Jordan Normal Form of Matrix Solve the following system of equations.
$$x_{n+1} = 2y_n − z_n$$
$$y_{n+1} = y_n$$
$$z_{n+1} = x_n − 2y_n + 2z_n$$
What is the solution in general for $x_0$, $y_0$, $z_0$ arbitrary?
It is intended to be solved using Jordan Normal Form of the Linear Algebra knowledge.
I have no idea how to start, can anyone give a hint?
Thank you!
 A: Hint. Note that for $n\geq 0$,
$$\begin{pmatrix}x_{n}\\y_{n}\\z_{n}\end{pmatrix}
=\begin{pmatrix}0&2&-1\\0&1&0\\1&-2&2\end{pmatrix}
\begin{pmatrix}x_{n-1}\\y_{n-1}\\z_{n-1}\end{pmatrix}=
\begin{pmatrix}0&2&-1\\0&1&0\\1&-2&2\end{pmatrix}^n
\begin{pmatrix}x_{0}\\y_{0}\\z_{0}\end{pmatrix}.$$
Now find the Jordan normal form $J$ of the matrix
$$M:=\begin{pmatrix}0&2&-1\\0&1&0\\1&-2&2\end{pmatrix}$$
and a matrix $P$ such that $M=PJP^{-1}$. Then $M^n=PJ^nP^{-1}$.
A: By the @Robert Z answer the jordan form of matrix $M$ is as follows:
$$
M=P\,J\,P^{-1}
$$
where the matices $P$ and $J$ are in the following form;
$$
P= \left( \begin {array}{ccc} -1&3&2\\0&1&1
\\ 1&0&0\end {array} \right)
\quad , \quad 
J=  \left( \begin {array}{ccc} 1&1&0\\ 0&1&0
\\ 0&0&1\end {array} \right)
$$
with the induction on $n$ you can prove that the $n$th power of matrix $J$ is as follows:
$$
J^n=
 \left( \begin {array}{ccc} 1&n&0\\ 0&1&0
\\ 0&0&1\end {array} \right)
$$
So, the $n$th power of matrix $M$ is in the following form:
$$
M^n=P\,J^n\,P^{-1}=
\left( \begin {array}{ccc} 1-n&2\,n&-n\\ 0&1&0
\\ n&-2\,n&1+n\end {array} \right)
$$
