Question

(M) Exercises 7-12 require MATLAB or other computational aid. In Exercises 7 and 8, use the power method with the \({{\bf{x}}_0}\) given. List \(\left\{ {{{\bf{x}}_k}} \right\}\) and \(\left\{ {{\mu _k}} \right\}\) for \(k = 1, \ldots .5.\) In Exercises 9 and 10, list \({\mu _5}\) and \({\mu _6}\).

9.\(A = \left( {\begin{aligned}{ {20}{r}}8&0&{12}\\1&{ - 2}&1\\0&3&0\end{aligned}} \right),{\rm{ }}{{\bf{x}}_0} = \left( {\begin{aligned}{ {20}{l}}1\\0\\0\end{aligned}} \right)\)

Short answer

The values are:

\(\begin{aligned}{l}{\mu _5} = 8.4233,\\{\mu _6} = 8.4246\end{aligned}\)

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

Use the power method for estimating a strictly dominant eigenvalue.

\({x_0} = \left( {\begin{aligned}{ {20}{l}}1\\0\\0\end{aligned}} \right)\) and \(A = \left( {\begin{aligned}{ {20}{c}}8&0&{12}\\1&{ - 2}&1\\0&3&0\end{aligned}} \right)\)

In MATLAB define \(x\) and \(A\), and use the given loop, which is based on the power method for estimating a strictly dominant eigenvalue:

For \({\rm{k}} = 0:6\)

\({\rm{y}} = {\rm{Ax}}\);

\(\left( {\max y,index} \right) = \max \left( {abs\left( y \right)} \right)\);

\(mu = \max ysign\left( {y\left( {index} \right)} \right)\)

x =(1/mu)*y

Note that we want to list \({\mu _k}\) for each \(k = 5\) and \(k = 6\), so sign ; is omitted from the end of the command row where \(\mu \) and \(x\) are calculated.

Therefore, \({\mu _5} = 8.4233\) and \({\mu _6} = 8.4246\).