Let V be the set of all complex square matrices. For $D \in V$ define $T_D : V \rightarrow V$ by $T_D(A) = DA$.
Using the inner product $=tr(XY^*)$, find the adjoint of $T_D$.
2. The adjoint $T_D^*$ of $T_D$ satisfies the condition $\langle T_D^*(A),B\rangle = \langle A,T_D(B)\rangle = \langle A,DB\rangle$ for all B, where the angled brackets denote the inner product. If you write this in terms of the trace then it becomes $\text{tr}(T_D^*(A)B^*) = \text{tr}(A(DB)^*) = \text{tr}(AB^*D^*)$. But the trace has the property that $\text{tr}(XY) = \text{tr}(YX)$. This means that $\text{tr}(T_D^*(A)B^*) = \text{tr}(D^*AB^*)$. From that you should be able to see what $T_D^*(A)$ is.