Does an isotropic vector always exist for an indefinite quadratic forms?

Question

I have some problem reading some paper. In that paper, the author proved that a quadratic form $Q$ has an isotropic vector(of course, nonzero) by showing $Q$ has both nonnegative and nonpositive values. Does an isotropic vector always exist for an indefinite quadratic forms? If true, why?

Answer 1

Yes. In a word: Continuity. Say $Q(u)\le0$ and $Q(v)\ge0$, with vectors $u$, $v$ both nonzero. If either of the inequalities is an equality, you have your isotropic vector already. Otherwise, note that $Q(tu+(1-t)v)$ is continuous with respect to $t$, and this function has opposite signs at the ends of $[0,1]$ – and hence is zero somewhere in the interval. Also, note that $tu+(1-t)v\ne0$ for $t\in(0,1)$, since otherwise, $Q(v)=(t/(1-t))^2Q(u)$ has the same sign as $Q(u)$. (You can make this even more explicit, since $Q(tu+(1-t)v)$ is a quadratic polynomial in $t$.)