If $M$ is $n$ by $n$ square matrix, and $\det(M+I) \ge 0$, show that $\det(M^5+I) \ge 0$

Question

If $M$ is $n$ by $n$ square matrix and $$\det(M+I) \ge 0$$ show that $$\det(M^5+I) \ge 0$$

I was trying with eigenvalues but I got stuck.

Answer 1

Hints.

• If $\det(M+I)$ is zero, $\det(M^5+I)$ must be zero too because $M+I$ is a factor of the polynomial $M^5+I$.
• If $\det(M+I)>0$, the number of real eigenvalues of $M$ that are smaller than $-1$ must be even.