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!
On
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:
$$ M^n=P\,J^n\,P^{-1}= \left( \begin {array}{ccc} 1-n&2\,n&-n\\ 0&1&0 \\ n&-2\,n&1+n\end {array} \right) $$
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 $M^n=PJ^nP^{-1}$.