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For a bilinear form $\varphi$, define $F_{\varphi }:V\to V^{\ast }$ by $F_{\varphi }(v)(u)=\varphi (u,v)$. Bilinearity makes $F_{\varphi }(v)$ a linear functional and makes $F_{\varphi }$ linear in $v$. Conversely, every linear $F:V\to V^{\ast }$ defines a bilinear form by ${\varphi }_F(u,v)=F(v)(u)$. These constructions undo each other, giving the desired bijective correspondence.

For a nondegenerate bilinear form, $F_{\varphi }$ has zero kernel and is an isomorphism because $V$ and its dual space have the same finite dimension. For the two forms, putIt is an isomorphism and satisfies ${\varphi }_1(u,\alpha v)=F_{{\varphi }_1}(\alpha v)(u)=F_{{\varphi }_2}(v)(u)={\varphi }_2(u,v)$ for all $u,v$. This is the representation of a bilinear form relative to a nondegenerate bilinear form.

If both forms are symmetric, thenNondegeneracy makes the adjoint with respect to ${\varphi }_1$ unique; this equality proves $\boxed{{\alpha }^{†}=\alpha }$ in that bilinear-form sense. No positivity or Hermitian conjugation is assumed over the arbitrary field.

A bilinear form $B$ on $V$ is nondegenerate whenIn finite dimensions this is equivalent to nondegeneracy in the first argument.

Define the linear mapNondegeneracy makes ${\beta }_1$ injective. Since $V$ and its dual space have the same finite dimension, ${\beta }_1$ is an isomorphism. Similarly define ${\beta }_2(w)=B_2(-,w)$ and setThen $\alpha$ is linear and

If $w\in \ker \alpha$, this identity gives $B_2(v,w)=0$ for every $v$. Conversely, if $B_2(v,w)=0$ for every $v$, then $B_1(v,\alpha w)=0$ for every $v$, so nondegeneracy gives $\alpha w=0$. HenceThis is the representation of a bilinear form relative to a nondegenerate bilinear form.