Let $M_n(\mathbb C)$ be the $n$ by $n$ complex matrices with the inner product $(A|B)=tr(AB^*)$ , let $Q \in M_n(\mathbb C)$ be a fixed matrix and $$T:M_n(\mathbb C) \rightarrow M_n(\mathbb C)$$$$T_Q(A)=QA-AQ$$ Find the adjoint of $T_Q$ and $T_Q$ is self-adjoint iff $Q=Q^*+\lambda I$ for some $\lambda\in\mathbb C$
I am pretty sure that this kind of questions have been asked before. But what confuses me is that;
Does not this$$(T_Q(A)|B)=(QA-AQ|B)=(QA|B)-(AQ|B)=tr(QAB^* )-tr(AQB^*)=0$$ For all $A,B\in M_n(\mathbb C)$ implies that $T_Q$ is zero mapping? Since we can pick $B=T_Q(A)$? And i conclude that $QA-AQ=0$ So $Q$ is central which implies that $Q$ is scalar matrice.
But is not the adjoint of a zero mapping is zero mapping again? Which is also self adjoint. I am missing something here i guess. Thanks for your help
EDIT: After @Daniel Fischer 's warning i spotted my mistake and solved it. Here is my solution $$(T_Q(A)|B)=(QA-AQ|B)=Tr((QA-AQ)B^*)=Tr(QAB^*-AQB^*)=Tr(AB^*Q-AQB^*)=Tr(A(B^*Q-QB^*))=Tr(A(Q^*B-BQ^*)^*)=(A|T_{Q^*}(B))$$ And $Q=Q^*+\lambda I$ for some $\lambda\in\mathbb C$ iff $T_Q(A)=QA-AQ=QA+\lambda A -\lambda A -AQ=(Q+\lambda I)A-A(\lambda I + Q)=Q^*A-AQ^*=T_{Q^*}(A)$