I have the following system of differnce equations:
$ \begin{bmatrix} Y_1(t+1)\\ Y_2(t+1) \end{bmatrix} = A \begin{bmatrix} Y_1(t)\\ Y_2(t) \end{bmatrix} + \begin{bmatrix} 0\\ G_0 \end{bmatrix} $
Where
$
A=\begin{bmatrix}
0 & 1\\
bk+b & -kb
\end{bmatrix},\quad 0<b<1,k>0
$,
is a real valued constant matrix.
I know the system is autonomous, and that the solution is given by: $ Y(n)=A^{n-n_0}y_0+\sum_{r=n_0}^{n-1}A^{n-r-1}g(r) $
Here $y(n_0)=y_0$, and i know that $g(r)=\begin{bmatrix} 0\\ G_0 \end{bmatrix}$ so the solution to the given system would be given accordingly:
$ Y(n)= \begin{bmatrix} 0 & 1\\ bk+b & -kb \end{bmatrix} ^{n-n_0}y_0+\sum_{r=n_0}^{n-1} \begin{bmatrix} 0 & 1\\ bk+b & -kb \end{bmatrix} ^{n-r-1}.\begin{bmatrix} 0\\ G_0 \end{bmatrix} $
Is this a satisfying solution?
Say that $n_0=0$ which then gives:
$ Y(n)= \begin{bmatrix} 0 & 1\\ bk+b & -kb \end{bmatrix} ^{n}y_0+\sum_{r=n_0}^{n-1} \begin{bmatrix} 0 & 1\\ bk+b & -kb \end{bmatrix} ^{n-r-1}.\begin{bmatrix} 0\\ G_0 \end{bmatrix} $
Is it then possible to find $A^n$ by Putzers Algorithm, I have tried myself, but whenever i put in real values for $b,k$ my $A^n$ returns complex numbers in the anti-diagonal.
Or would it be easier to approach the solution by The Jordan Form, or does that only work with homogenous difference equations?