Let \(V\) be the space of \(n \times n\) matrices over the complex numbers, with the inner product \((A, B) = \operatorname{tr}(AB^*)\). Let \(P\) be a fixed invertible matrix in \(V\), and let \(T_P\) be the linear operator on \(V\) defined by \(T_P(A) = P^{-1}AP\). Find the adjoint of \(T_P\).
Solution
The adjoint is \[ \boxed{ T_P^*(B)=(P^*)^{-1}BP^*.} \] Equivalently, \[ T_P^*=T_{P^*}. \]
1. Move the factors around the trace.
For arbitrary matrices \(A,B\), \[ \begin{aligned} (T_P(A),B) &=\operatorname{tr}\!\left(P^{-1}APB^*\right)\\ &=\operatorname{tr}\!\left(APB^*P^{-1}\right), \end{aligned} \] where the second equality uses cyclicity of trace [@HK71].
2. Identify the matrix whose adjoint appears in the second slot.
Set \[ S(B)=(P^*)^{-1}BP^*. \] Then \[ S(B)^* =P B^* P^{-1}, \] because \(((P^*)^{-1})^*=P^{-1}\) and \((P^*)^*=P\). Hence \[ \operatorname{tr}\!\left(APB^*P^{-1}\right) =\operatorname{tr}\!\left(A S(B)^*\right) =(A,S(B)). \] Thus \[ (T_P(A),B)=(A,S(B)) \] for every \(A,B\). By uniqueness of the adjoint, \[ T_P^*=S, \] which is the displayed formula.