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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.

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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.

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  • $\begingroup$ The 'easy way' of showing that $\det(A\pm iI)$ are conjugates is to look at $P_d(x) = \det(A+xI)$ and note that it's a polynomial in $x$, so $\overline{P_d(x)} = P_d(\bar{x})$; now just plug in $x=i$. $\endgroup$ – Steven Stadnicki Dec 3 '19 at 22:07
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If $A$ has eigenvalues $\lambda_1, \lambda_2, \lambda_3$, then $\det (A^2 + 1) = \prod (\lambda_i^2 + 1)$. (Either use the Jordan normal form, or note that it's clearly true if $A$ is diagonalizable and then conclude it for arbitrary $A$ by the fact that $\det$ is continuous.) If all $\lambda_i$ are real, then this product is clearly positive. Otherwise, assume without loss of generality that $\lambda_2 = \overline{\lambda_1}$ and $\lambda_3$ is real. Then we again have \begin{align*} \prod (\lambda_i^2 + 1) = (\lambda_1^2 + 1)(\overline{\lambda}{}_1^2 + 1)^*(\lambda_3 + 1) = \left|\lambda_1^2 + 1\right|^2 (\lambda_3^2 + 1) \geq 0. \end{align*}

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You can also check that it is true via the real Jordan normal form.

First if we have a complex eigenvalue $\alpha + i \beta$ and one real eigenvalue $\lambda$, then $A$ is (after conjugation) of the form $$ A = \begin{pmatrix} \alpha & \beta & 0 \\ -\beta & \alpha & 0 \\ 0 & 0 & \lambda \end{pmatrix} $$ and thus $$ det(A^2 + I) = det \begin{pmatrix} \alpha^2-\beta^2+1 & 2\alpha \beta & 0 \\ -2\alpha \beta & \alpha^2 - \beta^2+1 & 0 \\ 0 & 0 & \lambda^2 +1 \end{pmatrix} = (\lambda^2+1) \left( (\alpha^2 - \beta^2+1)^2 + 4\alpha^2 \beta^2 \right) $$

Now we consider the case when all the eigenvalues $\lambda, \mu, \nu\in \mathbb{R}$.

If they are all simple, then we get $det(A^2 + I) = (\lambda^2 +1) (\mu^2 +1) (\nu^2+1)$.

If one of the eigenvalues has multiplicity 2, then $A$ is (after conjugation) of the form $$ A = \begin{pmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \mu \end{pmatrix} $$ and thus $$ det(A^2 + I) = det \begin{pmatrix} \lambda^2+1 & 2\lambda & 0 \\ 0 & \lambda^2+1 & 0 \\ 0 & 0 & \mu^2 \end{pmatrix} = (\lambda^2+1)^2 (\mu^2+1) $$

Finally, we are left to consider the case when we have one eigenvalue of multiplicity 3, then $A$ is (after conjugation) of the form $$ A = \begin{pmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{pmatrix} $$ and thus $$ det(A^2 + I) = det \begin{pmatrix} \lambda^2+1 & 2\lambda & 1 \\ 0 & \lambda^2+1 & 2\lambda \\ 0 & 0 & \lambda^2 +1 \end{pmatrix} = (\lambda^2 + 1)^3 $$

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