Prove that $T_p$ is Self Adjoint

Question

Let $P \in M^{\Bbb C}_{n \times n}$ an invertible matrix. Let $T_p$ be a linear transformation $T_p: M^{\Bbb C}_{n \times n} \to M^{\Bbb C}_{n \times n}$ such that: $$T_p(X)=P^{-1}XP$$. for every $X \in M^{\Bbb C}_{n \times n}$. The scalar (inner) product is $<A,B>=tr(B^*A)$

Prove that if $P$ is self-adjoint, then $T_p$ is self-adjoint.

My question

Let $*$ be the self-adjoint operator. Should I prove $(T_p)^*=T_p$ OR $T_{p^*}=T_p$?

I think I should solve that with inner-product arithmetic but I don't know which term I have to prove.

Assuming I have to prove $(T_p)^*=T_p$, I am stuck here:

$$tr \left( B^*P^{-1}XP\right) =tr \left(P^{-1} X^*PB \right)$$

Thanks,

Alan

Answer 1

So starting from $\langle A,B\rangle = \operatorname{tr}(B^* A)$, we want to show that

$$ \langle T_p X,B \rangle = \langle X,T_p B\rangle \Longleftrightarrow \operatorname{tr}(B^* T_p X) = \operatorname{tr}((T_p B)^* X).$$

We have that $\operatorname{tr}((T_p B)^*X) = \operatorname{tr}((P^{-1}BP)^*X) = \operatorname{tr}(P^*B^* (P^{-1})^*X)$. But now $P^* = P$ and $(P^{-1})^* = P^{-1}$ so this reduces to $\operatorname{tr}(PB^*P^{-1}X)$. Now use the fact that traces are cyclic.

Answer 2

\begin{eqnarray} \langle X,T_p(Y) \rangle &=& \operatorname{tr}(X^* P^{-1} Y P ) \\ &=& \operatorname{tr}(PX^*P^{-1} Y) \\ &=& \operatorname{tr}((P^{-1}XP)^* Y) \\ &=& \langle P^{-1}XP,Y \rangle \end{eqnarray} Hence $(T_p)^* X = P^{-1}XP = T_p X$.