
\subsection{\bf Convergence of Conjugate Gradient in Diagonalized Form}

\begin{slide}

\begin{itemize}
\item We want $\tilde{r}_k \big|_i = 0$ for all $i$

\item But
\[
\alpha_k = \frac{\sum_i \left\lbrack \left( \tilde{r}^{(k-1)}
\right) \big|_i \right\rbrack^2}{\sum_i \lambda_i \left\lbrack
\tilde{r}^{(k-1)} \big|_i \right\rbrack^2}
\]

\item We order eigenvalues $\lambda_1 \geq \lambda_2 \ge \ldots$

\item If $\tilde{r}^{(k-1)} \big|_1$ non zero and reasonably big

\item Then
\begin{eqnarray*}
\alpha_k & \sim & \frac{1}{\lambda_1} \ \mbox{and} \\
\tilde{r}^{(k)}\big|_1 & = & \left( 1 - \frac{\lambda_1}{\lambda_1}
\right) \tilde{r}^{(k-1)}\big|_1 \\
& \sim & 0
\end{eqnarray*}

\item But other $\tilde{r}_k \big|_i$\ \ $i>1$\ \ are largely
unchanged if eigenvalues $\lambda_i \ll \lambda_1$ because
as $\tilde{r}^{(k-1)}\big|_i$ is multiplied by $\left( 1 -
\lambda_i / \lambda_1 \right) \sim 1$ (when $\lambda_i \ll \lambda_1$) on
going from iteration $(k-1)$ to $(k)$
\end{itemize}

\end{slide}

