A subspaces problem.

1. Oct 30, 2007

Evalesco

Hi, I've being going over some course material in preparation for my exams and I've come across an exercise which I can't seem to work out.

The exercise:
Let V be a vector space over a field F and let U and W be two subspaces of V.
Suppose V = U + W. Prove that V = U ⊕ W iff {(u,w)∈U×W : u + w = 0} = {(0,0)}

What I've done so far:
At first I was unsure of the operator ⊕ but from Wikipedia I discovered the following:
V = U ⊕ W ⇔ (V = U + W) ∧ (U ∩ W = ∅)
Which I guess is a consequence of the following: The dimension of V ⊕ W is equal to the sum of the dimensions of V and W.

So I thought ok that identity will make it quite easy to solve, I just need to make sure U and W share no common elements.

We are given {(u,w)∈U×W : u + w = 0} = {(0,0)}
Ok I thought, that uses the property that U and W are both closed under addition to ensure the two subspaces U and W have no non-zero elements in common.

But this is where I'm stuck I thought in order to solve it I would need U and V to share no common elements but 0 can be both an element of U and V.

I know this isn't a proof but I was just trying to formulate an idea of how I would go about solving it. Any insight would be welcome, Thanks.

2. Oct 30, 2007

OK, let V be the direct sum of U and W. It follows that U$\cap$W = {0}, so there cannot exist an element of (u, v) from U x W such that u + w = 0, because then the intersection of U and W would not be trivial.

For the other direction, assume {(0, 0)} = { (u, v) from U x W : u + w = 0}. Take an element from U$\cap$W, let's say x. Then x is in U and x is in W. Since the intersection is a subspace, there exists an inverse for the element x. It's easy to finish the proof now.

3. Oct 30, 2007

Evalesco

Thanks for the clarification, it helped me heaps, it looks like I had my definitions a little off lol.