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.


  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?
  • $\begingroup$ See this 3Blue1Brown video: youtube.com/watch?v=O85OWBJ2ayo $\endgroup$ Dec 29, 2021 at 21:05
  • $\begingroup$ If you plug the series into Wolfram Cloud you will get an equation for the result matrix. $\endgroup$ Dec 29, 2021 at 21:06
  • $\begingroup$ 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? $\endgroup$
    – Dirk
    Dec 29, 2021 at 21:19

3 Answers 3


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 $$\eqalign{ H\O &= \LR{e^{t\O} - I} \qiq H &= \LR{\frac{e^{iwt}-\o}{iw}}I \\ }$$


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