/*! 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 23 Give an example of a nonzero vec... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 23

Give an example of a nonzero vector \(\mathbf{x} \in \mathbb{R}^{2}\) for which $$\|\mathbf{x}\|_{\infty}=\|\mathbf{x}\|_{2}=\|\mathbf{x}\|_{1}$$

Short Answer

Expert verified
The vector \(\mathbf{x} = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix} \in \mathbb{R}^{2}\) satisfies the condition \(\|\mathbf{x}\|_{\infty}=\|\mathbf{x}\|_{2}=\|\mathbf{x}\|_{1}\).

Step by step solution

01

Define the norms

We first recall the definitions of the given norms: 1. The 1-norm, or Manhattan norm, of a vector \(\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \in \mathbb{R}^{2}\) is defined as \[\|\mathbf{x}\|_{1} = |x_1| + |x_2|\] 2. The 2-norm, or Euclidean norm, of a vector \(\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \in \mathbb{R}^{2}\) is defined as \[\|\mathbf{x}\|_{2} = \sqrt{x_1^2 + x_2^2}\] 3. The infinity norm, or maximum norm, of a vector \(\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \in \mathbb{R}^{2}\) is defined as \[\|\mathbf{x}\|_{\infty} = \max\{|x_1|, |x_2|\}\]
02

Set up the equalities

Since the norms must be equal, we set up the following equations: 1. \(\|\mathbf{x}\|_{1} = \|\mathbf{x}\|_{2}\): \[|x_1| + |x_2| = \sqrt{x_1^2 + x_2^2}\] 2. \(\|\mathbf{x}\|_{1} = \|\mathbf{x}\|_{\infty}\): \[|x_1| + |x_2| = \max\{|x_1|, |x_2|\}\] 3. \(\|\mathbf{x}\|_{2} = \|\mathbf{x}\|_{\infty}\): \[\sqrt{x_1^2 + x_2^2} = \max\{|x_1|, |x_2|\}\]
03

Consider cases for x1 and x2 signs

Now we should consider two cases: either both \(x_1\) and \(x_2\) are positive, or one is positive and the other is negative. Note that we can't have both negative, since the given vector is nonzero. Case 1: \(x_1\) and \(x_2\) are both positive. In this case, equation 2 reduces to \[x_1 + x_2 = \max\{x_1, x_2\}\]. Since both \(x_1\) and \(x_2\) are positive, this can only be true if one of them is equal to zero, but then the vector is not nonzero as required. Case 2: Assume, without loss of generality, that \(x_1\) is positive and \(x_2\) is negative. Then, we can write equation 2 as \[x_1 - x_2 = x_1\]. Thus, this implies that \(x_2 = 0\), but this contradicts our assumption that \(x_2\) is negative.
04

Find the appropriate vector

Since the third equation (\(\|\mathbf{x}\|_{2} = \|\mathbf{x}\|_{\infty}\)) encompasses both the first and second equations, we can focus on finding a nonzero vector \(\mathbf{x} \in \mathbb{R}^{2}\) that satisfies this equation. Let \(x_1 = x_2\). Then \(\sqrt{x_1^2 + x_2^2} = \sqrt{2x_1^2}\) and \(\max\{|x_1|, |x_2|\} = |x_1|\). If we set \(x_1 = x_2 = \frac{1}{\sqrt{2}}\), we have \[\sqrt{2x_1^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1\] and \[\max\{|x_1|, |x_2|\} = \frac{1}{\sqrt{2}}\] So, the vector that satisfies the given conditions is \[\mathbf{x} = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}\]

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 \(A\) be an \(m \times 2\) matrix. Show that if both the classical Gram- Schmidt process and the modified Gram-Schmidt process are applied to the column vectors of \(A\), then both algorithms will produce the exact same \(Q R\) factorization, even when the computations are carried out in finite-precision arithmetic (i.e., show that both algorithms will perform the exact same arithmetic computations)

Find the least squares solution of each of the following systems: (a) \(x_{1}+x_{2}=3\) \(2 x_{1}-3 x_{2}=1\) \(0 x_{1}+0 x_{2}=2\) (b) \(-x_{1}+x_{2}=10\) \(\begin{aligned} 2 x_{1}+x_{2} &=5 \\ x_{1}-2 x_{2} &=20 \end{aligned}\) (c) \(\quad x_{1}+x_{2}+x_{3}=4\) \(\begin{aligned}-x_{1}+x_{2}+x_{3} &=0 \\\\-x_{2}+x_{3} &=1 \\ x_{1} &+x_{3}=2 \end{aligned}\)

Sketch the set of points \(\left(x_{1}, x_{2}\right)=\mathbf{x}^{T}\) in \(\mathbb{R}^{2}\) such that (a) \(\|\mathbf{x}\|_{2}=1\) (b) \(\|\mathbf{x}\|_{1}=1\) (c) \(\|\mathbf{x}\|_{\infty}=1\)

Let \(p_{0}, p_{1}, \ldots\) be a sequence of orthogonal polynomials and let \(a_{n}\) denote the lead coefficient of \(p_{n}\) Prove that $$\left\|p_{n}\right\|^{2}=a_{n}\left\langle x^{n}, p_{n}\right\rangle$$

Use the Gram-Schmidt process to find an orthonormal basis for the subspace of \(\mathbb{R}^{4}\) spanned by \(\mathbf{x}_{1}=(4,2,2,1)^{T}, \mathbf{x}_{2}=(2,0,0,2)^{T}, \mathbf{x}_{3}=\) \((1,1,-1,1)^{T}\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

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.