Is $[Tr(AB)]^{*} = Tr(AB)^{\dagger} = Tr(B^{\dagger}A^{\dagger})$?
The second equal sign is trivial but the first equal sign is what I am puzzled about.
$*$ means the complex conjugate of a number.
$\dagger$ means the complex conjugate of a matrix.
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$\begingroup$In general $Tr(M^{\dagger}) = Tr(M)^*$, since $M^{\dagger} = (M^T)^*$, and $tr(M) = tr(M^T)$ and $tr(M^*) = tr(M)^*$, where $M^*$ denotes conjugating every element of the matrix.
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