# Linear Algebra: Subspace proof

1. Mar 2, 2009

### lockedup

1. The problem statement, all variables and given/known data:

Prove: A set U $$\subset$$ V = (V, $$\oplus$$, $$\odot$$) is a vector subspace of V if and only if ($$\forall$$u1, u2 $$\in$$ U) (1/2 $$\odot$$ (u1 $$\oplus$$ u2) $$\in$$ U) and ($$\forall$$u $$\in$$ U) ($$\forall$$t $$\in \mathbb{R}$$) (t $$\odot$$ u $$\in$$ U).

3. The attempt at a solution:

I don't have the first clue. To me, it seems that there is missing information. I know that for a subspace, it is sufficient to prove only closure under addition and scalar multiplication. Maybe he's defining a different sort of addition? Ugh, the whole proof thing is actually pretty new to me. I only started doing simple proofs last semester in Discrete Mathematics...

Last edited: Mar 2, 2009
2. Mar 2, 2009

### phreak

He's using $$\oplus$$ and $$\odot$$ as representatives of addition and multiplication, respectively, just with other symbols. In this proof, you're working with an arbitrary vector space, so that addition and multiplication may not be defined in the sense that you're used to. In fact, they're both defined arbitrarily, and we can define addition and multiplication however we want to, as long as they satisfy the axioms of a vector space. However, even though the vector space operations are unknown, we do know how multiplication works in R.

In any case, here is what you must prove: for each t in R and for each u in U, $$t \odot u \in U$$ and for all $$u_1, u_2 \in U, \, u_1 \oplus u_2 \in U$$. The first thing you need to prove has already been given to you. All you need to prove is closure under addition. I'll give you a hint, and then you're going to have to think a little bit about what to do. The hint is: $$2\in \mathbb{R}$$. Now you're going to have to use the vector space axioms and the first given to figure out why closure under addition holds.

Last edited: Mar 2, 2009
3. Mar 2, 2009

### lockedup

Thank you. Yeah, the next problem defines u $$\oplus$$ v (u, v in R3) as v X u. This one is simple in that v X u is not commutative and therefore R3 with this definition is not a vector space.