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Q7.4-19E

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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

In Exercises 17–24, \(A\) 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 \(m \times n\) “diagonal” matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

19. Show that the columns of \(V\) are eigenvectors of \({A^T}A\), the columns of \(U\) are eigenvectors of \(A{A^T}\), and the diagonal entries of \({\bf{\Sigma }}\) are the singular values of \(A\). (Hint: Use the SVD to compute \({A^T}A\) and \(A{A^T}\).)

It is verified that, \(V\) and \(U\) are eigenvectors of \({A^T}A\) and \(A{A^T}\) respectively, and the columns of \(U\) re eigenvectors of \(A{A^T}\) and the diagonal elements of \(\Sigma \) are the singular values of \(A\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Find the product of \({A^T}\) and \(A\)

As singular value decomposition of \(A\) is \(A = U\Sigma {V^T}\), then

\(\begin{array}{c}{A^T}A = {\left( {U\Sigma {V^T}} \right)^T}U\Sigma {V^T}\\ = V\mathop \Sigma \limits^T {U^T}\Sigma {V^T}\\ = V\left( {\mathop \Sigma \limits^T \Sigma } \right){V^T}\\ = V\left( {\mathop \Sigma \limits^T \Sigma {V^{ - 1}}} \right)\end{array}\)

Therefore, from the Diagonalization theorem. Singular values are the diagonal entries in \(\Sigma \) matrix. The columns of matrix \(V\) are the eigenvectors of \({A^T}A\).

Step 2: Find the product of \(A\) and \({A^T}\)

Now, find the product of \(A = U\Sigma {V^T}\) and its transpose.

\(\begin{array}{c}A{A^T} = U\Sigma {V^T}{\left( {U\Sigma {V^T}} \right)^T}\\ = U\mathop \Sigma \limits^T {V^T}V\mathop \Sigma \limits^T {U^T}\\ = U\left( {\Sigma \mathop \Sigma \limits^T } \right){U^T}\\ = U\left( {\Sigma \mathop \Sigma \limits^T } \right){U^{ - 1}}\end{array}\)

Thus, it can be observed that \(U\) diagonalizes \(A{A^T}\) and the columns of \(U\) must be eigenvectors of \(A{A^T}\).

Most popular questions for Math Textbooks

Let \(A = \left( {\begin{aligned}{{}}{\,\,\,2}&{ - 1}&{ - 1}\\{ - 1}&{\,\,\,2}&{ - 1}\\{ - 1}&{ - 1} &{\,\,\,2}\end{aligned}} \right)\),\({{\rm{v}}_1} = \left( {\begin{aligned}{{}}{ - 1}\\{\,\,\,0}\\{\,\,1}\end{aligned}} \right)\) and and\({{\rm{v}}_2} = \left( {\begin{aligned}{{}}{\,\,\,1}\\{\, - 1}\\{\,\,\,\,1}\end{aligned}} \right)\). Verify that\({{\rm{v}}_1}\), \({{\rm{v}}_2}\) an eigenvector of \(A\). Then orthogonally diagonalize \(A\).

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: 4. Let A be an \(n \times n\) symmetric matrix.

a. Show that \({({\rm{Col}}A)^ \bot } = {\rm{Nul}}A\). (Hint: See Section 6.1.)

b. Show that each y in \({\mathbb{R}^n}\) can be written in the form \(y = \hat y + z\), with \(\hat y\) in \({\rm{Col}}A\) and z in \({\rm{Nul}}A\).

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 \({\rm{u}} = {{\rm{x}}^ + }\).

Compute the quadratic form \({{\bf{x}}^T}A{\bf{x}}\), when \(A = \left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\) and

a. \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\)

b. \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\)

c. \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\)
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