"Let $H$ and $K$ be subspaces of a vector space $V$, and $H\cap K:=\{v\in V \vert v \in H \land v\in K\}$. Show that $H\cap K$ is a subspace of $V$."
The zero vector is in $H\cap K$, since $0 \in H$ and $0\in K$ ( They're both subspaces… ). Therefore $H\cap K \neq \emptyset$.
Since $H$ and $K$ are subspaces, we have that for all scalars $c$ : if $v\in H$ then $cv \in H$ and also if $v\in K$ then $cv \in K$. It follows that for all $v\in H\cap K$ and all scalars $c$ : $ cv \in H \cap K$. Thus $H\cap K$ is closed under multiplication by scalars.
How should I show that $H\cap K$ is closed under vector addition? That is, for all $v,u \in H\cap K$ we have that $v+u \in H\cap K$
Best Answer
Since $H\subseteq V$ and $K\subseteq V$ are linear subspaces, you may approach this problem by showing that $av + w$ belongs to $H\cap K\subseteq V$ whenever $v\in H\cap K$ and $w\in H\cap K$, where $a\in\textbf{F}$, the field associated to $V$.
Indeed, this is the case. If $v\in H\cap K$, then $v\in H$ and $v\in K$. Thus $av\in H$ and $av\in K$. Moreover, since $w\in H\cap K$, then $w\in H$ and $w\in K$. Since $H$ and $K$ are subspaces, we conclude that $av + w\in H$ and $av + w\in K$, that is to say, $av + w\in H\cap K$, and we are done.
BONUS
More generally, you can prove that the intersection of an arbitrary number of linear subspaces of $V$ is also a subspace. Can you prove it based on the previous argument?