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!
 A: Note that
$$A=P\cdot\begin{bmatrix} 2i&0\\0&-2i\end{bmatrix}\cdot P^{-1}$$
With $P=\begin{bmatrix} -1&-1\\-2i&2i\end{bmatrix}$. We have
$$e^A=P\cdot e^{D}\cdot P^{-1}$$
With $D$ the diagonal matrix above
A: By separating odd and even terms in the series, what you have found can be rewritten as:
$$
e^A=I \sum_{k=0}^\infty{(-1)^k4^k\over(2k)!}+A \sum_{k=0}^\infty{(-1)^k 4^k\over(2k+1)!}=I \cos2+A{1\over2}\sin2.
$$
A: 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, $$e^{At} =  \pmatrix{\cos 2t&\frac 12 \sin 2t\\-2\sin 2t&\cos 2t}.$$
A: For the general formula I get it to $e^A = I \cosh(\sqrt{\alpha\beta}) + A\sinh(\sqrt{\alpha\beta})/\sqrt{\alpha\beta}$ where the same square root is selected everywhere (the one under $\cosh$ doesn't matter due to symmetry).
You get this by checking the powers of $A$:


*

*$A^0 = I$

*$A^1 = A$

*$A^2 = (\alpha\beta)I$

*$A^3 = (\alpha\beta)A$


and so on. Expanding it results in
$e^A = I/0! + A/1! + (\alpha\beta)I/2! + (\alpha\beta)A/3! + ...$
suppose $\gamma^2 = \alpha\beta$ we get
$e^A = I\gamma^0/0! + (A/\gamma)\gamma^1/1! + I\gamma^2/2! + (A/\gamma)\gamma^3/3!+...$
then just separating the parts you get
$e^A = I\cosh(\gamma) + (A/\gamma)sinh(\gamma)$
