In article <MPG.email@example.com>, Kaba <firstname.lastname@example.org> 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.)