# 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 $\overrightarrow{0}=0v=(1+(-1))v=\mathrm{\dots}$). Then we can
calculate as follows:

$v+u$ | $=(v+u)+\overrightarrow{0}=(v+u)+[-(u+v)+(u+v)]$ | ||

$=[(v+u)+(-(u+v))]+(u+v)=[(v+u)+(-1)(u+v)]+(u+v)$ | |||

$=[(v+u)+((-1)u+(-1)v)]+(u+v)=[((v+u)+(-u))+(-v)]+(u+v)$ | |||

$=[(v+(u+(-u)))+(-v)]+(u+v)=[(v+\overrightarrow{0})+(-v)]+(u+v)$ | |||

$=[v+(-v)]+(u+v)=\overrightarrow{0}+(u+v)$ | |||

$=u+v$ |

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 |