Let \(g\) be a non-singular bilinear form on a finite-dimensional vector space \(V\). Suppose \(T\) is an invertible linear operator on \(V\) and that \(f\) is the bilinear form on \(V\) given by \(f(\alpha, \beta) = g(\alpha, T\beta)\). If \(U\) is a linear operator on \(V\), find necessary and sufficient conditions for \(U\) to preserve \(f\).
Solution
Because \(g\) is non-singular, every linear operator \(U\) has a unique adjoint \(U^\sharp\) relative to \(g\), characterized by \[ g(U\alpha,\gamma)=g(\alpha,U^\sharp\gamma) \] for all \(\alpha,\gamma\in V\).
1. Express preservation of \(f\) through the \(g\)-adjoint.
For arbitrary \(\alpha,\beta\), \[ \begin{aligned} f(U\alpha,U\beta) &=g(U\alpha,TU\beta)\\ &=g(\alpha,U^\sharp T U\beta). \end{aligned} \] On the other hand, \[ f(\alpha,\beta)=g(\alpha,T\beta). \] Therefore \(U\) preserves \(f\) exactly when \[ g(\alpha,(U^\sharp T U-T)\beta)=0 \] for all \(\alpha,\beta\).
2. Use non-singularity of \(g\) to obtain the operator criterion.
Since \(g\) is non-singular, the condition \[ g(\alpha,\gamma)=0 \] for every \(\alpha\) forces \(\gamma=0\). Hence the condition from step 1 is equivalent to \[ \boxed{U^\sharp T U=T.} \] Thus this equation is necessary and sufficient for \(U\) to preserve \(f\).
3. Give the equivalent inverse form of the condition.
If \[ U^\sharp T U=T, \] then the left-hand side is invertible because the right-hand side \(T\) is invertible. In particular \(U\) is invertible. Multiplying on the right by \(U^{-1}\) and then by \(T^{-1}\) gives \[ U^\sharp=T U^{-1}T^{-1}. \] Conversely this identity immediately implies \(U^\sharp T U=T\). Therefore one may equivalently state the answer as \[ \boxed{ U\text{ preserves }f \iff U^\sharp T U=T \iff U\in GL(V)\text{ and }U^\sharp=T U^{-1}T^{-1}.} \]