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