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Chapter 7: Q7.4-26E (page 395)

(M) Compute an SVD of each matrix in Exercises 26 and 27. Report the final matrix entries accurate to two decimal places. Use the method of Examples 3 and 4.

26. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{ - {\bf{18}}}&{{\bf{13}}}&{ - {\bf{4}}}&{\bf{4}}\\{\bf{2}}&{{\bf{19}}}&{ - {\bf{4}}}&{{\bf{12}}}\\{ - {\bf{14}}}&{{\bf{11}}}&{ - {\bf{12}}}&{\bf{8}}\\{ - {\bf{2}}}&{{\bf{21}}}&{\bf{4}}&{\bf{8}}\end{array}} \right)\)

Short Answer

Expert verified

The singular value decomposition of \(A\) is:\(A = \left( {\begin{array}{*{20}{c}}{.5}&{ - .5}&{ - .5}&{ - .5}\\{.5}&{.5}&{.5}&{ - .5}\\{.5}&{ - .5}&{.5}&{.5}\\{.5}&{.5}&{ - .5}&{.5}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{40}&0&0&0\\0&{20}&0&0\\0&0&{10}&0\\0&0&0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - .4}&{.8}&{ - .2}&{.4}\\{.8}&{.4}&{.4}&{.2}\\{.4}&{ - .2}&{ - .8}&{.4}\\{ - .2}&{ - .4}&{.4}&{.8}\end{array}} \right)\)

Step by step solution

01

Write the matrix 

Find the transpose of the matrix\(A\)by using MATLAB commands.

\( > > B = A';\)

\(B = \left( {\begin{array}{*{20}{c}}{ - 18}&2&{ - 14}&{ - 2}\\{13}&{19}&{11}&{21}\\{ - 4}&{ - 4}&{ - 12}&4\\4&{12}&8&8\end{array}} \right)\)

Find the product of \(A\) and \(B\).

\( > > {\rm{ }}C = B*A;\)

\(C = \left( {\begin{array}{*{20}{c}}{528}&{ - 392}&{224}&{ - 176}\\{ - 392}&{1092}&{ - 176}&{536}\\{224}&{ - 176}&{192}&{ - 128}\\{ - 176}&{536}&{ - 128}&{288}\end{array}} \right)\)

02

Find eigenvalues and eigenvectors of the matrix \(AB\)

Find the eigenvectors of the matrix \(AB\).

\( > > {\rm{ }}\left( {V{\rm{ }}B} \right) = eigs\left( C \right);\)

\(V = \left( {\begin{array}{*{20}{c}}{ - \frac{2}{5}}&{\frac{4}{5}}&{\frac{2}{5}}&{ - \frac{1}{5}}\\{\frac{4}{5}}&{\frac{2}{5}}&{ - \frac{1}{5}}&{ - \frac{2}{5}}\\{ - \frac{1}{5}}&{\frac{2}{5}}&{ - \frac{4}{5}}&{\frac{2}{5}}\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{2}{5}}&{\frac{2}{5}}\end{array}} \right)\)

\(\begin{array}{l} > > {\rm{ }}SIGMA = {\rm{ }}diag\left( {sqrt\left( E \right)} \right);\\ > > {\rm{ }}SS = sqrt\left( E \right);\end{array}\)

Then we have\(\sum \)the matrix.

\(\sum = \left( {\begin{array}{*{20}{c}}{40}&0&0&0\\0&{20}&0&0\\0&0&{10}&0\\0&0&0&0\end{array}} \right)\)

03

Find the singular value decomposition of\(A\)

Write the vectors of the\(U\)matrix:

\(\begin{array}{l} > > {\rm{ }}u1{\rm{ }} = {\rm{ }}\left( {1/SS\left( 1 \right)} \right)*A*V\left( {:,1} \right);\\ > > {\rm{ }}u2{\rm{ }} = {\rm{ }}\left( {1/SS\left( 2 \right)} \right)*A*V\left( {:,2} \right);\\ > > {\rm{ }}u3{\rm{ }} = {\rm{ }}\left( {1/SS\left( 3 \right)} \right)*A*V\left( {:,3} \right);\\ > > {\rm{ }}u4{\rm{ }} = {\rm{ }}(1/SS\left( 4 \right)*A*V\left( {:,4} \right);\\ > > {\rm{ }}U = \left( {u1;u2;u3;u4} \right);\end{array}\)

\(U = \left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}\\{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\end{array}} \right)\)

Therefore, the singular value decomposition of\(A\)is:

