Prove some statements about the spans of finite sets of vectors

Let $A$ and $B$ be finite sets of vectors in $\mathbb{R}^n$ . Let $L(A)$ and $L(B)$ denote the linear spans of $A$ and $B$ . Prove the following.

Proof. Let $X = L(A)$ . Then, there are vectors $\alpha_1, \ldots, \alpha_r \in A$ such that
$X = c_1 \alpha_1 + \cdots + c_r \alpha_r.$

But, $\alpha_i \in A$ implies $\alpha_i \in B$ since $A \subseteq B$ . Hence,

$X = c_1 \alpha_1 + \cdots + c_r \alpha_r$

for $\alpha_1, \ldots, \alpha_r \in B$ . Therefore, $X \in L(B). \qquad \blacksquare$
If $X \in L(A \cap B)$ then
$X = c_1 \alpha_1 + \cdots + c_r \alpha_r$

for $\alpha_i \in A$ and $\alpha_i \in B$ . But, $\alpha_i \in A$ implies $X \in L(A)$ and $\alpha_i \in B$ implies $X \in L(B)$ . Thus, $x \in L(A) \cap L(B). \qquad \blacksquare$
Let $A = \{ (0,1), (1,0) \}$ and let $B = \{ (1,1) \}$ . Then,
$L(A) = \mathbb{R}^2$, \qquad L(B) = \{ (a,a) \mid a \in \mathbb{R} \}.$

So, $L(A) \cap L(B) = \{ (a,a) \mid a \in \mathbb{R} \}$ . But, $A \cap B = \varnothing$ so $L (A \cap B) = \varnothing. \qquad \blacksquare$