# 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 $$$$A_{\omega}^4=-\omega^2 A_{\omega}^2 \tag{1}$$$$ but now how can we exploit this information to compute $$\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?
• See this 3Blue1Brown video: youtube.com/watch?v=O85OWBJ2ayo Dec 29, 2021 at 21:05
• If you plug the series into Wolfram Cloud you will get an equation for the result matrix. Dec 29, 2021 at 21:06
• This matrix seems to be diagonalizable. Why don't use that? If you want to use the characteristic polynomial: You already know that $A^4 = -\omega A^2$. How does that help you, to figure out all even powers of $A$? How can you get all odd powers?
– Dirk
Dec 29, 2021 at 21:19

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.

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.

$$\def\Cin#1{\in{\mathbb C}^{#1\times #1}} \def\D{\operatorname{Diag}} \def\v{\operatorname{vec}} \def\L{\left} \def\R{\right} \def\LR#1{\L(#1\R)} \def\p#1#2{\frac{\partial #1}{\partial #2}} \def\m#1{ \L[\begin{array}{c}#1\end{array}\R] } \def\mc#1{ \L[\begin{array}{c|c}#1\end{array}\R] } \def\qiq{\quad\implies\quad} \def\O{\Omega} \def\o{{\tt1}}$$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=iw\,I \\ &F = e^{0} = I,\qquad &G = e^{t\O} = e^{iwt}I \\ } This simplifies the equation such that it can be solved explicitly \eqalign{ H\O &= \LR{e^{t\O} - I} \qiq H &= \LR{\frac{e^{iwt}-\o}{iw}}I \\ }