Question

Let \(A = \left( {\begin{aligned}{ {20}{r}}{ - 2}&{ - 3}\\6&7\end{aligned}} \right)\). Repeat Exercise 5, using the following sequence \({\bf{x}},A{\bf{x}}, \ldots ,{A^5}{\bf{x}}\).

\(\left( {\begin{aligned}{ {20}{l}}1\\1\end{aligned}} \right),\left( {\begin{aligned}{ {20}{r}}{ - 5}\\{13}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{r}}{ - 29}\\{61}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{r}}{ - 125}\\{253}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{r}}{ - 509}\\{1021}\end{aligned}} \right),\left( {\begin{aligned}{ {20}{r}}{ - 2045}\\{4093}\end{aligned}} \right)\)

Short answer

The value is \(\lambda = 4.0024\).

Step-by-step solution

Definition of Eigenvector

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.

Find the Eigenvalue

The normalized form of \({A^5}x = \left( {\begin{aligned}{ {20}{c}}{ - 2045}\\{4093}\end{aligned}} \right)\) is:

\(\begin{aligned}{c}v = \frac{1}{{4093}}\left( {\begin{aligned}{ {20}{c}}{ - 2045}\\{4093}\end{aligned}} \right)\\ = \left( {\begin{aligned}{ {20}{c}}{ - .4996}\\1\end{aligned}} \right)\end{aligned}\)

Now, 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 \(Av\) :

\(\begin{aligned}{c}Av = \left( {\begin{aligned}{ {20}{c}}{ - 2}&{ - 3}\\6&7\end{aligned}} \right)\left( {\begin{aligned}{ {20}{c}}{ - .4996}\\1\end{aligned}} \right)\\ = \left( {\begin{aligned}{ {20}{c}}{ - 2.0008}\\{4.0024}\end{aligned}} \right)\end{aligned}\)

The largest entity is 4.0024. This means eigenvalue is \(\lambda = 4.0024\). The corresponding vector is \(v = \left( {\begin{aligned}{ {20}{c}}{ - .4996}\\1\end{aligned}} \right)\).