Diagonalise a matrix and show the formula I have diagonlised P to get
$$P=\left(\begin{matrix}
-1 &0 &0\\
0 &0 &0\\
0 &0 &1
\end{matrix}\right)$$
however am unsure on how to proceed, would appreciate any help!
By diagonalising P by a transformation of similarity, show that
$$    e^{Pt} = (I_3 - P^2) + P \sinh( t) + P^2 \cosh(t)$$
when 
$P$ is the matrix
$$P=\left(\begin{matrix}
0 &1 &0\\
0 &0 &1\\
0 &1 &0
\end{matrix}\right)$$
and where $I_3$ is the identity matrix of a $3\times3$
 A: Edit: You do not need to diagonalize, actually. Simply compute and observe that
$$
P^3=P.
$$
To figure this out, of course, one might need to compute the eigenvalues: $-1,0,1$. So actually, $P^3-P=0$ is nothing but Cayley-Hamilton.
Note that
$$
P^0=I_3\qquad P^{2n}=P^2\qquad P^{2n-1}=P\qquad\forall n\geq 1.
$$
So
$$
e^{tP}=\sum_{n\geq 0}\frac{t^nP^n}{n!}=I_3+\sum_{n\geq 1}\frac{t^{2n}}{(2n)!}P^2+\sum_{n\geq 1}\frac{t^{2n+1}}{(2n+1)!}P
$$
$$
=I_3 +(\cosh t-1)P^2+\sinh t P=I_3-P^2+\sinh tP+\cosh tP^2.
$$
Note: if you do absolutely need to diagonalize, go ahead and write
$$
P=AQA^{-1}
$$
where $Q$ is the diagonalized form you found.
Then all the previous argument holds for $Q$ and you find
$$
e^{tQ}=I_3 +(\cosh t-1)Q^2+\sinh t Q=I_3-Q^2+\sinh tQ+\cosh tQ^2.
$$
It only remains to see that
$$
AQ^nA^{-1}=(AQA^{-1})^n=P^n\qquad\forall n\geq 0
$$
to deduce that
$$
e^{tP}=Ae^{tQ}A^{-1}=I_3 +(\cosh t-1)P^2+\sinh t P=I_3-P^2+\sinh tP+\cosh tP^2.
$$
A: Given:
$$P=\left(\begin{matrix}
-1 &0 &0\\
0 &0 &0\\
0 &0 &1
\end{matrix}\right)$$
By diagonalising P by a transformation of similarity, show that
$$    e^{Pt} = (I_3 - P^2) + P \sinh( t) + P^2 \cosh(t)$$
If we diagonalize the matrix $P$, we arrive at:
$$P = S J S^{-1} = \begin{bmatrix} 1 & 1 & 1 \\ -1 & 0 & 1\\ 1 & 0 & 1\end{bmatrix} \cdot \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 0 & 1\end{bmatrix} \cdot \begin{bmatrix} 0 & -\frac{1}{2} & \frac{1}{2} \\ 1 & 0 & -1\\ 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix}$$
Note: $\cosh t = \frac{1}{2}(e^{-t} + e^{t}) ~ \text{and} ~ \sinh t = \frac{1}{2}(-e^{-t} + e^{t}) ~\text{and}~ P^{2} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$.
Using this diagonalization, the matrix exponential is given by:
$$\displaystyle e^{Pt} = e^{S J S^{-1} t} = \begin{bmatrix} 1 & -1 & 1 \\ -1 & 0 & 1\\ 1 & 0 & 1\end{bmatrix} \cdot \begin{bmatrix} e^{-1} & 0 & 0 \\ 0 & 0 & 0\\ 0 & 0 & e^{t}\end{bmatrix} \cdot \begin{bmatrix} 0 & -\frac{1}{2} & \frac{1}{2} \\ 1 & 0 & -1\\ 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix} + \begin{bmatrix} 0 & 0 & \frac{1}{2}(e^{-t}+e^{t}) \\ 0 & \frac{1}{2}(e^{-t}+e^{t}) & 0\\ 0 & 0 & \frac{1}{2}(e^{-t}+e^{t})\end{bmatrix} + \begin{bmatrix} 0 & \frac{1}{2}(-e^{-t}+e^{t}) & 0 \\ 0 & 0 & \frac{1}{2}(-e^{-t}+e^{t})\\ 0 & \frac{1}{2}(-e^{-t}+e^{t}) & 0\end{bmatrix} = (I_3 - P^2) + P \sinh( t) + P^2 \cosh(t) = \begin{bmatrix} 1 & \frac{1}{2}(-e^{-t}+e^{t}) & -1 + \frac{1}{2}(e^{-t}+e^{t}) \\ 0 & \frac{1}{2}(e^{-t}+e^{t}) & \frac{1}{2}(-e^{-t}+e^{t})\\ 0 & \frac{1}{2}(-e^{-t}+e^{t}) & \frac{1}{2}(e^{-t}+e^{t})\end{bmatrix}$$
