Proving conjugate property of Frobenius inner product over the complex field

Question

The Frobenius inner product defined by Inner product of A,B=trace(B*A), where , B*=conjugate of B transpose, over the complex field. I need help in proving the property that, Conjugate of Inner product of A,B = Inner product of B,A Can anyone please help me out in proving this particular property. Any help will be highly appreciated.

Answer 1

Note that if $A$ has entries $A_{ij}$ and $B%$ has entries $B_{ij}$, then $$\langle A,B \rangle = \sum_{i=1}^n \sum_{j=1}^n A_{ij}\overline{B_{ij}}\,.$$ That is, $\langle A,B \rangle$ is just as the dot product in $\mathbb F^N$ with $N=n^2$ (first you multiply each entry of the first vector with the conjugate of the respective entry of the second vector, then you add all). Can you finish from here?