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.

  • 1
    $\begingroup$ I'm not sure what part 2 has to do with part 1. The existence of $X$ is simply the definition of being finitely generated. $\endgroup$ – eyeballfrog May 26 '19 at 20:56
  • $\begingroup$ @eyeballfrog That's what I'm confused about!! $\endgroup$ – killpwnrang3 May 26 '19 at 21:12

(2) We are assuming that $V$ is a non-zero finitely generated vector space. What this means is that $V\neq\{0\}$ and that there is a finite set $X=\{v_1,\ldots,v_n\}$ which spans $V$. If $0\notin X$, 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 $v_{n+1}\in\operatorname{span}\bigl(\{v_1v_2,\ldots,v_n\}\bigr)$. 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.
  • $\begingroup$ For (2), I thought of the same thing, but it is asking to use (1) in the proof, and I'm not sure how to incorporate that into the proof $\endgroup$ – killpwnrang3 May 26 '19 at 21:10
  • $\begingroup$ Done. ${}{}{}{}$ $\endgroup$ – José Carlos Santos May 26 '19 at 21:18

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