/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q14SE Question: 14. Exercises 12鈥�14 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Q14SE (page 395)

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

Given any \({\rm{b}}\) in \({\mathbb{R}^m}\), adapt Exercise 13 to show that \({A^ + }{\rm{b}}\) is the least-squares solution of minimum length. [Hint: Consider the equation \(A{\rm{x}} = {\rm{b}}\), where \(\mathop {\rm{b}}\limits^\^ \) is the orthogonal projection of \({\rm{b}}\) onto Col \(A\).

Short Answer

Expert verified

The least-square solution of \(A{\rm{x}} = {\rm{b}}\) having minimum length is \({A^ + }\mathop {\bf{b}}\limits^\^ \).

Step by step solution

01

Describe the given information

It is given that \(\mathop {\bf{b}}\limits^\^ \) is the orthogonal projection of \({\bf{b}}\) onto Col\(A\), then the least square solution of \(A{\bf{x}} = {\bf{b}}\)can be said to be approximately same as that of \(A{\bf{x}} = \mathop {\bf{b}}\limits^\^ \).

02

Use the result of exercise 13

When the system\(A{\rm{x}} = {\rm{b}}\)is consistent and\({{\rm{x}}^ + } = {A^ + }{\rm{b}}\), then the minimum length solution of\(A{\bf{x}} = \mathop {\bf{b}}\limits^\^ \)is\({A^ + }\mathop {\bf{b}}\limits^\^ \), thus, the least-square solution of\(A{\rm{x}} = {\rm{b}}\), having minimum length, is\({A^ + }\mathop {\bf{b}}\limits^\^ \).

Although, it is known that \(A{A^ + }{\rm{y}}\) is the orthogonal projection of \({\rm{y}}\)onto \({\rm{Col}}\,\,A\), and \(A{A^ + }A = A\), \({A^ + }A{A^ + } = {A^ + }\). So, \({A^ + }\mathop {\rm{b}}\limits^\^ = {A^ + }{\bf{b}}\). Thus, it is confirmed that the least square solution of \(A{\rm{x}} = {\rm{b}}\)having minimum length is \({A^ + }\mathop {\bf{b}}\limits^\^ \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

18. Suppose \(A\) is square and invertible. Find a singular value decomposition of \({A^{ - 1}}\)

In Exercises 25 and 26, mark each statement True or False. Justify each answer.

26.

  1. There are symmetric matrices that are not orthogonally diagonizable.
  2. b. If \(B = PD{P^T}\), where \({P^T} = {P^{ - {\bf{1}}}}\) and D is a diagonal matrix, then B is a symmetric matrix.
  3. c. An orthogonal matrix is orthogonally diagonizable.
  4. d. The dimension of an eigenspace of a symmetric matrix is sometimes less than the multiplicity of the corresponding eigenvalue.

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.

20. \(\left( {\begin{aligned}{{}}5&8&{ - 4}\\8&5&{ - 4}\\{ - 4}&{ - 4}&{ - 1}\end{aligned}} \right)\)

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 \(A = {\sigma _1}{u_1}v_1^T + {\sigma _2}{u_2}v_2^T + ... + {\sigma _r}{u_r}v_r^T\).

Find the matrix of the quadratic form. Assume x is in \({\mathbb{R}^2}\) .

a. \(3x_1^2 - 4{x_1}{x_2} + 5x_2^2\) b. \(3x_1^2 + 2{x_1}{x_2}\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.