Proving The Linear Dependence Theorem: A Guide To Non-Trivial Linear Combination In Math.

{v1…vp} is linearly dependent if and only if

at least one of the vectors is in the span of the othersyou can remove at least one vector without shrinking the span

there exists a non-trivial linear combination of {v1…vp} that equals the zero vector.

To prove this statement, we must first define what it means for a set of vectors to be linearly dependent. A set of vectors {v1…vp} is said to be linearly dependent if there exist scalars c1, c2,…,cp not all equal to zero, such that c1v1 + c2v2 + … + cpvp = 0.

Now, we will prove the original statement:

Statement: {v1…vp} is linearly dependent if and only if there exists a non-trivial linear combination of {v1…vp} that equals the zero vector.

Proof:

Forward Direction: Assume that {v1…vp} is linearly dependent. This means that there exist scalars c1, c2,…,cp not all equal to zero, such that c1v1 + c2v2 + … + cpvp = 0. Without loss of generality, let c1 = 0 (if not, renumber the vectors). Then, we have:

c2v2 + c3v3 + … + cpvp = 0

Divide both sides by c2 (since c2 cannot be zero, as that would imply all c’s are zero), and we get:

v2 + (c3/c2)v3 + … + (cp/c2)vp = 0

Therefore, we have found a non-trivial linear combination of {v1…vp} that equals the zero vector.

Backward Direction: Assume that there exists a non-trivial linear combination of {v1…vp} that equals the zero vector. Without loss of generality, assume that v1 is one of the vectors in the linear combination (if not, renumber the vectors). Then, we have:

c1v1 + c2v2 + … + cpvp = 0

Divide both sides by c1 (since c1 cannot be zero, as that would imply all c’s are zero), and we get:

v1 + (c2/c1)v2 + … + (cp/c1)vp = 0

This means that {v1…vp} is linearly dependent, as we have found a non-trivial linear combination of {v1…vp} that equals the zero vector.

Therefore, we have proven that {v1…vp} is linearly dependent if and only if there exists a non-trivial linear combination of {v1…vp} that equals the zero vector.

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