Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 2 / 8B / Solution

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.