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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.

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Hmm I think the problem is this: $T_r\beta$ is orthogonal to $r$ for all $\beta$, and $T_q \alpha$ is orthogonal to $q$ for all $\alpha$, but that doesn't mean that $T_q\alpha$ is orthogonal to $T_r \beta$. For example think of $\mathbb{R}^3$ with the usual dot product. Then k is orthogonal to j and it is also orthogonal to i but it is not orthogonal to itself!

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thanks maxymoo, – 1yen Mar 17 '11 at 8:21

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