definition of vector space needs no commutativity

In the definition of vector space (http://planetmath.org/VectorSpace) one usually lists the needed properties of the vectoral addition and the multiplication of vectors by scalars as eight axioms, one of them the commutative law

u+v=v+u.

The latter is however not necessary, because it may be proved to be a consequence of the other seven axioms. The proof can be based on the fact that in defining the group (http://planetmath.org/Group), it suffices to postulate only the existence of a right identity element and the right inverses of the elements (see the article “redundancy of two-sidedness in definition of group (http://planetmath.org/RedundancyOfTwoSidednessInDefinitionOfGroup)”).

Now, suppose the validity of the seven other axioms (http://planetmath.org/VectorSpace), but not necessarily the above commutative law of addition. We will show that the commutative law is in force.

We need the identity $(-1)v=-v$ which is easily justified (we have $\vec{0}=0v=(1+(-1))v=\ldots$ ). Then we can calculate as follows:

	$\displaystyle v+u$	$\displaystyle=(v+u)+\vec{0}=(v+u)+[-(u+v)+(u+v)]$
		$\displaystyle=\;[(v+u)+(-(u+v))]+(u+v)=[(v+u)+(-1)(u+v)]+(u+v)$
		$\displaystyle=[(v+u)+((-1)u+(-1)v)]+(u+v)=[((v+u)+(-u))+(-v)]+(u+v)$
		$\displaystyle=[(v+(u+(-u)))+(-v)]+(u+v)=[(v+\vec{0})+(-v)]+(u+v)$
		$\displaystyle=[v+(-v)]+(u+v)=\vec{0}+(u+v)$
		$\displaystyle=u+v$

Q.E.D.

This proof by Y. Chemiavsky and A. Mouftakhov is found in the 2012 March issue of The American Mathematical Monthly.

Title	definition of vector space needs no commutativity
Canonical name	DefinitionOfVectorSpaceNeedsNoCommutativity
Date of creation	2015-01-25 12:26:14
Last modified on	2015-01-25 12:26:14
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	5
Author	pahio (2872)
Entry type	Feature