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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.

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    The diagonal of $D$ contains the eigenvalues of A. Compute $D^2$, $D^3$. What do you observe?2012-09-27
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    @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.+12012-09-27

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