Matrix exponential via Cayley-Hamilton

Problem

For any $t\in\mathbb{R}$ compute $\exp(A_\omega t)$, where \begin{equation*}A_\omega\triangleq\left[\begin{array}{c|c} 0_2 & I_2 \\ \hline 0_2 & \Omega \end{array}\right]\end{equation*} and

• $0_2$ is the $2\times2$ matrix whose entries are all zero;
• $I_2$ is the $2\times2$ identity matrix;
• $\omega$ is a given parameter and\begin{equation*}\Omega \triangleq \begin{bmatrix} 0 & -\omega\\ \omega & 0 \end{bmatrix}\end{equation*}

partial solution

Just to refresh my mind, I want to use the method (which I don't remember anymore) based on the Cayley-Hamilton theorem. First of all, the characteristic polynomial. Since $A_\omega$ is upper-triangular, holds \begin{equation*}\begin{aligned} \chi_{A_{\omega}}(s) &\triangleq \text{det}(sI_4-A_{\omega})\\ &=\text{det} (sI_2-0_2)\text{det} (sI_2-\Omega)\\ &=s^2(s^2+\omega^2)=s^4+\omega^2 s^2 \end{aligned}\end{equation*} Now the Cayley-Hamilton theorem says that \begin{equation*}\begin{aligned} \chi_{A_{\omega}}(A_{\omega}) = 0_4 \end{aligned}\end{equation*} or, more explicitly, \begin{equation*}\begin{aligned} A_{\omega}^4+\omega^2 A_{\omega}^2 = 0_4 \end{aligned}\end{equation*} so we know that \begin{equation}A_{\omega}^4=-\omega^2 A_{\omega}^2 \tag{1}\end{equation} but now how can we exploit this information to compute $\exp(A_{\omega}t)$? I don't remember very well.

Probably we can use $(1)$ to find a closed expression for $A_{\omega}^k$ that figures in the definition \begin{equation*}\exp(A_\omega t)\triangleq \sum_{k=0}^\infty \frac{(A_\omega t)^k}{k!}=\sum_{k=0}^\infty A_\omega^k \frac{t^k}{k!}\end{equation*} but honestly I don't remember how to do it.

questions

1. How can we use CH to compute $\exp(A_{\omega}t)$?
2. There is a more clever way to compute $\exp(A_{\omega}t)$? If yes, what is the procedure?

A couple ways to compute $\exp(tA_\omega)$ come to mind.

First, you can diagonalize $A_\omega = PDP^{-1}$ and then $\exp(tA_\omega) = P\exp(tD)P^{-1}$.

A second way: there is an embedding of $\mathbb{C} \hookrightarrow \mathbb{R}^{2 \times 2}$ where $1 \mapsto I_2$ and $i \mapsto \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.

So you can compute the matrix exponential of $\begin{pmatrix} 0 & 1 \\ 0 & \omega i \end{pmatrix}$ and then push it back through that embedding. This seems like the easiest approach to me.

Third, given that $A^4 = - \omega A^2$ by Cayley-Hamilton, we can compute the exponential by definition:

\begin{align*} \exp(tA) &= \left(I + \frac12A^2 + \frac1{4!}A^4 + \frac{1}{6!}A^6 + \cdots \right) + \left(A + \frac1{3!}A^3 + \frac1{5!}A^5 + \frac{1}{7!}A^7 + \cdots \right) \end{align*}

And then simplify using $A^4 = -\omega A^2$ and $A^6 = -\omega A^4 = \omega^2 A^2$ and $A^5 = -\omega A^3$ and so on.