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I am working through some test review questions and this one has me stumped. I am not sure I am interpreting the question correctly. I originally thought this was about spans of subspaces, but we have not covered that, so I am lost. I have bounced around ideas, but the proof I have right now is:

By the definition of the F-span of ⟨v1,...,vn⟩, we know there exists {λ1v1 + ···+ λnvn : λi ∈F for all 1 ≤i ≤n} such that any v in V can be represented as a linear combination. The definition of a spanning set is a system of vectors is spanning if any v in V can be represented as a linear combination, so clearly ⟨v1,...,vn⟩ spans V over F

This seems way too simple and I feel like I am missing something. Can someone explain?

Cole Tran
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  • ok thank you, I think I was overthinking it – Cole Tran Jan 23 '23 at 06:11
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    No @geetha290krm, what he should say is 'there exists $S\subset${λ1v1 + ···+ λnvn : λi ∈F for all 1 ≤i ≤n} such that any v in V can be represented as a linear combination of elements of $S$'. – Anne Bauval Jan 23 '23 at 06:11
  • In order to avoid writing down many linear combinations, the proof can be done by contradiction. Observe that for any vectors $w_1,w_2,\ldots, w_k\in \langle v_1,v_2,\ldots, v_n\rangle $ we have $\langle w_1,w_2,\ldots, w_k\rangle \subseteq \langle v_1,v_2,\ldots, v_n\rangle .$ Assume by contradiction $v_1,v_2,\ldots, v_n$ is not a spanning set. Then $\langle v_1,v_2,\ldots, v_n\rangle $ is a proper subspace of $V.$ By the remark made in the second sentence of this comment $\langle v_1,v_2,\ldots, v_n\rangle $ does not contain a spanning set. – Ryszard Szwarc Jan 23 '23 at 08:16
  • This answers your question: Span of a subset of a vector space is the smallest subspace containing that set because, denoting by $W$ the subspace of $V$ spanned by $v_1,\dots,v_n,$ what you are asking reduces to: if $S\subset W$ then $\operatorname{span}(S)\subset W.$ – Anne Bauval Jan 29 '23 at 14:08

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