TITLE: How to compute $e^{At}$ with $A=\left ( \begin{array}{cc} -3 & 4 \\ -4 & -3 \end{array} \right )$
QUESTION [5 upvotes]: If I wanted to calculate the matrix exponential of $\left ( \begin{array}{cc}  -3 & 4 \\ -4 & -3 \end{array} \right )$, I could do this by calculating the eigenvectors and values and then use $e^{At}=P\ \mbox{diag}(e^{\lambda t}) P^{-1}$ where $P$ is the matrix with the eigenvectors of $A$ as its columns and I get $$e^{At}= e^{-3t}\left ( \begin{array}{cc}    \cos(4t) & \sin(4t) \\ -\sin(4t) & \cos(4t) \end{array} \right )$$ but I was wondering how I do this using the definition, I'm sure it shouldn't be tough but I'm getting a bit lost; I have:
$e^{At}=\sum\limits^{\infty}_{m=0} \frac{1}{m!}(At)^m$ but then looking at the powers of this matrix I just get rubbish i think?
Thanks very much for any help.

REPLY [2 votes]: I know you explicitly wanted to use the definition, but another useful way to get $e^{tA}$ would be to use the Laplace transform.
$$
e^{tA}=\mathcal{L}^{-1}\left((sI-A)^{-1}\right)=
\mathcal{L}^{-1}\left(\begin{bmatrix}s+3 & -4 \\ 4 & s+3\end{bmatrix}^{-1}\right)
=\mathcal{L}^{-1}\left(\frac{1}{(s+3)^2+16}\begin{bmatrix}s+3 & 4 \\ -4 & s+3\end{bmatrix}\right)=
e^{-3t}\mathcal{L}^{-1}\left(\frac{1}{s^2+16}\begin{bmatrix}s & 4 \\ -4 & s\end{bmatrix}\right)
=e^{-3t}\begin{bmatrix}\cos{4t} & \sin{4t} \\ -\sin{4t} & \cos{4t}\end{bmatrix}.
$$