Prove or disprove : $\det(A^k + B^k) \geq 0$

This question came from here. As the OP hasn't edited his question and I really want the answer, I'm adding my thoughts.

Let $A, B$ be two real $n\times n$ matrices that commute and $\det(A + B)\ge 0$.
Prove or disprove : $\forall k\in \mathbb{N}^*$ $$\det(A^k + B^k) \geq 0$$

What I think about it
I think we can consider two cases : k even and k odd.
For k even, the result is pretty easy considering $$\det(A^{2k}+B^{2k})=\det(A^{k}+iB^{k}){\det(A^{k}-iB^{k})}=\det(A^{k}+iB^{k})\overline {\det(A^{k}+iB^{k})}\ge 0$$ For k odd, I thought about this equality but I don't know if it can help $$A^{2k+1}+B^{2k+1}=(A+B)\left(\sum_{i=0}^{2k} (-1)^i A^{2k-i}B^{i}\right)$$ so $$\det(A^{2k+1}+B^{2k+1})\ge 0 \iff \det\left(\sum_{i=0}^{2k} (-1)^i A^{2k-i}B^{i}\right)\ge 0$$

Solution 1:

Hint :

If $k=2p$ then $A^k+B^k = (A^p+iB^p)(A^p-iB^p)$ and use that $\det \overline{M}=\overline{\det M}$. In this case, the hypothesis $\det(A+B)\ge 0$ is useless.

If $k=2p+1$, use the factorization $A^k+B^k = (A+B)\prod_{m=1}^p (A+e^{\frac{2im\pi}{2p+1}}B)(A+e^{-\frac{2im\pi}{2p+1}}B)$ and then use the trick of the even case.

Solution 2:

Here is the proof for $det(B) \ne 0$. We have $$ det(A+B)=det(B)det(AB^{-1}+I)=det(B)det(PJ_{AB^{-1}}P^{-1}+I)=det(B)(\lambda _1 +1)\cdots (\lambda _n + 1) $$ and $$ det(A^k+B^k)=det(B^k)det(A^k B^{-k} + I)=det(B^k)det(PJ_{AB^{-1}}^kP^{-1}+I)=det(B)^k({\lambda _1}^k +1)\cdots ({\lambda _n}^k + 1), $$ where $\lambda _i$ are eighnvalues of $AB^{-1}$. There are four possible cases: $$1)det(B)\gt0,k~is~odd$$ $$1)det(B)\gt0,k~is~even$$ $$1)det(B)<0,k~is~odd$$ $$1)det(B)<0,k~is~even$$. It is easy to see that in all these cases $det(A^k+B^k)\ge0$.