Finding the zero vector

1. Feb 6, 2008

torquerotates

I came across an interesting problem. I'm trying to determine if the following is a vector space, x+y=xy, kx=x(risen to power k), and I came across an interesting result. I used ax. 4 to show x+y+1=(x+y)+1=(xy)+1=1(xy)=xy=x+y. Doesn't that just seem strange that 1 is the zero vector. 1 is not even a vector let alone 0. Is there something wrong with my thinking?

let a,b,c be vectors and V is a vector space, then
1)a&b is in V then a+b is in V
2)a+b=b+a
3)a+(b+c)=(a+b)+c
4)0+a=a+0=a
5)a+(-a)=(-a)+a=0
6)a is in V implies ka is in V
7)k(a+b)=ka+kb
8)(k+m)a=ka+ma
9)k(ma)=(km)a
10) 1a=a

2. Feb 6, 2008

Hurkyl

Staff Emeritus
What is the set of vectors? What is the field of scalars?

Why not?

3. Feb 6, 2008

torquerotates

Well, the problem said, the set of all positive real numbers such that, x+y=xy, kx=x(risen to power k). But if we designate this as set of vectors, not scalars( something can be both simulataneously). Then I would think that I arrived at a contradiction. Because 1 is obviously a scalar. We can designate it as a vector. I have no problems with that. But then in this situation, what would be a scalar? Obviously it can't be any number k since k is a multiple of one and one is a vector. The thing thats getting me is that I think whenever there is a set of vectors, there is a field of scalars associated with it. Is this true?

4. Feb 7, 2008

HallsofIvy

Staff Emeritus
You have completely confused yourself (well, at least me!). You say that you are using the set of real numbers as vectors (completely allowable) but then protest that 1 is a scalar not a vector! If you are thinking of the real numbers as a vector space over the real numbers, then any number, including 0 and 1, is both a scalar and a vector- you just have to keep track of which one you intend in a particular case. The real numbers, with ordinary addition as vector addition and ordinary multiplication as scalar multiplication, certainly does form a vector space over the real numbers. It has dimension 1 of course and isn't terribly interesting!

Now, back to this problem. Your set is the set of positive real numbers and you are defining vector addition, "x+ y", as xy, scalar multiplication, "kx", as xk. Yes, the number 1 is now the "vector 0", the additive identity. A vector space must have all of the "group" properties- in particular every member must have an additive inverse. What is the "additive inverse" of a "vector"? (And do you see why you need positive real numbers and not all real numbers? What would happen if 0 were in the set?)

I think the crucial point here is the "distributive law": if a is a scalar and u and v are vectors, then a(u+ v)= au+ av. What does that say in terms of the operations defined here? Is it true?

Actually, you can construct a simple isomorphism between this space and the vector space of real numbers with ordinary addition and multiplcation, based on the fact that ex+y= exey and ekx= (ex)k.

Last edited: Feb 7, 2008