For which $C$ is the linear map $\varphi_C : V \to V, A \mapsto CA$ selfadjoint with scalar product $\langle A, B \rangle = \operatorname{tr}(A^t B)$?

Question

Let $V=M_{n\times n}(\mathbb R)$.

By the definition of selfadjoint, I have to show for which $C$, $\langle \varphi_C(A), B\rangle = \langle A,\varphi_C (B)\rangle$ is true $\forall A,B \in V$. In other words, $\operatorname{tr}((CA)^t B) = \operatorname{tr}(A^t CB)$.

Using some cyclic permutations of trace, I cannot figure out how to do this. Having tried this, it seems to me like this cannot be true since I cannot reform the LHS to have some version of $CB$, instead it amounts to some variant of $BC$ inside the trace. How do I begin? Is this about showing $C$ must be symmetric?

Answer 1

The given condition implies that $\operatorname{tr}(BA^t(C-C^t))=0$ for all matrices $B$ and $A$. Thus $\operatorname{tr}(X(C-C^t))=0$ for every matrix $X$. Hence $C-C^t$ must be zero, i.e. $C$ must be symmetric.