
\subsection{\bf Diagonal Form of General}
\subsection{\bf Hermitean Matrix---II}

\begin{slide}

\begin{itemize}
\item But ANY vector
\[
| x > = \sum_{k=1}^n x_k | v_k >
\]
as $| v_k >$ must be complete as ORTHOnormal and $n$ of them

\item Then
\begin{eqnarray*}
\tilde{A} | x > & = & \sum_{k=1}^n x_k \tilde{A} | v_k > \\
& = & \sum_{k=1}^n x_k A | v_k > \\
& = & A | x > \\
\mbox{and\ we\ have\ proven}\ \tilde{A} & \equiv & A
\end{eqnarray*}
\item Thus, 
\[
A = U^+\ \mbox{Diag.}\  \left\lbrack \begin{array}{ccc}
\lambda_1 \\
& \ddots \\
&& \lambda_n \end{array} \right\rbrack U
\]
\end{itemize}

\end{slide}

