Talk:Vector space
From Academic Kids
For the sake of correctness. The * was used in the vector space axioms both as a map * : F x F -> F and as a map * : F x V -> V. I changed a*b to ab. What about the double used + : F x F -> F and + : V x V -> V? -- Georg Muntingh
- I wouldn't worry about it. Usually, each multiplication is written without a symbol, and each addition with "+". The reader has to infer from context what operation it is, and this is possible by looking at the elements operated on and asking where they came from. If this is too much for the reader to handle, they probably won't understand it anyway. (Sorry if I sound dismissive.) Otherwise, we would have 4 or 5 different symbols, (+, _, #, *, ...??) and this is just incredibly cluttered. Revolver 21:40, 9 Feb 2004 (UTC)
To say the same things in more technical language: we have a case here of operator overloading (see also overloading); which is not necessarily a bad thing when types can be inferred. Charles Matthews 19:13, 11 Feb 2004 (UTC)
How about using + for vector addition and + for field addition? —Daniel Brockman 07:16, Mar 8, 2004 (UTC)
The section on 'vectors in physics' really belongs at vector (spatial), rather than here.
Charles Matthews 09:24, 23 Feb 2004 (UTC)
I agree about the 'vectors in physics'... Is it really true what is atated "Note that property 5 (commutativity) actually follows from the other 9"? I am almost certain that we need to modify the 4:th axiom (exist -x: x + -x = 0) so that the invers of a vector is commutative in the following way: "exist -x: x + -x = -x + x = 0" for the statement to hold. So I will now modify the 4:th axiom myself.. I would be glad to see a proof of the statement if it was true in its original version. Dj, 14 Mar 2004
- The extra axiom you have added (-x + x = 0) is not needed, as it follows from the first four axioms. Any introductory book on group theory should have a proof, but here's one anyway. Suppose s is an idempotent (that is, s + s = s). Then s = s + 0 = s + (s + -s) = (s + s) + -s = s + -s = 0. So 0 is the only idempotent. For any x we have (-x + x) + (-x + x) = -x + (x + -x) + x = -x + 0 + x = -x + x, that is, -x + x is an idempotent, so -x + x = 0. --Zundark 09:09, 14 Mar 2004 (UTC)
Sirs:
You have a mathematical typesetting error in statement 4 of your formal definition of a vector space over a field F.
In the last part of that sentence, when you type v + w = 0, the 0 should be in BOLD FACE, because it is a vector 0, and not a scaler 0.
Regards, Harold
- You are correct; I've fixed it now. (You could have fixed it yourself.) --Zundark 19:07, 26 Jun 2004 (UTC)
How does property 5 follow from the other nine? I've added this as an open mathematical question, since it seems to have gone unanswered for several months. Prumpf 18:01, 15 Aug 2004 (UTC)
- (x+x)+(y+y) = (1+1)(x+y) = (x+y)+(x+y), so x+y = y+x. The same proof shows more generally that you can't have a "non-abelian module". --Zundark 20:01, 15 Aug 2004 (UTC)
Connection
Using the vector space R∞ and a ultrafilter U on the natural numbers we can creat a hyperreal field!
discussion at Wikipedia talk:WikiProject Mathematics/related articles
This article is part of a series of closely related articles for which I would like to clarify the interrelations. Please contribute your ideas at Wikipedia talk:WikiProject Mathematics/related articles. --MarSch 14:08, 12 Jun 2005 (UTC)
