Correct solution for this tensor - language-agnostic

Question

I'm implementing the system in this paper and I've come a little unstuck correctly implementing the radial tensor field.
All tensors in this system are of the form given on page 3, section 4
R [ cos(2t), sin(2t); sin(2t), -cos(2t) ]
The radial tensor field is defined as:
R [ yy - xx, -2xy; -2xy, -(yy-xx) ]
In my system I'm only storing R and Theta, since I can calculate the tensor based off just that information. This means I need to calculate R and Theta for the radial tensor. Unfortunately, my attempts at this have failed. Although it looks correct, my solution fails in the top left and bottom right quadrants.
Addendum: Following on from discussion in the comments about the image of the system not working, I'll put some hard numbers here too.
The entire tensor field is 800x480, the center point is at { 400, 240 }, and we're using the standard graphics coordinate system with a negative y axis (ie. origin in the top left).
At { 400, 240 }, the tensor is R = 0, T = 0
At { 200, 120 }, the tensor is R = 2.95936E+9, T = 2.111216
At { 600, 120 }, the tensor is R = 2.95936E+9, T = 1.03037679
I can easily sample any more points which you think may help.
The code I'm using to calculate values is:
float x = i - center.X;
float xSqr = x * x;
float y = j - center.Y;
float ySqr = y * y;
float r = (float)Math.Pow(xSqr + ySqr, 2);
float theta = (float)Math.Atan2((-2 * x * y), (ySqr - xSqr)) / 2;
if (theta < 0)
theta += MathHelper.Pi;

Answer 1

Evidently you are comparing formulas (1) and (2) of the paper. Note the scalar multiple l = || (u_x,u_y) || in formula (1), and identify that with R early in the section. This factor is implicit in formula (2), so to make them match we have to factor R out.
Formula (2) works with an offset from the "center" (x0,y0) of the radial map:
x = xp - x0
y = yp - y0
to form the given 2x2 matrix:
y^2 - x^2 -2xy
-2xy -(y^2 - x^2)
We need to factor out a scalar R from this matrix to get a traceless orthogonal 2x2 matrix as in formula (1):
cos(2t) sin(2t)
sin(2t) -cos(2t)
Since cos^2(2t) + sin^2(2t) = 1 the factor R can be identified as:
R = (y^2 - x^2)^2 + (-2xy)^2 = (x^2 + y^2)^2
leaving a traceless orthogonal 2x2 matrix:
C S
S -C
from which the angle 'tan(2t) = S/C` can be extracted by an inverse trig function.
Well, almost. As belisarius warns, we need to check that angle t is in the correct quadrant. The authors of the paper write at the beginning of Sec. 4 that their "t" (which refers to the tensor) depends on R >= 0 and theta (your t) lying in [0,2pi) according to the formula R [ cos(2t), sin(2t); sin(2t) -cos(2t) ].
Since sine and cosine have period 2pi, t (theta) is only uniquely determined up to an interval of length pi. I suspect the authors meant to write either that 2t lies in [0,2pi) or more simply that t lies in [0,pi). belisarius suggestion to use "the atan2 equivalent" will avoid any division by zero. We may (if the function returns a negative value) need to add pi so that t >= 0. This amounts to adding 2pi to 2t, so it doesn't affect the signs of the entries in the traceless orthogonal matrix (since 'R >= 0` the pattern of signs should agree in formulas (1) and (2) ).