Degenerate bilinear form ($\det A=0\text{in finite dimensions}$)

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 $A=\mathrm{diag}(1,0)$, $B((x,y),(s,t))=xs$. 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 $x=\pm 1$ as its unit level set. This is a lost direction, a stronger condition than being an isotropic vector: with $A=\mathrm{diag}(1,-1)$, $B((1,1),(1,1))=0$ but $B((1,1),(1,0))=1$.

For a symmetric bilinear form or an alternating bilinear form, the radical is the linear subspaceThese vectors are invisible to all pairings, and the bilinear form is nondegenerate exactly when this linear subspace is $\{0\}$. In coordinates, $\mathrm{rad}B=\ker A$ for both kinds of bilinear form, since $A^T=A$ or $A^T=-A$. Thus $\mathrm{diag}(1,0)$ has radical of a bilinear form $\mathrm{span}\{(0,1)\}$.

Passing to the quotient vector space $V/\mathrm{rad}B$ 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.^[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.

The left radical is $\{v\in V:B(v,w)=0\text{ for every }w\in V\}$. If $A$ is the matrix of a bilinear form, this condition is $v^{T}A=0$, so its coordinate vectors form $\ker A^T$. For $A=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, the bilinear form $B(v,w)=v_1{w}_2$ has left radical $\mathrm{span}\{(0,1)\}$.

The right radical is $\{w\in V:B(v,w)=0\text{ for every }v\in V\}$. If $A$ is the matrix of a bilinear form, this condition is $Aw=0$. For $B(v,w)=v_1{w}_2$ from the left radical example, the right radical is $\mathrm{span}\{(1,0)\}$. The left radical and right radical differ here because the bilinear form is neither symmetric nor alternating. For a symmetric bilinear form or an alternating bilinear form, reversing a zero pairing preserves zero and the two linear subspaces coincide.