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Chapter 7: Q7.4-23E (page 395)

In Exercises 17鈥�24, \(A\) is an \(m \times n\) matrix with a singular value decomposition \(A = U\Sigma {V^T}\) , where \(U\) is an \(m \times m\) orthogonal matrix, \({\bf{\Sigma }}\) is an \(m \times n\) 鈥渄iagonal鈥� matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

23. Let \(U = \left( {{u_1}...{u_m}} \right)\) and \(V = \left( {{v_1}...{v_n}} \right)\) where the \({{\bf{u}}_i}\) and \({{\bf{v}}_i}\) are in Theorem 10. Show that \(A = {\sigma _1}{u_1}v_1^T + {\sigma _2}{u_2}v_2^T + ... + {\sigma _r}{u_r}v_r^T\).

Short Answer

Expert verified

It is verified that \(A = {\sigma _1}{{\bf{u}}_{\bf{1}}}{\bf{v}}_{\bf{1}}^{\bf{T}} + {\sigma _2}{{\bf{u}}_2}{\bf{v}}_2^{\bf{T}} + ... + {\sigma _r}{{\bf{u}}_r}{\bf{v}}_r^{\bf{T}}\).

Step by step solution

01

Write the matrix 

It is given that, \(U = \left( {{{\bf{u}}_{\bf{1}}}...{{\bf{u}}_{\bf{m}}}} \right)\) and \(V = \left( {{{\bf{v}}_{\bf{1}}}...{{\bf{v}}_{\bf{n}}}} \right)\)

Therefore, \({V^T} = \left( {\begin{array}{*{20}{c}}{{\bf{v}}_1^T}\\{{\bf{v}}_n^T}\end{array}} \right)\)

02

Step 2: Compute the matrix A from SVD:

From theorem 10,\(U\sum = \left\{ {\left( {{\sigma _1}{{\bf{u}}_1}...{\sigma _r}{{\bf{u}}_r}} \right)} \right\}\)where\(r\)is the rank of the matrix\(A\), thus find\(A = U\sum {V^T}\).

\(\begin{array}{c}A = U\sum {V^T}\\ = \left\{ {\left( {{\sigma _1}{{\bf{u}}_1}...{\sigma _r}{{\bf{u}}_r}} \right)} \right\}\left( {\begin{array}{*{20}{c}}{{\bf{v}}_1^T}\\{{\bf{v}}_n^T}\end{array}} \right)\\ = {\sigma _1}{{\bf{u}}_{\bf{1}}}{\bf{v}}_{\bf{1}}^{\bf{T}} + {\sigma _2}{{\bf{u}}_2}{\bf{v}}_2^{\bf{T}} + ... + {\sigma _r}{{\bf{u}}_r}{\bf{v}}_r^{\bf{T}}\end{array}\)

Hence, \(A = {\sigma _1}{{\bf{u}}_{\bf{1}}}{\bf{v}}_{\bf{1}}^{\bf{T}} + {\sigma _2}{{\bf{u}}_2}{\bf{v}}_2^{\bf{T}} + ... + {\sigma _r}{{\bf{u}}_r}{\bf{v}}_r^{\bf{T}}\).

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Most popular questions from this chapter

Question: If A is \(m \times n\), then the matrix \(G = {A^T}A\) is called the Gram matrix of A. In this case, the entries of G are the inner products of the columns of A. (See Exercises 9 and 10).

9. Show that the Gram matrix of any matrix A is positive semidefinite, with the same rank as A. (See the Exercises in Section 6.5.)

Question: 5. Show that if v is an eigenvector of an \(n \times n\) matrix A and v corresponds to a nonzero eigenvalue of A, then v is in Col A. (Hint: Use the definition of an eigenvector.)

Classify the quadratic forms in Exercises 9鈥�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.

15. \( - {\bf{3}}x_{\bf{1}}^{\bf{2}} - {\bf{7}}x_{\bf{2}}^{\bf{2}} - {\bf{10}}x_{\bf{3}}^{\bf{2}} - {\bf{10}}x_{\bf{4}}^{\bf{2}} + {\bf{4}}{x_{\bf{1}}}{x_{\bf{2}}} + {\bf{4}}{x_{\bf{1}}}{x_{\bf{3}}} + {x_{\bf{1}}}{x_{\bf{4}}} + {\bf{6}}{x_{\bf{3}}}{x_{\bf{4}}}\)

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

9. \(\left[ {\begin{aligned}{{}}{ - 4/5}&{\,\,\,3/5}\\{3/5}&{\,\,4/5}\end{aligned}} \right]\)

Orthogonally diagonalize the matrices in Exercises 13鈥�22, giving an orthogonal matrix\(P\)and a diagonal matrix\(D\). To save you time, the eigenvalues in Exercises 17鈥�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.

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