Matrix Exponential using the Cayley-Hamilton theorem For the matrix $$P=\left( \begin{matrix} 0&1&0 \\ 0&0&1 \\ 0&1&0 \end{matrix} \right)$$ how do you find $e^{Pt}$ using the Cayley-Hamilton theorem?
I have found it by diagonalising $P$, but the question states to use the Cayley-Hamilton theorem.

If you calculate the characteristic polynomial $\chi_A(x)=x^3-x$, you know from Cayley-Hamilton theorem that $P^3-P=0$, i.e. $P^3=P$.
This implies that $P=P^3=P^5=\dots$ and $P^2=P^4=P^6=\dots$. Hence
$$
\begin{align}
e^{Pt} 
&= I+Pt + \frac{P^2t^2}2 + \frac{P^3t^3}{3!} + \dots = \\
&= I + P\left(t+\frac{t^3}{3!}+\frac{t^5}{5!}+\dots\right) + P^2\left(\frac{t^2}2+\frac{t^4}{4!}+\frac{t^6}{6!}+\dots\right) = \\
&= I+ P \frac{e^t-e^{-t}}2 + P^2 \left(\frac{e^t+e^{-t}}2-1\right) = \\
&= I+ P\sinh t + P^2(\cosh t-1)
\end{align}
$$
Using
$P=\begin{pmatrix}0&1&0\\0&0&1\\0&1&0\end{pmatrix}$
and
$P^2=\begin{pmatrix}0&0&1\\0&1&0\\0&0&1\end{pmatrix}$
I get
$$e^{Pt}=
\begin{pmatrix}
1&\sinh t&\cosh t-1\\
0&\cosh t&\sinh t\\
0&\sinh t&\cosh t
\end{pmatrix}.$$
If I did not miss something, this seems to be the same result as WolframAlpha returns. (Up to some algebraic manipulation.)
