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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 analytically?

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:

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

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

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

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

I tried asking on MSE: https://math.stackexchange.com/questions/4576505/proving-an-implication-of-two-dimensional-matrix

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    $\begingroup$ Have you tried using Lagrange multipliers? Might be possible to use them to prove the slightly stronger statement that under your circumstances $\det(AB+A+I) + \det(BA+B+I) \geq 0$. For your matrices, if I've done it correctly, the eigenvalues for $A$ are given by $\frac{x}{2} \pm \left(\frac{x^2}{4} - y\right)^{\frac{1}{2}},$ and a symmetric formula for $B$. $\endgroup$
    – JoshuaZ
    Nov 25, 2022 at 13:41
  • $\begingroup$ Can you please elaborate? I could not understand your comment. What is the stringer version? Lagrange multiplier using which function? $\endgroup$
    – BAYMAX
    Nov 26, 2022 at 1:06
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    $\begingroup$ Your constraint would be on the eigenvalues of $A^2B$ and $AB^2$. You would then be trying to find the minimum of the function given by $\det(AB+A+I) + \det(BA+B+I)$. If that function is at least 0, then one cannot have both terms negative. That approach might work. $\endgroup$
    – JoshuaZ
    Nov 26, 2022 at 1:11
  • $\begingroup$ This is a nice idea but I have not used Lagrange multipliers before. Would you like to frame it as an answer? $\endgroup$
    – BAYMAX
    Nov 26, 2022 at 3:18
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    $\begingroup$ Related: mathoverflow.net/questions/426170 $\endgroup$
    – Fred Hucht
    Dec 2, 2022 at 17:03

3 Answers 3

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Using Groebner bases and inequalities on $\mathbb{R}$, we prove that the OP's conjecture is true.

The real polynomial $x^2-sx+p$ admits roots of modulus $<1$ iff $p<1,p>-1,p>-s-1,p>s-1,s>-2$. -We use more inequalities than necessary to speed up the algorithm-.

Let $s_1=tr(AB^2),p_1=\det(AB^2),s_2=tr(A^2B),p_2=\det(A^2B)$ and

$d_1=\det(AB+A+I_2),d_2=\det(BA+B+I_2)$.

Here we show

$\textbf{Proposition 1.}$ The set of $\{(x,y,z;w)\}$ satisfying the following conditions is empty

$\{d_1<0,d_2<0,p_1<1,p_1>-1,p_1>-s_1-1,p_1>s_1-1,s_1>-2,p_2<1,p_2>-1,p_2>-s_2-1,p_2>s_2-1,s_2>-2\}$.

$\textbf{Proof.}$ To do that, we use the computer algebra software Raglib which is a patch in Maple.

with(Fgb); with(RAG);

PointsPerComponents([d1 < 0, d2 < 0, p1 > -1, p1 < 1, p1 > -s1-1, p1 > s1-1, p2 > -1, p2 < 1, p2 > -s2-1, p2 > s2-1, s1 > -2, s2 > -2], [x, y, z, w]);

The duration of the calculation is 5h40'. Below is a screenshot of the work:

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  • $\begingroup$ I did not downvote... Nice idea but i dont understand the output files of the code.. Is the proof already above Or there are more arguments to be put? $\endgroup$
    – BAYMAX
    Dec 2, 2022 at 5:47
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    $\begingroup$ @loupblanc Feel free to revert my edit if this isn't the form you want. Still, my impression is that the output is a bit more readable when formatted as code compared to the original version. $\endgroup$ Dec 2, 2022 at 8:47
  • $\begingroup$ @Martin Sleziak , thanks. $\endgroup$
    – loup blanc
    Dec 2, 2022 at 10:54
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    $\begingroup$ @Max Alekseyev , I consider real and imaginary roots. The set of considered $(s,p)$ is the interior of the triangle $A(-2,1),B(2,1)C(0,-1)$; $AC$ and $BC$ are tangent to the parabola $P$ of equation $s^2=4p$. Above $P$: the imaginary roots. Below $P$: the real roots. $\endgroup$
    – loup blanc
    Dec 2, 2022 at 14:52
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    $\begingroup$ Despitemy admiration for the solution, I support @BAYMAX' request for a proof which can be done by hand. $\endgroup$ Dec 20, 2022 at 21:45
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By considering cases of real/complex eigenvalues, the problem is reduced to a system of polynomial inequalities in each of the four cases.

