Talk:Definite bilinear form

Definite bilinear forms: symmetric only?

I added the word symmetric, because definite is only defined over symmetric bilinear forms. Eulermatroid (talk) 10:48, 1 February 2012 (UTC)[reply]

Is this necessarily the case? Any bilinear form can be separated into symmetric and antisymmetric parts:

B (x, y) = B + (x, y) + B - (x, y)

where

B + (x, y) = (B (x, y) + B (y, x))/2

and

B - (x, y) = (B (x, y) - B (y, x))/2

. The associated quadratic form

Q (x) = B (x, x) = B + (x, x)

depends only on the symmetric part

B +

, but that does not mean that it is only defined when

B = B +

(i.e. when

B

is symmetric), and that it would be a matter of definition, not of necessity. Do you have a reference to substantiate your assertion? — Quondum ^☏✎ 11:55, 1 February 2012 (UTC)[reply]

A square matrices of reell numbers is pos. definite if 1) A = A ^ t and at the same time 2) for every x <> 0; x ^ t * A * x> 0

Prof. Veselic, Fernuniversität Hagen, Methoden der Physik Prof. Luise Unger , Fernuniversität Hagen, Lineare Algebra-2, Eulermatroid (talk) 15:44, 2 February 2012 (UTC)[reply]

I wouldn't have guessed that definiteness is only defined for symmetric forms, but a brief browse through google books makes it seem the case. I don't have a counterexample of an asymmetric pos def matrix. I think the main goal is to say "a quadratic form is definite" when the outputs have fixed sign, and quadratic forms are associated with symmetric bilinear forms. Rschwieb (talk) 18:14, 2 February 2012 (UTC)[reply]

Review my logic above. The counterexample of an asymmetric pos-def matrix is trivially easy: take any positive-definite symmetric matrix, and to it add any antisymmetric matrix, and you have it. Put another way: Any square matrix (symmetric or not) with only positive eigenvalues is positive-definite. See [1] for an example of not restricting the definition to symmetric bilinear forms. — Quondum ^☏✎ 19:49, 2 February 2012 (UTC)[reply]

The fact that a given unsymmetric matrice satisfies condition 2) [ x ^ t * A * x> 0]; does not mean that it is pos. definite. Every professor should know that, ask one. 93.199.170.224 (talk) 08:42, 3 February 2012 (UTC)[reply]

I think the definition should be given for symmetric forms only. For a non-symmetric form to be "definite" in the article's sense, a necessary and sufficient condition is that its symmetric part be definite. If there are any sources that substantiate this viewpoint, then it can be added later in the article. But the majority of sources (in my experience) assume symmetry. Perhaps the reason for this is twofold. First, many theorems hold for the symmetric case that otherwise do not in the asymmetric. Second, the asymmetric case isn't actually a useful generalization anyway, for the reason I indicated. Sławomir Biały (talk) 11:31, 3 February 2012 (UTC)[reply]

I think that the term "positive definite" has been introduced primarily for quadratic forms. Therefore it is a pity to have two different articles for Definite bilinear form and Definite quadratic form. I propose to merge these two pages, which should not be difficult, as the latter contains only 4 lines. If the merge is done, this discussion would be almost meaningless.

By the way the former is also a stub as it should contain, at least, the two following important results.

* Over a vector space of finite dimension, all positive definite quadratic forms are equivalent in the sense that they have the same matrix on well chosen bases. It follows that a Euclidean vector space is simply a vector space equipped with a positive definite quadratic form.

* If f is a two times differentiable function of several real variables and x is a value of its variable corresponding to a local minimum of f, then the gradient of f is null at x and the part of order 2 of its Taylor expansion (associated to its Hessian matrix) is a positive semidefinite quadratic form. Conversely, if the gradient is null and the Hessian is positive definite, then x is a local minimum.

D.Lazard (talk) 17:18, 3 February 2012 (UTC)[reply]

I would support a merge into Definite quadratic form, which should show how the associated symmetric bilinear form is determined by the quadratic form. — Quondum ^☏✎ 17:49, 3 February 2012 (UTC)[reply]