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

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Paul Sperry

Posts: 1,392
Registered: 12/6/04
Re: Bilinear forms and reflexivity
Posted: Dec 26, 2011 11:30 PM
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In article <MPG.29631200b93dc448989a07@news.cc.tut.fi>, Kaba
<kaba@nowhere.com> wrote:

> Paul Sperry wrote:
> > > Since B is non-degenerate, 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 non-zero vector. Use non-degeneracy 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.

>
> Yes, the proof seems acceptable to me.
>

> > [...]
> >

> > > 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 skew-symmetric is the zero form:

> >
> > I suspect the author of the wiki article just forgot to add
> > "non-degenerate". I can't recall seeing the result elsewhere without
> > "non-degenerate".

>
> That would explain it. Thanks.
>
> Regarding the Wikipedia page, since I don't have a reference, I will
> just create a mention in the discussion page about this.


In the book "Groups and Characters" (Prop 1.7.6 page 24) Larry C Grove
claims to have proved the result without assuming non-degeneracy. If
the proof is correct, I don't follow it. The book is searchable at
Amazon. (Use "reflexive" for the search.)

--
Paul Sperry
Columbia, SC (USA)



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