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Chapter 7: Q7.4-27E (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.

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

Expert verified

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

01

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\(B\).

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

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

02

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.

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

03

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

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Most popular questions from this chapter

Question: 11. Given multivariate data \({X_1},................,{X_N}\) (in \({\mathbb{R}^p}\)) in mean deviation form, let \(P\) be a \(p \times p\) matrix, and define \({Y_k} = {P^T}{X_k}{\rm{ for }}k = 1,......,N\).

  1. Show that \({Y_1},................,{Y_N}\) are in mean-deviation form. (Hint: Let \(w\) be the vector in \({\mathbb{R}^N}\) with a 1 in each entry. Then \(\left( {{X_1},................,{X_N}} \right)w = 0\) (the zero vector in \({\mathbb{R}^p}\)).)
  2. Show that if the covariance matrix of \({X_1},................,{X_N}\) is \(S\), then the covariance matrix of \({Y_1},................,{Y_N}\) is \({P^T}SP\).

Question: 13. Exercises 12–14 concern an \(m \times n\) matrix \(A\) with a reduced singular value decomposition, \(A = {U_r}D{V_r}^T\), and the pseudoinverse \({A^ + } = {U_r}{D^{ - 1}}{V_r}^T\).

Suppose the equation\(A{\rm{x}} = {\rm{b}}\)is consistent, and let\({{\rm{x}}^ + } = {A^ + }{\rm{b}}\). By Exercise 23 in Section 6.3, there is exactly one vector\({\rm{p}}\)in Row\(A\)such that\(A{\rm{p}} = {\rm{b}}\). The following steps prove that\({{\rm{x}}^ + } = {\rm{p}}\)and\({{\rm{x}}^ + }\)is the minimum length solution of\(A{\rm{x}} = {\rm{b}}\).

  1. Show that \({{\rm{x}}^ + }\) is in Row \(A\). (Hint: Write \({\rm{b}}\) as \(A{\rm{x}}\) for some \({\rm{x}}\), and use Exercise 12.)
  2. Show that\({{\rm{x}}^ + }\)is a solution of\(A{\rm{x}} = {\rm{b}}\).
  3. Show that if \({\rm{u}}\) is any solution of \(A{\rm{x}} = {\rm{b}}\), then \(\left\| {{{\rm{x}}^ + }} \right\| \le \left\| {\rm{u}} \right\|\), with equality only if \({\rm{u}} = {{\rm{x}}^ + }\).

Determine which of the matrices in Exercises 1–6 are symmetric.

4. \(\left( {\begin{aligned}{{}}0&8&3\\8&0&{ - 4}\\3&2&0\end{aligned}} \right)\)

Determine which of the matrices in Exercises 7–12 are orthogonal. If orthogonal, find the inverse.

7. \(\left( {\begin{aligned}{{}{}}{.6}&{\,\,\,.8}\\{.8}&{ - .6}\end{aligned}} \right)\)

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.

22. \(\left( {\begin{aligned}{{}}4&0&1&0\\0&4&0&1\\1&0&4&0\\0&1&0&4\end{aligned}} \right)\)
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