I have tried to research more about this following problem, and I have tried to come up with a proof myself, but unfortunately I can’t think of anything that makes sense. I am having trouble with part (2) and (4) of this problem. Would really appreciate any help or clarification on how to begin solving this problem.

**Question:**

Prove that every finitely generated vector space has a

basis. In fact, every vector space has a basis, but the proof of that is beyond the scope of this course. Let V be a non-zero finitely generated vector space.

**(1) Let w, v1,…,vn be in V . Prove w is in Span(v1,…, vn) if and only if Span(w, v1,…, vn) = Span(v1,…, vn).**

**(2) Prove, using part (1) that there exists a finite set X with nonzero members such that V = SpanX.**

**(3) Suppose V is spanned by a finite set of cardinality n. By part (2) we can assume 0 not in X. Our claim is V has a basis. We will use induction on n. Prove that the induction base holds. That is,**

if n = 1, then V has a basis.

**(4) Induction hypothesis: suppose that V has a basis whenever it is spanned by a set of cardinality n. Now suppose V is spanned by a set of cardinality n + 1. Prove that V has a basis.**

(2) We are assuming that $V$ is a non-zero finitely generated vector space. What this *means* is that $Vneq{0}$ and that there is a finite set $X={v_1,ldots,v_n}$ which spans $V$. If $0notin X$, you’re done. Otherwise, if, say, $v_1=0$, then, by (1), $V=operatorname{span}bigl({v_2,ldots,v_n}bigr)$ and ${v_2,ldots,v_n}$ is a set of non-zero vectors.

(4) Suppose that $V$ is spanned by ${v_1v_2,ldots,v_{n+1}}$. There are two possibilities:

- the set ${v_1v_2,ldots,v_{n+1}}$ is linearly dependent. Then one of the vectors belongs to the span of the others. You can assume without loss of generality that $v_{n+1}inoperatorname{span}bigl({v_1v_2,ldots,v_n}bigr)$. Therefore, $V=operatorname{span}bigl({v_1v_2,ldots,v_n}bigr)$ But then, by the induction hypothesis, $V$ has a basis.
- the set ${v_1v_2,ldots,v_{n+1}}$ is linearly independent. But then it is a basis.