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Re: Bilinear forms and reflexivity
Posted:
Dec 25, 2011 3:02 AM
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In article <MPG.29609fe9d59cd5a9989a01@news.cc.tut.fi>, Kaba
<kaba@nowhere.com> wrote:
> Hi,
>
> Let B : V times V --> F be linear in both arguments, where V is a vector
> space over F. Let F = RR, the reals, for concreteness.
>
> Let
> B is reflexive :<=> forall v, w in V: B(v, w) = 0 <=> B(w, v) = 0
> B is symmetric :<=> forall v, w in V: B(v, w) = B(w, v)
> B is alternating :<=> forall v, w in V: B(v, w) = -B(w, v)
>
> The Wikipedia page for the bilinear form states:
>
> "One can prove that B is reflexive if and only if it is either:
> symmetric ... or alternating ..."
>
> http://en.wikipedia.org/wiki/Bilinear_form
>
> If B is symmetric, then B(v, w) = 0 <=> B(w, v) = 0, and thus B is
> reflexive.
>
> If B is alternating, then B is skew-symmetric since
>
> B(v + w, v + w) = B(v, v) + B(v, w) + B(w, v) + B(w, w)
> = B(v, w) + B(w, v)
> = 0
>
> <=> B(v, w) = -B(w, v)
>
> If B is skew-symmetric, then B(v, w) = 0 <=> -B(w, v) = 0 = B(w, v), and
> thus B is reflexive.
>
> 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 anti-symmetric.
As far as I know this is not trivial. A Google search turns up
<www.maths.qmul.ac.uk/~pjc/class_gps/ch3.pdf>
--
Paul Sperry
Columbia, SC (USA)
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