Numerical Analysis/Inverse iteration exercises

Consider the matrix, $$ A=\left[\begin{array}{c c c}6 & 2 & -1 \\2 & 5 & 1 \\-1 & 1 & 4 \end{array} \right] $$, and the vector, $$ \textbf{x}^{(0)} = \left[\begin{array}{c}1 \\1 \\ 1 \\\end{array} \right]. $$

The LU decomposition of A is $$ L=\left[\begin{array}{c c c} 1&0&0\\0.3333&1&0\\-0.167&0.3077&1\end{array}\right], U=\left[\begin{array}{c c c}6&2&-1\\0&4.333&1.333\\0&0&3.423\end{array}\right]$$.

By hand, use the LU decomposition to do two iterations of the inverse power method (without shift) starting with $$ \textbf{x}^{(0)}. $$

Solution, first iteration: $$ LU\textbf{y}^{(1)}=\textbf{x}^{(0)} \Rightarrow \textbf{y}^{(1)}=\left[\begin{array}{c}0.1910\\0.0674\\0.2809\end{array}\right]$$. The estimated eigenvalue is $$ \mu_1 = \frac{1}{0.2809}=3.56 $$ and the estimated eigenvector is $$ \textbf{x}^{(1)}= \frac{\textbf{y}^{(1)}}{0.2809} = \left[ \begin{array}{c} 0.6800 \\ 0.2400 \\ 1.0000 \end{array} \right] $$.

Solution, second iteration: $$ LU\textbf{y}^{(2)}=\textbf{x}^{(1)} \Rightarrow \textbf{y}^{(2)}=\left[\begin{array}{c}0.1996\\-0.0966\\0.3240\end{array}\right]$$. The estimated eigenvalue is $$ \mu_2 = \frac{1}{0.3240}=3.0864 $$ and the estimated eigenvector is $$ \textbf{x}^{(2)} = \frac{\textbf{y}^{(2)}}{0.3240} = \left[ \begin{array}{c} 0.6158 \\ -0.2982 \\ 1.0000 \end{array} \right] $$.

Use a computational software package to do 50 iterations. What was the result? Solution, 50 iterations: The estimated eigenvalue after 50 iterations is $$ \mu_{50} = \frac{1}{0.4375}=2.2855 $$ and the estimated eigenvector is $$ \textbf{x}^{(50)} = \left[ \begin{array}{c} 0.7750 \\ -0.9394 \\ 1.0000 \end{array} \right] $$.