Any quicker method to find matrix exponent for a scaled permutation matrix? I am trying to find the eigen value of matrix $e^A$ such that the matrix A is given as 
$$ 
\begin{bmatrix}
a & 0 & 0\\
0 & 0 & a\\
0 & a & 0
\end{bmatrix}
$$
I found $A^2 = \begin{bmatrix} a^2 & 0 & 0 \\ 0 & a^2 & 0 \\ 0 & 0 & a^2 \end{bmatrix}$, $A^3 = \begin{bmatrix} a^3 & 0 & 0 \\ 0 & 0 & a^3 \\ 0 & a^3 & 0 \end{bmatrix}$ and so forth...
So that I can find, $e^A$ by substituting above matrices in its power series expansion
\begin{equation}
e^A = 1 + A + \frac{1}{2!} A^2+\dots 
\end{equation}
which gives
\begin{equation}
e^A =
\begin{bmatrix}
1+a+\frac{1}{2}a^2+\frac{1}{6}a^3+\dots & 0 & 0 \\
0 & 1+\frac{1}{2}a^2+\frac{1}{24}a^4+\dots & a+\frac{1}{6}a^3+\frac{1}{120}a^5\dots \\
0 & a+\frac{1}{6}a^3+\frac{1}{120}a^5\dots & 1+\frac{1}{2}a^2+\frac{1}{24}a^4+\dots
\end{bmatrix}
\end{equation}
which can be written as 
\begin{equation}
e^A =
\begin{bmatrix}
e^a & 0 & 0 \\
0 & \frac{e^a}{2}+\frac{e^{-a}}{2} & \frac{e^a}{2}-\frac{e^{-a}}{2} \\
0 & \frac{e^a}{2}-\frac{e^{-a}}{2} & \frac{e^a}{2}+\frac{e^{-a}}{2}
\end{bmatrix}
\end{equation}
I am able to solve this, but I am wondering if there is a quicker way or a trick that I can use to find $e^A$ quickly that this method?

By diagonalization,
$$\exp\begin{pmatrix} 0 & a \\ a & 0 \end{pmatrix} = \begin{pmatrix} \cosh(a) & \sinh(a) \\ \sinh(a) & \cosh(a)\end{pmatrix} $$
is just De Moivre/Euler's formula $e^{i\alpha}=\cos(\alpha)+i\sin(\alpha)$ written in a equivalent way.
