Theorem 1.9. If a vector space V is generated by a finite set S, then some subset of S is a basis for V. Hence V has a finite basis.

Proof. If S = or S = {0 }, then V = {0 } and is a subset of S that is a basis for V. Otherwise S contains a nonzero vector u1. By item 2 on page 37, {u1} is a linearly independent set. Continue, if possible, choosing vectors u2,...,uk in S such that {u1, u2,...,uk} is linearly independent. Since S is a finite set, we must eventually reach a stage at which = {u1, u2,...,uk} is a linearly independent subset of S, but adjoining to any vector in S not in produces a linearly dependent set. We claim that is a basis for V. Because is linearly independent by construction, it suffices to show that spans V.

By Theorem 1.5 (p. 30) we need to show that S span(). Let v S. If v , then clearly v span(). Otherwise, if v / , then the preceding construction shows that {v} is linearly dependent. So v span() by Theorem 1.7 (p. 39). Thus S span().

Hi there: Iam working on the proof of this question, but I don't know why we need to prove S span() as I marked above? Before this part, we already proved that is a basis for V, correct?

here are the Theorem:

Theorem 1.7. Let S be a linearly independent subset of a vector space

V, and let v be a vector in V that is not in S. Then S {v} is linearly dependent if and only if v span(S).

Theorem 1.5. The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S.