Eigenvector and eigenvalue for exponential matrix $X$ 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 $e^X$'.
 A: 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
$e^X v = e^a v$.
A: Without restricting the generality, we can suppose that the ground field is algebraically closed. There exists $P$ such that $B=PXP^{-1}$ is a sup triangular matrix, and $P(v)$ is an eigenvector of $B$ associated to $a$, $exp(B)=Pexp(A)P^{-1}$, $P(v)$ is an eigenvector of $exp(B)$ associtated to $e^a$. This implies that $v$ is an eigenvector of $exp(X)$ associated to $e^a$.
A: You don't need to assume that $X$ is diagonalisable to use the "Taylor expansion". By definition, 
$$
\exp(X) = \sum_{k=0}^\infty \frac{1}{k!} X^k
$$
Also, if $Xv = av$, then $X^2v = X(Xv) = aXv = a^2v$, etc. By induction, $X^n v = a^n v$. 
Hence
$$
\exp(X)v 
= \left(\sum_{k=0}^\infty \frac{1}{k!} X^k \right)v 
= \sum_{k=0}^\infty \frac{1}{k!} X^k v 
= \sum_{k=0}^\infty \frac{1}{k!} a^k v 
= \left( \sum_{k=0}^\infty \frac{1}{k!} a^k \right) v = e^a v. 
$$
