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Let $S$ be a subset of a vector space. I conjecture that $S$ is linearly dependent if and only if there exists a proper subset $S' \subset S$ such that $\operatorname{span}(S')=\operatorname{span}(S)$. Is this the case?

If this is the case, I see an analogy between this and the definition of infinite set as a set which is bijective with a proper subset of itself. Could someone elaborate on this if it is true?

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    $\begingroup$ Yes. Say you have $x \in S - S'.$ Then $x$ can be expressed as a finite linear combination of elements in $S'$ (since $\text{span} (S)=\text{span}(S')$ and $x\in S.$ $\endgroup$ – matt biesecker May 16 '15 at 4:23
  • $\begingroup$ There is an analogy between cardinality and dimension. Both measure size. Any infinite cardinal has a proper subset of the same size, and an infinite-dimensional vector space has a proper subspace with the same size. (Not sure where choice axioms come into play here.) $\endgroup$ – anon May 16 '15 at 4:31
  • $\begingroup$ @anon: There is no mention of cardinality here, just a strict subset. $\endgroup$ – copper.hat May 16 '15 at 4:40
  • $\begingroup$ @copper.hat OP does mention cardinality ("the definition of an infinite set as ..."). That is what I was responding to. $\endgroup$ – anon May 16 '15 at 4:56
  • $\begingroup$ What I was getting at is that with the definitions of both linear dependence and infinite cardinality we say that a set $X$ has property $P$ if there is a proper subset $X' \subset X$ such that $\{X, X'\}$ has property $Q$. In the former case, property $P$ is linear dependence and property $Q$ is both elements have the same span. In the latter case property $P$ is infinite cardinality and property $Q$ is there exists a bijection between both elements. $\endgroup$ – agcha May 16 '15 at 9:55
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If $S$ is linearly dependent, then we can write $\sum_k \alpha_k x_k =0$ for some $\alpha_k$ not all zero and $x_k \in S$. Without loss of generality, we can assume that $\alpha_1 \neq 0$, and so we have $x_1 \in \operatorname{sp} \{x_2,...\}$. Then let $S' = S \setminus \{x_1\}$, which is a proper subset, and clearly $\operatorname{sp} S = \operatorname{sp} S'$.

For the other direction, if $\operatorname{sp} S = \operatorname{sp} S'$ with $S'$ being a proper subset of $S$, then pick $x \in S \setminus S'$.

If $x=0$, then $S' \cup \{0\}$ is linearly dependent and hence so is $S$, otherwise we have $x =\sum_k \alpha_k x_k$ for some $x_k \in S'$. Hence $\sum_k \alpha_k x_k + (-1) x = 0$ and so $S$ is linearly dependent.

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