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!

• If you have Maple software you can first define your matrix and after that run package with(MTM) and in continue with running code jordan(matrix name) see the jordan form of your matrix. – Amin235 Jun 7 '17 at 13:06

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 $M^n=PJ^nP^{-1}$.

• Thank you so much! It really makes sense. Then should I consider the jordan form of matrix n-square since it then can be written in terms of $x_0$,$y_0$,$z_0$? – Castalia520 Jun 7 '17 at 12:49
• @Castalia520 Find the the Jordan form $J$ of $M$ and a matrix $P$ such that $M=PJP^{-1}$. – Robert Z Jun 7 '17 at 12:52
• Hint: $M = PJP^{-1}$ and $M^n = PJ^nP^{-1}$ – user392395 Jun 7 '17 at 12:52

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:

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