Let \(V\) be a finite-dimensional vector space and let \(f\) be a non-degenerate symmetric bilinear form on \(V\). Show that for each linear operator \(T\) on \(V\) there is a unique linear operator \(T'\) on \(V\) such that \[ f(T\alpha,\beta)=f(\alpha,T'\beta) \] for all \(\alpha,\beta\in V\). Also show that \[ (T')'=T. \] How much of the above remains valid without the assumption that \(f\) is non-degenerate?
Solution
Define \[ L_f:V\longrightarrow V^*, \qquad L_f(\gamma)(\alpha)=f(\alpha,\gamma). \] Because \(f\) is non-degenerate and \(V\) is finite-dimensional, \(L_f\) is an isomorphism.
1. Every operator has a unique adjoint relative to \(f\).
Fix \(\beta\in V\). The map \[ \alpha\longmapsto f(T\alpha,\beta) \] is a linear functional on \(V\). Since \(L_f\) is onto, there is a unique vector, denoted \(T'\beta\), such that \[ f(T\alpha,\beta)=f(\alpha,T'\beta) \] for every \(\alpha\).
To see that \(T'\) is linear, let \(a,b\) be scalars. 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 gives \[ T'(a\beta+b\gamma)=aT'\beta+bT'\gamma. \]
If \(S\) also satisfies the adjoint identity, then \[ f(\alpha,(S-T')\beta)=0 \] for all \(\alpha\), so non-degeneracy implies \((S-T')\beta=0\) for every \(\beta\). Thus \(S=T'\).
2. The adjoint operation is an involution.
Symmetry gives, for all \(\alpha,\beta\), \[ \begin{aligned} f(T'\beta,\alpha) &=f(\alpha,T'\beta)\\ &=f(T\alpha,\beta)\\ &=f(\beta,T\alpha). \end{aligned} \] Thus \(T\) is an adjoint of \(T'\). By uniqueness from step 1, \[ \boxed{(T')'=T.} \]
3. Describe the degenerate case.
Let \[ R=V^\perp=\{r\in V:f(r,v)=0\text{ for every }v\in V\} \] be the radical of the symmetric form.
An adjoint of \(T\) exists if and only if \[ T(R)\subseteq R. \] Indeed, the image of \(L_f\) is the annihilator \[ R^0=\{\lambda\in V^*: \lambda(r)=0\text{ for all }r\in R\}. \] For fixed \(\beta\), the functional \(\alpha\mapsto f(T\alpha,\beta)\) lies in \(R^0\) exactly when \[ f(Tr,\beta)=0 \] for every \(r\in R\). Requiring this for all \(\beta\) is exactly \(Tr\in R\) for all \(r\in R\).
When an adjoint exists it need not be unique. If \(S_1,S_2\) are adjoints, then \[ f(\alpha,(S_1-S_2)\beta)=0 \] for all \(\alpha,\beta\), so \[ \operatorname{im}(S_1-S_2)\subseteq R. \] Conversely, adding any operator with image in \(R\) to an adjoint gives another adjoint.
Finally, symmetry still shows that if \(S\) is an adjoint of \(T\), then \(T\) is an adjoint of \(S\). Thus the adjoint relation remains symmetric, but without non-degeneracy there is generally neither existence for every \(T\) nor uniqueness, so one cannot define a single-valued involution \(T\mapsto T'\) on all operators.