Non-degenerate quadratic forms have zero kernel

Question

Let $V$ be a vector space over $F$, we say $Q:V\to F$ is a quadratic form if

1. $Q(\lambda v)=\lambda^2Q(v)$, and
2. the map $B:V\times V\to F$ defined by $B(u,v)=Q(u+v)-Q(u)-Q(v)$ is bilinear

I'd like to show that if $B$ is non-degenerate, then $\{x\in V\mid Q(x)=0\}=\{0\}$, but I'm kind of stuck. The problem is that $B$ is not necessarily an inner product.

Note: The bilinear map $B$ is called non-degenerate if $B(u,V)=0\implies u=0$

Answer 1

$V=\mathbb{F}^2$, $Q(\vec{x})=x_1x_2$, $B(\vec{x},\vec{y})=x_1y_2+x_2y_1$ is a counterexample, where $\vec{x}= {{x_1}\choose{x_2}}, \vec{y}= {{y_1}\choose{y_2}}.$

Answer 2

On $\mathbf C$ considered as an $\mathbf R$-vector space, the standard inner product $z\cdot z'=xx'+yy'$ ($z=x+iy$, $z'=x'+iy'$) is non-degenerate, but the set of vectors $\;z=t(1\pm i),\enspace t\in\mathbf R$, satisfy $q(z)=0$.

(The set of vectors $v$ such that $q(v)=0$ is called the isotropic cone of the bilinear form associated to the quadratic form)

It is the reason why on complex vector spaces, one uses hermitian forms.