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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:Here $A$ is the matrix of a bilinear form, and $V^{\ast }$ 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.^[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 $A=\mathrm{diag}(1,-1)$, 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.

Nondegeneracy of $b$ means both radicals are zero: if $b(x,y)=0$ for every $y$, then $x=0$, and the analogous implication holds in the other variable. Define linear maps $B_U:U\to V^{\ast }$ and $B_V:V\to U^{\ast }$ by $B_U(x)(y)=b(x,y)$ and $B_V(y)(x)=b(x,y)$. Both are injective. Finite dimensionality gives $\dim U\le \dim V$ and $\dim V\le \dim U$, so they are isomorphisms.

Similarly define $C_U(x)(y)=c(x,y)$ and $C_V(y)(x)=c(x,y)$, which are linear by bilinearity. SetThen $S$ and $T$ are linear endomorphisms of $U$ and $V$, respectively, and their definitions give $b(Sx,y)=c(x,y)=b(x,Ty)$ for every pair. These dual isomorphisms induced by a nondegenerate pairing also prove uniqueness. No symmetry of $b$, and no identification of the two vector spaces, is required.