Why is quadratic form defined via a symmetric bilinear form?

Question

A typical definition of quadratic form goes like this:

Let $B:V\times V \to F$ be a symmetric bilinear form. A function $Q : V → F$ defined by $Q(v) = B(v, v)$ is called a quadratic form.

Why do we need $B$ to be symmetric if the way $Q$ is defined doesn't use it?

Edit: And is that related to the reason why we ask for a symmetric bilinear form in the definition of a positive definite bilinear form even though the definition itself $(B(v,v)>0)$ again doesn't directly use it?

Answer 1

Every bilinear form gives a quadratic form by $Q(v)=B(v,v)$, but the quadratic form doesn't depend on the antisymmetric part of $B$. So if you change $B$ by adding some antisymmetric form, then $Q$ won't change.

Every bilinear form can be written as a sum of a symmetric and antisymmetric form in a unique way:

$$B=\frac{B+B^T}{2}+\frac{B-B^T}{2}$$

This means that every quadratic form arising from an arbitrary form $B$ also arises from a symmetric form $\frac{B+B^T}{2}$. So it doesn't really matter if we put the word "symmetric" in the definition or not.

EDIT: If you're working over a field of characteristic $2$ then the above doesn't work (because you can't divide by 2) and the notions of "quadratic form generated by an arbitrary bilinear form" and "quadratic form generated by a symmetric bilinear form" are genuinely different (for example $x_1^2+x_1x_2+x_2^2$ is of the first kind but not the second). I don't know what definition people in this area tend to use, but my guess would be the first.

Answer 2

It is so because quadratic forms are closely related to norms and inner products on a vector space. It's all the more so as one can get back the bilinear form from the quadratic, through one of the polarisations identities, if $\operatorname{char} F\neq 2$, e.g.: $$B(u,v)=\tfrac 12 \bigl(Q(u+v)-Q(u)-Q(v)\bigr)$$

Answer 3

To see the reason why we need $B$ to be symmetric let's assume that $V$ is finite dimensional space, so we know that $Q$ takes the form

$$Q(x)=x^tAx$$ and then we have

$$Q(x)=x^t\left(\frac12(A+A^t)+\frac12(A-A^t)\right)x=\frac12x^t(A+A^t)x$$ since $Q(x)$ is real and then $(Q(x))^t=Q(x)$. In other words the quadratic form $Q$ can only capture the information contained in the symmetric part $\frac12 (A + A^t )$, but nothing in the skew-symmetric part $\frac12 (A−A^t)$.