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I posted a similar question here

But as the title suggests, I'm trying to show that the converse is true.

Here's what I have:
let $y \in span\{S\} \cap span\{T\}$, then $y=c_1v_1+c_2v_2+...+c_nv_n$, where $c$ is in a field, and $\forall v \in S \cap T$.

Am I correct so far?

From there I'm not too sure how to show that $y \in span\{S \cap T \}$

Any suggestions would be appreciated. Thanks.

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    What you are trying to prove is not true. For example, let $S = \{ (1,0), (0,1)\}$ and $T = \{(1,1), (1,-1)\}$. Then $S \cap T = \emptyset$, but each of $S$ and $T$ span all of $\mathbb R^2$. Do you see what's wrong? – 2017-02-06
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    Could we say it's false because $\mathbb{R}^2$ is not a subset of the zero vector? Also, is this always false? – 2017-02-06
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    It's not even true in $1$ dimension. In $\mathbb R$ we have $span (\{1\})=span (\{2\})=\mathbb R$ but $span (\{1\}\cap \{2\})=span (\phi)=\phi. $The inclusion should be the other way around. – 2017-02-06

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