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Q7.4-26E

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Linear Algebra and its Applications
Found in: Page 395
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

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

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

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: 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 \(B\).

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

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

Step 2: 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.

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

Step 3: 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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