# 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 DefinitionOfVectorSpaceNeedsNoCommutativity 2015-01-25 12:26:14 2015-01-25 12:26:14 pahio (2872) pahio (2872) 5 pahio (2872) Feature