# Proof that $\operatorname{span}(S_1\cap S_2) \subseteq \operatorname{span}(S_1) \cap \operatorname{span}(S_2)$

An exercise in my book says :

Prove that $\operatorname{span}(S_1\cap S_2) \subseteq \operatorname{span}(S_1) \cap \operatorname{span}(S_2)$. Give an example in which $\operatorname{span}(S_1 \cap S_2)$ and $\operatorname{span}(S_1) \cap \operatorname{span}(S_2)$ are equal and one in which they are unequal.

Why are those two not necessarily equal?

Thanks a lot.

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1-st part: Can you show that $A\subseteq B$ $\Rightarrow$ $span(A)\subseteq span(B)$? If yes, the rest of the proof is just elementary set theory. 2-nd part: What do you get for $S_1=\{e_1,e_2\}$ and $S_2=\{2e_2,e_3\}$, where $e_1=(1,0,0)$, $e_2=(0,1,0)$ and $e_3=(0,0,1)$? –  Martin Sleziak Sep 29 '12 at 13:53
@MartinSleziak 1st part: like so often, I think it's somewhat obvious that if A is a subset of B, then $span(A)$ won't contain any elements not in $span(B)$. Still I'm not sure how to 'prove' it in math-terms. –  foaly Sep 29 '12 at 13:57
@MartinSleziak ok since $S_1 \cap S_2 \subseteq S_1, S_2$, $span(S_1\cap S_2) \subseteq span(S_1), span(S_2)$, thus $span(S_1\cap S_2) \subseteq span(S_1)\cap span(S_2)$. That itself makes sense to me so far. –  foaly Sep 29 '12 at 14:05
no, the last step doesn't. How can I be sure that $span(S_1\cap S_2)$ does not always equal $span(S_1)\cap span(S_2)$, if I only know $span(S_1\cap S_2)$ is contained in $S_1$ and $S_2$? –  foaly Sep 29 '12 at 14:14
Martin's example in the first comment answers the question. –  Shankara Pailoor Sep 29 '12 at 14:28

I will write $[A]$ instead of $\operatorname{span}(A)$ for the sake of brevity.

Of course, the proofs depend on your definition of linear span. E.g., we can define the span of the set $A$ as the set of all linear combinations, i.e. $$[A]=\{c_1\vec{a}_1+\dots+c_n\vec{a}_n; n\in\mathbb N; c_1,\dots,c_n\in F, a_1,\dots,a_n\in A\},$$ where $F$ is the fields we are working with.

Observation 1. $A\subseteq B$ implies $[A]\subseteq[B]$.

The proof is basically just observing that every linear combination of elements of $A$ is a linear combination of elements of $B$.

Observation 2. $[S_1\cap S_2] \subseteq [S_1]\cap[S_2]$.

From observation 1 we get $[S_1\cap S_2]\subseteq [S_1]$ and $[S_1\cap S_2]\subseteq [S_2]$, which yields $[S_1\cap S_2] \subseteq [S_1]\cap[S_2]$.

Example 3. We want to find an example where the inclusion is strict. Let us try $S_1=\{e_1,e_2\}$ and $S_2=\{2e_1,e_3\}$, where $e_1=(1,0,0)$, $e_2=(0,1,0)$ and $e_3=(0,0,1)$. We can see that:

• $S_1\cap S_2=\emptyset$ and thus $[S_1\cap S_2]=\{0\}$
• $e_1\in[S_1]\cap [S_2] \subseteq [\{e_1\}]$ and $e_1\notin [S_1\cap S_2]$. So $[S_1\cap S_2]\subsetneqq [S_1]\cap[S_2]$ (In fact, we have $[S_1]\cap [S_2] = [\{e_1\}]$.)
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Thank you. Observation 1 and Example 3 seem obvious to me. I at first had problems to see why the first part of Observation 2 yields the second. Now I'm fine :) –  foaly Sep 29 '12 at 14:39