Week 8 Tutorial Solutions

  1. (12,12)(12,0)(0,12)(12,12)
  2. (13,13)(0,23)(23,23)(0,13)(13,13)
  3. By multiplying the matrix A with [xy] and then reducing the entries mod one, one gets T(x,y).
  4. The matrix A has determinant det(A)=1 so areas of polygons are not changed when mapped by A1. Thus R and A1(R) have the same area.
  5. Let D be the collection of all Cartesian rectangles in X together with . This is a π-system so it suffices to check that R and T1(R) have the same area. But R and A1(R) have the same area, and reducing modulo one does not cause any overlap, so T1(R) has the same area as R.
  6. Set μ=13(δ(1/2,1/2)+δ(1/2,0)+δ(0,1/2)) which is T invariant because T cannot change the cardinalty of {(12,12),(12,0),(0,12)}E and {(12,12),(12,0),(0,12)}T1(E) have the same cardinality for all E[0,1)2.
  7. The eigenvalues of A are the roots of its characteristic polynomial λ23λ+1. These are λ1=3+52λ2=352 with v1=[λ11]v2=[λ21] the respective eigenvalues.
  8. 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.
  9. We will be able to prove the existence of points with dense orbit after developing some more theory.