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

*

*How can we use CH to compute $\exp(A_{\omega}t)$?

*There is a more clever way to compute $\exp(A_{\omega}t)$? If yes, what is the procedure?

 A: 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.
A: $
\def\Cin#1{\in{\mathbb C}^{#1\times #1}}
\def\D{\operatorname{Diag}}
\def\v{\operatorname{vec}}
\def\L{\left}
\def\R{\right}
\def\LR#1{\L(#1\R)}
\def\p#1#2{\frac{\partial #1}{\partial #2}}
\def\m#1{ \L[\begin{array}{c}#1\end{array}\R] }
\def\mc#1{ \L[\begin{array}{c|c}#1\end{array}\R] }
\def\qiq{\quad\implies\quad}
\def\O{\Omega}
$Apply the analytic function $f$ to the matrices $A$ and $B$
$$\eqalign{
F &= f(A),\qquad G = f(B) \\
}$$
and thence to the block upper triangular matrix $X$
$$\eqalign{
X &= \m{A&C\\0&B},\qquad\quad &Y = f(X) = \m{F&H\\0&G} \\
}$$
Since a matrix will commute with any analytic function (i.e. power series) of itself
$$\eqalign{
YX &= XY \qiq &(AH - HB) = (FC - CG) \\
}$$
This is a Sylvester Equation which must be solved for the unknown block $H$.
In the current problem
$f(x)=e^{tx},\:A=0,\:B=\O,\:C=I,\:$ and
$$\eqalign{
&\O^2 = -w^2I \qiq
  &\O^{-1} = -\LR{\frac{\O}{w^2}} \\
&F = e^{0} = I,\qquad
  &G = e^{t\O} = I\cos(wt)+\O\LR{\frac{\sin(wt)}{w}} \\
}$$
This simplifies the equation such that it can be solved explicitly
$$\eqalign{
H\O &= \LR{e^{t\O} - I} \qiq
H &= \frac{\O-\O e^{t\O}}{w^2} \\
}$$
A: I believe the technique that you are trying to remember is Hermite-Sylvester interpolation.
Any analytic function of $A\in{\mathbb C}^{n\times n}$ can be written as a power series, which can be divided by the characteristic polynomial of degree $n$ leaving a polynomial of degree $(n-1)$ as the remainder, i.e.
$$\def\l{\lambda}
f(A) = c_0I + c_1A + \cdots + c_{n-1}A^{n-1}$$
If $\l$ is an eigenvalue of $A$ then $f(\l)$ is an eigenvalue of $f(A)$. Since the eigenvalues satisfy the characteristic polynomial, they also satisfy the interpolation equation.
$$f(\l) = c_0 + c_1\l + \cdots + c_{n-1}\l^{n-1}$$
This produces a system of $n$ equations (one for each eigenvalue) in $n$ unknowns (the $c_k$ coefficients).
If the eigenvalues are distinct then you are good, but for a duplicate eigenvalue you lose one equation, which must be replaced by the derivative, i.e.
$$f'(\l) = c_1 + 2c_2 + \cdots + (n-1)c_{n-1}\l^{n-2}$$
If an eigenvalue has multiplicity $m$, then you must use up to
the $(m-1)^{th}$ derivative.
You have already calculated the characteristic polynomial, from which the eigenvalues are seen to be $\{0,\pm iw\}$, with the zero eigenvalue having multiplicity $m=2$.
Given the functions
$$f(\l) = e^{t\l},\qquad f'(\l) = te^{t\l}$$
all you need to do is setup $4$ interpolation equations and solve for the $4$ coefficients.
