How to solve a discrete dynamical system

Question

I am working through an example to solve the following system but I believe there might be typing errors in the book I will type it exactly as in the book

$x(n+1)= \begin{bmatrix}2 & 1\\0 & 2 &\end{bmatrix}x(n)+\begin{bmatrix}2\\-1 &\end{bmatrix}$

Then the fixed point is given by: $x_*= \begin{bmatrix}2 & 1\\0 & 2 &\end{bmatrix}x_*+\begin{bmatrix}-2\\-1 &\end{bmatrix}$

Therefore $\begin{bmatrix} 1& 1\\0 & 1 &\end{bmatrix}x(n)+\begin{bmatrix}-2\\-1 &\end{bmatrix}=0$

and hence $x_*=\begin{bmatrix}1\\1 &\end{bmatrix}$

$X(n)=x(n)-x_*$

then $X(n+1)= \begin{bmatrix}2 & 1\\0 & 2 &\end{bmatrix}X(n)$

the eigenvalue equation is $(\lambda-2)^2=0$ with $\lambda_1=2$

let $K_1=\begin{bmatrix}1\\0 &\end{bmatrix}$

I am not going to type the rest but I am sure there are errors. can anyone refer me to a proper worked example of a similar system.

Answer 1

Correction:

$x(n+1)= \begin{bmatrix}2 & 1\\0 & 2 &\end{bmatrix}x(n)+\begin{bmatrix}2\\-1 &\end{bmatrix}\ \ \ \ $ (Eq. 1)

Then the fixed point is given by ($x(n+1)=x(n)=x_*$):

$x_*= \begin{bmatrix}2 & 1\\0 & 2 &\end{bmatrix}x_*+\begin{bmatrix}2\\-1 &\end{bmatrix}$

Therefore $\begin{bmatrix} 1& 1\\0 & 1 &\end{bmatrix}x_*=\begin{bmatrix}-2\\1 &\end{bmatrix}$

and hence $x_*=\begin{bmatrix}-3\\1 &\end{bmatrix}$

Redefine zero: $X(n)=x(n)-x_*$ (substitute in Eq. 1)

then $X(n+1)= \begin{bmatrix}2 & 1\\0 & 2 &\end{bmatrix}X(n)$

The eigenvalue equation is $(\lambda-2)^2=0$ with $\lambda_{1,2}=2$

let $K_1=\begin{bmatrix}1\\0 &\end{bmatrix}$

I do not know what the rest of the exercise is.