Let $A,B\in\text{Mat}_n(\mathbb{C})$ if $A$, $B$ commute, then $\det(A^2+B^2)\geq 0$

Question

Let $A,B\in\text{Mat}_n(\mathbb{C})$. If $A$, $B$ commute, then $\det\left(A^2+B^2\right)\geq 0$.

We know: $A^2+B^2=(A+iB)(A-iB)$ so $$\det\left(A^2+B^2\right)=\det(A+iB)\det(A-iB)=\det(A+iB)\,\overline{\det(A+iB)}$$

I'm stuck here, can someone help me?

Is it correct to say $\det(A+iB)\,\overline{\det(A+iB)}=\bigl(\det(A+iB)\bigr)^2$?

Answer 1

If $A=0$ and $B=diag(1,i)$ (the diagonal matrix with entries $1,i$) then the result is incorrect. Maybe you needed something else?

If I am not wrong this is true if we assume that the matrixes are real. Try using that if two matrixes commute then they can be diagonalized (EDIT:assuming they can be diagonalized in the first place) by the same matrix, meaning that there exists real $P$ such that $PAP^{-1}=diagonal$ and $PBP^{-1}=diagonal$.