$X$ 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 $e^X$'.
Best Answer
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 $$ \exp(X)v = \left(\sum_{k=0}^\infty \frac{1}{k!} X^k \right)v = \sum_{k=0}^\infty \frac{1}{k!} X^k v = \sum_{k=0}^\infty \frac{1}{k!} a^k v = \left( \sum_{k=0}^\infty \frac{1}{k!} a^k \right) v = e^a v. $$