Pedantic question about symmetric/non-symmetric bilinear forms: is symmetry of the bilinear form redundant? A quadratic form $q$ is defined in terms a symmetric bilinear form $f$ by $q(\mathbf{v}):=f(\mathbf{v},\mathbf{v}).$ However, this definition seems too restrictive. Why must $f$ be symmetric? Even if $f$ were a non-symmetric bilinear form, $q$ could still be defined in terms of $f$ and would still have the defining property of quadratic forms, namely that,

In mathematics, a quadratic form is a polynomial with terms all of degree two.

Any clarifications are welcomed, thank you.
 A: Actually it's the opposite: the non-symmetric part of the bilinear form is redundant. Any bilinear form $B(u, v)$ has a canonical decomposition (over a field of characteristic $\neq 2$) into a symmetric part and a skew-symmetric part
$$B(u, v) = \frac{B(u, v) + B(v, u)}{2} + \frac{B(u, v) - B(v, u)}{2}$$
and the skew-symmetric part of $B$ does not contribute to $Q(v) = B(v, v)$ at all, so can be safely ignored. That is, if we write $S(u, v)$ for the symmetric part, then $B(v, v) = S(v, v)$. This is why we define quadratic forms using only symmetric bilinear forms. The symmetric part can be recovered from the quadratic form $Q$ as the polarization $\frac{Q(u + v) - Q(u) - Q(v)}{2}$ so the quadratic form and the symmetric bilinear form are two different ways of talking about the same thing.
You can work this out quite explicitly for two variables, for example: here the vector space of quadratic forms is $3$-dimensional with basis $\{ x^2, xy, y^2 \}$ and the vector space of symmetric bilinear forms is also $3$-dimensional and can be identified with the vector space of symmetric matrices, with basis
$$\left\{ \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right] \right\}$$
and these correspond to the quadratic forms $x^2, 2xy, y^2$ respectively. Meanwhile the vector space of skew-symmetric forms is $1$-dimensional with basis $\left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right]$ and the corresponding quadratic form is identically zero.
(There are additional subtleties if the field has characteristic $2$ because we can no longer perform the division by $2$ necessary for the formulas above, but on a first pass we can safely ignore them. Most people will never need to deal with them.)
