- #1
ythamsten
- 16
- 0
Homework Statement
Hey PF, I'm here again asking about linear transformations, ha ha.
Let C={(x,y) [itex]\in[/itex] [itex]\mathbb R[/itex]2 | x²+y²≤1} a circle of radius 1 and consider the linear transform
T:[itex]\mathbb R[/itex]2→[itex]\mathbb R[/itex]2
(x,y) [itex]\mapsto[/itex] ([itex]\frac{5x+3y}{4}[/itex],[itex]\frac{3x+5y}{4}[/itex])
Find all values of a natural n for which Tn(C), the image of C after n applications of T, contains at least 2013 points (a,b) with coordinates a, b [itex]\in[/itex] [itex]\mathbb Z[/itex].(x,y) [itex]\mapsto[/itex] ([itex]\frac{5x+3y}{4}[/itex],[itex]\frac{3x+5y}{4}[/itex])
Homework Equations
N/A
The Attempt at a Solution
At first I've fixed for both input and output basis for the map ε = {e1,e2} (i.e. the canonic basis) writing the linear transformation in a matrix form:
[tex]\begin{pmatrix}
T\end{pmatrix}^{ε}_{ε}\begin{pmatrix}
x\\
y
\end{pmatrix}_{ε}
=\begin{pmatrix}
\frac{5}{4} & \frac{3}{4}\\
\frac{3}{4} & \frac{5}{4}
\end{pmatrix}\begin{pmatrix}
x\\
y
\end{pmatrix}_ε[/tex]
Then, knowing I'm going to be applying the linear transformation n times, I thought would be a wise choice to diagonalize it, taking advantage of it's nice form to do so:
[tex]\begin{pmatrix}
T\end{pmatrix}^{ε}_{ε} = -\frac{1}{2}\begin{pmatrix}
1 & 1\\
1 & -1
\end{pmatrix}\begin{pmatrix}
2 & 0\\
0 & \frac{1}{2}
\end{pmatrix}\begin{pmatrix}
-1 & -1\\
-1 & 1
\end{pmatrix}[/tex]
But right now, I'm having a little bit of trouble to figure out how to count the number of points on the maps that has both coordinates belonging to [itex]\mathbb Z[/itex]. Hope for some help, thanks in advance guys.