Exponential Matrix 2 - complex eigenvalues Euler

Question

Given a matrix $A=\begin{pmatrix} \sigma & \omega \\ -\omega & \sigma \end{pmatrix}$ with two complex eigenvalues $\sigma\pm i\omega$, using the Euler formula $e^{i \omega t}=\cos(\omega t)+i \sin(\omega t)$ my objective is to show that $e^{At}=e^{\sigma t}\begin{pmatrix} \cos(\omega t) & \sin(\omega t)\\ -\sin(\omega t)& \cos(\omega t) \end{pmatrix}$.
It should be done without using Taylor expansion.

Answer 1

The eigen values of the matrix are :

λ=α+iω and λ¯=α−iω. The corresponding eigenvectors are $v=\begin{bmatrix}1 \\ i\end{bmatrix} = \begin{bmatrix}1 \\ 0\end{bmatrix} + i\begin{bmatrix}0\\ 1\end{bmatrix},$ and $v¯= \begin{bmatrix}1 \\ −i\end{bmatrix} = \begin{bmatrix}1 \\ 0\end{bmatrix} − i\begin{bmatrix}0 \\ 1\end{bmatrix}$

Write, $A = αI + ωJ, \: I = \begin{bmatrix}1 & 0\\ 0& 1\end{bmatrix},J = \begin{bmatrix}0 &−1 \\ 1& 0\end{bmatrix}$

The matrices I and J commute (check), so $e^{tA}=e^{αtI + ωtJ} = e^{αtI} e^{ωtJ}$.

We have,

$e^{αtI} = \begin{bmatrix}e^{αt} & 0 \\ 0 & e^{αt}\end{bmatrix}, \: e^{ωtJ} = \begin{bmatrix}cosωt & −sinωt \\ sinωt & cosωt\end{bmatrix}$

which gives the desired result.

Check that this holds true $e^{ωtJ} = \begin{bmatrix}cosωt & −sinωt \\ sinωt & cosωt\end{bmatrix}$ The definition of matrix exponential (refer to any standard book) is based on the converging Taylor series, but since you are not allowed to use that definition to conclude the above, you may instead use the Lemma for the derivative of a matrix exponential to prove that the above expression holds true.

For any matrix $A \in \mathbb{R}^{n \times n}$ and $t \in \mathbb{R}$, we have,

$ \frac{d}{dt} e^{At} = A e^{At} $

Also, note that the above only holds for scalar matrix $A$ and not when A is also a function of time. Using this lemma, you can substitute the expression for $e^{\omega tJ} $ above (i.e. for $A = \omega J$, in this case) , differentiate with respect to $t$ and show that it holds to complete the proof.