Computing the exponential matrix with complex eigenvalues

Question

Let $a,b\in\mathbb{R}$ two fixed parameters, and consider the following matrix $$
A:=\left(\begin{matrix} 0 & -a & 0 & 0 \\ a & 0 & 0 & 0 \\ 0 &0&0&b \\ 0&0&b&0\end{matrix}\right).
$$
I am trying to compute the exponential matrix $e^{Ax}$ associated with $A$ above. In order to do it, I started by computing its eigenvalues, from where I obtained $\{\pm b,\pm ia\}$, and hence $A$ is diagonalizable. Here is where everything became dark for me, specifically, the fact that I have two complex eigenvalues confuses me. For example, in order to obtain the eigenspace associated with $\pm b$, I know that I just have to solved the following system $$
(A\mp bI)\vec{v}=\vec{0},
$$
from where I obtained (respectively): $$
\left(\begin{matrix} 0 \\ 0 \\ 1 \\ 1\end{matrix}\right) \qquad \hbox{and}\qquad \left( \begin{matrix} 0 \\ 0 \\ 1 \\ -1\end{matrix}\right).
$$
However, as I said before, when trying to mimic the previous computations with the eigenvalues $\pm ia$, I got confused and I blocked. For example, if I pick $+ia$, the first two equations associated with $(A-iaI)\vec{v}=\vec{0}$ writes $$
ia v_1+av_2=0 \quad \hbox{ and } \quad av_1-iav_2=0,
$$
from where I cannot solve for $v_1$ or $v_2$. Does anyone knows how to follow?

Answer 1

Noticing that $$ A^{2n}={\rm diag}\{(-1)^n a^{2n},(-1)^n a^{2n}, b^{2n}, b^{2n}\} $$ and $$ A^{2n+1}= \begin{pmatrix} 0 & (-1)^{n+1} a^{2n+1}& 0 & 0 \\ (-1)^{n} a^{2n+1} & 0 & 0 \\ 0 & 0 & 0 & b^{2n+1}\\ 0 & 0 & b^{2n+1} & 0 \end{pmatrix} $$ You can separate $e^{x A}$ into two parts $$ e^{x A} = \sum_{n=0}^\infty \frac{A^{2n}}{(2n)!}x^{2n}+ \sum_{n=0}^\infty \frac{A^{2n+1}}{(2n+1)!}x^{2n+1} $$ The first and second summation are separately $$ \begin{pmatrix} \cos (a x) & 0 & 0 & 0 \\ 0 & \cos (a x) & 0 & 0 \\ 0 & 0 & \cosh (b x) & 0 \\ 0 & 0 & 0 & \cosh (b x) \\ \end{pmatrix} ,\qquad \begin{pmatrix} 0 & -\sin (a x) & 0 & 0 \\ \sin (a x) & 0 & 0 & 0 \\ 0 & 0 & 0 & \sinh (b x) \\ 0 & 0 & \sinh (b x) & 0 \\ \end{pmatrix} $$