Let $V= M_n(\mathbb{C})$ be an inner product space defined by $\langle A, B\rangle = Tr(AB^*)$
Let $T\in Hom(V,V)$ be a linear operator defined by $T(A)=P^{-1}AP$ for $P \in M_n(\mathbb{C})$
Find $T^*$
I tried to approach it in two differnt ways :
First is to use the fact that for every matrix $A$: $A^* = \overline{A^T}$
I found that the inner product $\langle A, B\rangle = Tr(AB^*)$ can be written as $\sum_{i=1}^{n} v_i\overline u_i$ where as $v_i$ is a row of A and $\overline u_i$ is a row of $B^*$
I don't know how to proccceed from there.
The second way I tried was to use the formula $\langle T(A), B\rangle = \langle A, T^*(B)\rangle$ but couldn't find a breakthrough there as well.
questions:
1 - Which way do you recommend ?
2- How do I procceed in that way?
From $\langle A,B \rangle =Tr(AB^*)$ we get \begin{align} \langle T(A),B \rangle & = Tr(T(A)\, B^*)\\ &=Tr(P^{-1}AP \, B^*) \\ &=Tr(A\,\,\color{red}{PB^* P^{-1}}) \ \ \textrm{ (since $Tr(MN)=tr(NM)$ for any $N,M$ in $V$) } \end{align}
But we know that $$ \langle T(A), \, B\rangle =\langle A, \,T^*(B)\rangle=Tr(A \, \color{red}{(T^*(B))^*})$$ Thus, the adjoint operator $T^*$ must satisfy $$ (T^*(B))^* = P B^* P^{-1} \implies \boxed{T^*(B) = (P^{-1})^* B P^*}. $$