\(\begin{array}{c}A = U\sum {V^T}\\ = \left( {\begin{array}{*{20}{c}}{\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}\\{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{40}&0&0&0\\0&{20}&0&0\\0&0&{10}&0\\0&0&0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - \frac{2}{5}}&{\frac{4}{5}}&{\frac{2}{5}}&{ - \frac{1}{5}}\\{\frac{4}{5}}&{\frac{2}{5}}&{ - \frac{1}{5}}&{ - \frac{2}{5}}\\{ - \frac{1}{5}}&{\frac{2}{5}}&{ - \frac{4}{5}}&{\frac{2}{5}}\\{\frac{2}{5}}&{\frac{1}{5}}&{\frac{2}{5}}&{\frac{2}{5}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{.5}&{ - .5}&{ - .5}&{ - .5}\\{.5}&{.5}&{.5}&{ - .5}\\{.5}&{ - .5}&{.5}&{.5}\\{.5}&{.5}&{ - .5}&{.5}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{40}&0&0&0\\0&{20}&0&0\\0&0&{10}&0\\0&0&0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - .4}&{.8}&{ - .2}&{.4}\\{.8}&{.4}&{.4}&{.2}\\{.4}&{ - .2}&{ - .8}&{.4}\\{ - .2}&{ - .4}&{.4}&{.8}\end{array}} \right)\end{array}\)

Thus, \(A = \left( {\begin{array}{*{20}{c}}{.5}&{ - .5}&{ - .5}&{ - .5}\\{.5}&{.5}&{.5}&{ - .5}\\{.5}&{ - .5}&{.5}&{.5}\\{.5}&{.5}&{ - .5}&{.5}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{40}&0&0&0\\0&{20}&0&0\\0&0&{10}&0\\0&0&0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - .4}&{.8}&{ - .2}&{.4}\\{.8}&{.4}&{.4}&{.2}\\{.4}&{ - .2}&{ - .8}&{.4}\\{ - .2}&{ - .4}&{.4}&{.8}\end{array}} \right)\).

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Most popular questions from this chapter

Orthogonally diagonalize the matrices in Exercises 13–22, giving an orthogonal matrix \(P\) and a diagonal matrix \(D\). To save you time, the eigenvalues in Exercises 17–22 are: (17) \( - {\bf{4}}\), 4, 7; (18) \( - {\bf{3}}\), \( - {\bf{6}}\), 9; (19) \( - {\bf{2}}\), 7; (20) \( - {\bf{3}}\), 15; (21) 1, 5, 9; (22) 3, 5.

15. \(\left( {\begin{aligned}{{}}{\,3}&4\\4&9\end{aligned}} \right)\)

Make a change of variable, \({\bf{x}} = P{\bf{y}}\), that transforms the quadratic form \(x_{\bf{1}}^{\bf{2}} + {\bf{10}}{x_{\bf{1}}}{x_{\bf{2}}} + x_{\bf{2}}^{\bf{2}}\) into a quadratic form with no cross-product term. Give P and the new quadratic form.

Question: 4. Let A be an \(n \times n\) symmetric matrix.

a. Show that \({({\rm{Col}}A)^ \bot } = {\rm{Nul}}A\). (Hint: See Section 6.1.)

b. Show that each y in \({\mathbb{R}^n}\) can be written in the form \(y = \hat y + z\), with \(\hat y\) in \({\rm{Col}}A\) and z in \({\rm{Nul}}A\).

Orthogonally diagonalize the matrices in Exercises 13–22, giving an orthogonal matrix\(P\)and a diagonal matrix\(D\). To save you time, the eigenvalues in Exercises 17–22 are: (17)\( - {\bf{4}}\), 4, 7; (18)\( - {\bf{3}}\),\( - {\bf{6}}\), 9; (19)\( - {\bf{2}}\), 7; (20)\( - {\bf{3}}\), 15; (21) 1, 5, 9; (22) 3, 5.

18. \(\left( {\begin{aligned}{{}}1&{ - 6}&4\\{ - 6}&2&{ - 2}\\4&{ - 2}&{ - 3}\end{aligned}} \right)\)

Classify the quadratic forms in Exercises 9–18. Then make a change of variable, \({\bf{x}} = P{\bf{y}}\), that transforms the quadratic form into one with no cross-product term. Write the new quadratic form. Construct \(P\) using the methods of Section 7.1.

14. \({\bf{3}}x_{\bf{1}}^{\bf{2}} + {\bf{4}}{x_{\bf{1}}}{x_{\bf{2}}}\)
See all solutions

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