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To reiterate:

Suppose $V$ is a vector space generated by a subset $S$, and $U$ is a proper subspace of $V$. Might one also say that $S$ generates $U$, since $U \subsetneq \text{Span}(S)$? Or, is it the fact that $U \neq \text{Span}(S)$ that implies we cannot say this?

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2 Answers 2

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Most certainly not. For example, take $S=\{(1,0,0), (0,1,0)\}$, and $U=\{(x, 0, 0)|x\in\mathbb R\}$. Then, clearly, $U$ is the subspace of $V=\{(x,y,0)|x,y\in\mathbb R\}$, however, $S$ does not generate $U$.

In fact, it is impossible for $U$ to be generated by $S$. By definition, the vector space, generated by $S$, is the smallest vector space that contains all vectors contained in $S$. Therefore, by definition, there does not exist any smaller vector space that also contains all of $S$.

In other words, if there exists some proper vector subspace $U$ such that $S\subset U$, then $V$ is, by definition, not generated by $S$.

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We have $V= Span(S)$. If we also have that $U= Span(S)$, then $U=V$.

Hence, if $U$ is a proper subspace of $V$, then $ U \subsetneq Span(S).$

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