Reflexive bilinear forms.

Question

Let $V$ be a vector space and $B: V \times V \to \Bbb R$ be a bilinear form.

Usually, I see books defining that if $B$ is symmetric, vectors ${\bf u},{\bf v} \in V$ are $B$-orthogonal if $B({\bf u},{\bf v}) = 0$, and there is no problem in this, since $B({\bf u},{\bf v}) = B({\bf v},{\bf u})$.

Then, asking for symmetry of $B$ seems a lot, after seeing the definition: $B$ is reflexive if for any ${\bf u},{\bf v} \in V$, $B({\bf u},{\bf v}) = 0 \implies B({\bf v},{\bf u}) = 0 $.

We can define orthogonality for reflexive bilinear forms. Clearly anti-symmetric forms are reflexive too.

Question: can you give me an example of a reflexive bilinear form which is not symmetric or anti-symmetric?

Answer 1

$$f((x,y); (s,t)) = ys + 2 xt.$$ So $$f((s,t);(x,y) ) = tx + 2 sy.$$ Because the matrix $$ \left( \begin{array}{rr} 0 & 2 \\ 1 & 0 \end{array} \right) $$ is not symmetric or antisymmetric.

For second version of the question, from books by Larry C. Grove, either Groups and Characters (1997) or Classical Groups and Geometric Algebra (2002), a bilinear form is reflexive if and only if it is either symmetric or "alternating," where alternating means that $B(v,v) = 0$ for every $v.$ Meanwhile, every alternating form is also skew symmetric, just expand $B(u+v,u+v).$ So there is no example such as was requested.

Meanwhile, in characteristic 2, there are symmetric (same as skew symmetric now) forms that are not alternating, which are then also not reflexive. Go Figure.

Answer 2

The answer to your question is "no", there is no such bilinear form. In fact, we have that $$B \text{ reflexive} \iff B \text{ symmetric or }B\text{ alternating}.$$ The statement holds over any field, so no exception for characteristic 2 here.

This result deserves a wider announcement in my opinion, as it motivates the study of symmetric and alternating bilinear forms. In my freshman course the choice of the investigated bilinear forms (in particular the alternating ones) felt a bit arbitrary to me.

For a proof, see for example Steven Roman, Advanced Linear Algebra, Springer, Graduate Texts in Mathematics 135. In the first edition 1995, it is Theorem 11.5, and in the second edition 2005 and the third edition 2008, it is Theorem 11.4.

As the proof is not completely trivial, it would be interesting to know the earliest source...