$\langle A,B\rangle = \operatorname{tr}(B^*A)$

Question

"define the inner product of two matrices $A$ and $B$ in $M_{n\times n}(F)$ by $$\langle A,B \rangle = \operatorname{tr}(B^*A), $$ where the {conjugate transpose} (or {adjoint}) $B^*$ of a matrix $B$ is defined by $B^*_{ij} = \overline{B_{ji}}.$

Prove that $\langle B,A \rangle = \overline{\langle A,B \rangle}$

This is left to the reader in my linear algebra text book and can't seem to work out a solution. Any help would be much appreciated.

Answer 1

This is trivial once you realize that $\mathrm{tr}(AB)=\mathrm{tr}(BA)$, as you have: $$\langle B,A\rangle=\mathrm{tr}(A^*B)=\mathrm{tr}(BA^*)=\langle A^*,B^*\rangle=\overline{\langle A,B\rangle}$$

Answer 2

Hint: $\operatorname{tr}(AB)=\operatorname{tr}(BA)$.

Answer 3

The trace is invariant under transposition, and the complex conjugate of the trace is the trace of the complex conjugate; also $A^*=\overline A^T =\overline{A^T}$, thus: $$\langle B,A\rangle = \operatorname{tr}(A^*B) = \operatorname{tr}(A^*B)^T = \operatorname{tr}B^T\overline A = \operatorname{tr}\overline{(B^*A)} =\overline{\operatorname{tr}(B^*A)}=\overline{\langle A,B\rangle}$$

After having written that, I noticed that there's a deleted answer which essentially says the same; I wonder why the owner deleted it, since it's the right answer.