Content deleted Content added
m Maintain {{WPBS}}: 1 WikiProject template. Remove 1 deprecated parameter: field. Tag: |
|||
| (10 intermediate revisions by 5 users not shown) | |||
Line 1:
{{Old AfD multi |date=16 February 2022 |result='''keep''' |page=Examples of vector spaces}}
{{WikiProject banner shell|class=Start|
{{WikiProject Mathematics| priority = Low}}
}}
==Unbounded sequences==
Line 20 ⟶ 25:
:The [[vector space]] article already discusses [[module (mathematics)|module]]s in the section on generalizations. The link is somewhat out of place here. If there were an article on [[examples of modules]], it might be appropriate to link it from here (I'm not claiming such an article should exist). -- [[User:Fropuff|Fropuff]] 02:41, 12 January 2006 (UTC)
A vector space is a module over a division ring -- not over a field. This article errs because the set of quaternions is a division ring but not a field. To speak of a vector space over the quaternions you can't define vector spaces only over fields. [[User:Jalongi|Jalongi]] ([[User talk:Jalongi|talk]]) 07:11, 19 July 2009 (UTC)
:Maybe I'm misunderstanding what you're saying or I missed something in the article but it seems correct to me. The only time the article mentions quaternions is as a vector space over the reals which is a field. The sources I've checked all define a vector space to be over a field, do you have a reference defining it over a division ring? Off the top of my head I don't see that any of the theory would break if you defined it that way but maybe there just aren't a lot of applications.--[[User:RDBury|RDBury]] ([[User talk:RDBury|talk]]) 22:37, 19 July 2009 (UTC)
::Hungerford's Algebra (which seems to be a popular reference and graduate text) defines a vector space to be a unitary module over a division ring. I misread the article's example involving the quaternions. The quaternions are vector space over the real numbers, but according to Hungerford's definition we could use quaternions as scalars and still have a vector space. One part of the theory that breaks down under the more general definition is that a matrix over a division ring does not have a well-defined rank. It has a left-rank and a right-rank, I think. We would just have to be careful about commutativity in proofs. Thanks for catching my error.[[User:Jalongi|Jalongi]] ([[User talk:Jalongi|talk]]) 15:59, 20 July 2009 (UTC)
:::You forgot to mention that Hungerford's Algebra has a yellow cover, pretty much all you need to say really. Somewhat ironically, Springer's website is one of the sources I mentioned above. It may be worthwhile to add Hungerford's definition to the Vector Space article in the Generalizations section. It's probably not a good idea to add it near the beginning since most of the article should be at a college freshman level of accessibility.--[[User:RDBury|RDBury]] ([[User talk:RDBury|talk]]) 19:33, 20 July 2009 (UTC)
== Function spaces and generalized coordinate spaces ==
Line 38 ⟶ 48:
:I'm new to adding comments to Wikipedia, my apologies if I screw anything up. Here's my question...There seems to be an ambiguous statement in the Field vector space example. The article states that the basis of a field is the identity element, but it does not specify whether it is the multiplicitive identity element, or the additive identity element. Later in the article the basis is expressed as the set {0), but I don't see how a linear combination of zeros could result in any non-zero element of the field. Shouldn't the basis be {1}?
::It's more or less implicit when you're talking about fields that identity element means multiplicative identity, but the article should be more specific. I couldn't find where {0} was given as a basis, but if it is then it's incorrect since no set containing 0 can be a basis. The basis of {0} is the empty set.--[[User:RDBury|RDBury]] ([[User talk:RDBury|talk]]) 19:43, 20 July 2009 (UTC)
| |||