Let \(V\) be a finite-dimensional vector space over a subfield \(F\) of \(\CC\), and let \(f\) be a non-degenerate skew-symmetric bilinear form on \(V\). Prove the following skew-symmetric analogue of the adjoint theorem.
For each linear operator \(T\) on \(V\), there is a unique linear operator \(T'\) such that \[ f(T\alpha,\beta)=f(\alpha,T'\beta) \] for all \(\alpha,\beta\in V\). Moreover, \[ (T_1T_2)'=T_2'T_1',\qquad (c_1T_1+c_2T_2)'=c_1T_1'+c_2T_2',\qquad (T')'=T. \]
Solution
Define the second-slot identification \[ \Phi_f:V\longrightarrow V^*, \qquad \Phi_f(\alpha)(\beta)=f(\beta,\alpha). \] Because \(f\) is non-degenerate and \(V\) is finite-dimensional, \(\Phi_f\) is an isomorphism.
1. Every operator \(T\) has a unique adjoint \(T'\) relative to \(f\).
For fixed \(\beta\in V\), the map \[ \alpha\longmapsto f(T\alpha,\beta) \] is a linear functional on \(V\). Since \(\Phi_f\) is onto, there is a unique vector \(T'\beta\) such that \[ f(T\alpha,\beta)=f(\alpha,T'\beta) \] for every \(\alpha\).
It remains to see that \(\beta\mapsto T'\beta\) is linear. Let \(a,b\in F\). For all \(\alpha\), \[ \begin{aligned} f(\alpha,T'(a\beta+b\gamma)) &=f(T\alpha,a\beta+b\gamma)\\ &=a f(T\alpha,\beta)+b f(T\alpha,\gamma)\\ &=f(\alpha,aT'\beta+bT'\gamma). \end{aligned} \] Non-degeneracy in the second variable therefore gives \[ T'(a\beta+b\gamma)=aT'\beta+bT'\gamma. \] Thus \(T'\) is linear. Uniqueness follows from the same argument: if \(S\) also satisfies the adjoint identity, then \[ f(\alpha,(T'-S)\beta)=0 \] for all \(\alpha\), hence \((T'-S)\beta=0\) for every \(\beta\).
2. Taking adjoints reverses products.
For all \(\alpha,\beta\), \[ \begin{aligned} f(T_1T_2\alpha,\beta) &=f(T_2\alpha,T_1'\beta)\\ &=f(\alpha,T_2'T_1'\beta). \end{aligned} \] By uniqueness from step 1, \[ \boxed{(T_1T_2)'=T_2'T_1'.} \]
3. Taking adjoints is linear.
For scalars \(c_1,c_2\) and all \(\alpha,\beta\), \[ \begin{aligned} f((c_1T_1+c_2T_2)\alpha,\beta) &=c_1f(T_1\alpha,\beta)+c_2f(T_2\alpha,\beta)\\ &=f(\alpha,(c_1T_1'+c_2T_2')\beta). \end{aligned} \] Hence uniqueness gives \[ \boxed{(c_1T_1+c_2T_2)'=c_1T_1'+c_2T_2'.} \]
4. The adjoint operation is an involution.
Apply the defining identity to \(T'\): \[ f(T'\beta,\alpha)=f(\beta,T''\alpha). \] Using skew-symmetry and then the defining identity for \(T\), \[ \begin{aligned} f(T'\beta,\alpha) &=-f(\alpha,T'\beta)\\ &=-f(T\alpha,\beta)\\ &=f(\beta,T\alpha). \end{aligned} \] Therefore \[ f(\beta,T''\alpha)=f(\beta,T\alpha) \] for every \(\alpha,\beta\). Non-degeneracy implies \[ T''\alpha=T\alpha \] for every \(\alpha\), so \[ \boxed{(T')'=T.} \]