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If $A = \begin{bmatrix} x & 1\\ y & 0\end{bmatrix}, B = \begin{bmatrix} z & 1\\ w & 0\end{bmatrix}$, for $x,y,z,w \in \Bbb{R}$.

I have observed by considering many examples of $x,y,z,w$ that:

If all the eigen values of $A^2B$ and $AB^2$ are less than one in absolute value $\implies$ $\det(AB+A+I)<0$ and $\det(BA+B+I)<0$ is not possible.

Any way how to prove it actually?

I am thinking if $\det(AB+A+I)<0$ and $\det(BA+B+I)<0$, then perhaps it would violate certain assumptions on the eigenvalues of $A^2B, AB^2$?

Explicit forms of matrices:

$A^2B = \begin{bmatrix} z(x^2+y)+xw & x^2+y\\ xyz+wy & xy\end{bmatrix}$

$AB^2 = \begin{bmatrix} x(w+z^2)+wz & xz+w\\ y(z^2+w) & yz\end{bmatrix}$

$AB +A+I = \begin{bmatrix} xz+w+x+1& x+1\\ yz+y & y+1\end{bmatrix}$

$BA+B+I = \begin{bmatrix} xz+y+z+1 & z+1\\ xw+w & w+1\end{bmatrix}$

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    $\begingroup$ You mean that the determinants cannot be simultaneously negative ? $\endgroup$ Commented Nov 14, 2022 at 16:00
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    $\begingroup$ Just a quick idea, suppose on the contrary it was true. Then the product of two negative numbers is positive and using the fact that determinate is multiplicative might be good idea. If not, try to factor a matrix out of the product inside and view that determinant as the characteristic polynomial of a matrix evaluated at point. $\endgroup$ Commented Nov 14, 2022 at 16:15
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    $\begingroup$ Another idea : using Cayley-Hamilton theorem : $A^2$ can be replaced by $xA+yI$ and $B^2$ by $zB+wI$... $\endgroup$ Commented Nov 14, 2022 at 21:36
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    $\begingroup$ A limit case : $A=\pmatrix{a&1\\a&0}$ with $B=\pmatrix{-1&1\\-1&0}$ $\endgroup$ Commented Nov 14, 2022 at 23:50
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    $\begingroup$ I've checked by computer that if $y = -x^2/4$ and $w = -z^2/4$ (this is the case when neither $A$ nor $B$ is diagonalizable over $\mathbb{C}$) then your conjecture is correct. $\endgroup$ Commented Nov 17, 2022 at 2:14

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This has been proven on MathOverflow. Making a Community Wiki answer so that it can stop showing up in the unanswered list while avoiding me "taking credit" for someone else's solution.

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  • $\begingroup$ The conjecture is known to be true via software and numerical searches but a formal mathematical proof is still not there.. Before posting this question I have mentioned that after trying a lot of examples, I have derived this conjecture but the answers have shown that using softwares the conjecture is true.. And i dont see a formal. Proof. So the question still remains open. $\endgroup$ Commented Dec 8, 2022 at 0:58
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    $\begingroup$ There's a big difference between checking a lot of examples (leading to suspecting the statement might be true) vs. using a computer to check cases as part of a full proof. The MO post is a full proof of the statement you asked for, and thus it fully answers the question you've asked here. If you don't want to accept their answer, you should probably open a new question that links to the old one but adds "I want a simpler proof" or "I want a proof that can be fully checked by hand with no computer" or whatever your true condition is. $\endgroup$ Commented Dec 8, 2022 at 1:08

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