Adjoint of linear transformation $T: \mathbb{M_n(C)} \rightarrow \mathbb{M_n(C)}$ Let V = ${M_n(\mathbb C)}$ with
inner product $\langle A, B\rangle = \text{tr}\,(B^*A)$, $A, B \in V$. 
Let $M \in {M_n(\mathbb C)}$, Define $T: V \rightarrow V$ by $T(A) = MA$. What is adjoint of $T$?  ($T^*(A)=$?)
Here is my approach but not successful,
By definition of adjoint, 
$$\langle T(A), B\rangle  = \langle A, T^*(B)\rangle, \ \ \ \forall A,B \in {M_n(\mathbb C)}.$$
$$\Rightarrow\langle MA, B\rangle =\langle A, T^*(B)\rangle $$
$$\text{tr}\,(B^*(MA))=\text{tr}\,((T^*(B))^* A).$$
How to proceed?? suppose if I equate (May I equate?) $B^*(MA)=(T^*(B))^* A$ How to get $T^*$?
 A: In general, we don't solve an equation to find $T^*B$, but rather, we know that $T^*$ exists (from general theory) and it should satisfy a certain property (namely $\langle TA,B\rangle=\langle A,T^*B\rangle$). From this property we should infer some new properties of $T^*$ which point us at a suitable candidate for $T^*$.
As in your last line, we want that $tr(B^*MA)=tr(T^*(B)^*A)$ for all $A$ and $B$. This will be satisfied if $B^*MA=T^*(B)^*A$ for all $A$ and $B$ (this is not and "if and only if"). In particular, for $A=I$, we would get something of the form $B^*M=T^*(B)^*$, or equivalently $T^*B=M^*B$.
The previous paragraph shows that the map $S(B)=M^*B$ satisfies, for all $A$ and $B$, $\langle TA,B\rangle=\langle A,SB\rangle$. But this is precisely the property which determines $T^*$, so $T^*=S$, i.e., $T^*(B)=S^*(B)=M^*B$ for all $B$.
(I am being rather pedantic here but I think this should make things as clear as possible.)
A: From $$\text{Tr}\,([B^*M-(T^*(B))^*]A),$$ you get that $T^*(B)^*=B^*M$ (because $A$ is arbitrary and so we can take it to be the adjoint of the expression between square brackets). 
Taking adjoints, $$T^*(B)=M^*B.$$
