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Short context on how the inequality problem I have was derived :

After solving an exercise regarding dynamical systems and a discrete time linear system, I yielded the conclusion that the eigenvalues of the matrix

$$A=\begin{bmatrix} 0 & 1 \\ a & b\end{bmatrix}$$

should lie inside the unit circle, which in short means :

$$\det(A-λI) =0 \Rightarrow\dots \Rightarrow λ^2-bλ - a=0$$

Now, we have that $D=b^2 + 4a$. If $D>0$ then we have two non-equal real eigenvalues, which are given by :

$$λ_{1,2} = \frac{b \pm \sqrt{b^2 + 4a}}{2}$$

As mentioned, we need them to lie within the unit circle, which means :

$$|λ_{1,2}|<1\Rightarrow \bigg|\frac{b \pm \sqrt{b^2 + 4a}}{2} \bigg| <1 \Rightarrow |b \pm \sqrt{b^2 + 4a}| < 2$$

Joining these two inequalities with the initial one to have the existence of two non-equal real solutions, we derive the system, which I found to be stuck while solving :

Question part :

$$\begin{cases}|b + \sqrt{b^2 + 4a}| < 2\\|b - \sqrt{b^2 + 4a}| < 2 \\ b^2 + 4a \quad \quad \quad \space > 0\end{cases}$$

How would one proceed with solving this system of inequalities ? Seems there would be many cases due to the absolute values. I looked it up quickly on Wolfram Alpha and the domain of solutions is a weird-curved triangle.

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    $\begingroup$ if $$b^2+4a<0$$ then we get no Solutions, the radicand is negative, or is this a typo? $\endgroup$ Dec 18 '17 at 19:40
  • $\begingroup$ @Dr.SonnhardGraubner Typo ! Thanks for pointing it out. $\endgroup$
    – Rebellos
    Dec 18 '17 at 19:47
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The system of inequations is invariant under $\,b \mapsto -b\,$, so it can be assumed WLOG that $\,b \ge 0\,$ (then at the end the solution has to be "symmetrized" back $b \mapsto \pm b$ of course).

For $\,b \ge 0\,$ the second inequality is redundant because $|b - \sqrt{b^2 + 4a}| \le |b + \sqrt{b^2 + 4a}|$ so the system reduces to:

$$ \begin{cases} \begin{align} b + \sqrt{b^2 + 4a} &\lt 2 \\ b^2 + 4a &\gt 0 \end{align} \end{cases} $$

The first inequality requires $b \lt 2\,$, so $0 \le b \lt 2\,$, then after rearranging and squaring:

$$ b^2 + 4a \lt (2-b)^2 = b^2-4b+4 \quad\iff\quad 4a < 4 - 4b \quad\iff\quad b < 1-a $$

Combining with the second inequality, and depending on the sign of $a$:

  • if $a \gt 0$ then the second inequality is automatically satisfied, so the solution set is given by $b \lt 1 - a$ with $0 \le b \lt 2$, equivalent to $\;\boxed{0 \lt a \lt 1, \; 0 \le b \lt 1-a}\,$

  • if $a \le 0$ then the second inequality gives $-a \lt b^2 / 4 \iff b \gt 2 \sqrt{-a}\,$, and the condition $0 \le b \lt 2$ requires $2 \sqrt{-a} \lt 2 \iff -1 \lt a \le 0\,$, so in the end the solution set is $\;\boxed{-1 \lt a \le 0, \;2 \sqrt{-a} \lt b \lt 1-a }\,$

To recover the negative values of $b$, too, just replace $b \mapsto |b|$ in the boxed solutions above.

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    $\begingroup$ Couldn't have asked for a better explanation and answer, amazing presentation right here ! Appreciate it a lot ! $\endgroup$
    – Rebellos
    Dec 19 '17 at 0:17

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