# Determining if set is a real vector space

1. Jun 21, 2009

### mlarson9000

1. The problem statement, all variables and given/known data
The set R^2 with addition defined by <x,y>+<a,b>=<x+a+1,y+b>, and scalar multiplication defined by r<x,y>= <rx+r-1,ry>.

The answer in the back of the book says it is a vector space, but I am having trouble proving that 0+v=v and v+(-v)=0

2. Relevant equations

3. The attempt at a solution
My guess is that you take v+(-1)v=0, applying the above definition of scalar multiplication to (-1)v, which gives 0=<-1,0>. Using that definition of 0, then 0+v=v, but I'm not sure if any of that is correct.

2. Jun 21, 2009

### slider142

The vector 0 is defined as a vector z such that z + v = v for each v in V. It is an axiom that this vector exists, so in showing that a space is a vector space, you must show that there exists such a vector. The easiest way is to directly use your formula for the sum of two vectors:
<x, y> + <z1, z2> = <x + z1 + 1, y + z2> = <x, y>
and solve for z, which must not depend on x or y.

3. Jun 21, 2009

### dx

Argh! Sorry, I should read more carefully..

4. Jun 21, 2009

### HallsofIvy

Staff Emeritus
In this vector space, the "0 vector" is not <0, 0>.

Any easy way of determining what the 0 vector, $\vec{0}$, is is to use the fact, true in any vector space, that $0\vec{v}= \vec{0}$. (That follows from the "distributive law": $a\vec{v}= (0+ a)\vec{v}= 0\vec{v}+ a\vec{v}$.)

Here, r<x,y>= <rx+r-1,ry>. What do you get if r= 0?