Prove that $\text{span}(S_{1}\cap S_{2})\subseteq \text{span}(S_{1})\cap\text{span}(S_{2})$.

Question

Let $S_{1}$ and $S_{2}$ be subsets of a vector space $V$. Prove that $\text{span}(S_{1}\cap S_{2})\subseteq \text{span}(S_{1})\cap\text{span}(S_{2})$. Give an example in which $\text{span}(S_{1}\cap S_{2})$ and $\text{span}(S_{1})\cap\text{span}(S_{2})$ are equal and one in which they are unequal.

MY ATTEMPT

If $v\in\text{span}(S_{1}\cap S_{2})$, then $v$ can be expressed as the linear combination of vectors common to $S_{1}$ and $S_{2}$. Consequently, it is the linear combination of vectors from $S_{1}$ as well as $S_{2}$, that is to say, $\text{span}(S_{1}\cap S_{2}) \subseteq \text{span}(S_{1})\cap\text{span}(S_{2})$.

As to the second part, it suffices to consider $S_{1} = \{e_{1},e_{2}\}\subseteq\textbf{R}^{3}$ and $S_{2} = \{e_{1},e_{3}\}\subseteq\textbf{R}^{3}$.

In such case, one has that

$$\text{span}(S_{1}\cap S_{2}) = \text{span}(\{e_{1}\}) = \text{span}(\{e_{1},e_{2}\})\cap\text{span}(\{e_{1},e_{3}\}) = \text{span}(S_{1})\cap\text{span}(S_{2})$$

As to the last question, I am having trouble to answer it properly.

Can someone help me out with it?

Answer 1

Hint

$$\text{span}(\{e_{1},e_{2}\}) = \text{span}(\{e_{1}+e_2,e_{2}\})$$ and $$ \{e_{1}+e_2,e_{2}\}\cap \{e_{1},e_{3}\}= \emptyset$$

If you do not want to work with the emptyset, work instead in $\mathbb R^2$ and add $e_4$ to both spanning sets.

Answer 2

This follows from the definition on Span. Indeed, let $$x \in \text{Span}(S_{1} \cap S_{2}) ,$$ then by definition of Span, $$ x = a_1u_1 + a_2u_2 + \dots + a_n u_n.$$ Then since \begin{align} &u_1, \dots ,u_n \in \text{Span}(S_{1}), \implies x \in \text{Span}(S_{1}) \\ \quad \text{and } \quad &u_1, \dots , u_n \in \text{Span} (S_2) \implies x \in \text{Span}(S_2) \end{align} $$ \therefore x\in \text{Span}(S_{1}) \cap \text{Span}(S_{2}) $$ We conclude that $\text{Span}(S_{1}\cap S_{2}) \subseteq \text{Span}(S_{1}) \cap \text{Span}(S_{2})$.

Equality

If $S_{1}= S_{2} \equiv S \implies S_{1} \cap S_{2} = S \implies \text{Span}(S) = \text{Span}(S) \cap \text{Span}(S) = \text{Span}(S) $.

Answer 3

Another approach to solve this is using the alternative definition of $\operatorname{span}$, you only need to recall that the intersection of an arbitrary number of linear subspaces is also a linear subspace:

Definition. Let $S$ be a subset of a vector space $V$. Consider the family $$\mathcal{F}_S = \{ H \subseteq V :\, H \textrm{ is a subspace of $V$ and $S \subseteq H$} \}$$ which is non-empty since it contains $V$. Then, define $\operatorname{span}(S) := \bigcap \mathcal{F}_S$.

With this definition, if $S_1$ and $S_2$ are subsets of a vector space $V$ such that $S_1 \subseteq S_2$, then $\mathcal{F}_{S_2} \subseteq \mathcal{F}_{S_1}$ and then $\operatorname{span}(S_1) \subseteq \operatorname{span}(S_2)$.

Finally, the proof of your problem is as goes:

$\quad$Proof. Since $S_1 \cap S_2$ is contained in $S_1$ and $S_2$, then $\operatorname{span}(S_1 \cap S_2)$ is contained in $\operatorname{span}(S_1)$ and $\operatorname{span}(S_2)$, which is the same as saying that $\text{span}(S_{1}\cap S_{2})\subseteq \text{span}(S_{1})\cap\text{span}(S_{2})$. $\blacksquare$

Also, for the last asking, you don't have to think in a too complicated example, just consider $S_1 = \{e_1\}$ and $S_2 = \{-e_1\}$. Observe that, with the definition that I gave you, we have $\operatorname{span}(\varnothing) = \{\textit0\,\}$.