# Serie of a matrix

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 $A$?

-
The diagonal of $D$ contains the eigenvalues of A. Compute $D^2$, $D^3$. What do you observe? – Lucien Sep 27 '12 at 17:51
@Lucien, $D^k$ have the powers of each eigenvalue of course. But I was thinking that my problem was $P$, because it is not in terms of eigenvalues. But still each column vector P depends on each eigenvalue, then I think there is no problem. Thank you.+1 – Hiperion Sep 27 '12 at 17:58

You are there. You are correct that the diagonal elements of $D$ are the eigenvalues of $A$. A power of a diagonal matrix is that power of each element on the diagonal, so you just exponentiate each diagonal element of $D$. You can write $e^A=Pe^DP^{-1}$
If $$D=\left(\begin{matrix}d_1&0&\cdots&0\\0&d_2&\cdots & 0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&d_n\end{matrix}\right),$$ then $$e^D=\left(\begin{matrix}e^{d_1}&0&\cdots&0\\0&e^{d_2}&\cdots & 0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&e^{d_n}\end{matrix}\right)$$ because the powers of $D$ simply have the powers of the $d_i$ as diagonal entries, hence you obtain the usual exponential series for $e^{d_i}$ in the end.