Linear algebra question Subspaces

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The discussion revolves around proving that the set S = { (a,b) | b > 0 } is a vector space under newly defined operations of addition and scalar multiplication. Participants emphasize the need to demonstrate closure under addition and scalar multiplication, noting that if b1 > 0 and b2 > 0, then b1*b2 > 0 ensures closure. It is clarified that, unlike subspaces of existing vector spaces, all vector space axioms must be verified due to the unique definitions provided. Key axioms include associativity, commutativity of addition, existence of a zero vector, and additive inverses. The original poster expresses gratitude after receiving guidance on how to approach the problem.
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Hey guys, new to the forum here, and its midterm time and I am working through a few questions and I can't seem to figure this one out.

Homework Statement



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.

Homework Equations



None

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
 
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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?
 
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.
 
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?
 
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.
 
Are you supposed to prove that V is a vector space, or that V is a subspace of R^2?
 
Says to prove that V is a vector space over R
 
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.
 
Okay, thanks guys! I just sort of needed a kick-start to get going. I figured it out, so thanks again!
 

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