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 \({\mu _6}\).

8.\(A = \left( {\begin{aligned}{ {20}{l}}2&1\\4&5\end{aligned}} \right),{{\bf{x}}_0} = \left( {\begin{aligned}{ {20}{l}}1\\0\end{aligned}} \right)\)

Short answer

The values are shown below:

\(\begin{aligned}{l}{\mu _1} = 4\\{\mu _2} = 7\\{\mu _3} = 6.1429\\{\mu _4} = 6.0233\\{\mu _5} = 6.0039\end{aligned}\)

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.

Consider \({x_0} = \left( {\begin{aligned}{ {20}{l}}1\\0\end{aligned}} \right)\) and \(A = \left( {\begin{aligned}{ {20}{l}}2&1\\4&5\end{aligned}} \right)\).

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 \({\rm{k}} = 1:5\);

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

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

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

\(x = \left( {{1 \mathord{\left/

{\vphantom {1 {mu}}} \right.

\kern-\nulldelimiterspace} {mu}}} \right) y\)

end

Note that we want to list \({x_k}\) and \({\mu _k}\) for each \(k = 1, \ldots ,5\), so sign ; is omitted from end of the command row where \(\mu \) and \(x\) are calculated.

List of \({\mu _k}\) and \({x_k}\) is:

\(\begin{aligned}{c}{x_1} = \left( {\begin{aligned}{ {20}{c}}{0.5}\\1\end{aligned}} \right)\\{x_2} = \left( {\begin{aligned}{ {20}{c}}{0.2857}\\1\end{aligned}} \right)\\{x_3} = \left( {\begin{aligned}{ {20}{c}}{0.2558}\\1\end{aligned}} \right)\\{x_4} = \left( {\begin{aligned}{ {20}{c}}{0.251}\\1\end{aligned}} \right)\\{x_5} = \left( {\begin{aligned}{ {20}{c}}{0.2502}\\1\end{aligned}} \right)\end{aligned}\)

And,

\(\begin{aligned}{l}{\mu _1} = 4\\{\mu _2} = 7\\{\mu _3} = 6.1429\\{\mu _4} = 6.0233\\{\mu _5} = 6.0039\end{aligned}\)

Thus, the values are listed as shown below:

\(\begin{aligned}{l}{\mu _1} = 4\\{\mu _2} = 7\\{\mu _3} = 6.1429\\{\mu _4} = 6.0233\\{\mu _5} = 6.0039\end{aligned}\)