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Let $A\in\mathbb{R}^{2\times 2}$ and assume that $|{\rm tr}A|<4$. Prove or disprove that there exist $a_1,a_2,a_3,a_4\in\mathbb{C}$ such that $|a_1|<1$, $|a_2|<1$, $|a_3|<1$, $|a_4|<1$, $p(x)\triangleq\prod_{i=1}^{4} (x-a_i)$ is a polynomial with real coefficients, and

\begin{equation} b_1I_2+b_2A+b_3A^2+A^3=0, \end{equation}

where for all $i\in\{1,2,3\}$, \begin{gather} b_i\triangleq(-1)^i\mspace{-40mu}\sum_{j_1,\ldots,j_{4-i}\in\{1,\ldots,4\}}\mspace{-40mu}a_{j_1}\times \cdots\times a_{j_{4-i}}, \end{gather} that is,

\begin{align} b_1&\triangleq-a_1a_2a_3-a_1a_2a_4-a_2a_3a_4-a_1a_3a_4,\\ b_2&\triangleq a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4,\\ b_3&\triangleq -a_1-a_2-a_3-a_4. \end{align}

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    $\begingroup$ It seems as though the $b_i$ coefficients have been defined so that $$ p(x) = x^4 + b_1 x + b_2x^2 + b_3 x + a_1a_2a_3a_4 $$ $\endgroup$ – Ben Grossmann May 29 at 1:39
  • $\begingroup$ That's right. Thanks $\endgroup$ – CyberX May 29 at 1:41
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    $\begingroup$ It would be helpful if you could share some of your thoughts on the problem: Where did you encounter the problem; do we have a reason to suspect that this problem has a nice solution? What have you tried? How far did you get before you got stuck? Are there any results/techniques that you suspect would be applicable here? $\endgroup$ – Ben Grossmann May 29 at 1:54
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    $\begingroup$ Some of my initial thoughts on the problem: assume such coefficients exist. Let $p$ denote the polynomial $p(x) = (x-a_1) \cdots (x-a_4)$. The given equation tells us that for the polynomial $q(x) = \frac{p(x) - p(0)}{x}$, we have $q(A) = 0$. That tells us that any eigenvalue $\lambda$ of $A$ must satisfy $q(\lambda) = 0$, which means that they must satisfy $p(\lambda) = p(0)$. The trace, of course, is the sum of the eigenvalues of $A$, so the statement is equivalent to saying that the solutions to $p(x) = p(0)$ must have magnitude less than $2$, given $|a_i| < 1$ for all $i$. $\endgroup$ – Ben Grossmann May 29 at 1:57
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    $\begingroup$ If $A$ has a repeated eigenvalue, we may assume that $A$ is not diagonalisable. So, we can always assume that the characteristic polynomial $x^2-tx+d$ of $A$ is the minimal polynomial of $A$. Now you are essentially seeking two real numbers $c$ and $r$ such that all zeros of $p(x)=x(x^2-tx+d)(x-c)+r$ lie inside the open unit disc, given that $|t|<4$. $\endgroup$ – user1551 May 29 at 13:35

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