Is V a Vector Space? A Quick Question on Axiom 8 Explained

[kwal0203]

Homework Statement

Determine whether this set equipped with the given operations is a vector space. For those that are not vector spaces identify the axiom that fails.

Set = V = all pairs of real real numbers of the form (x,y) where x>=0, with the standard operations on R^2.

The Attempt at a Solution

This set is not a vector space because it is not closed under scalar multiplication I.e. -1*(1,1)=(-1,-1) which is not in V as x<0 and because there is not always a vector in V such that u+(-u)=(-u)+u=0 I.e. when u=(1,1) then -u=(-1,-1) which is not in V as again x<0.

My question is why does axiom 8 hold which states:

(K+m)u=ku+km

I.e. if k=-1, m=-1, u=(1,1) ----> (-1+-1)u=(-1,-1)+(-1,-1)=(-2,-2) which is not in V as x<0.

Does axiom 8 not require the solution to be in the set V?

 

[micromass]

You showed that the space is not closed under scalar multiplication, so it's not a vector space. But also, since it's not closed under scalar multiplication, all the axioms which use scalar multiplication make no sense. It makes no sense to asl $(\alpha + \beta) v = \alpha v + \beta v$, since scalar multiplication is not well-defined.

so I wouldn't say that Axiom 8 holds in this case. I would rather say that it makes no sense.

 

[kwal0203]

You showed that the space is not closed under scalar multiplication, so it's not a vector space. But also, since it's not closed under scalar multiplication, all the axioms which use scalar multiplication make no sense. It makes no sense to asl $(\alpha + \beta) v = \alpha v + \beta v$, since scalar multiplication is not well-defined.

so I wouldn't say that Axiom 8 holds in this case. I would rather say that it makes no sense.

Oh I see, but it could be the case that a set is closed under scalar multiplication but one of the axioms that depend on that, such as axiom 8, do not hold. Just not vice versa.

Thanks!