Matrix exponential, computation. What is $e^A$, where$$A = \begin{pmatrix} 0 & 1 \\ 4 & 0 \end{pmatrix}?$$
 A: Hint:
It is not so difficult to show  that if a $2\times 2$ matrix  $A$ has null trace than the exponential is:
$$
e^A=I \cos \theta +A \dfrac{\sin \theta}{\theta}
$$
where $\theta= \sqrt{\det(A)}$.
This comes from the definition of exponential as a series:
$$e^A=\sum_{n=0}^{\infty}\frac{A^n}{n!}$$
 and from the fact that, for such a matrix, we have $A^2=-\det(A) I$
In your case, since $\det A <0$ use:
$$
\cos (i|\theta|)=\cosh |\theta| \qquad \dfrac{\sin (i|\theta|)}{i|\theta|}=\dfrac{\sinh |\theta|}{|\theta|} 
$$
A: You first have to find a basis in which the matrix has the form $B= D+N$, where $D$ is a diagonal matrix, $N$ is nilpotents and $DN=ND$. Then, since $D$ and $N$ commute, 
$$\mathrm e^{B}=\mathrm e^{D}\mathrm e^{N}$$
The exponential of a diagonal matrix $D$ is the diagonal matrix with the exponentials of the diagonal elements of $D$ on the diagonal. The exponential of a nilpotent matrix can be computed with the series expansion of the exponential, which is really a finite sum for a nilpotent matrix.
Finally, let $P$ the change of basis matrix from the canonical basis to the new basis. We have $A=PBP^{-1}$, whence
$$\mathrm e^{A}=P\mathrm e^{B}P^{-1}.$$
Here $A$ is diagonalisable: one finds the eigenvalues are $\pm 2$ and a basis of eigenvectors is $\;\bigl\{(1,2),(1,-2)\bigr\}$, hence
$$P=\begin{bmatrix}1&1\\2&-2\end{bmatrix},\quad P^{-1}=\frac14\begin{bmatrix}2&1\\2&-1\end{bmatrix}$$
and finally
$$\mathrm e^{A}=\begin{bmatrix}1&1\\2&-2\end{bmatrix}\begin{bmatrix}\mathrm e^2&0\\0&\mathrm e^{-2}\end{bmatrix}\cdot\frac14\begin{bmatrix}2&1\\2&-1\end{bmatrix} =\begin{bmatrix}\cosh 2&\frac12\sinh 2\\2\sinh2&\cosh2\end{bmatrix}.$$
