% m211oh84.tex (prev.: *oh77.*) - Solving discrete linear systems
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\begin{center}\tcr{\rmH Solving Discrete Linear Systems
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\begin{description} 
\item[\tcr{Theorem 15:}]
The discrete linear system \tcg{with constant coefficients} 
\begin{equation} 
\tcb{\label{5.9d1}
\vec{u}_{n+\scr1}=A\vec{u}_n+\vec{b}_n, \;\;\; n=0,1,\dots} 
\end{equation} has the (unique) \tcr{solution} 
\begin{eqnarray}
\label{5.9d2}
\tcb{\vec{u}_n }&\tcb{=} &
\tcb{A^n\vec{u}_{\scr0}+A^{n-\scr1}\vec{b}_{\scr0}+A^{n-\scr2}\vec{b}_{\scr1}+\dots} \\
& & \tcb{\ldots + A\vec{b}_{n-\scr2}+\vec{b}_{n-\scr1}.} \nonumber
\end{eqnarray} 
In particular, if the discrete linear system 
\sref{5.9d1} is \tcg{homogeneous} (i.e.\ \mc{\vec{b}_n=\vec{0}} for all \mc{n}), then 
\begin{equation}
\tcb{\label{5.9d3}
\vec{u}_n=A^n\vec{u}_{\scr0}.}
\end{equation}

\item[\tcr{Method} \tcg{for solving discrete linear systems:}]\hspace*{10pt}\vspace{5pt}
 
\tcr{Step 1:} Use \tcg{Theorem 12} to write down the \tcg{general formula} for \mc{\vec{u}_n} in terms of  powers of \mc{A}. 
\vspace{5pt}

\tcr{Step 2:} Derive an \tcr{explicit formula} for \mc{A^n}. (Use the theory of \tcg{matrix polynomials}.)

\item[\tcr{Note:}] To solve a \tcr{difference equation}, first transform it into a \tcg{discrete linear system} and then then \tcr{solve} this system by using the above method.

\end{description}
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