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 $3\times3$

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:
$$\displaystyle e^{Pt} = e^{S J S^{-1} t} = \begin{bmatrix} 1 & -1 & 1 \\ -1 & 0 & 1\\ 1 & 0 & 1\end{bmatrix} \cdot \begin{bmatrix} e^{-1} & 0 & 0 \\ 0 & 0 & 0\\ 0 & 0 & e^{t}\end{bmatrix} \cdot \begin{bmatrix} 0 & -\frac{1}{2} & \frac{1}{2} \\ 1 & 0 & -1\\ 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix} + \begin{bmatrix} 0 & 0 & \frac{1}{2}(e^{-t}+e^{t}) \\ 0 & \frac{1}{2}(e^{-t}+e^{t}) & 0\\ 0 & 0 & \frac{1}{2}(e^{-t}+e^{t})\end{bmatrix} + \begin{bmatrix} 0 & \frac{1}{2}(-e^{-t}+e^{t}) & 0 \\ 0 & 0 & \frac{1}{2}(-e^{-t}+e^{t})\\ 0 & \frac{1}{2}(-e^{-t}+e^{t}) & 0\end{bmatrix} = (I_3 - P^2) + P \sinh( t) + P^2 \cosh(t) = \begin{bmatrix} 1 & \frac{1}{2}(-e^{-t}+e^{t}) & -1 + \frac{1}{2}(e^{-t}+e^{t}) \\ 0 & \frac{1}{2}(e^{-t}+e^{t}) & \frac{1}{2}(-e^{-t}+e^{t})\\ 0 & \frac{1}{2}(-e^{-t}+e^{t}) & \frac{1}{2}(e^{-t}+e^{t})\end{bmatrix}$$
