/*! 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 55 Give an example of a nonlinear m... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 55

Give an example of a nonlinear map \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) such that \(F^{-1}(0)=\\{0\\}\) but \(F\) is not one-to-one.

Short Answer

Expert verified
The nonlinear map that satisfies the given conditions is \(F(x, y) = (x^3, y^3)\). \(F\) is not linear, has the inverse image \(F^{-1}(0) = \{0\}\), and is not one-to-one.

Step by step solution

01

1. Confirm Non-linearity

We need to check that the map \(F\) is not linear. A map is linear if it satisfies the following conditions: 1. \(F(x_1 + x_2, y_1 + y_2) = F(x_1, y_1) + F(x_2, y_2)\) for all \(x_1, x_2, y_1, y_2 \in \mathbf{R}\). 2. \(F(ax, ay) = aF(x, y)\) for all \(x, y \in \mathbf{R}\) and \(a \in \mathbf{R}\). For \(F(x, y) = (x^3, y^3)\), let's test the conditions: \(F((x_1+x_2), (y_1+y_2)) = ((x_1+x_2)^3, (y_1+y_2)^3) \neq (x_1^3+x_2^3, y_1^3+y_2^3) = F(x_1, y_1) + F(x_2, y_2)\), so the map is not linear.
02

2. Check Inverse Image

Let's find the inverse image of 0, meaning the set of points that map to (0,0). We have: \[F(x, y) = (0, 0)\] \[(x^3, y^3) = (0, 0)\] This implies that \(x^3 = 0\) and \(y^3 = 0\), so \(x = 0\) and \(y = 0\). Thus, the inverse image of \(0\) is \((0, 0)\), i.e., \(F^{-1}(0) = \{0\}\).
03

3. Confirming Not One-to-One

In order for a map to be one-to-one, each point in the domain must map to a unique point in the codomain. We need to show that our map \(F\) does not satisfy this condition. Let's take two different points \((x_1,y_1)\) and \((x_2,y_2)\) in \(\mathbf{R}^{2}\): \((x_1,y_1) = (0,0)\) and \((x_2,y_2) = (-1,1)\) When we evaluate the map \(F\) on these points, we get: \[F(0, 0) = (0^3, 0^3) = (0, 0)\] \[F(-1, 1) = ((-1)^3, 1^3) = (-1, 1)\] So, we see that two different points in the domain map to two different points in the codomain, showing that the map \(F\) is not one-to-one. Considering all the above points, the nonlinear map \(F(x, y) = (x^3, y^3)\) satisfies all the given conditions.

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 \(T\) be a linear operator on a finite-dimensional vector space \(V\), and let \(\beta\) be an ordered basis for \(\mathrm{V}\). Prove that \(\lambda\) is an eigenvalue of \(\mathrm{T}\) if and only if \(\lambda\) is an eigenvalue of \([\mathrm{T}]_{\beta}\).

Prove: (a) The zero mapping \(0,\) defined by \(\mathbf{0}(v)=0 \in U\) for every \(v \in V,\) is the zero element of \(\operatorname{Hom}(V, U) .\) (b) The negative of \(F \in \operatorname{Hom}(V, U)\) is the mapping \((-1) F,\) that is, \(-F=(-1) F\)

Consider the linear map \(G: \mathbf{R}^{3} \rightarrow \mathbf{R}^{3}\) defined by \(G(x, y, z)=(x+y+z, y-2 z, y-3 z)\) and the unit sphere \(S_{2}\) in \(\mathbf{R}^{3},\) which consists of the points satisfying \(x^{2}+y^{2}+z^{2}=1 .\) Find \((\mathrm{a}) G\left(S_{2}\right),\) (b) \(G^{-1}\left(S_{2}\right)\)

Consider linear mappings \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}, \quad G: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}, H: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y, z)=(x+y+z, x+y), \quad G(x, y, z)=(2 x+z, x+y), \quad H(x, y, z)=(2 y, x)\) Show that \(\left.F, G, H \text { are linearly independent [as elements of } \operatorname{Hom}\left(\mathbf{R}^{3}, \mathbf{R}^{2}\right)\right]\)

Consider the matrix mapping \(A: \mathbf{R}^{4} \rightarrow \mathbf{R}^{3},\) where \(A=\left[\begin{array}{rrrr}1 & 2 & 3 & 1 \\ 1 & 3 & 5 & -2 \\ 3 & 8 & 13 & -3\end{array}\right] .\) Find a basis and the dimension of (a) the image of \(A,(\mathrm{b})\) the kernel of \(A.\)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Statistics

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.