Trace of product of three matrices

Question

I have an expression of the form \begin{equation} i\operatorname{tr}(ABC - B^\dagger AC), \end{equation} where $A$ and $C$ are Hermitian matrices, but $B$ is not. $B-B^\dagger := i D$, where $D$ is also a Hermitian matrix. It can be seen that the above is real by cyclic permutation of the second term inside the trace, $B^\dagger AC \to CB^\dagger A$: $$ i\operatorname{tr}(ABC - CB^\dagger A) = i \underbrace{\operatorname{tr}(ABC - (ABC)^\dagger)}_{2 i \operatorname{Im} (\operatorname{tr}(ABC))}=-2\operatorname{Im}(\operatorname{tr}(ABC))\in\mathbb{R}. $$ However, for reasons concerning the specific form of the matrices, I would like to rewrite the first expression like so: \begin{align} i\operatorname{tr}(ABC - B^\dagger AC) &= i\operatorname{tr}(ABC - B^\dagger CA + B^\dagger \left[C,A\right])\\ &= i \operatorname{tr}(ABC - A B^\dagger C) - i\operatorname{tr}(B^\dagger \left[A,C\right])\\ &= - \operatorname{tr}(ADC) - i \operatorname{tr}(B^\dagger \left[A,C\right]), \end{align} where $\left[A,C\right]$ is the commutator of $A$ and $C$. My (perhaps minor) issue is that it is not readily apparent (to me at least) that the final expression is also real. Is there a way of seeing explicitly that the final expression is real, without knowing the specifics of A, B and C and without just rewriting it to its original form?

Answer 1

The Hermitian conjugate is the composition of the complex conjugate and the transpose $$\eqalign{ \def\d{\dagger} &Z^\d \;\equiv\; (Z^*)^T \;\equiv\; (Z^T)^* \\ }$$ Hermitian matrices satisfy $$\eqalign{ &A^\d = A \quad\implies\quad A^T=A^* \\ &C^\d = C \quad\implies\quad C^T=C^* \\ }$$ The double-dot product $(:)$ is a convenient notation for the trace and has these properties $$\eqalign{ \def\a{\alpha} \def\b{\beta} \def\l{\lambda} \def\tr{\operatorname{tr}} X:Y \;&\equiv\; \sum_{i=1}^m\sum_{k=1}^n X_{ik}Y_{ik} \;\equiv\; \tr(X^TY) \\ X:Y &= Y:X \;=\; Y^T:X^T \\ PQ:Y &= P:(YQ^T) \;=\; Q:(P^TY) \\ }$$ Use the above notations to analyze the first expression $$\eqalign{ &\tr(ABC) = AB:C^T = B:A^TC^T = B:(AC)^* &\equiv \l \\ &\tr(B^\d AC) = B^*:AC = \Big(B:(AC)^*\Big)^* &\equiv \l^* \\ &\l \;\equiv\; \a+i\b \quad\iff\quad \l^* = \a-i\b \\ &\therefore \;i\tr(ABC-B^\d AC) \;=\; i(\l-\l^*) \;\equiv\; -2\b \\ }$$ and (the negative of) the second $$\eqalign{ &\tr(ADC) = D^T:CA = D:(AC)^* = iB^\d:(AC)^*-iB:(AC)^* \\ &\tr(ADC) \equiv i\mu-i\l \\ &i\tr(B^\d AC-B^\d CA) = iB^*:AC - iB^\d:(AC)^* \;\equiv\; i\l^* - i\mu \\ &\therefore\;\tr(ADC)+i\tr(B^\d[AC]) = i(\l^*-\l) \;\equiv\; 2\b \\ }$$ So your first and second expressions are equal and are real.