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

\begin{slide}

\begin{itemize}
\item Consider transformed matrix $\tilde{A}$ using $U$ defined
previously
\[
\tilde{A} = U^* \left\lbrack \begin{array}{cccc}
\lambda_1 \\
& \lambda_2 & 0 \\
0 \\
& & & \lambda_n \end{array} \right\rbrack\ U
\]

\begin{center}
$\uparrow$ \\
Diagonal \\
\end{center}

\item By inspection
\[
\tilde{A} | v_k > = \lambda_k | v_k >
\]
i.e., $\tilde{A}$ and $A$ give same result on each eigenvector.
\end{itemize}

\end{slide}

