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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.

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