# Dropping a rule in vector spaces

1. Oct 25, 2012

### eric_h22

What happen if i drop the
"V is a vector spaces. If k,l \in \mathbb{F}, u \in V, (k+l)u=ku + lu"
rule in vector spaces?

2. Oct 25, 2012

### HallsofIvy

Staff Emeritus
Well, the first thing that goes is the whole notion of scalar multiplication and representation in terms of bases or as n-tuples of numbers!

If lu+ ku is not equal to (l+ k)v then scalar multiplication no longer has the usual properties. And while we may still be able to write $u= a_1e_1+ a_2e_2$ and $v= b_1e_1+ b_2e_2$ we could no longer say $u+ v= (a_1+ b_1)e_1+ (a_2+ b_2)e_2$.

3. Oct 27, 2012

### Erland

We should also ask ourselves if this axiom is dependent of the other, that is, if it can be proved from the other vector space axioms, and thus being redundant.

The answer is no, it is in fact independent of the other axioms. To see this, we construct a structure which satisfies all vector space axioms except the one in question here.

To do this, we use the real numbers both as our vectors and our field of scalars (with its ordinary field structure).

We define addition of vectors as ordinary addition of reals, but we define multiplication (we denote it with x here) of a vector v by a scalar r by

r x v = v if r>0, r x v = -v if r<0, and r x v = 0 if r=0.

It is then easy show that all the vector space axioms are satisfied for this structure, except the axiom in question here, which is false here, since (1+1) x 1 = 2 x 1 = 1, while 1 x 1 + 1 x 1 = 1 + 1 = 2.

4. Oct 27, 2012

### Erland

A more general problem is to decide which vector space axioms are independent and which can be proved from the other axioms. We found here that the distributivity rule discussed here is independent. In another tread we found that the existence of additive inverses is also independent, although it can be replaced with the rule 0v=v, for all vectors v.

The following I know:

Dependent (can be proved from the other axioms)

Independent (cannot be proved from the other axioms):