Bilinear and quadratic forms
Degenerate bilinear form has a non-trivial kernel: there exist some non-zero x in V such that $B(x,y)=0$ for all $y\in V$, it’s orthogonal to V.
Bilinear form is degenerate if and only if the determinant of the associated matrix is zero.
A nondegenerate (or nonsingular) form — $f(x,y)=0$ for all $y \in V$ implies that $x = 0$.
Find kernel of bilinear form with matrix $B$.
$\operatorname{Ker} \beta(x, y) = { y \vert \forall x \beta(x, y) = 0 }$
So we have $x^t B y = 0$ for all x. We can’t just get rid of $x^t$.
But we can choose $x = By$, then we have $(By)^t By = 0$, which is dot product, this means $By=0$.