Euler equation, Jordan normal form

Question

My question is: Prove the following:

$$e^{ \begin{bmatrix} \sigma_{i}t & \omega_{i}t \\ -\omega_{i}t & \sigma_{i}t \\ \end{bmatrix}} = e^{\sigma_{i}t} {\begin{bmatrix} \cos(\omega_{i}t) & \sin (\omega_{i}t) \\ -\sin(\omega_{i}t) & \cos(\omega_{i}t) \\ \end{bmatrix}} $$

Thanks in advance, wonderful community of stackexchange.

Answer 1

Here is an direct answer. Just follow the definitions as @Kwin mentions.

Answer 2

Hint: Write $$ \pmatrix{ \sigma_{i}t & \omega_{i}t \\ -\omega_{i}t & \sigma_{i}t } = \underbrace{\sigma_{i}t \pmatrix{1 &0\\0&1}}_A + \underbrace{\omega_{i}t \pmatrix{0&1\\-1&0}}_B $$ Now, note that $AB = BA$, which means that $\exp(A + B) = \exp(A) \exp(B)$. $\exp(A)$ and $\exp(B)$ can both be nicely computed using the power series definition.