Subspace of Vectors Let $W_1$ and $W_2$ be subspaces of a vector space $V$,


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*Show that $W_1\cap W_2$ is a subspace of $V$.

*Let $W_1+W_2=\{w_1+w_2 \mid w_1 ∈ W1, w_2 ∈ W_2\}$. Show that $W_1+W_2$ is also a subspace of $V$.


Any suggestions where to start welcome.
 A: I dont know what you mean $W_1 W_2$. However, to show $W_1 + W_2 \leq V$ is fairly routine. In fact, let $u,v \in W_1$ and $u'v' \in W_2$. We have that $u+v \in W_1$ and $u'+v' \in W_2$. It follows that 
$$ (u + v) + (u' + v') \in W_1 + W_2 $$
Let $\alpha \in \mathbb{K}$ and $u,u' \in W_1,W_2$ respectively. Hence, $\alpha u \in W_1 $ and $\alpha u' \in W_2$. It follows that 
$$ \alpha u + \alpha u' \in W_1 + W_2 $$
A: To show that $W_1 \cap W_2$ is a subspace, first note that the $0$ element is in the intersection, since every subspace of $V$ contains zero. Then we must show that this set is closed under vector addition and scalar multiplication. 
Let $u,v \in W_1 \cap W_2$. It follows that $u,v \in W_1$, and $W_1$ is, by definition, closed under addition. Hence $u+v \in W_1$. The same is similarly true for $W_2$, and we can conclude that $u+v \in W_1 \cap W_2$. 
Now let $\alpha$ be a scalar. The argument for closure of scalar multiplication is similar. For $u \in W_1 \cap W_2$, we will have, by virtue of $W_1$ and $W_2$ being subspaces, $\alpha u \in W_1$ and $\alpha u \in W_2$. Hence $\alpha u \in W_1 \cap W_2$.
So we have closure of vector addition, closure of scalar multiplication, and the zero vector. These are the three things we need to prove something is a subspace. 
