Let $A$ a matrix $n\times n$.
Define $e^A=\sum ^{\infty}_{n=0} \frac{A^n}{n!}$ (also you can see this question).
If $A$ is a diagonalizable matrix, find $e^A$ in terms of eigenvalues of $A$.
I was trying this:
If $A$ is diagonalizable, exists an inversible matrix $P$ such that: $D=P^{-1}AP$ With $D$ a diagonal matrix. Then: $e^A=\sum ^{\infty}_{n=0} \frac{(PDP^{-1})^n}{n!}$ $=P\left(\sum ^{\infty}_{n=0} \frac{D^n}{n!} \right)P^{-1}$
Because $D$ is diagonal matrix, the eigenvalues of $A$ are the diagonal elements of $D$, but I think that this is not enough.
How can I find $e^A$ in terms of eigenvalues of $A$?
Thanks for your help.