Jump to content

Talk:Quotient space (linear algebra)

Page contents not supported in other languages.
From Wikipedia, the free encyclopedia

Untitled

[edit]

what's the significance of Quotient Space?

They're used everywhere. If you want a concrete example, you can use quotient space to define tensor products (let V and W be two finite dimensional vector spaces, let X be (infinite dimensional) free vector space on pairs (v,w), and let X_0 be the subspace spanned by (v,w) + (v,z) - (v,w+z) and (v,sw) - s(v,w) for any v,w,z and scalars s (and likewise for the left component of the pair). The quotient X/X_0 is the (finite dimensional!) tensor product. You get the wedge product if you include (v,w) + (w,v) as well (of course in this case we need V = W). 66.117.135.137 (talk) 09:44, 28 April 2008 (UTC)[reply]


Noting an error in the article

[edit]

Quote : "Let C[0,1] denote the Banach space of continuous real-valued functions on the interval [0,1]". This is an error, the continuous real-valued functions on the interval [0,1] do form a normed vector space, but do not form a Banach space (since the space is not closed). However I am afraid to fix it myself since I don't know enough on the subject. zermalo (talk) 16:31, 27 April 2009 (UTC)[reply]

It depends on the norm. With the L^1 norm it is not complete, but with the supremum norm it is. —Preceding unsigned comment added by MikeRumex (talkcontribs) 09:57, 29 October 2009 (UTC)[reply]

Another error

[edit]

"Another example is the quotient of Rn by the subspace spanned by the first m standard basis vectors. The space Rn consists of all n-tuples of real numbers (x1,…,xn). The subspace, identified with Rm, consists of all n-tuples such that the last n-m entries are zero: (x1,…,xm,0,0,…,0)."

I think it should be "The subspace, identified with Rm, consists of all n-tuples such that the last n-m entries are NOT zero: (x1,…,xn-m,0,0,…,0)."

I did not correct it since I am not an expert in the area. — Preceding unsigned comment added by Tagib (talkcontribs) 19:26, 7 March 2013 (UTC)[reply]

I think that's correct. Only the first m entries will be non-zero, correctly defining the space. 194.204.105.6 (talk) 12:53, 18 April 2023 (UTC)[reply]

Norm on the quotient space

[edit]

I think that it worth to explicitly note that the infimum in

is not necessarily attained. However I can't find such an example. Can someone add such an example to the article? bungalo (talk) 15:06, 13 November 2010 (UTC)[reply]

Functorial?

[edit]

Isn't 'making the quotient' functorial? The article only highlights one side of this natural construction and completely ignores the appropriate operation on linear maps, hence its functorial nature.

This is a big problem in many mathematical wikipedia articles. Way to little care is take for the appropriate morphims and hence the categorical background. This is ancient mathematical thinking!

equivalence classes "over" or "of"?

[edit]

@Wuffuwwuf Hi, by comparison of the result of this query

https://wiki.riteme.site/w/index.php?search=%22equivalence+classes+of%22&title=Special:Search&profile=advanced&fulltext=1&ns0=1

and this query

https://wiki.riteme.site/w/index.php?search=%22equivalence+classes+over%22&title=Special:Search&profile=advanced&fulltext=1&ns0=1

I believe that "of" is the correct proposition. Additionally, in my opinion, semantically, "of" is more appropriate for this sentence:

The quotient space V/N is then defined as V/~, the set of all equivalence classes over V by ~.

equivalence class belongs to V by ~, we can use the verb define by "over", but the latter part says that "it belongs" and therefore the correct proposition is "of". Hooman Mallahzadeh (talk) 05:21, 5 July 2022 (UTC)[reply]

I don't think that's right. In the results listed at the first link, you will notice that "equivalence classes of" is used when what follows the "of" is elements of the set V being partitioned into equivalence classes, (i.e., the elements of those classes,) while "equivalence classes over" or "equivalence classes on" is used when what follows "over" is the set V itself. Wuffuwwuf (talk) 05:27, 5 July 2022 (UTC)[reply]
So for example the equivalence classes induced by the relation of being congruent mod 3 are classes of integers, but over the set Z of integers. Wuffuwwuf (talk) 05:29, 5 July 2022 (UTC)[reply]
@Wuffuwwuf I'v done a search and find that the proposition "on" is more frequent than "over". Do you agree with this sentence?

The quotient space V/N is then defined as V/~, the set of all equivalence classes ««on»» V by ~.

Hooman Mallahzadeh (talk) 05:35, 5 July 2022 (UTC)[reply]
Yes, I would agree with that. I also think "by ~" is a little vague—"induced by ~" or even "relative to ~" would be better. Thanks for asking! Wuffuwwuf (talk) 05:38, 5 July 2022 (UTC)[reply]