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(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

Expert verified

The middle eigenvalue is \ (3.3212\ ).

Step by step solution

01

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

02

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 \(3.3212\).

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Most popular questions from this chapter

Exercises 19–23 concern the polynomial \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + ... + {a_{n - {\bf{1}}}}{t^{n - {\bf{1}}}} + {t^n}\) and \(n \times n\) matrix \({C_p}\) called the companion matrix of \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

21. Use mathematical induction to prove that for \(n \ge {\bf{2}}\),\(\begin{aligned}{c}det\left( {{C_p} - \lambda I} \right) = {\left( { - {\bf{1}}} \right)^n}\left( {{a_{\bf{0}}} + {a_{\bf{1}}}\lambda + ... + {a_{n - {\bf{1}}}}{\lambda ^{n - {\bf{1}}}} + {\lambda ^n}} \right)\\ = {\left( { - {\bf{1}}} \right)^n}p\left( \lambda \right)\end{aligned}\)

(Hint: Expanding by cofactors down the first column, show that \(det\left( {{C_p} - \lambda I} \right)\) has the form \(\left( { - \lambda B} \right) + {\left( { - {\bf{1}}} \right)^n}{a_{\bf{0}}}\) where \(B\) is a certain polynomial (by the induction assumption).)

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

2. \({\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,\,\,{\mathop{\rm and}\nolimits} \,\,\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}\)

Question: In Exercises \({\bf{5}}\) and \({\bf{6}}\), the matrix \(A\) is factored in the form \(PD{P^{ - {\bf{1}}}}\). Use the Diagonalization Theorem to find the eigenvalues of \(A\) and a basis for each eigenspace.

6. \(\left( {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{0}}&{{\bf{ - 2}}}\\{\bf{2}}&{\bf{5}}&{\bf{4}}\\{\bf{0}}&{\bf{0}}&{\bf{5}}\end{array}} \right){\bf{ = }}\left( {\begin{array}{*{20}{c}}{{\bf{ - 2}}}&{\bf{0}}&{{\bf{ - 1}}}\\{\bf{0}}&{\bf{1}}&{\bf{2}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{5}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\bf{0}}&{\bf{0}}&{\bf{1}}\\{\bf{2}}&{\bf{1}}&{\bf{4}}\\{{\bf{ - 1}}}&{\bf{0}}&{{\bf{ - 2}}}\end{array}} \right)\)

Assume the mapping\(T:{{\rm P}_2} \to {{\rm P}_{\bf{2}}}\)defined by \(T\left( {{a_0} + {a_1}t + {a_2}{t^2}} \right) = 3{a_0} + \left( {5{a_0} - 2{a_1}} \right)t + \left( {4{a_1} + {a_2}} \right){t^2}\) is linear. Find the matrix representation of\(T\) relative to the bases \(B = \left\{ {1,t,{t^2}} \right\}\).

Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

3. \(\left[ {\begin{array}{*{20}{c}}3&-2\\1&-1\end{array}} \right]\)

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