z-transforms of a system of coupled difference equations I have the following system of difference equations 
$$\mathbf{x}_{n+1}=A\mathbf{x}_{n},\qquad A=\frac{1}{3}\begin{bmatrix}
1&5&2\\
2&0&0\\
0&2&0
\end{bmatrix}.$$
$L$ is diagonalisable, and so via the spectral decomposition determining a general $\mathbf{x}_{n}$ is easy. Although I am required to implement z-transforms on the original coupled system, this is what I'm having issues with. Some hints on where to start with this would be great. (I am given $\mathbf{x}_0=\begin{bmatrix} 
6 & 0& 0\end{bmatrix}^T$).
Thank you
 A: We start from
$$
\left\{ \matrix{
  {\bf x}_{\,n}  = 0\quad \left| {\;n < 0} \right. \hfill \cr 
  {\bf x}_{\,0}  = {\bf x}_{\,0}  \hfill \cr 
  {\bf x}_{\,n + 1}  = {\bf x}_{\,n} \,A \hfill \cr}  \right.
$$
we put 2nd and 3rd into a single identity
$$
{\bf x}_{\,n} I = {\bf x}_{\,n - 1} \,A + [0 = n]\,{\bf x}_{\,0} I
$$
where $[P]$ denotes the Iverson bracket
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
   1 & {P = TRUE}  \\
   0 & {P = FALSE}  \\
 \end{array} } \right.
$$
Define the vectorial function of $z$
$$
{\bf F}(z) = \left( {f_{\,1} (z),f_{\,2} (z), \cdots ,f_{\,q} (z)} \right) = \sum\limits_{0\, \le \,n} {{\bf x}_{\,n} \,z^{\,n} } 
$$
then the recurrence becomes
$$
\eqalign{
  & {\bf F}(z)I = \sum\limits_{0\, \le \,n} {{\bf x}_{\,n} \,z^{\,n} I}  = \sum\limits_{0\, \le \,n} {{\bf x}_{\,n - 1} z^{\,n} } \,A + \sum\limits_{0\, \le \,n} {[0 = n]\,z^{\,n} {\bf x}_{\,0} } I =   \cr 
  &  = {\bf x}_{\, - 1} z^{\,0}  + \sum\limits_{1\, \le \,n} {{\bf x}_{\,n - 1} z^{\,n} } \,A + z^{\,0} {\bf x}_{\,0} I =   \cr 
  &  = \sum\limits_{1\, \le \,n} {{\bf x}_{\,n - 1} z^{\,n} } \,A + {\bf x}_{\,0} I =   \cr 
  &  = \sum\limits_{0\, \le \,n - 1} {{\bf x}_{\,n - 1} z^{\,n - 1 + 1} } \,A + {\bf x}_{\,0} I =   \cr 
  &  = z\sum\limits_{0\, \le \,n - 1} {{\bf x}_{\,n - 1} z^{\,n - 1} } \,A + {\bf x}_{\,0} I =   \cr 
  &  = z{\bf F}(z)\,A + {\bf x}_{\,0} I \cr} 
$$
i.e.
$$
\eqalign{
  & {\bf F}(z)I = z{\bf F}(z)\,A + {\bf x}_{\,0} I\quad  \Rightarrow \quad {\bf F}(z)\left( {I - zA} \right) = {\bf x}_{\,0} I\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad {\bf F}(z) = {\bf x}_{\,0} \left( {I - zA} \right)^{\, - 1}  \cr} 
$$