For example, the case of real eigenvalues of $A^2B$ and non-real ones of $AB^2$ leads to the system: $$ \begin{cases} \det(AB+A+I)<0,\\ \det(BA+B+I)<0,\\ \mathrm{tr}(A^2B)^2 - 4\det(A^2B)\geq 0,\\ \det(A^2B) \pm \mathrm{tr}(A^2B) + 1 > 0,\\ \mathrm{tr}(AB^2)^2 - 4\det(AB^2) < 0, \\ \det(AB^2) < 1, \end{cases} $$ where each inequality is polynomial in $x,y,z,w$.

Essentially the question asks to show that neither of such four systems has a real solution.

There exist a few computational methods for solving systems of polynomial inequalities (eg. see this paper). I'm checking if those implemented in QEPCAD software can help here.

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  • $\begingroup$ Good idea!it might give us a hint of a mechanism behind the proof. But why we need real eigenvalues of A^2B and non real eigenvalues of AB^2 from the start? $\endgroup$
    – BAYMAX
    Dec 2, 2022 at 5:44
  • $\begingroup$ It is one of the four cases shown as an example. Another case is when both pairs of eigenvalues are real, etc. $\endgroup$ Dec 2, 2022 at 11:45
  • $\begingroup$ Right, can it be shown analytically [idea gained from the software or QEPCAD calculations] that the above system of inequalities does not have real solutions? then the proof would have been complete? $\endgroup$
    – BAYMAX
    Dec 2, 2022 at 12:17
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    $\begingroup$ I have even checked by considering a lot of examples computationally that the above conjecture is true for $n \times n$ matrices of the form in the question for $n \geq 2$ $\endgroup$
    – BAYMAX
    Dec 2, 2022 at 12:20
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    $\begingroup$ Oh ok, for $n>2$: Suppose $A,B \in M_{n}(\Bbb{R})$ such that $A = \left[C_{1}\middle|\frac{I}{0\dots0}\right], B= \left[C_{2}\middle|\frac{I}{0\dots0}\right]$ , where $A$ and $B$ have different first columns (represented as $C_{1}, C_{2}$). $\endgroup$
    – BAYMAX
    Dec 2, 2022 at 14:07
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Here is a slight generalization of the above Proposition 1 where the hypotheses relate to the matrices as a whole and no longer to the entries of the matrices.

$\textbf{Proposition 2.}$ Let $A,B\in M_2(\mathbb{R})$ be s.t. $A-B$ is singular, $\rho(A^2B)<1,\rho(AB^2)<1$.

Then the following $2$ inequalities cannot be simultaneously true $\det(AB+A+I)<0,\det(BA+B+I)<0$.

$\textbf{Proof.}$ If $\dim(\ker(A-B))>1$, then $A=B$ and the required result is easy to show.

Then $\dim(\ker(A-B))=1$ and there is $v\in\mathbb{R}^2\setminus\{0\}$ s.t. $Av=Bv$.

$\bullet$ If $v$ is not an eigenvector of $A$, then $\{u=Av,v\}$ is a basis of $\mathbb{R}^2$. In this basis, $A,B$ are in the form chosen by the OP and we conclude with the Proposition 1.

$\bullet$ If $v$ is an eigenvector of $A$, then $Av=Bv=a v$, where $a\in\mathbb{R}$. Then the spectra of $A,B$ are real.

Then we may assume that $A,B$ are in the form $A=\begin{pmatrix}a&b\\0&c\end{pmatrix},B=\begin{pmatrix}a&d\\0&f\end{pmatrix}$.

The inequalities relate only to the variables $a,c,f$. Using the above software, we obtain quickly the required result. $\square$

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