TITLE: Calculating matrix exponential
QUESTION [10 upvotes]: Given matrix $$M = \begin{pmatrix} 7i& -6-2i\\6-2i&-7i\end{pmatrix}$$ how do I calculate matrix exponential $e^M$?


I know I can use that $e^A=Pe^DP^{-1}$ where $D=P^{-1}AP$. I computed the characteristic polynomial of the above matrix as
$$P(\lambda)=\lambda^2+89$$
Is there an easier way to do this than trying to compute the diagonalized matrix?

REPLY [4 votes]: Via Cayley-Hamilton,
$${\rm M}^2 = - 89 \, {\rm I}_2 = \left( i \sqrt{89} \right)^2 {\rm I}_2$$
Hence, matrix ${\rm A} := \frac{{\rm M}}{i \sqrt{89}}$ is involutory, i.e., ${\rm A}^2 = {\rm I}_2$. Using Euler's formula,
$$\begin{aligned} \exp({\rm M}) = \exp \left( i \sqrt{89} {\rm A} \right) &= \cos \left( \sqrt{89} {\rm A} \right) + i \sin \left( \sqrt{89} {\rm A} \right)\\ &= \cos \left( \sqrt{89} \right) {\rm I}_2 + i \sin \left( \sqrt{89} \right) {\rm A}\\ &= \color{blue}{\cos \left( \sqrt{89} \right) {\rm I}_2 + \frac{\sin \left( \sqrt{89} \right)}{\sqrt{89}} {\rm M}}\end{aligned}$$
where $\cos \left( \sqrt{89} {\rm A} \right) = \cos \left( \sqrt{89} \right) {\rm I}_2$ and $\sin \left( \sqrt{89} {\rm A} \right)=  \sin \left( \sqrt{89} \right) {\rm A}$ because ${\rm A}$ is involutory.