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Put $b(v,w)=\phi (w)(v)$. Each $\phi (w)$ belongs to the dual space $V^{\prime }$, so it is a linear functional. For real numbers $a,b_0$ and vectors $v_1,v_2,w$, this givesThe linearity of $\phi$ gives linearity in the second argument as well:Thus $b$ is a bilinear form.

Its right radical of a bilinear form is exactly $\ker \phi$, becauseIf $b$ is a nondegenerate bilinear form, this kernel is zero. Since $V$ is a finite-dimensional vector space and $\dim V^{\prime }=\dim V$, the rank-nullity theorem makes $\phi$ surjective.

Conversely, if $\phi$ is surjective, the same dimension equality makes it injective, so the right radical is zero. If $v$ lies in the left radical, then $f(v)=0$ for every $f\in V^{\prime }$, because every $f$ is some $\phi (w)$. A nonzero $v$ can be included in a basis and detected by its coordinate functional, so this forces $v=0$. Both radicals vanish, proving

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.