Proving H intersect K is subspace of H Suppose $H$ and $ K$ are subspaces of vector space $V$.
How to prove $H\cap K$ is subspace of $H$?
 A: To show that something is a subspace you have to show three things:


*

*It's closed under scalar multiplication.

*It's closed under addition

*It's non-empty


I'll do (1) so that you can see how it goes, you'll have to do (2) and for (3) show that it contains $0$.
To show that $H \cap K$ is closed under scalar multiplication assume $v \in H \cap K$ is a vector and $r$ is a scalar.  I need to show that $rv$ is contained in $H \cap K$.
As $v \in H \cap K$ and $H \cap K$ is exactly the set of vectors contained in both $H$ and $K$ we know that $v$ is contained in both $H$ and $K$.  We are told that $H$ is a subspace and subspaces are closed under scalar multiplication so $v \in H$ implies $rv \in H$.  With the same kind of argument you can get $rv \in K$ as well.  Now $rv$ is contained in both $H$ and $K$ so by definition $rv \in H \cap K$.  This proves that $H \cap K$ is closed under scalar multiplication.
A: Treat them as groups, and show that they intersection must be closed under addition, inverses. Then show that this subgroup is closed under linear combination.
