# 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.

## 3 Answers

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.

• 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 $x=i$. – Steven Stadnicki Dec 3 '19 at 22:07

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*}

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 $$det(A^2 + I) = det \begin{pmatrix} \lambda^2+1 & 2\lambda & 1 \\ 0 & \lambda^2+1 & 2\lambda \\ 0 & 0 & \lambda^2 +1 \end{pmatrix} = (\lambda^2 + 1)^3$$