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\begin{center}\tcr{\rmH Difference Equations as \vspace{10pt} \\Discrete Linear Systems} \end{center} %\vspace{0pt}
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\begin{description}  

\item[\tcr{Definition:}]\hspace*{5pt} A \tcr{difference equation of order} \mc{k} is a sequence of equations of the form
\end{description}
\beq
\begin{array}[t]{r}
\tcb{a_{\scr n+k} = c_{\scr0}a_{\scr n} + c_{\scr1}a_{\scr n+1}
+ \ldots + c_{\scr k-1}a_{\scr n+k-1} + b_{\scr n},}\\ \tcb{n\ge0,}
\end{array} \vspace{-12pt} \label{1}
\eeq 
\begin{description}
\item[\hspace*{22pt}] where \mc{c_{\scr0}, c_{\scr1}, \ldots, c_{\scr k-1}} and \mc{b_{\scr0}, b_{\scr1},\ldots,b_{\scr n},\ldots} are \tcg{given} and we want to \tcg{solve} for the \mc{a_{\scr n}}'s.

\item[\tcr{Theorem 14:}] Given the \tcg{difference equation} \sref{1}, put
$$
\tcb{\vec{u}_{\scr n} = 
\left(\begin{array}{c}
a_{\scr n}\\ a_{\scr n+1} \\ \vdots \\ a_{\scr n+k-1}
\end{array}\right),\quad 
A = \left(\begin{array}{ccccc}
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots &  & \ddts & \ddts & \vdots \\
0 & 0 & \cdots & 0 & 1 \\
c_{\scr0} & c_{\scr1} & \cdots & c_{\scr k-2} & c_{\scr k-1}
\end{array}\right), %\vec{b}_{\scr n} =
} $$
and \mc{\vec{b}_n = (0, \ldots, 0, b_n)^t}. Then we have the \tcg{discrete linear system}
\beq
\tcb{\vec{u}_{\scr n+1} = A\vec{u}_{\scr n} + \vec{b}_{\scr n}}.
\label{2}
\eeq 
\item[\tcr{Remark:}] The above matrix \mc{A} has \tcg{characteristic polynomial}\vspace{-7pt}
$$
\tcb{\mbox{ch}_{\scr A}(t) = t^{\scr k} - c_{\scr k-1}t^{\scr k-1} - \ldots
-c_{\scr1}t - c_{\scr0}.}
$$   
\end{description}
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