Determine $T^*$ such that $T(A):M_2(K) \rightarrow AM-MA$ and inner product $tr(B^tA) = \langle A,B\rangle$

Question

Well $tr(B^tA)$ is clearly an inner product, now we have to find $T^*$ such that $\langle T(A),B\rangle=\langle A,T^*(B)\rangle $

Where M =$\begin{bmatrix}1 & 0\\-1& 2\end{bmatrix}$

I think I'm not doing the right approach simply by setting the inner products equal and yielding a system of equations, since that in this case T* would depende on A and B and not only on the transformation T. Can someome give me a hint on how to proceed? Thanks in advance.

Answer 1

Hint: $tr(AB) = tr(BA)$ you will have $$\left\langle T(A), B\right\rangle = \left\langle A, BM^t-M^tB \right\rangle$$