= Radical of a bilinear form
{title2=$\operatorname{rad}B$}
For a <symmetric bilinear form> or an <alternating bilinear form>, the radical is the <linear subspace>
$$
\operatorname{rad}B=\{v\in V:B(v,w)=0\text{ for every }w\in V\}.
$$
These <vectors> are invisible to all pairings, and the <bilinear form> is <nondegenerate> exactly when this <linear subspace> is $\{0\}$. In <coordinates>, $\operatorname{rad}B=\ker A$ for both kinds of <bilinear form>, since $A^T=A$ or $A^T=-A$. Thus $\operatorname{diag}(1,0)$ has <radical of a bilinear form> $\operatorname{span}\{(0,1)\}$.
Passing to the <quotient vector space> $V/\operatorname{rad}B$ removes precisely the invisible directions. The rule $\overline B([v],[w])=B(v,w)$ is well-defined because adding a <vector> in the <radical of a bilinear form> to either input changes no pairing, and the resulting <bilinear form> is <nondegenerate>. For $B((x,y),(s,t))=xs$, this <quotient vector space> retains just the $x$ direction and its form is ordinary multiplication.https://www.math.ucla.edu/~rse/oldbook/Kniga.pdf[Elman, Karpenko and Merkurjev, The Algebraic and Geometric Theory of Quadratic Forms, §1: radical and nondegenerate quotient]{ref}
For a general <bilinear form>, distinguish the <left radical of a bilinear form> from the <right radical of a bilinear form>; they can be different <linear subspaces> even though their <dimensions> agree in finite dimensions.
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