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I have a subspace $S$ and a vector set $V$, on what conditions is $V$ not a basis for $S$?

I know to be a basis, $\text{span } V = S$ and $V$ must be linearly independent.

So to not be a basis for $S$, $V$ must be either linearly dependent or $\text{span } V \ne S$.

Question: Say $V$ is linearly dependent (hence not a basis), could $\text{span } V = S$, still be a possibility?

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    Yes, in fact if $B$ is a basis, let $V = B \cup \{0\}$. Then $V$'s span is $S$ but $V$ is not a basis. – 2017-01-20
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    The conditions for $V$ to not be a basis for $S$, is either that it is linearly independent, or that it doesn't have the same cardinality as the dimension of $S$ (if it is finite). The answer can sometimes be specific to $S$ and $V$, For example, see the Muntz- Szasz theorem, that tells you **exactly** when a set of powers of $x$ span $C[0,1]$ with uniform norm, – 2017-01-20
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    @OpenBall - simpler example: Let $V = S$. – 2017-01-20

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