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Chapter 7: Q7.4-22E (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.

22. Show that if \(A\) is an \(n \times n\) positive definite matrix, then an orthogonal diagonalization \(A = PD{P^T}\) is a singular value decomposition of \(A\).

Short Answer

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Step by step solution

01

Show that vectors orthogonal to \({{\bf{v}}_{\bf{1}}}\)

Since the matrix\(P\)is a square and orthogonal matrix then we have,

\(\begin{array}{c}P{P^T} = I\\{P^T} = {P^{ - 1}}\\{({P^T})^{ - 1}} = {({P^{ - 1}})^{ - 1}}\\ = P\end{array}\)

02

Show that \(A = PD{P^T}\) is a singular value decomposition of \(A\)

Simplify\(P{P^T}\).

\(\begin{array}{c}P{P^T} = I\\{P^T} = {P^{ - 1}}\\{({P^T})^T} = {({P^{ - 1}})^T}\\ = {\left( {{P^T}} \right)^{ - 1}}\\ = P\end{array}\)

Therefore,\({P^T}\)is an orthogonal matrix, and the diagonal matrix becomes the\(\sum \)matrix.

Thus, the factorization \(A = PD{P^{ - 1}}\) satisfies the properties which make it singular value decomposition.

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Most popular questions from this chapter

Let A be the matrix of the quadratic form

\({\bf{9}}x_{\bf{1}}^{\bf{2}} + {\bf{7}}x_{\bf{2}}^{\bf{2}} + {\bf{11}}x_{\bf{3}}^{\bf{2}} - {\bf{8}}{x_{\bf{1}}}{x_{\bf{2}}} + {\bf{8}}{x_{\bf{1}}}{x_{\bf{3}}}\)

It can be shown that the eigenvalues of A are 3,9, and 15. Find an orthogonal matrix P such that the change of variable \({\bf{x}} = P{\bf{y}}\) transforms \({{\bf{x}}^T}A{\bf{x}}\) into a quadratic form which no cross-product term. Give P and the new quadratic form.

Orhtogonally diagonalize the matrices in Exercises 37-40. To practice the methods of this section, do not use an eigenvector routine from your matrix program. Instead, use the program to find the eigenvalues, and for each eigenvalue \(\lambda \), find an orthogonal basis for \({\bf{Nul}}\left( {A - \lambda I} \right)\), as in Examples 2 and 3.

40. \(\left( {\begin{aligned}{{}}{\bf{8}}&{\bf{2}}&{\bf{2}}&{ - {\bf{6}}}&{\bf{9}}\\{\bf{2}}&{\bf{8}}&{\bf{2}}&{ - {\bf{6}}}&{\bf{9}}\\{\bf{2}}&{\bf{2}}&{\bf{8}}&{ - {\bf{6}}}&{\bf{9}}\\{ - {\bf{6}}}&{ - {\bf{6}}}&{ - {\bf{6}}}&{{\bf{24}}}&{\bf{9}}\\{\bf{9}}&{\bf{9}}&{\bf{9}}&{\bf{9}}&{ - {\bf{21}}}\end{aligned}} \right)\)

Question: Compute the singular values of the \({\bf{4 \times 4}}\) matrix in Exercise 9 in Section 2.3, and compute the condition number \(\frac{{{\sigma _1}}}{{{\sigma _4}}}\).

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

Question: Let \({x_1}\,,{x_2}\) denote the variables for the two-dimensional data in Exercise 1. Find a new variable \({y_1}\) of the form \({y_1} = {c_1}{x_1} + {c_2}{x_2}\), with\(c_1^2 + c_2^2 = 1\), such that \({y_1}\) has maximum possible variance over the given data. How much of the variance in the data is explained by \({y_1}\)?
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