I am trying to prove all the results regarding linear algebra in my ODE class. I have already convinced myself that if I have a matrix $T$ which has an eigenvalue $\lambda = a + ib$ and an associated eigenvector $v = u + iw$ I can always "almost diagonalize" $T$ using $T = PDP^{-1}$ where $P = [u | v]$ is the real canonical base and $D = \begin{pmatrix} a & b\\ -b & a \end{pmatrix}$
When it comes to linear systems of ODE's, using matrix exponentials, I just don't get why $e^{Dt} = e^{at}\begin{pmatrix} \cos bt & \sin bt\\ -\sin bt & \cos at\end{pmatrix}$
As far as I know, you can transform $D$ in sines and cosines forms factoring out $\rho = \sqrt{a^2 + b^2}$ and getting:
$D = \rho \begin{pmatrix} \cos \theta & \sin \theta\\ -\sin \theta & \cos \theta\end{pmatrix}$, where $\tan \theta = \frac{b}{a}$.
I can't see the connexion between these two forms and how one would get rid $\theta$ for example, when calculating the exponential via the infinite series form. Any help is highly appreciated! Thanks in advance!
Write $D=aI+bJ$, $I=$identity matrix, and $J=\begin{bmatrix}0 & 1\\ -1 & 0 \end{bmatrix}$. \begin{align*} \exp tD & =\exp\left(taI\right)\exp\left(tbJ\right)\\ & =\begin{bmatrix}e^{ta} & 0\\ 0 & e^{ta} \end{bmatrix}\begin{bmatrix}\cos bt & \sin bt\\ -\sin bt & \cos bt \end{bmatrix}\\ & =e^{ta}\begin{bmatrix}\cos bt & \sin bt\\ -\sin bt & \cos bt \end{bmatrix} \end{align*}