# Linear independence proof

Mr Davis 97

## Homework Statement

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

## The Attempt at a Solution

I think that that it would be easier to prove the logically equivalent statement: Prove that a set S of vectors is linearly dependent if and only if there exists a finite subset of S that is linearly dependent.

First assume that a set S is linearly dependent. Then there exists a linear dependence relation among its vectors ##a_1s_1 + \cdots + a_ns_n = 0##, such that not all ##a_1,..., a_n## are zero. Since S is a subset of itself, there must exist a finite subset of S that is linearly dependent.

Second, assume that S has a finite subset S' that is linearly dependent. Since S' is a subset of S, this means that there exists a linear dependence relation among the vectors in S. Thus, S is also linearly dependent.

Does this prove the original statement? It seemed a little too easy.

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## Homework Statement

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

## The Attempt at a Solution

I think that that it would be easier to prove the logically equivalent statement: Prove that a set S of vectors is linearly dependent if and only if there exists a finite subset of S that is linearly dependent.

First assume that a set S is linearly dependent. Then there exists a linear dependence relation among its vectors ##a_1s_1 + \cdots + a_ns_n = 0##, such that not all ##a_1,..., a_n## are zero. Since S is a subset of itself, there must exist a finite subset of S that is linearly dependent.

Second, assume that S has a finite subset S' that is linearly dependent. Since S' is a subset of S, this means that there exists a linear dependence relation among the vectors in S. Thus, S is also linearly dependent.

Does this prove the original statement? It seemed a little too easy.

I think the result is wrong: consider ##S = \{ (1,0,0), (0,1,0), (0,0,1), (1,1,1)\}.## Any three of the vectors in ##S## are linearly independent, but ##S## itself is not.

Mr Davis 97
I think the result is wrong: consider ##S = \{ (1,0,0), (0,1,0), (0,0,1), (1,1,1)\}.## Any three of the vectors in ##S## are linearly independent, but ##S## itself is not.
But S is a subset of itself

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But S is a subset of itself

Of course I understand that; but if the question meant "including itself" then it is trivial and pointless.

Mr Davis 97
Of course I understand that; but if the question meant "including itself" then it is trivial and pointless.
The problem is from Friedberg, a widely used textbook. So I don't think the statement could be wrong. Maybe it is just a trivial exercise...