I am having a little trouble understanding the notion of non-degeneracy for bilinear maps.
A bilinear map $\psi : V \times V \mapsto \mathbb{R}$ always has a matrix representation, so $\psi(x,y)=x^TAy$.
A bilinear map is non-degenerate if and only if $A$ is full rank. This is also equivalent to stating that $\psi(x,y)=0 \; \forall y \in V \Rightarrow x = 0$.
Consider now the following symmetric bilinear form $r^TQ^{-1}q = 0$ ($Q$ is symmetric).
Define $W_q = \{ q \}$ and $W_r = \{ r \}$. We can define the b-orthogonal subspaces as
$W_q^{\perp} = \{x \; | \; x=Qy, \; y \in \mathcal{N}(q^T)\}$, where $\mathcal{N}(q^T)$ is the null-space of $q$. In particular, we can express $y$ as $y=T_q\alpha$, with columns of $T_q$ spanning the null-space.
Similarly, $W_r^{\perp} = \{x \; | \; x=Qy, \; y \in \mathcal{N}(r^T)\}$, with $y=T_r\beta$.
If $r \in W_q^{\perp}$ and $q \in W_r^{\perp}$, then
$0 = r^TQ^{-1}q = \alpha^TT_q^TQT_r\beta$, and this should hold for all $\alpha$ and $\beta$.
However, the matrix $T_q^TQT_r$ is full-rank, and the transformed bilinear map is non-degenerate suggesting this equality can not hold. I feel like I am missing an important argumentation step. Any help would be much appreciated.