Let \(f\) be a non-degenerate form on a finite-dimensional space \(V\). Show that each linear operator \(S\) has an ‘adjoint relative to \(f\),’ i.e., an operator \(S'\) such that \(f(S\alpha, \beta) = f(\alpha, S'\beta)\) for all \(\alpha, \beta\).
Solution
Fix a linear operator \(S\) on \(V\).
1. For each \(\beta\), construct a unique vector \(S'\beta\).
For fixed \(\beta\in V\), define \[ L_\beta(\alpha)=f(S\alpha,\beta). \] Because \(S\) is linear and \(f\) is linear in its first variable, \(L_\beta\) is a linear functional on \(V\). By Exercise 8, since \(f\) is non-degenerate, there is a unique vector, which we denote by \(S'\beta\), such that \[ L_\beta(\alpha)=f(\alpha,S'\beta) \] for every \(\alpha\). Therefore \[ f(S\alpha,\beta)=f(\alpha,S'\beta) \] for all \(\alpha,\beta\).
2. The assignment \(\beta\mapsto S'\beta\) is linear.
Let \(c\) be a scalar and let \(\beta,\gamma\in V\). Since a sesquilinear form is conjugate-linear in its second variable, \[ \begin{aligned} f(S\alpha,c\beta+\gamma) &=\overline c\,f(S\alpha,\beta)+f(S\alpha,\gamma)\\ &=\overline c\,f(\alpha,S'\beta)+f(\alpha,S'\gamma)\\ &=f(\alpha,cS'\beta+S'\gamma). \end{aligned} \] But by the defining property of \(S'\) applied to \(c\beta+\gamma\), \[ f(S\alpha,c\beta+\gamma)=f(\alpha,S'(c\beta+\gamma)). \] Thus \[ f\bigl(\alpha,S'(c\beta+\gamma)-cS'\beta-S'\gamma\bigr)=0 \] for every \(\alpha\). Right non-degeneracy implies \[ S'(c\beta+\gamma)=cS'\beta+S'\gamma. \] Hence \(S'\) is a linear operator.
Thus every linear operator \(S\) has a relative adjoint \(S'\) satisfying \[ \boxed{f(S\alpha,\beta)=f(\alpha,S'\beta)} \] for all \(\alpha,\beta\in V\).