Let $V$ be a finite-dimensional real vector space, and let $\langle \,\cdot\,,\,\cdot\, \rangle : V \times V \to \mathbb{R}$ be a nondegenerate symmetric bilinear form on $V$. For $u, v \in V$, let us define $$ Q(u,v) = \langle u,u \rangle \langle v,v \rangle - \langle u,v \rangle^2 \in \mathbb{R}. $$ Let $W$ be a $2$-dimensional subspace of $V$. I want to show that the following three are equivalent.
(1) The restriction $\langle \,\cdot\,,\,\cdot\, \rangle|_W$ is nondegenerate.
(2) $Q(u,v) \ne 0$ for any basis $\{ u,v \}$ of $W$.
(3) $Q(u,v) \ne 0$ for some basis $\{ u,v \}$ of $W$.
Could you help me? Thank you.
(1) $\Rightarrow$ (3). Let $\{u, v\}$ be an orthonormal basis of $W$ with respect to $\langle \,\cdot\,,\,\cdot\, \rangle|_W$ (it exists as it is non-degenerate), you have that $$Q(u, v) = \|u\|^2\|v\|^2 \neq 0.$$
(2) $\Rightarrow$ (3). Trivial.
(2) $\Rightarrow$ (1). By contradiction, let us assume that there exists $x \neq 0$ and $\langle \,x\,,\,z\, \rangle = 0$ for all $z \in W$. As $x \neq 0$ we can find a basis $\{x, u\}$ of $W$. By taking $z = u$, we deduce that $Q(x, u) = 0$.
(3) $\Rightarrow $ (2). Left as an exercise (it's not too hard, just take any basis of $W$ and develop it in term of the basis you have in (3)).