= Nondegenerate bilinear form
{title2=$\operatorname{rad}_L B=\operatorname{rad}_R B=\{0\}$}
= Nondegenerate
{synonym}
A <bilinear form> $B$ is nondegenerate when its <left radical> and <right radical> are both zero: every nonzero <vector> can be detected by some pairing, whichever input slot it occupies. On a <finite-dimensional vector space>, either condition implies the other, and the following tests are equivalent:
$$
\begin{aligned}
B(v,w)=0\text{ for every }v&\quad\Longrightarrow\quad w=0,\\
V\longrightarrow V^*,\quad w\longmapsto B(-,w)&\quad\text{is an isomorphism},\\
\det A&\ne0.
\end{aligned}
$$
Here $A$ is the <matrix of a bilinear form>, and $V^*$ is the <dual space>. The <linear map> $w\mapsto B(-,w)$ has <matrix> $A$ in the chosen <basis> and its <dual basis>, so its <kernel> is the <right radical>. The map $v\mapsto B(v,-)$ has <matrix> $A^T$. An <invertible matrix> has zero <kernel> on both sides, proving the finite-dimensional equivalence. In infinite dimensions, zero <left radical> and <right radical> imply these maps are <injective>, but do not generally make them <surjective>.<Bilinear Forms by Keith Conrad>[§2: the two maps to the dual space]{ref}
For the real <symmetric bilinear form> $B((x,y),(s,t))=xs-yt$, the <determinant> is $-1$. Its <isotropic vector> $(1,1)$ has zero self-pairing but pairs to $1$ with $(1,0)$, so it is detected. The <alternating bilinear form> with <matrix> $J$ has <determinant> $1$ and detects every nonzero <vector> even though all self-pairings vanish. Neither zero self-pairings nor the absence of <positive definiteness> implies degeneracy.
Nondegeneracy is useful because it turns a <linear functional> into a unique <vector>. For $A=\operatorname{diag}(1,-1)$, the <linear functional> $\ell(s,t)=2s+3t$ is $B((2,-3),(s,t))$; solving for that representing <vector> amounts to solving an invertible <system of linear equations>. The analogous identification underlies <bilinear orthogonal complements> and the <dual isomorphisms induced by a nondegenerate pairing>.
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