Question

(M) Let \(A\) be as in Exercise 9. Use the inverse power method with \({{\bf{x}}_{\bf{0}}}{\bf{ = }}\left( {{\bf{1,0,0}}} \right)\) to estimate the eigenvalue of \(A\) near \(\alpha {\bf{ = - 1}}{\bf{.4}}\), with an accuracy to four decimal places.

Short answer

The middle eigenvalue is \( - 1.4244\).

Step-by-step solution

Write the function of MATLAB

Consider \(A = \left( {\begin{aligned}{{20}{c}}8&0&{12}\\1&{ - 2}&1\\0&3&0\end{aligned}} \right)\), \({{\bf{x}}_0} = \left( {\begin{aligned}{{20}{c}}1\\0\\0\end{aligned}} \right)\)

Now estimate the middle eigenvalue of the matrix \(A\) near about \( - 1.4\).

Write the required function of MATLAB

\({\rm{function}}\left( {{\rm{v,lambda}}} \right){\rm{ = IPM}}\left( {{\rm{B,tol}}} \right)\)

tic;

\({\rm{A}} = {\rm{inv}}\left( {\rm{B}} \right){\rm{;}}\)

\({\rm{n}} = {\rm{size}}\left( {{\rm{A}},1} \right);\)

\({\rm{v}} = {\rm{rand}}\left( {{\rm{n}},1} \right);\)

\({\rm{v}} = {\rm{v}}/{\rm{norm}}\left( {\rm{v}} \right);\)

\({\rm{res}} = 1;\)

\({\rm{while}}\left( {{\rm{rse}} > {\rm{tol}}} \right)\)

\({\rm{W}} = {\rm{A}}*{\rm{v}};\)

\({\rm{lambda}} = {\rm{max}}\left( {{\rm{abs}}\left( {\rm{W}} \right)} \right);\)

\({\rm{V}} = {\rm{W}}/{\rm{lamda}};\)

\({\rm{res}} = {\rm{norm}}\left( {{\rm{A}}*{\rm{v}} - {\rm{lambda}}*{\rm{v}}} \right);\)

toc

end

Find the middle eigenvalue

Enter the Matrix\(B\)in MATLAB:

\(B = \left( {\begin{aligned}{ {20}{c}}{8\;2\;12;}&{1\; - 2\;1;}&{0\;3\;0}\end{aligned}} \right)\)

Enter the\({x_0}\)in MATLAB:

\({x_0} = \left( {\begin{aligned}{ {20}{c}}1&0&0\end{aligned}} \right)'\)

Now find the eigenvector.

\({\rm{IPM}}\left( {{\rm{B,tol}}} \right)\)

Construct the data in the table shown below:

\(k\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)
\({{\bf{x}}_k}\)	\(\left( {\begin{aligned}{ {20}{c}}1\\0\\0\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}1\\{.3646}\\{ - .7813}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}1\\{.3734}\\{ - .7854}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}1\\{.3729}\\{ - .7854}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}1\\{.3734}\\{ - .7854}\end{aligned}} \right)\)
\({{\bf{y}}_k}\)	\(\left( {\begin{aligned}{ {20}{c}}{40}\\{14.5833}\\{ - 31.25}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}{ - 38.125}\\{ - 14.2361}\\{29.9479}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}{ - 41.1134}\\{ - 15.3300}\\{32.2888}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}{ - 40.9243}\\{ - 15.2608}\\{32.1407}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}{ - 40.9358}\\{ - 15.2650}\\{32.1497}\end{aligned}} \right)\)
\({\mu _k}\)	\(40\)	\( - 38.125\)	\( - 41.1134\)	\( - 40.9243\)	\( - 40.9358\)
\({v_k}\)	\( - 1.375\)	\( - 1.42623\)	\( - 1.42432\)	\( - 1.42444\)	\( - 1.42443\)

Thus, the middle eigenvalue is \( - 1.4244\).