# Linear algebra question Subspaces

1. Nov 3, 2009

### Technique101

Hey guys, new to the forum here, and its midterm time and Im working through a few questions and I can't seem to figure this one out.

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

Let S = { (a,b) | b > 0 } and define addition by (a,b) + (c,d) = (a*d + a*c, b*d) and define scalar multiplication by k(a,b) = ( k*a*b^(k-1) , b^k ).
Prove that S is a vector space of R.

2. Relevant equations

None

3. The attempt at a solution

I'm just confused! I want to prove that it's closed under addition, scalar multiplication, but I don't know how to start for this one.

Thanks

2. Nov 3, 2009

### Staff: Mentor

Let (a1, b) and (a2, b2) be two elements of set S.

Show that (a1, b) + (a2, b2) is also in set S.
Show that k(a2, b2) is in set S.

How can you tell if a pair (u, v) is in S?

3. Nov 3, 2009

### Dick

You'll need to prove a lot more than just closure, but you can certainly start with that. If b>0 and d>0 does (a*d + a*c, b*d) satisfy the condition that b*d>0? That's additive closure isn't it? Is it closed? Just take the properties one at a time.

4. Nov 4, 2009

### Technique101

Okay, so i state that since b1 > 0 and b2 > 0, therefore b1*b2 > 0, and therefore is closed under addition and therefore is in the set.

I do the same for scalar multiplication.

For a question like this, do I really need to prove all 10 axioms, or is there a more simplified way to prove that it is a vector space?

5. Nov 4, 2009

### Dick

Ok, you've got closure. But no, you aren't done. The other axioms are important. Like I said, take them one at a time.

6. Nov 4, 2009

### Staff: Mentor

Are you supposed to prove that V is a vector space, or that V is a subspace of R^2?

7. Nov 4, 2009

### Technique101

Says to prove that V is a vector space over R

8. Nov 4, 2009

### HallsofIvy

Staff Emeritus
The reason Mark44 was confused was that you titled this "linear algebra question... subspaces" and originally said "Prove V is a vector space of R".

If you were given a vector space V and asked to show that U is a subspace of V, then you would only have to prove that U is closed under addition and scalar multiplication because all the other properties, associativity and commutativity of addition, etc. follow from the fact that V is a vector space.

But here you are given new definitions of addition and scalar multiplication so you have to prove that addition is associative and commutative, that there is a "zero" vector, that every vector has an additive inverse, etc.

9. Nov 4, 2009

### Technique101

Okay, thanks guys! I just sorta needed a kick-start to get going. I figured it out, so thanks again!