Question

(M) Use the inverse power method to estimate the middle eigenvalue of the \(A\) in Example 3, with accuracy to four decimal places. Set \({{\bf{x}}_{\bf{0}}}{\bf{ = }}\left( {{\bf{1,0,0}}} \right)\).

Short answer

The middle eigenvalue is \ (3.3212\ ).

Step-by-step solution

Write the function of MATLAB

Consider \(A = \left( {\begin{aligned}{{20}{c}}{10}&{ - 8}&{ - 4}\\{ - 8}&{13}&4\\{ - 4}&5&4\end{aligned}} \right)\), \({{\bf{x}}_0} = \left

( {\begin{aligned}{{20}{c}}1\\0\\0\end{aligned}} \right)\)

Now calculate the middle eigenvalue of the matrix \(A\).

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:

\(B = \left( {10{\rm{ }} - 8{\rm{ }} - 4; - 8{\rm{ }}13{\rm{ }}4;{\rm{ }} - 4{\rm{ }}5{\rm{ }}4} \right)\)

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

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

Now find the eigenvector.

IPM(B,tol)

Construct the data in the table shown below:

\(k\)	\(0\)	\(1\)	\(2\)
\({{\bf{x}}_k}\)	\(\left( {\begin{aligned}{*{20}{c}}1\\0\\0\end{aligned}} \right)\)	\(\left( {\begin{aligned}{*{20}{c}}1\\{.7873}\\{.0908}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{*{20}{c}}1\\{.7870}\\{.0957}\end{aligned}} \right)\)
\({{\bf{y}}_k}\)	\(\left( {\begin{aligned}{*{20}{c}}{26.0552}\\{20.5128}\\{2.3669}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{*{20}{c}}{47.1975}\\{37.1436}\\{4.5187}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{*{20}{c}}{47.1233}\\{37.0866}\\{4.5083}\end{aligned}} \right)\)
\({\mu _k}\)	\(26.0552\)	\(47.1975\)	\(47.1233\)
\({v_k}\)	\(3.3384\)	\(3.32119\)	\(3.3212209\)

Thus, the middle eigenvalue is \(3.3212\).