From: David G Radcliffe Subject: Re: generalized axioms for Vector Spaces? Date: 25 Jan 2001 03:39:00 GMT Newsgroups: sci.math Summary: Independence of axioms for vector spaces mathdiva@my-deja.com wrote: : Hello hello. I have a question about something : extraordinarily basic that has me a bit confused... : most linear algebra texts state the commutativity of : addition (i.e. u+v=v+u) as one of the axioms of a Vector : Space. However, a friend of mine claims that this axiom : follows from the rest and need not be included in the list : of axioms. I'm a bit stumped. Can anyone verify this? (u + v) + (-v + -u) = 0 by associativity of addition. Therefore u + v = -(-v + -u) = (-1)((-1)v + (-1)u) = (-1)(-1)v + (-1)(-1)u = v + u. ============================================================================== From: mareg@mimosa.csv.warwick.ac.uk () Subject: Re: generalized axioms for Vector Spaces? Date: 25 Jan 2001 09:04:10 GMT Newsgroups: sci.math In article <94o45k$dfa$1@nnrp1.deja.com>, mathdiva@my-deja.com writes: >Hello hello. I have a question about something >extraordinarily basic that has me a bit confused... >most linear algebra texts state the commutativity of >addition (i.e. u+v=v+u) as one of the axioms of a Vector >Space. However, a friend of mine claims that this axiom >follows from the rest and need not be included in the list >of axioms. I'm a bit stumped. Can anyone verify this? > This is a nice exercise - I have not seen it before. First prove that 0v = 0 for all vectors v. Then show (-1)v = -v for all vectors v. The proofs are the same as in a standard vector space. Then for all vectors v and w, (-w) + (-v) = -(v + w) = (-1)(v + w) = (-1)v + (-1)w = (-v) + (-w) Derek Holt. ============================================================================== From: Fred W. Helenius Subject: Re: generalized axioms for Vector Spaces? Date: Thu, 25 Jan 2001 13:45:38 -0500 Newsgroups: sci.math mareg@mimosa.csv.warwick.ac.uk () wrote: >In article <94o45k$dfa$1@nnrp1.deja.com>, > mathdiva@my-deja.com writes: >>most linear algebra texts state the commutativity of >>addition (i.e. u+v=v+u) as one of the axioms of a Vector >>Space. However, a friend of mine claims that this axiom >>follows from the rest and need not be included in the list >>of axioms. I'm a bit stumped. Can anyone verify this? >This is a nice exercise - I have not seen it before. >First prove that 0v = 0 for all vectors v. >Then show (-1)v = -v for all vectors v. >The proofs are the same as in a standard vector space. >Then for all vectors v and w, >(-w) + (-v) = -(v + w) = (-1)(v + w) = (-1)v + (-1)w = (-v) + (-w) There is a slick solution that does without the lemmas which I learned from Herstein's _Topics in Algebra_ (he applies it to rings with unity, not vector spaces, but the idea is the same). Expand (1 + 1)(u + v) using the distributive laws in both orders, yielding: (1 + 1)u + (1 + 1)v = 1(u + v) + 1(u + v), or u + u + v + v = u + v + u + v. The existence of additive inverses allows the outer terms to be cancelled, leaving u + v = v + u. -- Fred W. Helenius