Talk:Degenerate bilinear form

1

Could we make this more general?

futurebird 22:46, 21 June 2006 (UTC)[reply]

2

It would be much more useful if the article starts with describing the Non-degenerate forms rather than degenerate forms. Non-degenerate forms are much more used and more general. Degeneracy in mathematics is not very important. If nobody objects can we change the title and corresponding article to better reflect the actual mathematics.

nirax (talk) 16:26, 12 August 2009 (UTC)[reply]

3

What is meant by $v\mapsto f(-,v)$ ? What is the function of the minus sign as an apparent argument in the bilinear function $f(-,v)$ ? It would be helpful if the author of this article could provide a link to explain the meaning of $v\mapsto f(-,v)$ .

Howard McCay (talk) 23:37, 28 January 2010 (UTC)[reply]

It's not a minus sign, it's a placeholder for an omitted argument. f(-,v) means the function that maps x to f(x,v). Algebraist 00:42, 29 January 2010 (UTC)[reply]

Which dual space?

In the infinite-dimensional case, doesn't the article need to explain which dual space the notation V* refers to? It it's the algebraic dual, then all forms are degenerate because V* is too large to be isomorphic to V.

I assume therefore that it is meant to be the continuous dual. But then the article could use a hint that "degeneracy" may depend on the topology of V that one is working with. –Henning Makholm (talk) 20:49, 15 January 2011 (UTC)[reply]

This is very confusing and unclear. If anyone wants to sort it out, the same problem is present on the Inner product space page.Hiiiiiiiiiiiiiiiiiiiii (talk) 03:48, 28 September 2011 (UTC)[reply]

Ungrammatical wording: If and only if

or equivalently in finite dimensions, if and only if

What is the significance of this "if and only if"? The thought behind it seems to be "A symmetric bilinear form is nondegenerate if and only if..." Perhaps it's a relic of a time when the paragraph began like that. Should it be deleted? Fomentalist (talk) 01:19, 31 October 2011 (UTC)[reply]

No, it's ok. The phrase continues on the next line … --84.75.62.94 (talk) 22:10, 15 February 2012 (UTC)[reply]

Infinite Dimensions

This section is extremely unclear. First of all, I gather that it's meant to be an example of how in infinite dimensions a bilinear form can satisfy f(x,y)=0 for all y only if x=0 without being an isomorphism, and therefore doesn't satisfy the criteria for a non-degenerate form, but it doesn't state very clearly that this is meant to be the purpose. More importantly, though, as a physicist and not a mathematician, some of the details are unclear to me, and I think need to be clarified for a less specialized audience. For example, does the space of continuous functions need to include the dirac-delta function, since it is only the limit of a series of functionals and not a regular function, per-se? If this is the only counter example to the non-degeneracy, does that actually prove anything? Second of all, what is meant when it says that it is in the space of the dual but "not of the required form"? What is the required form to have the dirac delta function as its dual? This is very confusing to me. Mpalenik (talk) 21:36, 27 March 2015 (UTC)[reply]

Proposed merge of Nondegenerate form with Degenerate bilinear form

The articles Degenerate bilinear form and Nondegenerate form are duals ("A nondegenerate form is one that is not degenerate") and therefore mostly redundant. Given that Nondegenerate form is quite short it makes sense to merge these articles. Tea2min (talk) 09:08, 13 February 2020 (UTC)[reply]

Support. It does not generally make sense to have separate articles about a class and the complement of that class, and that is definitely the case here. —Quondum 19:18, 19 April 2020 (UTC)[reply]

Merger complete. Klbrain (talk) 21:51, 6 January 2021 (UTC)[reply]

Examples

The Examples section just repeats the last paragraph of Nondegenerate forms. Maybe some concrete examples could be added? Angulon (talk) 19:45, 5 January 2023 (UTC)[reply]