<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] By separating odd and even terms in the series, what you have found can be rewritten as:
$$
e^A=I \sum_{k=0}^\infty{(-1)^k4^k\over(2k)!}+A \sum_{k=0}^\infty{(-1)^k 4^k\over(2k+1)!}=I \cos2+A{1\over2}\sin2.
$$
</s>