ATRACTOR

Take \(A=\begin{pmatrix}-1 & 2\\ 1 & -1 \end{pmatrix}\) and \(Q=\left\{ (x,y)\in\mathbb{R}^{2}:x,y>0\right\}\).

Diagonalizing the matrix \(A\), we get: \[\begin{array}{cll} A & = & \begin{pmatrix}-1 & 2\\ 1 & -1 \end{pmatrix}\\ & = & \begin{pmatrix}-\sqrt{2} & \sqrt{2}\\ 1 & 1 \end{pmatrix}\begin{pmatrix}-1-\sqrt{2} & 0\\ 0 & -1+\sqrt{2} \end{pmatrix}\begin{pmatrix}-\sqrt{2} & \sqrt{2}\\ 1 & 1 \end{pmatrix}^{-1} \end{array}\]

So, the successive powers of \(A\) are given by: \[\begin{array}{cll} A^{n} & = & \begin{pmatrix}-\sqrt{2} & \sqrt{2}\\ 1 & 1 \end{pmatrix}\begin{pmatrix}-1-\sqrt{2} & 0\\ 0 & -1+\sqrt{2} \end{pmatrix}^{n}\begin{pmatrix}-\sqrt{2} & \sqrt{2}\\ 1 & 1 \end{pmatrix}^{-1}\\ & = & \begin{pmatrix}-\sqrt{2} & \sqrt{2}\\ 1 & 1 \end{pmatrix}\begin{pmatrix}\left(-1-\sqrt{2}\right)^{n} & 0\\ 0 & \left(-1+\sqrt{2}\right)^{n} \end{pmatrix}\begin{pmatrix}-\frac{1}{2\sqrt{2}} & \frac{1}{2}\\ \frac{1}{2\sqrt{2}} & \frac{1}{2} \end{pmatrix}\\ & = & \begin{pmatrix}\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2} & \frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\sqrt{2}}\\ \frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}} & \frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2} \end{pmatrix} \end{array}\]

Therefore, we get: \[\begin{array}{cll} P_{n} & = & A^{n}P_{0}\\ & = & \begin{pmatrix}\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2} & \frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\sqrt{2}}\\ \frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}} & \frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2} \end{pmatrix}\begin{pmatrix}x_{0}\\ y_{0} \end{pmatrix}\\ & = & \begin{pmatrix}\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}x_{0}+\frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\sqrt{2}}y_{0}\\\frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}x_{0}+\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}y_{0} \end{pmatrix} \end{array}\]

If \(P_{n}\in\mathbb{Q}\), we have: \[\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}x_{0}+\frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\sqrt{2}}y_{0}>0\] \[\frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}x_{0}+\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}y_{0}>0\]

If \(n\) is even, then \[\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}=\frac{\left(1+\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}>0\] and \[\frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}=\frac{-\left(1+\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}<0,\] because \(1+\sqrt{2}>-1+\sqrt{2}>0\Rightarrow\left(1+\sqrt{2}\right)^{n}>\left(-1+\sqrt{2}\right)^{n}>0\). So, we get:
\[\begin{array}{cl} & \frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}x_{0}+\frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\sqrt{2}}y_{0}>0\\ \Leftrightarrow & x_{0}>\frac{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}\sqrt{2}y_{0}\end{array}\]
\[\begin{array}{cl} & \frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}x_{0}+\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}y_{0}>0\\ \Leftrightarrow & x_{0}<\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}}\sqrt{2}y_{0}\end{array}\]
\[\begin{array}{ccl} \frac{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}} & = & \frac{1-\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}{1+\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}\\ & = & \frac{1-\left(2\sqrt{2}-3\right)^{n}}{1+\left(2\sqrt{2}-3\right)^{n}}\\ & = & \frac{1-\left(3-2\sqrt{2}\right)^{n}}{1+\left(3-2\sqrt{2}\right)^{n}}\\ & = & \frac{2}{1+\left(3-2\sqrt{2}\right)^{n}}-1 \end{array}\]
\[\begin{array}{ccl} \frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}} & = & \frac{1+\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}{1-\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}\\ & = & \frac{1+\left(2\sqrt{2}-3\right)^{n}}{1-\left(2\sqrt{2}-3\right)^{n}}\\ & = & \frac{1+\left(3-2\sqrt{2}\right)^{n}}{1-\left(3-2\sqrt{2}\right)^{n}}\\ & = & \frac{2}{1-\left(3-2\sqrt{2}\right)^{n}}-1 \end{array}\]

So, we have: \[P_{n}\in\mathbb{Q}\Leftrightarrow\left(\frac{2}{1+\left(3-2\sqrt{2}\right)^{n}}-1\right)\sqrt{2}y_{0}<x_{0}<\left(\frac{2}{1-\left(3-2\sqrt{2}\right)^{n}}-1\right)\sqrt{2}y_{0}\]

If \(n\) is odd, then \[\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}=\frac{-\left(1+\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}<0\] and \[\frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}>0,\] because \(1+\sqrt{2}>-1+\sqrt{2}>0\Rightarrow\left(1+\sqrt{2}\right)^{n}> \left(-1+\sqrt{2}\right)^{n}>0\).

