### Dynamic II

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?