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.

2. \(A = \left( {\begin{aligned}{ {20}{c}}{1.8}&{ - .8}\\{ - 3.2}&{4.2}\end{aligned}} \right]\)

\(\left( {\begin{aligned}{ {20}{c}}1\\0\end{aligned}} \right],{\rm{ }}\left( {\begin{aligned}{ {20}{c}}{ - .5625}\\1\end{aligned}} \right],{\rm{ }}\left( {\begin{aligned}{ {20}{c}}{ - .3021}\\1\end{aligned}} \right],{\rm{ }}\left( {\begin{aligned}{ {20}{c}}{ - .2601}\\1\end{aligned}} \right],{\rm{ }}\left( {\begin{aligned}{ {20}{c}}{ - .2520}\\1\end{aligned}} \right]\)

Short answer

The largest eigenvalue of A is 5.0064, and the corresponding eigenvector is \(\left( {\begin{aligned}{ {20}{c}}{ - .2560}\\1\end{aligned}} \right]\).

Step-by-step solution

Given information

A matrix \(A = \left( {\begin{aligned}{ {20}{l}}{1.8}&{ - .8}\\{ - 3.2}&{4.2}\end{aligned}} \right]\). A sequence \(\left\{ {{x_k}} \right\}\).

Find the Eigenvalue

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

\(A{x_0} = \left( {\begin{aligned}{ {20}{c}}{1.8}&{ - .8}\\{ - 3.2}&{4.2}\end{aligned}} \right]\left( {\begin{aligned}{ {20}{l}}1\\0\end{aligned}} \right] = \left( {\begin{aligned}{ {20}{c}}{1.8}\\{ - 3.2}\end{aligned}} \right]\),\({\mu _0} = - 3.2\)

\(A{x_1} = \left( {\begin{aligned}{ {20}{c}}{1.8}&{ - .8}\\{ - 3.2}&{4.2}\end{aligned}} \right]\left( {\begin{aligned}{ {20}{c}}{ - .5625}\\1\end{aligned}} \right] = \left( {\begin{aligned}{ {20}{c}}{ - 1.8125}\\{4.7625}\end{aligned}} \right]\),\({\mu _1} = 4.7625\)

\(A{x_2} = \left( {\begin{aligned}{ {20}{c}}{1.8}&{ - .8}\\{ - 3.2}&{4.2}\end{aligned}} \right]\left( {\begin{aligned}{ {20}{c}}{ - .3021}\\1\end{aligned}} \right] = \left( {\begin{aligned}{ {20}{c}}{ - 1.34378}\\{5.16672}\end{aligned}} \right]\),\({\mu _2} = 5.16672\)

\(A{x_3} = \left( {\begin{aligned}{ {20}{c}}{1.8}&{ - .8}\\{ - 3.2}&{4.2}\end{aligned}} \right]\left( {\begin{aligned}{ {20}{c}}{ - .2601}\\1\end{aligned}} \right] = \left( {\begin{aligned}{ {20}{c}}{ - 1.26818}\\{5.03232}\end{aligned}} \right]\),\({\mu _3} = 5.03232\)

\(A{x_3} = \left( {\begin{aligned}{ {20}{c}}{1.8}&{ - .8}\\{ - 3.2}&{4.2}\end{aligned}} \right]\left( {\begin{aligned}{ {20}{c}}{ - .2520}\\1\end{aligned}} \right] = \left( {\begin{aligned}{ {20}{c}}{ - 1.2536}\\{5.0064}\end{aligned}} \right]\),\({\mu _4} = 5.0064\)

Hence, the largest absolute entry is 5.0064. So, the eigenvalue is equal to 4.9978. The corresponding eigenvector is \(\left( {\begin{aligned}{ {20}{c}}{ - .2520}\\1\end{aligned}} \right]\).