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Suppose I have a general vector space $V$ (could be possible uncountably infinite dimensional) and a list of vectors $(e_s)_{s \in S}$ that is linearly independent but doesn't span the whole of $V$. Is it true that I will always have a subspace $U \subset V$ such that $V= \langle e_s \rangle_{s \in S} \oplus U$?

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up vote 2 down vote accepted

Yes, extend $\{e_s\}_{s\in S}$ to a basis of $V$, call it $B$. Letting $U =B \setminus\{e_s\}$ gives a complementary subspace. so that $V = \langle e_s\rangle _{s\in S}\oplus \langle U\rangle $

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Yes. You can always complete your linearly independent set to a basis, say $(e_s)_{s\in S}\cup(f_t)_{t\in T}$, and take $W=\text{span}\,\{f_t\}_{t\in T}$.

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The worst obstacle might be that the Axiom of Choice might be used. – Hagen von Eitzen Nov 21 '12 at 15:50
Hard to do much about infinite-dimensional spaces without it ;) – Martin Argerami Nov 21 '12 at 15:56
Oh, I see. Thank you. This is to do with the existence of the Hamel basis which uses Zorn's lemma which is equivalent to the axiom of choice, right? – chango Nov 21 '12 at 16:01
Indeed.${\ \ \ \ }$ – Martin Argerami Nov 21 '12 at 16:06

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