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Topic:
Bilinear forms and reflexivity
Replies:
24
Last Post:
Jan 1, 2012 5:21 AM




Re: Bilinear forms and reflexivity
Posted:
Dec 26, 2011 12:11 AM


In article <MPG.2961f642dd362374989a02@news.cc.tut.fi>, Kaba <kaba@nowhere.com> wrote:
> Paul Sperry wrote: > > > But how about the other direction, that reflexive implies symmetric or > > > alternating? > > > > Reflexive only implies symmetric if the field has characteristic 2  so > > your assumption that the field is the reals means you would show > > reflexive implies antisymmetric. > > It think this is not true. See below.
I think you are correct. I have no idea what I had in mind.
> > A far as I know this is not trivial. A Google search turns up > > <www.maths.qmul.ac.uk/~pjc/class_gps/ch3.pdf> > > The Theorem 3.3 of that paper gives ideas for a proof, although I think > the proof itself is misleading if not flawed. > > Claim >  > > Let B : V times V > F be a bilinear form, where V is a vector space > over the field F. If B is reflexive _and_ nondegenerate, then B is > symmetric. (Note that these are the same conditions as in the paper) > > Proof >  > > Denote the bilinear form B(u, v) by u * v. It trivially holds forall u, > v in V that > > (u * u)(u * v)  (u * u)(u * v) = 0. > > This can be modified by bilinearity to > > u * (u * v)u  u * (u * u)v = 0 > <=> > u * ((u * v)u  (u * u)v) = 0. > > By reflexivity > > ((u * v)u  (u * u)v) * u = 0. > > By bilinearity again > > (u * u)((u * v)  (v * u)) = 0. > > Since B is nondegenerate, u * u != 0, forall u != 0. Thus > > forall u != 0, v: u * v = v * u.
For _some_ v u*v != 0.
The cited article now proceeds like this:
Call u good if there is a v such that u*v = v*u != 0.
Use the above equation to show that if u is good then u*w = w*u for all w. This commutivity implies that if u is good and u*w != 0 then w is also good.
Let u be good and w any nonzero vector. Use nondegeneracy to show that there is a v such that u*v != 0 and w*v != 0. Since u is good v is good; since v is good w is good; so, if there is at least one good vector, _all_ vectors are good and B is symmetric.
If there are _no_ good vectors set w = u in the equation and get u*u = 0 for all u and B is alternating.
[...]
> This still leaves open the case of degenerate reflexive bilinear forms. > Let us assume we are away from 2. The only bilinear form that is both > symmetric and skewsymmetric is the zero form:
I suspect the author of the wiki article just forgot to add "nondegenerate". I can't recall seeing the result elsewhere without "nondegenerate".
> u * v = v * u = u * v > ==> > 2 (u * v) = 0 > ==> > u * v = 0 > > Thus a skewsymmetric bilinear form must be degenerate (since skew > symmetry implies reflexivity and the zero form is degenerate): > > B is skewsymmetric ==> B is reflexive and degenerate.
It is true that skewsymmetric implies reflexive if char(F) != 2 but reflexive does _not_ imply both skewsymmetric _and_ alternating. There is at least one well known skewsymmetric nondegenerate form. > Does someone see a way forward?
 Paul Sperry Columbia, SC (USA)



