= Degenerate bilinear form
{title2=$\det A=0\quad\text{in finite dimensions}$}
A <bilinear form> is degenerate if some nonzero <vector> pairs to zero with every <vector> on at least one side. For a <symmetric bilinear form> or an <alternating bilinear form>, this says its <radical of a bilinear form> contains a nonzero <vector>. On a <finite-dimensional vector space>, it is equivalent to a singular <matrix of a bilinear form>, or $\det A=0$.
For $A=\operatorname{diag}(1,0)$, $B((x,y),(s,t))=xs$. The nonzero <vector> $(0,1)$ is invisible to every pairing, and changing $y$ never changes $B((x,y),w)$. The <quadratic form> $q=x^2$ therefore has two parallel lines $x=\pm1$ as its unit <level set>. This is a lost direction, a stronger condition than being an <isotropic vector>: with $A=\operatorname{diag}(1,-1)$, $B((1,1),(1,1))=0$ but $B((1,1),(1,0))=1$.
Back to article page
