Linear algebra question Subspaces

In summary, the student is having trouble understanding how to start proving that a set is a vector space. He is given new definitions of addition and scalar multiplication and has to prove that these operations are associative and commutative.
  • #1
Technique101
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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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  • #2
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
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
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
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
Are you supposed to prove that V is a vector space, or that V is a subspace of R^2?
 
  • #7
Says to prove that V is a vector space over R
 
  • #8
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
Okay, thanks guys! I just sort of needed a kick-start to get going. I figured it out, so thanks again!
 

FAQ: Linear algebra question Subspaces

What is a subspace in linear algebra?

A subspace in linear algebra is a subset of a vector space that maintains the properties of a vector space. This means that it is closed under addition and scalar multiplication, contains the zero vector, and is closed under linear combinations.

How do you determine if a set is a subspace?

To determine if a set is a subspace, you need to check if it satisfies the three properties of a vector space: closure under addition, closure under scalar multiplication, and contains the zero vector. If it satisfies all three, then it is a subspace.

Can a subspace contain the zero vector?

Yes, a subspace must contain the zero vector. This is one of the three properties of a vector space that a subspace must satisfy. The zero vector is essential for the closure under addition and scalar multiplication properties.

How are subspaces related to linear independence?

A set of vectors is linearly independent if none of the vectors can be written as a linear combination of the other vectors. Subspaces are made up of linearly independent vectors, as any linear combination of the vectors in the subspace will also be in the subspace.

Can a set of vectors that span a subspace be infinite?

Yes, a set of vectors that span a subspace can be infinite. As long as the vectors are linearly independent and satisfy the properties of a vector space, they can span a subspace regardless of the number of vectors in the set.

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