Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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

2 Answers 2

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 $e^A=Pe^DP^{-1}$

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.