Question about the definition of a vector space

1. Aug 5, 2009

AxiomOfChoice

Suppose $V$ is a vector space over a field $F$ that has multiplicative identity 1. Do we have to take, as an axiom, that $1\vec{v} = \vec v 1= \vec{v}$ for every $\vec v\in V$, or is this a direct consequence of other, more rudimentary vector space axioms?

2. Aug 5, 2009

g_edgar

I think that will depend on what the other axioms are. In any case, this vector $u = v-1v$ has the property: for every scalar $a$ we have $au = 0$ ... If you postulate that such a vector $u$ must be zero, that will be enough.

3. Aug 5, 2009

JazzFusion

Part of the definition of a vector space is the existance a multiplicative identity. In other words, if 1 is the multiplicative identity, 1v = v for every v in V, or V is not a vector space.

Vector spaces, however, are not required to exhibit commutivity under multiplication, so there is nothing inherent in the definition of a Vector Space to require that v1 = v.

4. Aug 9, 2009

Ravid

The expression v1 is not defined in a typical (i.e. left) vector space. The fact that 1v=v follows from the properties of 1 in the field, since $$1v=w\Rightarrow v=1w=1v$$. The real question is, why do you care? You know that it's true, so whether or not you have to take it as an axiom is a triviality.

5. Aug 10, 2009

g_edgar

If you leave out the axiom $1v=v$ I do not see how $1v=w\Rightarrow v=1w$ although I do see that $1w=1v$

6. Aug 10, 2009

CaffeineJunky

Dummit and Foote requires all R-modules (R is a ring with identity) to have the axiom 1m = m "to avoid 'pathologies' such as having rm = 0 for all r in R and m in M". Since all vector spaces are K-modules (where K is the field acting on the vector space), it is safe to assume that 1v = v is required to be an axiom to avoid problems.

The condition is probably redundant considering (rs)v = r(sv) is an axiom, and you can take s = r^{-1}, which is guaranteed to exist because K is a field to produce an identity 1, and then take r = 1 to show that 1v = v. Nevertheless, there's no harm in taking or neglecting it as a vector space axiom.

7. Aug 10, 2009

Rasalhague

In their Introduction to Vectors and Tensors, Vol. 1, Bowen and Wang define a vector space as a 3-tuple (V,F,f) consisting of an additive abelian group V, a field F, and a function $$f : F \times V \rightarrow V$$, called scalar multiplication, such that, for all $$\lambda$$ and $$\mu$$ in F, and for all u, v in V

(1) $$f\left(\lambda, f\left(\mu,\mathbf{v} \right) \right) = f\left(\lambda\mu, \mathbf{v} \right)$$

(2) $$f\left(\lambda + \mu,\mathbf{v} \right) = f\left(\lambda,\mathbf{v} \right) + f\left(\mu,\mathbf{v} \right)$$

(3) $$f\left(\lambda,\mathbf{u} + \mathbf{v} \right) = f\left(\lambda,\mathbf{u} \right) + f\left(\lambda,\mathbf{v} \right)$$

(4) $$f\left(1,\mathbf{v} \right) = \mathbf{v}$$

http://repository.tamu.edu/handle/1969.1/2502

They add that it's customary to use the following simplified notation for the scalar multiplication function:

$$f\left(\lambda,\mathbf{v} \right) = \lambda \mathbf{v}$$

and "we shall [...] also regard $$\lambda \mathbf{v}$$ and $$\mathbf{v} \lambda$$ to be identical." Thus, for Bowen and Wang, the existence of a scalar identity element for scalar multiplication is one of the (possible) axioms, and "communtativity" of scalar multiplication simply a matter of notation, not part of the formal definition. This seems close to what Ravid wrote about v1 being not defined, except that they've defined it as just an alternative way of denoting 1v = f(1,v).

They also note that their fourth axiom can be replaced with

(4) $$\lambda \mathbf{u} = \mathbf{0} \Leftrightarrow \lambda = 0 \text{ or } \mathbf{u} = \mathbf{0}$$

and that their second axiom is redundant, proofs of both these statements being left as exercises.

8. Aug 11, 2009

Rasalhague

Hmm, I can see how this alternative axiom 4 could be derived from the original set of axioms:

Let $$\lambda \mathbf{u} = \mathbf{0}$$.

Then $$\lambda \mathbf{u} + \lambda \mathbf{u} = \lambda \mathbf{u}$$

$$\Rightarrow \lambda \left( \mathbf{u} + \mathbf{u} \right) = \lambda \mathbf{u}$$

Divide both sides by $$\lambda$$.

$$\text{If } \lambda \neq 0, \text{ then } \mathbf{u} + \mathbf{u} = \mathbf{u}$$.

Subtracting u from each side, u = 0.

But I don't know if a similar arument in reverse would ensure that the original axiom 4 was true, given this alternative version.

9. Aug 11, 2009

Rasalhague

Oh, actually I see they prove this properly in their Theorem 8.1, also covering the case where lambda = 0.

(a) $$\left(a \right) \quad 0\mathbf{u} = \mathbf{0}$$
because $$0 = \lambda - \lambda = \lambda\mathbf{u} - \lambda\mathbf{u} = \mathbf{0}$$

(b) $$\mathbf{0} = \lambda \mathbf{0}$$
because $$\lambda\mathbf{u} = \lambda \left(\mathbf{u} + \mathbf{0} \right)$$, so

$$\lambda\mathbf{u} = \lambda\mathbf{u} + \lambda\mathbf{0}$$

$$\lambda\mathbf{u} - \lambda\mathbf{u} = \lambda\mathbf{u} - \lambda\mathbf{u} + \lambda\mathbf{0}$$.

(c) $$\lambda\mathbf{u} = \mathbf{0} \Rightarrow \lambda = 0 \text{ or } \mathbf{u} = \mathbf{0}$$.

Proof. Let $$\lambda\mathbf{u} = \mathbf{0}$$

By (a), this equation is satisfied if lambda = 0. If lambda does not equal 0, u must = 0 because

$$\mathbf{u} = 1\mathbf{u} = \lambda \left(\frac{1}{\lambda} \right)\mathbf{u} = \frac{1}{\lambda} \left(\lambda \mathbf{u}\right) = \frac{1}{\lambda} \left(\mathbf{0} \right) = \mathbf{0}$$

Can we infer that u = 1u, from this final statement, given (c)?

10. Aug 11, 2009

Ravid

I was using $$1^{-1}\cdot 1v=v$$ but now that you mention it I suppose that requires the axiom.