Is there some theorem that says that if the number of vectors of a base $S$ is less than the dimension of a vector space $V$ than it cannot span the vector space?
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$S$ can not be called a base if its cardinality is less than the dimension of your vector space. – Suzet Aug 01 '18 at 10:49
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It follows from the fact that all bases must have the same cardinality. In fact, if $S$ spans $V$, then there is a subset of $S$ that is a basis of $V$, by taking a linearly independent subset that is maximal by inclusion. But since the dimension of $V$ is strictly larger, then it has a basis of that larger cardinality, which is a contradiction. – Aug 01 '18 at 10:49
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How about the definition of dimension? – amd Aug 01 '18 at 20:11
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@amd: While the fact of equal cardinality of bases is essential to the definition of dimension, it is a theorem/proposition that needs proof that "dimension" is well-defined in this context. – hardmath Aug 02 '18 at 15:56
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@hardmath If you’re throwing around a term like “dimension,” there’d better be a definition of it already. That definition usually involves the minimal cardinality of a spanning set. This question puts the cart before the horse. – amd Aug 02 '18 at 20:30
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I suuppose that you're after this:
Theorem: If $V$ is a vector space, if $S\subset V$ and if $\#S<\dim V$, then $\operatorname{span}(S)\neq V$.
Yes, this is a standard Linear Algebra theorem.
Proof: I will assume that $\dim V<+\infty$. If $S\subset V$ and $\operatorname{span}(S)=V$, then, if $S'$ is a maximal linearly independent subset of $S$ such that $\operatorname(S')=V$ (it must exist, since $S$ is finite), $S'$ is a basis of $V$. Therefore$$\#S\geqslant\#S'=\dim V.$$
José Carlos Santos
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