Linear Algebra and Its Applications, 5th Edition

Linear Algebra and Its Applications, 5th Edition

Authors: David C. Lay, Steven R. Lay, Judi J. McDonald

ISBN-13: 978-0321982384

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See our solution for Question 21E from Chapter 7.4 from Lay's Linear Algebra and Its Applications, 5th Edition.

Problem 21E

Chapter:
Problem:
0

Step-by-Step Solution

Given Information
We have to justify the statement that the second singular value of a matrix $A$ is the maximum of $\| A \mathbf { x } \|$ as $\mathbf { x }$ varies over all unit vectors orthogonal to $\mathbf { v } _ { 1 } ,$ with $\mathbf { v } _ { 1 }$ a right singular vector corresponding to the first singular value of $A .$

Step-1:
We know that the right singular vector $v _ { 1 }$ is an eigenvector for the largest eigenvalue $\lambda _ { 1 }$ of $A ^ { T } A$. Also, the second largest eigenvalue $\lambda _ { 2 }$ is the maximum of $x ^ { T } \left( A ^ { T } A \right)x$ overall unit vectors orthogonal to $\mathbf { v } _ { 1 }$ .\[\begin{array}{l}{X^T}\left( {{A^T}A} \right)X = {X^T}{A^T}(AX)\\ = {(AX)^T}(AX)\\ = {\left| {\left| {AX} \right|} \right|^2}\end{array}\]Thus,

The square root of $\lambda _ { 2 } ,$ is the maximum of $\| A X \|$ overall unit vectors orthogonal to $v _ { 1 }$ .