How to compute $e^{At}$ with $A=\left ( \begin{array}{cc} -3 & 4 \\ -4 & -3 \end{array} \right )$ 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.
 A: 
For a solution to the second version of the question, see below. 

This applies to the first version of the question, where $\color{red}{A=\begin{pmatrix}3 & 4 \\ -4 & -3\end{pmatrix}}$. Since $\text{tr}(A)=0$ and $\det(A)=7$, Cayley-Hamilton theorem yields $A^2=-7I$. Hence, for every $n\geqslant0$,  $A^{2n}=(-7)^n\cdot I$ and $A^{2n+1}=(-7)^n\cdot A$. This yields
$$
\mathrm e^{tA}=\sum\limits_{n=0}^{+\infty}\frac{t^n}{n!}\cdot A^n=\sum\limits_{n=0}^{+\infty}(-1)^n7^n\frac{t^{2n}}{(2n)!}\cdot I+\sum\limits_{n=0}^{+\infty}(-1)^n7^n\frac{t^{2n+1}}{(2n+1)!}\cdot A,
$$
that is,
$$
\color{red}{\mathrm e^{tA}=\cos(\sqrt7t)I+\frac{\sin(\sqrt7t)}{\sqrt7}A}.
$$

The second version of the question, where $\color{blue}{A=\begin{pmatrix}-3 & 4 \\ -4 & -3\end{pmatrix}}$, can be solved even more easily. Since $A+3I=4J$ with $J=\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$, one has $\mathrm e^{tA}=\mathrm e^{-3t}\cdot\mathrm e^{4tJ}$. Note that $J^2=-I$ (and/or that $J$ is nothing but the matrix of the clockwise rotation by $90°$), hence, for every $n\geqslant0$, $J^{2n}=(-1)^nI$ and $J^{2n+1}=(-1)^nJ$.
This yields
$$
\mathrm e^{tJ}=\sum\limits_{n=0}^{+\infty}\frac{t^n}{n!}\cdot J^n=\sum\limits_{n=0}^{+\infty}(-1)^n\frac{t^{2n}}{(2n)!}\cdot I+\sum\limits_{n=0}^{+\infty}(-1)^n\frac{t^{2n+1}}{(2n+1)!}\cdot J,
$$
that is,
$$
\mathrm e^{tJ}=\cos(t)I+\sin(t)J.
$$
Coming back to $A$ and using the formula for $\mathrm e^{4tJ}$, one gets
$$
\color{blue}{\mathrm e^{tA}=\mathrm e^{-3t}\cdot\left(\cos(4t)I+\sin(4t)J\right)}.
$$
A: Post problem change; i.e $A=\begin{bmatrix}-3&4\\-4&-3\end{bmatrix}$. $A^2+6A+25I=0$ therefore $A^n=a_nA+b_nI$ for some $a_n$ and $b_n$. Note that
$$
\begin{align}
a_nA+b_n
&=(a_{n-1}A+b_{n-1}I)A\\
&=a_{n-1}(-6A-25I)+b_{n-1}A\tag{1}
\end{align}
$$
Therefore, $a_n=-6a_{n-1}+b_{n-1}$ and $b_n=-25a_{n-1}$; that is,
$$
A^n=a_nA-25a_{n-1}I\tag{2}
$$
where
$$
a_n=-6a_{n-1}-25a_{n-2}\tag{3}
$$
Solving $(3)$ with $a_1=1$ and $a_0=0$ yields
$$
a_n=\frac{(-3+4i)^n-(-3-4i)^n}{8i}\tag{4}
$$
Now
$$
\begin{align}
e^{tA}
&=\sum_{n=0}^\infty\frac{1}{n!}t^nA^n\\
&=\sum_{n=0}^\infty\frac{t^n}{n!}\left(\frac{(-3+4i)^n-(-3-4i)^n}{8i}A-25\frac{(-3+4i)^{n-1}-(-3-4i)^{n-1}}{8i}I\right)\\
&=\frac14e^{-3t}\sin(4t)A-\frac{1}{4}e^{-3t}(-3\sin(4t)-4\cos(4t))I\\
&=e^{-3t}\begin{bmatrix}\cos(4t)&\sin(4t)\\-\sin(4t)&\cos(4t)\end{bmatrix}\tag{5}
\end{align}
$$
A: I know you explicitly wanted to use the definition, but another useful way to get $e^{tA}$ would be to use the Laplace transform.
$$
e^{tA}=\mathcal{L}^{-1}\left((sI-A)^{-1}\right)=
\mathcal{L}^{-1}\left(\begin{bmatrix}s+3 & -4 \\ 4 & s+3\end{bmatrix}^{-1}\right)
=\mathcal{L}^{-1}\left(\frac{1}{(s+3)^2+16}\begin{bmatrix}s+3 & 4 \\ -4 & s+3\end{bmatrix}\right)=
e^{-3t}\mathcal{L}^{-1}\left(\frac{1}{s^2+16}\begin{bmatrix}s & 4 \\ -4 & s\end{bmatrix}\right)
=e^{-3t}\begin{bmatrix}\cos{4t} & \sin{4t} \\ -\sin{4t} & \cos{4t}\end{bmatrix}.
$$
