<s>[INST] Matrix exponential: $\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$ It is asked to calculate $e^A$, where
$$A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$$
I begin evaluating some powers of A:
$A^0= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\; ; A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix} \; ; A^2 = \begin{pmatrix} -4 & 0 \\ 0 & -4\end{pmatrix} \; ; A^3 = \begin{pmatrix} 0 & -4 \\ 16 & 0\end{pmatrix}\; ; $
$ A^4=\begin{pmatrix} 16 & 0 \\ 0 & 16\end{pmatrix},\; \ldots$
I've noted that, since
$$e^A = \sum_{k=0}^\infty \frac{A^k}{k!}$$
we will have the cosine series at the principal diagonal for $\cos(2)$. But  couldnt get what we will have in $(e^A)_{12}$ and $(e^A)_{21}$.
Also, we know that if $B=\begin{pmatrix} 0 & \alpha \\ -\alpha & 0 \end{pmatrix}$, then $e^B = \begin{pmatrix} \cos(\alpha) & \sin(\alpha) \\ -\sin(\alpha) & \cos(\alpha) \end{pmatrix} $. Is there a general formula for
$$B=\begin{pmatrix} 0& \alpha \\ \beta & 0 \end{pmatrix}$$?
Thanks!
 [/INST] here is another way to find $e^A.$  we will use the interpretation that $x = e^{At}x_0$ is the unique solution of $$\frac{dx}{dt} = Ax, x(0) = x_0.$$ 
for $$\frac{d}{dt}\pmatrix{x\\y} = \pmatrix{0&1\\-4&0}\pmatrix{x\\y}$$ which in component form is $$\dot x = y, \dot y = -4x \to \ddot x=-4x, \ddot y= -4y \text{ and } x(0) = x_0, y(0) = y_0.$$  the solutions are $$x = x_0\cos 2t+\frac 12y_0 \sin 2t, y = \dot x = -2x_0\sin 2t + y_0\cos 2t$$ 
writing the last equation in matrix form, we have $$ \pmatrix{x\\y} = \pmatrix{\cos 2t&\frac 12 \sin 2t\\-2\sin 2t&\cos 2t}\pmatrix{x_0\\y_0}.$$  therefore, $$e^{At} =  \pmatrix{\cos 2t&\frac 12 \sin 2t\\-2\sin 2t&\cos 2t}.$$
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