I want to prove the following:
$\langle A,B \rangle = \overline{\langle B,A \rangle}$
where
$\langle A,B \rangle := tr(AB^{*})\,\, and \,\, A,B \in \mathbb{C}^{n \times n} $
Note:
The bar denotes the complex conjugate of all entries of a matrix and the "*" the adjoint matrix.
Edit:
$\langle A,B \rangle = tr(AB^{*}) = tr(\overline{(BA^{*})^T}) = tr(\overline{BA^{*}}) = \overline{tr(BA^*)} = \overline{\langle B,A \rangle}$
The third equality holds since A and B are square matrices.
Best Answer
Just observe that $(AB^*)^*=B^{**}A^*=BA^*$ and therefore the matrices $AB^*$ and $BA^*$ are complex conjugate of each other. Their diagonal elements are thus conjugate and their sums are as well, that is, $tr(AB^*)=\overline{tr(BA^*)}$