So, we have:
\[\begin{array}{cl} & \frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}x_{0}+\frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\sqrt{2}}y_{0}> 0\\ \Leftrightarrow & x_{0}<\frac{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}\sqrt{2}y_{0}\end{array}\]
\[\begin{array}{cl} & \frac{-\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2\sqrt{2}}x_{0}+\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{2}y_{0}>0\\ \Leftrightarrow & x_{0}>\frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}}\sqrt{2}y_{0}\end{array}\]
\[\begin{array}{ccl} \frac{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}} & = & \frac{1-\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}{1+\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}\\ & = & \frac{1-\left(2\sqrt{2}-3\right)^{n}}{1+\left(2\sqrt{2}-3\right)^{n}}\\ & = & \frac{1+\left(3-2\sqrt{2}\right)^{n}}{1-\left(3-2\sqrt{2}\right)^{n}}\\ & = & \frac{2}{1-\left(3-2\sqrt{2}\right)^{n}}-1 \end{array}\]
\[\begin{array}{ccl} \frac{\left(-1-\sqrt{2}\right)^{n}+\left(-1+\sqrt{2}\right)^{n}}{\left(-1-\sqrt{2}\right)^{n}-\left(-1+\sqrt{2}\right)^{n}} & = & \frac{1+\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}{1-\left(\frac{-1+\sqrt{2}}{-1-\sqrt{2}}\right)^{n}}\\ & = & \frac{1+\left(2\sqrt{2}-3\right)^{n}}{1-\left(2\sqrt{2}-3\right)^{n}}\\ & = & \frac{1-\left(3-2\sqrt{2}\right)^{n}}{1+\left(3-2\sqrt{2}\right)^{n}}\\ & = & \frac{2}{1+\left(3-2\sqrt{2}\right)^{n}}-1 \end{array}\]

Therefore, we get: \[P_{n}\in\mathbb{Q}\Leftrightarrow\left(\frac{2}{1+\left(3-2\sqrt{2}\right)^{n}}-1\right)\sqrt{2}y_{0}<x_{0}<\left(\frac{2}{1-\left(3-2\sqrt{2}\right)^{n}}-1\right)\sqrt{2}y_{0}\]

In both cases, we have: \(P_{n}\in\mathbb{Q}\) if and only if \(x_{0}\) lies between \(\left(\frac{2}{1+\left(3-2\sqrt{2}\right)^{n}}-1\right)\sqrt{2}y_{0}\) and \(\left(\frac{2}{1-\left(3-2\sqrt{2}\right)^{n}}-1\right)\sqrt{2}y_{0}\), com \(y_{0}>0\)

Note that, since \(0<3-2\sqrt{2}<1\), \(\left(\frac{2}{1+\left(3-2\sqrt{2}\right)^{n}}-1\right)_{n\in\mathbb{N}}\) is a strictly increasing sequence whose limit is \[\lim\left(\frac{2}{1+\left(3-2\sqrt{2}\right)^{n}}-1\right)=\frac{2}{1+\lim\left(3-2\sqrt{2}\right)^{n}}-1=\frac{2}{1+0}-1=1\]

Analogously, \(\left(\frac{2}{1-\left(3-2\sqrt{2}\right)^{n}}-1\right)_{n\in\mathbb{N}}\) is a strictly decreasing sequence which converges to \(1\) (indeed, it is the inverse of the previous sequence).

So, \(P_{n}\in\mathbb{Q}\) for all \(n\in\mathbb{N}\) if and only if \(x_{0}=\sqrt{2}y_{0}\) and \(y_{0}>0\). We can therefore conclude that the successive iterates of the initial point belong to the first quadrant only when the initial point belongs to the line \(x=\sqrt{2}y\) and to the first quadrant, that is, its coordinates correspond to the length of the diagonal and the side of a square.

Assume now that, instead of the matrix \(A\), we take its inverse. What happens with the successive images of a point in the plane by applying that matrix?

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