/*! 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 3 Assume \(A \in L(X . Y)\) and \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 3

Assume \(A \in L(X . Y)\) and \(A x=0\) only when \(x=0\). Prove that \(A\) is then \(1-1\)

Short Answer

Expert verified
Question: Prove that a linear operator A is injective if and only if A acts on some vector x in such a way that the result is the zero vector only when x is the zero vector. Answer: We have proven that if A acts on some vector x in such a way that the result is the zero vector only when x is the zero vector, then A is an injective function. To complete the proof, we need to show the converse, i.e., if A is injective, then A acts on some vector x in such a way that the result is the zero vector only when x is the zero vector. Proof: Suppose A is an injective function. Let x 鈭� X, and assume that A(x) = 0. Since A is injective, if A(x) = A(y), then x = y. In our case, A(x) = 0 = A(0), and thus, x = 0. This confirms that A acts on some vector x in such a way that the result is the zero vector only when x is the zero vector. Therefore, a linear operator A is injective if and only if A acts on some vector x in such a way that the result is the zero vector only when x is the zero vector.

Step by step solution

01

Recall the definition of an injective function

An injective function (also called a 1-1 function) is a function for which every element in its domain maps to a unique element in its codomain. In other words, if f(x) = f(y), then x = y for all x and y in the domain.
02

State the assumption

We are given that \(A \in L(X,Y)\) is a linear operator and that \(Ax = 0\) only when \(x = 0\). We will use this property to prove that A is an injective function.
03

Start the proof by assuming that A(x) = A(y)

Suppose that for some vectors \(x, y \in X\), we have \(Ax = Ay\). Our task is to show that under this assumption, \(x\) and \(y\) must be equal.
04

Use linearity property to rewrite the equation

The linearity property of A states that for any vectors \(u, v \in X\) and scalars \(c\), A(cu + v) = cA(u) + A(v). In our case, we can use linearity to rewrite our equation: \(A(x) - A(y) = A(x-y)\)
05

Use the assumption to conclude that x - y = 0

Since \(A(x) = A(y)\), the left side of our equation simplifies to the zero vector: \(0 = A(x - y)\) According to the given assumption, this is only possible if the vector \(x - y\) itself is equal to the zero vector: \(x - y = 0\)
06

Add y to both sides to obtain x = y

Solving for x, we add y to both sides of the equation: \(x - y + y = 0 + y\) \(x = y\)
07

Conclude that A is an injective function

We have shown that under our assumption \(A(x) = A(y)\), we must have \(x = y\). This confirms that A is an injective function, as required by the problem statement.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear algebra
Linear algebra is a foundational branch of mathematics that deals with vector spaces and linear mappings between these spaces. It includes the study of lines, planes, and subspaces, but is also concerned with properties that are common to all vector spaces. A major aspect of linear algebra is the study of linear transformations or operators, which are functions that preserve vector addition and scalar multiplication.

The rules of linear algebra are essential for solving systems of linear equations, which appear across various fields such as engineering, computer science, and economics. Understanding the concepts of linear algebra helps us manipulate and solve for unknown variables within a given problem, whether it be simple equations or more complex functions involving multi-dimensional spaces.
1-1 function
A 1-1 function, also known as an injective function, has a distinct characteristic: each element of the function's domain (the input) is mapped to a unique element of its codomain (the output). This means that no two distinct elements in the domain correspond to the same element in the codomain.

This concept is crucial because it establishes a one-to-one relationship between input and output, ensuring that the function preserves distinctiveness. When you have an injective function, you can say with certainty that if the function outputs are the same, then the inputs must have been the same as well. This property is especially important in fields like cryptography and computer science, where preserving unique mappings between elements can be essential.
Linearity property
The linearity property of a function tells us that the function behaves predictably with respect to addition and scalar multiplication. Specifically, for a linear operator 'A', the linearity property is outlined by two conditions: A(u + v) = A(u) + A(v) and A(cv) = cA(v), for any vectors u, v in the vector space, and any scalar c.

This property allows us to manipulate equations in ways that reveal underlying relationships between variables and simplifies the process of analysing and solving linear equations. It plays a pivotal role in various applications such as physics, where it ensures that the principles governing a system add up linearly under different conditions, a key concept known as superposition.
Injective function proof
Proving a function is injective involves showing that if two inputs produce the same output, then the inputs must be identical. In the context of a linear operator, this can often be accomplished by leveraging the operator's linearity property.

Proof Outline

First, assume that A(x) = A(y). Then, using the linear property, show that A(x) - A(y) can be rewritten as A(x - y). Since A(x) = A(y), it follows that A(x - y) must be the zero vector.

Given that A(x - y) = 0, and by the assumption that A maps only the zero vector to itself, it implies x - y must be the zero vector. Hence, x must be equal to y. We have thus demonstrated that the operator A is injective, as it preserves the uniqueness of the inputs through to the outputs.

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

Fix two real numbers \(a\) and \(b, 0

Take \(n=m=1\) in the implicit function theorem, and interpret the theorem (as well as its proof) graphically.

Suppose \(\mathrm{f}\) is a differentiable mapping of \(R^{1}\) into \(R^{3}\) such that \(|\mathbf{f}(t)|=1\) for every \(t\). Prove that \(f^{\prime}(t) \cdot f(t)=0\). Interpret this result geometrically.

If \(f\) is a real function defined in a convex open set \(E \subset R^{n}\), such that \(\left(D_{1} f\right)(\mathbf{x})=0\) for every \(\mathbf{x} \in E\), prove that \(f(\mathbf{x})\) depends only on \(x_{2}, \ldots, x_{n}\). Show that the convexity of \(E\) can be replaced by a weaker condition, but that some condition is required. For example, if \(n=2\) and \(E\) is shaped like a horseshoe, the statement may be false.

Show that the system of equations $$ \begin{array}{r} 3 x+y-z+u^{2}=0 \\ x-y+2 z+u=0 \\ 2 x+2 y-3 z+2 u=0 \end{array} $$ can be solved for \(x, y, u\) in terms of \(z ;\) for \(x, z, u\) in terms of \(y ;\) for \(y, z, u\) in terms of \(x\); but not for \(x, y, z\) in terms of \(u\).
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

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.