Let \(A = \left( {\begin{aligned}{ {20}{r}}{15}&{16}\\{ - 20}&{ - 21}\end{aligned}} \right)\). The vectors \({\bf{x}}, \ldots ,{A^5}{\bf{x}}\) are \(\left( {\begin{aligned}{ {20}{l}}1\\1\end{aligned}} \right)\),

\(\left( {\begin{aligned}{ {20}{r}}{31}\\{ - 41}\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{r}}{ - 191}\\{241}\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{r}}{991}\\{ - 1241}\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{r}}{ - 4991}\\{6241}\end{aligned}} \right),{\rm{ }}\left( {\begin{aligned}{ {20}{r}}{24991}\\{ - 31241}\end{aligned}} \right)\).

Find a vector with a 1 in the second entry that is close to an eigenvector of \(A\). Use four decimal places. Check your estimate, and give an estimate for the dominant eigenvalue of \(A\).

Eigenvectors, also known as characteristic vectors, appropriate vectors, or latent vectors, are a specific collection of vectors associated with a linear system of equations. Each eigenvector is associated with an eigenvalue.

The normalized form of \({A^5}x = \left( {\begin{aligned}{ {20}{c}}{24991}\\{ - 31241}\end{aligned}} \right)\) is:

\(\begin{aligned}{c}v = - \frac{1}{{31241}}\left( {\begin{aligned}{ {20}{c}}{24991}\\{ - 31241}\end{aligned}} \right)\\ = \left( {\begin{aligned}{ {20}{c}}{ - .7999}\\1\end{aligned}} \right)\end{aligned}\)

This is a vector with 1 in a second entry that is an approximation of eigenvector of \(A\).

To estimate the eigenvalue of \(A\), compute \(Av\):

\(\begin{aligned}{c}Av = \left( {\begin{aligned}{ {20}{c}}{15}&{16}\\{ - 20}&{ - 21}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}{ - .7999}\\1\end{aligned}} \right)\\ = \left( {\begin{aligned}{ {20}{c}}{4.0015}\\{ - 5.002}\end{aligned}} \right)\end{aligned}\)

The largest entity is 5.002. This means eigenvalue is \(\lambda = - 5.002\). The corresponding vector is \(v = \left( {\begin{aligned}{ {20}{c}}{ - .7999}\\1\end{aligned}} \right)\).