
\subsection{\bf Diagonalization of Quadratic Form}

\begin{slide}

\begin{itemize}
\item Note that for any vector $\underline{x}$
\begin{eqnarray*}
&& \frac{1}{2} \underline{x}^T A \underline{x} - \underline{b}^T
\cdot \underline{x} \\
&& \\
& = & \frac{1}{2} \underbrace{\normalsize\underline{x}^T P^{-1}}_{x^T P^T}
\Lambda P\underline{x} - \underbrace{\underline{b}^T
P^{-1}}_{\normalsize(P\underline{b})^T} P\underline{x} \\
&& \\
& = & \frac{1}{2} ( P\underline{x} )^T \Lambda P\underline{x} - (
P\underline{b} )^T P\underline{x} \\
&& \\
& = & \frac{1}{2} \underline{\tilde{x}}^T \Lambda
\underline{\tilde{x}} - \tilde{b}^T \underline{\tilde{x}}
\end{eqnarray*}

\item Note this transformation is correct but not useful as a
solution technique as it takes longer to find $\Lambda$ and $P$ than
to solve equations $A\underline{x} = \underline{b}$.

\item In transformed space denoted by $\widetilde{}$ superscript
\begin{center}
\begin{tabular}{|ccc|}
\hline
&& \\
$\tilde{x}\big|_{i\ {\rm compt}}$ & = & $\tilde{b}_i / \lambda_i$ \\
&& \\
\hline
\end{tabular}
\end{center}
\end{itemize}

\end{slide}

