# Cayley-Hamilton theorem to compute this.

$A = \pmatrix{0&-3&0\\3&0&0\\0&0&-1}$

Compute the $e^{At}$. Well, the first problem of this is to calculate the inverse of $A$ using Cayley-Hamilton theorem. But for this second problem, I don't know how to solve it, should I use the Cayley-Hamilton theorem?

• Why do you need to invert $A$ in order to compute its exponential?
– amd
Mar 10 '18 at 18:46
• He didn't said he needed to. Mar 10 '18 at 18:48
• Is $t$ a scalar? Mar 10 '18 at 18:49
• What exactly is your question, then?
– amd
Mar 10 '18 at 19:40
• Your book covers this, no? Mar 10 '18 at 23:17

Note that the powers of $A$ are very regular, so calculating the exponential explicitly is not hard.

You can use Cayley-Hamilton to compute this exponential by noting that the characteristic polynomial of $A$ is a cubic, so that any polynomial in $A$ can be reduced to the quadratic remainder after dividing by $A$’s characteristic polynomial. This also extends to $f(A)$, where $f$ is an analytic function. It’s not too difficult to show that if $R$ is this remainder polynomial, then for any eigenvalue $\lambda_i$ of $A$, $f(\lambda_i)=R(\lambda_i)$. Therefore, $$e^{At}=\alpha_0 I+\alpha_1 A+\alpha_2 A^2$$ for some unknown coefficients $\alpha_i$ that can be determined from the equations $$e^{\lambda_i t}=\alpha_0+\alpha_1 \lambda_i+\alpha_2 \lambda_i^2,$$ which is a system of linear equations in the unknown coefficients $\alpha_i$. (If there’s a repeated eigenvalue, these equations need a small tweak, but that’s not the case for this matrix.)

Explicitly calculating the characteristic polynomial yields $$p_A(x) = -(x^2+9)(x+1) = -(x-3i)(x+3i)(x+1)$$ so $\sigma(A) = \{3i, -3i, -1\}$.

Therefore, $A$ is diagonalizable and after a bit of computation we find:

$$A = PDP^{-1} = \frac12\pmatrix{i & -i & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1}\pmatrix{3i & 0 & 0 \\ 0 & -3i & 0 \\ 0 & 0 & -1}\pmatrix{-i & 1 & 0 \\ i & 1 & 0 \\ 0 & 0 & 2}$$

Therefore, assuming that $t \in \mathbb{C}$ is a scalar, we get:

\begin{align} e^{At} = \frac12\pmatrix{i & -i & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1}\pmatrix{e^{3it} & 0 & 0 \\ 0 & e^{-3it} & 0 \\ 0 & 0 & e^{-t}}\pmatrix{-i & 1 & 0 \\ i & 1 & 0 \\ 0 & 0 & 2} = \pmatrix{\frac{e^{3it}+e^{-3it}}{2} & -\frac{e^{3it}-e^{-3it}}{2i} & 0 \\ \frac{e^{3it}-e^{-3it}}{2i} & \frac{e^{3it}+e^{-3it}}{2} & 0 \\ 0 & 0 & e^{-t}}\\ \end{align}

This can be written as

$$e^{At} = \pmatrix{\cos 3t & -\sin 3t & 0 \\ \sin 3t & \cos 3t & 0 \\ 0 & 0 & e^{-t}}$$

which is a real matrix if $t \in \mathbb{R}$.