Proving a determinant inequality Let $A$ be a square matrix in $M_n(\mathbb R)$. Prove that:
$$det(A^2+I_n) \ge 0$$
I wrote  $A^2+I_n=A^2 I_n+I_n=I_n(A^2+1)$:
$$det(I_n)\cdot det(A^2+1)=det(A^2+1)$$
How can I prove that is $\ge 0$ ? Thank you.

The problem in the OP approach is that we don't know what $A^2+1$ is. 
It's better to use complex numbers and the relatioships:


*

*$A^2+I_n=(A-iI_n)(A+iI_n)$, and 

*$\det(\overline B)=\overline{\det B}$ for any complex square matrix $B$.

