/*! 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} Problem 42 Let \(A\) be an \(n \times n\) m... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 42

Let \(A\) be an \(n \times n\) matrix and let \(Q\) and \(V\) be \(n \times n\) orthogonal matrices. Show that (a) \(\|Q A\|_{2}=\|A\|_{2}\) (b) \(\|A V\|_{2}=\|A\|_{2}\) (c) \(\|Q A V\|_{2}=\|A\|_{2}\)

Short Answer

Expert verified
To summarize, we have shown that for an \(n \times n\) matrix \(A\) and orthogonal matrices \(Q\) and \(V\), the following properties hold true: \(\|QA\|_{2}=\|A\|_{2}\), \(\|AV\|_{2}=\|A\|_{2}\), and \(\|QAV\|_{2}=\|A\|_{2}\). This was done by computing expressions involving the transpose of the product with orthogonal matrices, and making use of the property that the spectral norm is the square root of the maximum eigenvalue of the matrix product with its transpose, along with useful properties of orthogonal matrices.

Step by step solution

01

(a) Proof that \(\|Q A\|_{2}=\|A\|_{2}\)

First, recall the definition of the spectral norm (2-norm): \(\|A\|_{2} = \sqrt{\lambda_\text{max}(A^TA)}\) where \(\lambda_\text{max}(A^TA)\) is the largest eigenvalue of the matrix \(A^TA\). We are going to show that \(\|QA\|_2 = \|A\|_2\). To prove this, let's compute \( (QA)^T(QA)\): \((QA)^T(QA) = A^T(Q^TQ)A = A^TIA = A^TA\) Now, since \((QA)^T(QA) = A^TA\), they have the same eigenvalues. Thus, \(\lambda_\text{max}((QA)^T(QA)) = \lambda_\text{max}(A^TA)\), and so \(\|QA\|_{2} = \sqrt{\lambda_\text{max}((QA)^T(QA))} = \sqrt{\lambda_\text{max}(A^TA)} = \|A\|_{2}\).
02

(b) Proof that \(\|A V\|_{2}=\|A\|_{2}\)

Proceeding in a similar manner as in part (a), we will now compute \((AV)^T(AV)\): \((AV)^T(AV) = V^T(A^TAV) = V^TA^TV\) Since \(V^TV = I\), we have: \((AV)^T(AV) = V^TA^TV = A^TA\) Again, \((AV)^T(AV)\) and \(A^TA\) have the same eigenvalues, so \(\lambda_\text{max}((AV)^T(AV)) = \lambda_\text{max}(A^TA)\), and \(\|AV\|_{2} = \sqrt{\lambda_\text{max}((AV)^T(AV))} = \sqrt{\lambda_\text{max}(A^TA)} = \|A\|_{2}\).
03

(c) Proof that \(\|QAV\|_{2}=\|A\|_{2}\)

Finally, let's compute \((QAV)^T(QAV)\): \((QAV)^T(QAV) = V^T(A^TQ^TQAV) = V^TA^TAV\) Now we can use the result of step (b) and obtain: \((QAV)^T(QAV) = V^TA^TAV = A^TA\) Again, \((QAV)^T(QAV)\) and \(A^TA\) have the same eigenvalues, so \(\lambda_\text{max}((QAV)^T(QAV)) = \lambda_\text{max}(A^TA)\), and \(\|QAV\|_{2} = \sqrt{\lambda_\text{max}((QAV)^T(QAV))} = \sqrt{\lambda_\text{max}(A^TA)} = \|A\|_{2}\). From steps (a), (b) and (c), we have shown that \(\|QA\|_{2}=\|A\|_{2}\), \(\|AV\|_{2}=\|A\|_{2}\), and \(\|QAV\|_{2}=\|A\|_{2}\), as needed.

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

Let \(R_{1}\) and \(R_{2}\) be two \(2 \times 2\) rotation matrices and let \(G_{1}\) and \(G_{2}\) be two \(2 \times 2\) Givens transformations. What type of transformations are each of the following? (a) \(R_{1} R_{2}\) (b) \(G_{1} G_{2}\) (c) \(R_{1} G_{1}\) (d) \(G_{1} R_{1}\)

Show each of the following: (a) \(\left(A^{+}\right)^{+}=A\) (b) \(\left(A A^{+}\right)^{2}=A A^{+}\) (c) \(\left(A^{+} A\right)^{2}=A^{+} A\)

Find the three-digit decimal floating-point representation of each of the following numbers: (a) 2312 (b) 32.56 (c) 0.01277 (d) 82,431

If \(A\) is a \(5 \times 3\) matrix with \(\|A\|_{2}=8, \operatorname{cond}_{2}(A)=2\) and \(\|A\|_{F}=12,\) determine the singular values of \(A\)

Let \\[ A=\left(\begin{array}{lll} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 2 \end{array}\right) \quad \text { and } \quad \mathbf{u}_{0}=\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right) \\] (a) Apply the power method to \(A\) to compute \(\mathbf{v}_{1}\) \(\mathbf{u}_{1}, \mathbf{v}_{2}, \mathbf{u}_{2},\) and \(\mathbf{v}_{3} .\) (Round off to two decimal places.) (b) Determine an approximation \(\lambda_{1}^{\prime}\) to the largest eigenvalue of \(A\) from the coordinates of \(\mathbf{v}_{3}\) Determine the exact value of \(\lambda_{1}\) and compare it with \(\lambda_{1}^{\prime} .\) What is the relative error?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

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.