True/false: $\det(A^2+I)\ge 0$ for every $3 \times 3$ matrix with real entries and rank $>0$ I have the following proposition about which I have to say whether it is true or false.

$\det(A^2+I)\ge 0$ for every $3 \times 3$ matrix with real entries and rank $>0$. $I$ is the identity matrix.

I tried brutal ways (taking a generic matrix, evaluating its square and adding $I$), but there are too many calculations and I feel that manner will not lead me to anything of interesting. I also tried to construct a counterexample, but nothing. I am not able to prove nor confute this assertion.

If you factor $A^2 + I = (A+iI)(A-iI)$ you get $\det(A^2 + I) = \det(A+iI)\cdot\det(A-iI)$. Now I didn't check it till the end, but it seems like $\det(A\pm iI)$ are complex conjugates, so you get something non-negative.
$$
\begin{vmatrix}
a \pm i & b & c \\
d & e \pm i & f \\
g & h & j \pm i
\end{vmatrix} = (a \pm i)(e \pm i)(j \pm i) + \dots = ((ae - 1) \pm i(a+e))(j\pm i) + \dots = (j(ae - 1) - a - e \pm i(ae - 1 + ja + je)) + \dots
$$
After the dots it seems trivial.
