Let $V$ be a real vector space and let $q: V \to \mathbb R$ be a real quadratic form,
Prove that if the set $L = \{v \in V | q(v) \ge 0\}$ forms a subspace of $V$ then q is definite (meaning $q$ is positive definite, positive semidefinite, negative definite or negative semidefinite)
Suppose by contradiction that there exists $x$ and $y$ such that $q(x) < 0 < q(y)$. Then for any real $\lambda$, $q(x + \lambda y) = q(x) + \lambda^2 q(y) + 2\lambda B(x,y)$ where $B$ is the bilinear form associated to $q$. So when $\lambda$ is large, $q(x + \lambda y)$ is positive, so $x + \lambda y \in L$. Hence $x = (x + \lambda y) - \lambda y$ is in $L$, which is a contradiction.
Hint: You want to show that if $L=\{x \mid q(x)\geq 0\}$ is a subspace, then $L=0$ or $L=V$