Is {(x,y): x>/0; y>/0} a Vector Space in R^2?

[012983]

Homework Statement

Just started learning vector spaces... not as fun as matrices. Anyway, I have a problem here, and I just want to make sure I'm understanding it correctly.

It states: "The set {(x,y): x>/0; y>/0} with the standard operations in R^2." It asks me to prove whether or not it's a vector space.

Homework Equations

The Attempt at a Solution

Does that mean I pick an arbitrary set u, let's say (4,6), and figure out if it forms a vector space using the ten axioms? I don't understand how that point would fit in R^2, so I'm missing something here.

Thanks for any help.

Edit: would it fail under axiom six (cU is in V)? Since if c is a negative number, U would be negative, which wouldn't satisfy the >/ requirement? Or am I learning this incorrectly?

 

[Mark44]

Homework Statement

Just started learning vector spaces... not as fun as matrices. Anyway, I have a problem here, and I just want to make sure I'm understanding it correctly.

It states: "The set {(x,y): x>/0; y>/0} with the standard operations in R^2." It asks me to prove whether or not it's a vector space.

What does this notation - x>/0 - mean?
Is it supposed to say x >= 0? That's the usual notation.

Homework Equations

The Attempt at a Solution

Does that mean I pick an arbitrary set u, let's say (4,6),

That's neither arbitrary nor a set. That is a specific vector in R² (i.e., in the plane.

and figure out if it forms a vector space using the ten axioms? I don't understand how that point would fit in R^2, so I'm missing something here.

Yes. Assuming you have to verify that all the axioms are true for this set, start with a couple of arbitrary vectors, such as u = <u₁, u₂> and v = <v₁, v₂>, and an arbitrary scalar.

Inasmuch as R² is a vector space and you're dealing with a subset of it, perhaps all you need to do is show that your set is a subspace of R². In that case, all you need to do is show that 1) <0, 0> is in your set, 2) the set is closed under addition, and 3) the set is closed under scalar multiplication.

Thanks for any help.

Edit: would it fail under axiom six (cU is in V)? Since if c is a negative number, U would be negative, which wouldn't satisfy the >/ requirement? Or am I learning this incorrectly?

I'm not sure what the ">/" requirement is.

 

[012983]

What does this notation - x>/0 - mean?
Is it supposed to say x >= 0? That's the usual notation.
That's neither arbitrary nor a set. That is a specific vector in R² (i.e., in the plane.
Yes. Assuming you have to verify that all the axioms are true for this set, start with a couple of arbitrary vectors, such as u = <u₁, u₂> and v = <v₁, v₂>, and an arbitrary scalar.

Inasmuch as R² is a vector space and you're dealing with a subset of it, perhaps all you need to do is show that your set is a subspace of R². In that case, all you need to do is show that 1) <0, 0> is in your set, 2) the set is closed under addition, and 3) the set is closed under scalar multiplication.

I'm not sure what the ">/" requirement is.

Sorry, had a brain fart there and forgot what the standard notation was... indeed I meant >=, or rather ≥.

Could you explain what it means to be closed under addition and scalar multiplication? For the former, does it mean that whenever two vectors within vector space V are added they will fall within vector space V? And similarly for scalar multiplication, that whenever a vector in vector space V is multiplied by a scalar it will fall within vector space V?

 

[Mark44]

Sorry, had a brain fart there and forgot what the standard notation was... indeed I meant >=, or rather ≥.

Could you explain what it means to be closed under addition and scalar multiplication? For the former, does it mean that whenever two vectors within vector space V are added they will fall within vector space V?

Yes. If u and v are in the space, then u + v is also in the space.

And similarly for scalar multiplication, that whenever a vector in vector space V is multiplied by a scalar it will fall within vector space V?

Right.

 

[012983]

Yes. If u and v are in the space, then u + v is also in the space.

Right.

So if u(4,2) is a vector within R^2, is it not a vector space since (I think) it fails the axiom 'cU is within V?' Given that x>=0 and y>=0 as defined in the problem statement.

 

[Mark44]

So if u(4,2) is a vector within R^2, is it not a vector space since (I think) it fails the axiom 'cU is within V?' Given that x>=0 and y>=0 as defined in the problem statement.

A better way to say that is - <4, 2> is a vector in the set, but -1*<4, 2> = <-4, -2> is not in the set, so the set is not closed under scalar multiplication.

 

[012983]

A better way to say that is - <4, 2> is a vector in the set, but -1*<4, 2> = <-4, -2> is not in the set, so the set is not closed under scalar multiplication.

Thanks, Mark44. Appreciate all your help.