How to Prove that the Span of k-1 Vectors is also the Span of V?

[oxlade15]

Homework Statement

1. If V is spanned by {v1,v2, ..., vk} and one of these vectors can be written as a linear combination of the other k-1 vectors, prove that the span of these k-1 vectors is also V.

Homework Equations

A set S = {v1,v2, ..., vk}, k >= 2 is linearly dependent if and only if at least one of the vectors vj can be written as a linear combination of the other vectors in S.

The Attempt at a Solution

Since one of the vectors can be written as a linear combination of the other k-1 vectors, this means that the set of vectors is linearly dependent. Also, since V is spanned by {v1,v2, ..., vk}, then span(S) = {c1v1 + c2v2 + ...+ ckvk : c1, c2, ..., ck are real numbers} by the definition of the span of a set. In addition, by the definition of linearly dependent, there exists a nontrivial solution to the c1v1 + c2v2 +...+ ckvk = 0. These are all the pieces of information that I have deduced from the information given in the problem. From here, I am unsure as to how to proceed in this proof.

 

[hamsterman]

Here's what I suggest. Say V₁ is the span of {v₁,...v_k} and V₂ is the span of {v₁,...v_k-1}. Now, assuming that V₁ and V₂ are different, there exists a vector x in V₂, not in V₁ or a vector y in V₁, but not in V₂.

Since x is in V₂, there exists a sum c₁v₁ + c₂v₂ + ... + c_k-1v_k-1 = x. We can write a sum c₁v₁ + c₂v₂ + ... + c_k-1v_k-1 + 0v_k = x + 0 = x, expressing x with a base of V₁, so x is in V₁.

Now y. It is given that v_k is in V₂, so we'll write it as b₁v₁ + ... b_k-1v_k-1.
y = a₁v₁ + ... + a_k-1v_k-1 + a_kv_k = a₁v₁ + ... + a_k-1v_k-1 + a_k( b₁v₁ + ... b_k-1v_k-1 ) = (a₁+a_kb₁)v₁ + ... + (a_k-1 + a_kb_k-1)v_k-1, so y is in V₂.

Thus, since V₁/V₂ = V₂/V₁ = ∅, V₁ = V₂.