Question

(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.

27. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{6}}&{ - {\bf{8}}}&{ - {\bf{4}}}&{\bf{5}}&{ - {\bf{4}}}\\{\bf{2}}&{\bf{7}}&{ - {\bf{5}}}&{ - {\bf{6}}}&{\bf{4}}\\{\bf{0}}&{ - {\bf{1}}}&{ - {\bf{8}}}&{\bf{2}}&{\bf{2}}\\{ - {\bf{1}}}&{ - {\bf{2}}}&{\bf{4}}&{\bf{4}}&{ - {\bf{8}}}\end{array}} \right)\)

Short answer

The singular value decomposition of \(A\) is:\(A = \left( {\begin{array}{*{20}{c}}{ - .57}&{ - .65}&{ - .42}&{.27}\\{.63}&{ - .24}&{ - .68}&{ - .29}\\{.07}&{ - .63}&{.53}&{ - .56}\\{ - .51}&{.34}&{ - .29}&{ - .73}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{16.46}&0&0&0&0\\0&{12.16}&0&0&0\\0&0&{4.87}&0&0\\0&0&0&{4.31}&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 10}&{.61}&{ - .21}&{ - .52}&{.55}\\{ - .39}&{.29}&{.84}&{ - 1.4}&{ - .19}\\{ - .74}&{ - .27}&{ - .07}&{.38}&{.49}\\{.41}&{ - .50}&{.45}&{ - .23}&{.58}\\{ - .36}&{ - .48}&{ - .19}&{ - .72}&{ - .29}\end{array}} \right)\)

Step-by-step solution

Write the matrix 

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

\( > > {\rm{ }}A = \left( {6{\rm{ }} - 8{\rm{ }} - 4{\rm{ }}5{\rm{ }} - 4;{\rm{ }}2{\rm{ }}7{\rm{ }} - 5{\rm{ }} - 6{\rm{ }}4;{\rm{ }}0{\rm{ }} - 1{\rm{ }} - 8{\rm{ }}2{\rm{ }}2;{\rm{ }} - 1{\rm{ }} - 2{\rm{ }}4{\rm{ }}4{\rm{ }} - 8} \right);\)

Find transpose of the matrix:

\( > > B = A';\)

\(B = \left( {\begin{array}{*{20}{c}}6&2&0&{ - 1}\\{ - 8}&7&{ - 1}&{ - 2}\\{ - 4}&{ - 5}&{ - 8}&4\\5&{ - 6}&2&4\\{ - 4}&4&2&{ - 8}\end{array}} \right)\)

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

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

\(C = \left( {\begin{array}{*{20}{c}}{41}&{ - 32}&{ - 38}&{14}&{ - 8}\\{ - 32}&{118}&{ - 3}&{ - 92}&{74}\\{ - 38}&{ - 3}&{121}&{10}&{ - 52}\\{14}&{ - 92}&{10}&{81}&{ - 72}\\{ - 8}&{74}&{ - 52}&{ - 72}&{100}\end{array}} \right)\)

Find eigenvalues and eigenvectors of matrix \(AB\)

Find the eigenvalues of matrix \(AB\).

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

\(E = \left( {\begin{array}{*{20}{c}}{270.87}\\{147.85}\\{23.73}\\{18.55}\\0\end{array}} \right)\)

Find eigenvectors.

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

\(V = \left( {\begin{array}{*{20}{c}}{ - 10}&{ - .39}&{ - .74}&{.41}&{ - .36}\\{.61}&{.29}&{ - .27}&{ - .50}&{ - .48}\\{ - .21}&{.84}&{ - .07}&{.45}&{ - .19}\\{ - .52}&{ - .14}&{.38}&{ - .23}&{ - .72}\\{.55}&{ - .19}&{.49}&{.58}&{ - .29}\end{array}} \right)\)

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

Then we have\(\sum \)matrix.

\(\sum = \left( {\begin{array}{*{20}{c}}{16.46}&0&0&0&0\\0&{12.16}&0&0&0\\0&0&{4.87}&0&0\\0&0&0&{4.31}&0\end{array}} \right)\)

Find the singular value decomposition \(A\)

Write the vectors of\(U\)matrix:

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

\(U = \left( {\begin{array}{*{20}{c}}{ - .57}&{ - .65}&{ - .42}&{.27}\\{.63}&{ - .24}&{ - .68}&{ - .29}\\{.07}&{ - .63}&{.53}&{ - .56}\\{ - .51}&{.34}&{ - .29}&{ - .73}\end{array}} \right)\)

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

\(\begin{array}{c}A = U\sum {V^T}\\ = \left( {\begin{array}{*{20}{c}}{ - .57}&{ - .65}&{ - .42}&{.27}\\{.63}&{ - .24}&{ - .68}&{ - .29}\\{.07}&{ - .63}&{.53}&{ - .56}\\{ - .51}&{.34}&{ - .29}&{ - .73}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{16.46}&0&0&0&0\\0&{12.16}&0&0&0\\0&0&{4.87}&0&0\\0&0&0&{4.31}&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 10}&{ - .39}&{ - .74}&{.41}&{ - .36}\\{.61}&{.29}&{ - .27}&{ - .50}&{ - .48}\\{ - .21}&{.84}&{ - .07}&{.45}&{ - .19}\\{ - .52}&{ - .14}&{.38}&{ - .23}&{ - .72}\\{.55}&{ - .19}&{.49}&{.58}&{ - .29}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - .57}&{ - .65}&{ - .42}&{.27}\\{.63}&{ - .24}&{ - .68}&{ - .29}\\{.07}&{ - .63}&{.53}&{ - .56}\\{ - .51}&{.34}&{ - .29}&{ - .73}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{16.46}&0&0&0&0\\0&{12.16}&0&0&0\\0&0&{4.87}&0&0\\0&0&0&{4.31}&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 10}&{.61}&{ - .21}&{ - .52}&{.55}\\{ - .39}&{.29}&{.84}&{ - 1.4}&{ - .19}\\{ - .74}&{ - .27}&{ - .07}&{.38}&{.49}\\{.41}&{ - .50}&{.45}&{ - .23}&{.58}\\{ - .36}&{ - .48}&{ - .19}&{ - .72}&{ - .29}\end{array}} \right)\end{array}\)

Thus, \(A = \left( {\begin{array}{*{20}{c}}{ - .57}&{ - .65}&{ - .42}&{.27}\\{.63}&{ - .24}&{ - .68}&{ - .29}\\{.07}&{ - .63}&{.53}&{ - .56}\\{ - .51}&{.34}&{ - .29}&{ - .73}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{16.46}&0&0&0&0\\0&{12.16}&0&0&0\\0&0&{4.87}&0&0\\0&0&0&{4.31}&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 10}&{.61}&{ - .21}&{ - .52}&{.55}\\{ - .39}&{.29}&{.84}&{ - 1.4}&{ - .19}\\{ - .74}&{ - .27}&{ - .07}&{.38}&{.49}\\{.41}&{ - .50}&{.45}&{ - .23}&{.58}\\{ - .36}&{ - .48}&{ - .19}&{ - .72}&{ - .29}\end{array}} \right)\)