Question

In Exercises 1–4, the matrix A is followed by a sequence \(\left\{ {{{\bf{x}}_k}} \right\}\) produced by the power method. Use these data to estimate the largest eigenvalue of A, and give a corresponding eigenvector.

3. \(A = \left( {\begin{aligned}{ {20}{c}}{.5}&{.2}\\{.4}&{.7}\end{aligned}} \right)\)

\(\left( {\begin{aligned}{ {20}{c}}1\\0\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{c}}1\\{.8}\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{c}}{.6875}\\1\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{c}}{.5577}\\1\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{c}}{.5188}\\1\end{aligned}} \right)\)

Short answer

The largest eigenvalue of A is 0.9075, and the corresponding eigenvector is \(\left( {\begin{aligned}{ {20}{c}}{.5188}\\1\end{aligned}} \right)\).

Step-by-step solution

Given information

A matrix \(A = \left( {\begin{aligned}{ {20}{l}}{.5}&{.2}\\{.4}&{.7}\end{aligned}} \right)\). A sequence \(\left\{ {{x_k}} \right\}\).

Find the Eigenvalue

Compute the value of\(A{x_k}\)and identify the largest entry as follows:

\(A{{\bf{x}}_0} = \left( {\begin{aligned}{ {20}{l}}{0.5}&{0.2}\\{0.4}&{0.7}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{l}}1\\0\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{l}}{0.5}\\{0.4}\end{aligned}} \right)\)Which implies\({\mu _0} = 0.5{\rm{ }}\)

\(A{{\bf{x}}_1} = \left( {\begin{aligned}{ {20}{l}}{0.5}&{0.2}\\{0.4}&{0.7}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}1\\{0.8}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{l}}{0.66}\\{0.96}\end{aligned}} \right)\)Which implies\({\mu _1} = 0.96{\rm{ }}\)

\(A{{\bf{x}}_2} = \left( {\begin{aligned}{ {20}{l}}{0.5}&{0.2}\\{0.4}&{0.7}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}{.6875}\\1\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}{0.54375}\\{0.975}\end{aligned}} \right)\)Which implies\({\mu _2} = 0.975{\rm{ }}\)

\(A{{\bf{x}}_3} = \left( {\begin{aligned}{ {20}{l}}{0.5}&{0.2}\\{0.4}&{0.7}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}{.5577}\\1\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{l}}{0.47885}\\{0.92308}\end{aligned}} \right)\)Which implies\({\mu _3} = 0.92308\)

\(A{{\bf{x}}_4} = \left( {\begin{aligned}{ {20}{l}}{0.5}&{0.2}\\{0.4}&{0.7}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}{.5188}\\1\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}{0.4594}\\{0.90752}\end{aligned}} \right)\)Which implies\({\mu _4} = 0.90752\)

Hence, the largest absolute entry is 0.9075. So, the eigenvalue is equal to 0.9075. The corresponding eigenvector is \(\left( {\begin{aligned}{ {20}{c}}{.5188}\\1\end{aligned}} \right)\).