I can't seem to find a reference which confirms this for me.
I know that the standard inner product for $A,B \in M_{n \times n}(\mathbb{R})$ is given by $\langle A,B\rangle = \operatorname{Tr}(B^{T}A)$.
Would the standard inner product for $A,B \in M_{n \times n}(\mathbb{C})$ be given by $\langle A,B\rangle = \operatorname{Tr}(B^{\dagger}A)$?
This is the only thing that makes sense, but I can't find anything about the matter online. I think this would give a valid "hermitian inner product" in the sense that we have $\langle A,B\rangle = \overline{\langle B,A\rangle}$.