ATRACTOR

In the case of the square, note that, in terms of matrices, the sequence of points is given by: \[P_{n}=AP_{n-1},\;\forall n\in\mathbb{N},\] where \[P_{n}=\begin{pmatrix}x_{n}\\ y_{n} \end{pmatrix}\] and \[A=\begin{pmatrix}-1 & 2\\ 1 & -1 \end{pmatrix},\] that is, \[P_{n}=A^{n}P_{0},\;\forall n\in\mathbb{N}\]

The matrix \(A\) has eigenvalues \(-1-\sqrt{2}\) and \(-1+\sqrt{2}\), associated to the eigenspaces spanned by the vectors \((-\sqrt{2},1)\) and \((\sqrt{2},1)\), respectively. Since these are one-dimensional subspaces, they are represented geometrically by lines going through the origin, whose equations are \(x=-\sqrt{2}y\) and \(x=\sqrt{2}y\). When the initial point is different from the origin and lies on one of these lines, it corresponds to an eigenvector of the matrix and, therefore, all its iterates are also eigenvectors, so they all lie on the same line. However, what happens to nonzero points belonging to the line \(x=\sqrt{2}y\) is very different from what happens to nonzero points belonging to the line \(x=-\sqrt{2}y\). Indeed, since the eigenvalue associated to the line \(x=\sqrt{2}y\) is, in absolute value, smaller than \(1\), the successive iterates will get closer and closer to the origin, without ever reaching it. On the other hand, since the eigenvalue associated to the line \(x=-\sqrt{2}y\) is, in absolute value, larger than \(1\), the successive iterates will get farther and farther away from the origin. If the initial point belongs to neither of the two lines, then its iterates will also not belong to those lines, approaching the line \(x=-\sqrt{2}y\) and moving away from the line \(x=\sqrt{2}y\). Check that this is so in the app below, where the line \(x=-\sqrt{2}y\) is shown in blue and the line \(x=\sqrt{2}y\) is shown in yellow. You can also check that something similar happens in the case of the regular hexagon and the pentagon, with only the lines corresponding to eigenspaces being different.

Instructions:
Click on the red dot to change the initial point; you can also choose if you want to see the lines joining the iterates shown in the graphic, highlight the \(n\)-th iterate for \(n\) between \(0\) and \(5\), restrict the initial points to the lines corresponding to eigenspaces or change the regular polygon on whose geometric proof of incommensurability this dynamic process is based.

Bearing this in mind, can we conclude again that the diagonal and the side of the square are incommensurable?

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