# Vector Space Axioms

1. May 13, 2008

### pjallen58

I am trying to shorten and generalize the the definition of a vector space to redefine it in such a way that only four axioms are required. The axioms must hold for all vectors u, v and w are in V and all scalars c and d.

I believe the four would be:

1. u + v is in V,
2. u + 0 = u
3. u + -u = 0
4. cu is in V

I believe 1 and 2 can be used to satisfy:

u + v = v + u
(u + v) + w = u + (v + w)

and 3 and 4 can be used to satisfy:

c(u + v) = cu + cv
(c + d)u = cu + du
c(du) = (cd)u
1u = u

Not sure if I am on the right track here so any suggestions or corrections would be appreciated. Thanks to all who reply.

2. May 13, 2008

### CompuChip

Actually I think that all the axioms are necessary, and if you leave out for example the commutativity or associativity axiom you don't get what people would ordinarily call a vector space.
If you think that you would, you should prove for example that "u + v = v + u" indeed follow from the four axioms you have, as you claim, though I wouldn't see how that could be done. In fact, I don't even see how to prove something as simple as 0 + u = u without using at least associativity ((u + v) + w = u + (v + w)) and -(-u) = u.

3. May 13, 2008

### CRGreathouse

A model for axioms 1 and 2 would be:
$$V:=\mathbb{Z}$$
$$u\mathbf{+}v:=u-v$$
$$\mathbf{0}:=0$$

where 1 holds by the closure of integers under subtraction and 2 holds by the additive identity of integers. But in this model $u+v\neq v+u$ for most u and v.

4. May 13, 2008

### CRGreathouse

Ah! If you define $-v:=v$ and $cv:=v$ in the above model, you can see that 1, 2, 3, and 4 hold but commutativity still fails in general, as does (u + v) + w = u + (v + w). (It doesn't matter here, but let c be drawn from the reals.) With an appropriate step function instead for scalar multiplication (say cv := 0 for c = 0 and v = 1 and cv := v otherwise) you can make the scalar distribution properties fail as well.

Last edited: May 13, 2008