Linear Algebra Proof 1.?

Question

Let U and V be subspaces of a vector space W. Prove that their intersection U (intersection) V is also a subspace of W.

I don't understand. If U lies in W, how could the intersection EVER be outside of W?

Answer 1

There is a distinction in linear algebra between a subset and a subspace. To show that the intersection is a subspace, you must show that it has the properties of a vector space - namely, that it is closed under linear combinations. This is not hard, just use that U and V are closed under linear combinations to show that any combination of vectors in both U and V is in U, then use the same logic to show that it is in V, and conclude that it is in U ∩ V.

Answer 2

Pascal outlined it fairly nicely. The sufficient condition that a subset of W be a subspace is that every linear combination of 2 vectors in the set is in W. Call this "the condition."

So any pair of vectors in U ∩ V are in U, and since U is a subspace of W, they satisify the condition. That makes U ∩ V a subspace. Likewise the same pair of vectors would be in V, and since V is a subspace of W.....

Almost trivial, isn't it?