenvector of $X$ with corresponding eigenvalue $a$. Show that $v$ is also an eigenvector of $e^{X}$ with eigenvalue $e^{a}$
If $X$ is diagonalizable, then we can start writing out terms using Taylor expansion of $e^{X}$ but I can't seem to get anywhere. 
Thanks for the help
Edit: Corrected question to read 'Let $v$ be an eigenvector of $X$' instead of 'Let $v$ be an eigenvector of $e^X$'.

REPLY [6 votes]: If we let $ \Phi(t) = e^{t X}$, we see that
$\Phi$ satisfies the (matrix) equation $\dot{Y} = Y X$ subject to the initial condition $Y(0) = I$.
Let $\xi(t) = \Phi(t) v$, where $Xv = a v$, then we see that
$\dot{\xi}(t) = \Phi(t) X v = a \Phi(t)v= a \xi(t)$, and
so $\xi(t) = e^{a t} v$. Taking $t=1$ we get
$e^X v = e^a v$.