Question about linear transformations

ythamsten
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Homework Statement


Hey PF, I'm here again asking about linear transformations, ha ha.
Let C={(x,y) \in \mathbb R2 | x²+y²≤1} a circle of radius 1 and consider the linear transform
T:\mathbb R2→\mathbb R2
(x,y) \mapsto (\frac{5x+3y}{4},\frac{3x+5y}{4})​
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 \in \mathbb Z.

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:
\begin{pmatrix}<br /> T\end{pmatrix}^{ε}_{ε}\begin{pmatrix}<br /> x\\<br /> y<br /> \end{pmatrix}_{ε}<br /> =\begin{pmatrix}<br /> \frac{5}{4} &amp; \frac{3}{4}\\<br /> \frac{3}{4} &amp; \frac{5}{4}<br /> \end{pmatrix}\begin{pmatrix}<br /> x\\<br /> y<br /> \end{pmatrix}_ε
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:
\begin{pmatrix}<br /> T\end{pmatrix}^{ε}_{ε} = -\frac{1}{2}\begin{pmatrix}<br /> 1 &amp; 1\\<br /> 1 &amp; -1<br /> \end{pmatrix}\begin{pmatrix}<br /> 2 &amp; 0\\<br /> 0 &amp; \frac{1}{2}<br /> \end{pmatrix}\begin{pmatrix}<br /> -1 &amp; -1\\<br /> -1 &amp; 1<br /> \end{pmatrix}
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 \mathbb Z. Hope for some help, thanks in advance guys.
 
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I will respond, but I am not sure how much help I can offer. I have two questions, at least.

1. What does the image of a circle under a linear map look like?

2. We have spoken before about how the the determinant of the matrix associated with a linear map will tell you how area scales. So what is then the area of the transformed region after ##n## applications of the mapping?

After answering these questions, do you have any thoughts on it?
 
Well, I can see that after you apply the transformation on the circle, we expand vectors in the (1,1) by a factor of 2, and contract the ones in the (1,-1) direction by a factor of two either. Then, the image of the circle after applying T n times is limited by an ellipse centered at the origin, with semi-axis sized 2^n, in the (1,1) direction, and 1/(2^n), in the (1,-1) direction.
Since det(T)=1, applying the transformation doesn't actually change the value of the area.
My thoughts on it would be to count only the (c,c); c \in \mathbb Z, since I'm sure that the number of those kind of points is growing with each time I apply T in the region C. Is this a good way to go?
 
Right. It would seem to me as well that as ##n## grows larger the image looks more and more like the line ##y=x##.

Maybe you can use your diagonalization to determine for which ##n## we reach points like ##(1000,1000)##. You have ##T^n \mathbf{x} = P^{-1}D^nP \mathbf{x}##, right?

Do you have access to an answer to the question?
 
Hm, I'm going to try that indeed... I guess the right answer would be n≥11. I think I would be able to find somewhere an answer model, since it was a question in a Math Olympiad. If you are interested let me now.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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