\[
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\newcommand{\R}{\mathbb{R}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\d}{\mathsf{d}}
\newcommand{\g}{>}
\newcommand{\l}{<}
\newcommand{\intd}{\,\mathsf{d}}
\newcommand{\Re}{\mathsf{Re}}
\newcommand{\area}{\mathop{\mathsf{Area}}}
\newcommand{\met}{\mathop{\mathsf{d}}}
\newcommand{\emptyset}{\varnothing}
\DeclareMathOperator{\borel}{\mathsf{Bor}}
\]
Week 8 Tutorial Solutions
- $(\tfrac{1}{2},\tfrac{1}{2}) \mapsto (\tfrac{1}{2},0) \mapsto (0,\tfrac{1}{2}) \mapsto (\tfrac{1}{2},\tfrac{1}{2})$
- $(\tfrac{1}{3},\tfrac{1}{3}) \mapsto (0,\tfrac{2}{3}) \mapsto (\tfrac{2}{3},\tfrac{2}{3}) \mapsto (0,\tfrac{1}{3}) \mapsto (\tfrac{1}{3},\tfrac{1}{3})$
- By multiplying the matrix $A$ with $[\begin{smallmatrix} x \\ y \end{smallmatrix}]$ and then reducing the entries mod one, one gets $T(x,y)$.
- The matrix $A$ has determinant $\mathsf{det}(A) = 1$ so areas of polygons are not changed when mapped by $A^{-1}$. Thus $R$ and $A^{-1}(R)$ have the same area.
- Let $\mathcal{D}$ be the collection of all Cartesian rectangles in $X$ together with $\emptyset$. This is a π-system so it suffices to check that $R$ and $T^{-1}(R)$ have the same area. But $R$ and $A^{-1}(R)$ have the same area, and reducing modulo one does not cause any overlap, so $T^{-1}(R)$ has the same area as $R$.
- Set
\[
\mu = \dfrac{1}{3} \left( \delta_{(1/2,1/2)} + \delta_{(1/2,0)} + \delta_{(0,1/2)} \right)
\]
which is $T$ invariant because $T$ cannot change the cardinalty of
\[
\{ (\tfrac{1}{2},\tfrac{1}{2}), (\tfrac{1}{2},0), (0,\tfrac{1}{2}) \} \cap E
\]
and
\[
\{ (\tfrac{1}{2},\tfrac{1}{2}), (\tfrac{1}{2},0), (0,\tfrac{1}{2}) \} \cap T^{-1}(E)
\]
have the same cardinality for all $E \subset [0,1)^2$.
- The eigenvalues of $A$ are the roots of its characteristic polynomial $\lambda^2 - 3\lambda + 1$. These are
\[
\lambda_1 = \dfrac{3 + \sqrt{5}}{2}
\qquad
\lambda_2 = \dfrac{3 - \sqrt{5}}{2}
\]
with
\[
v_1 = \begin{bmatrix} \lambda_1 \\ 1 \end{bmatrix}
\qquad
v_2 = \begin{bmatrix} \lambda_2 \\ 1 \end{bmatrix}
\]
the respective eigenvalues.
- As each eigenspace wraps around the unit square, its consecutive intersections with the bottom of the square correspond to the orbit of an irration rotation. As such orbits are dense in the unit interval, and as the slope does not change, we get that the eigenspace is dense in $[0,1)^2$.
- We will be able to prove the existence of points with dense orbit after developing some more theory.