/*! 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 16 Prove that a set \(S\) of vector... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 16

Prove that a set \(S\) of vectors is linearly independent if and only if each finite subset of \(S\) is linearly independent.

Short Answer

Expert verified
To prove that a set $S$ of vectors is linearly independent if and only if each finite subset of $S$ is linearly independent, we demonstrate the following: 1. If $S$ is a set of linearly independent vectors, then each finite subset of $S$ is linearly independent: Given a finite subset $A$ of $S$, all vectors in $A$ are also in $S$. Since $S$ is linearly independent, any linear combination of the vectors in $A$ equals 0 only when all coefficients are zero, making $A$ linearly independent. 2. If each finite subset of $S$ is linearly independent, then the set $S$ of vectors is linearly independent: Consider any finite collection of vectors from $S$. As every finite subset of $S$ is linearly independent, any linear combination of the vectors that equals the zero vector implies all coefficients are zero. Thus, we conclude that a set $S$ of vectors is linearly independent if and only if each finite subset of $S$ is linearly independent.

Step by step solution

01

Part 1: If a set S is linearly independent, then each finite subset of S is linearly independent.

Suppose S is a set of linearly independent vectors. Let A be a finite subset of S, and let's assume that \(v_1, v_2, ..., v_n\) are the vectors in A. To prove that A is linearly independent, we need to show that every linear combination of the vectors in A equals the zero vector only when all the coefficients are zero. Formally, we have to show that for any scalar coefficients \(a_1, a_2, ..., a_n\): \(a_1v_1 + a_2v_2 + ... + a_nv_n = 0\) implies \(a_1 = a_2 = ... = a_n = 0\) Since S is linearly independent and all the vectors in A are also in S, by the definition of linear independence, we can say that this condition holds. Thus, each finite subset A of S is also linearly independent.
02

Part 2: If every finite subset of S is linearly independent, then the set S is linearly independent.

Suppose each finite subset of S is linearly independent. Now, we need to show that S is linearly independent. Let \(\{v_1, v_2, ..., v_n\}\) be a finite collection of vectors from S. Since every finite subset of S is linearly independent, the set \(\{v_1, v_2, ..., v_n\}\) is also linearly independent. To prove that S is linearly independent, we need to show that for any given finite collection of vectors from S, any linear combination that equals the zero vector implies that all the coefficients are zero. Formally, we have to show that for any scalar coefficients \(a_1, a_2, ..., a_n\): \(a_1v_1 + a_2v_2 + ... + a_nv_n = 0\) implies \(a_1 = a_2 = ... = a_n = 0\) Since every finite subset of S is linearly independent, this condition holds. Hence, the set S of vectors is linearly independent. Combining both parts, we have proven that a set S of vectors is linearly independent if and only if each finite subset of S is linearly independent.

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

Prove that a subset \(W\) of a vector space \(V\) is a subspace of \(V\) if and only if \(0 \in \mathrm{W}\) and \(a x+y \in \mathrm{W}\) whenever \(a \in F\) and $x, y \in W$.

Show that the set of convergent sequences is an infinite-dimensional subspace of the vector space of all sequences of real numbers. (See Exercise 21 in Section 1.3.)

The set of all \(n \times n\) matrices having trace equal to zero is a subspace \(W\) of \(M_{n \times n}(F)\) (see Example 4 of Section 1.3). Find a basis for W. What is the dimension of W?

Let \(u=(1,-2,4), v=(3,5,1), w=(2,1,-3) .\) Find: (a) \(3 u-2 v\) (b) \(5 u+3 v-4 w\) \(\begin{array}{llll}\text { (c) } u \cdot v, & u \cdot w, & v \cdot w ;\end{array}\) (d) \(\|u\|,\|v\|\) (e) \(\cos \theta,\) where \(\theta\) is the angle between \(u\) and \(v\) \((\mathrm{f}) \quad d(u, v)\) \((g) \quad \operatorname{proj}(u, v)\)

Prove: For any complex numbers \(z, w \in \mathbf{C},(\text { i) } \overline{z+w}=\bar{z}+\bar{w}\) (ii) \(\overline{z w}=\bar{z} \bar{w}\) (iii) \(\bar{z}=z\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Geometry

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.