How to "rotate" a function? Or, how to write a function which has a known, rotational symmetry with respect to another function?

Question

EDIT 2: I've posted my "real" question here: https://mathematica.stackexchange.com/questions/115766/finding-closed-form-eigenvalues-of-a-particular-matrix

I have posed my question formally in LaTeX here:

Formalized problem statement.

But in words, here is my question. Suppose I have two real-valued functions in the real plane, $f$ and $g$. Moreover, I know that $g$ is given exactly by $f$, with its argument rotated by $2\pi/3$. In other words, given knowledge of $f$, we simply rotate the output by $2\pi/3$ to get $g$ everywhere.

How can I formalize this additional information? I used a 2D rotation matrix to find the corresponding argument of $\{x,y\}$ rotated by $2\pi/3$, but I still do not know how to write function $g$ in terms of $f$.

In my full problem, I have a system of nonlinear equations which involve both $f$ and $g$ and I wish to incorporate the rotational symmetry information to reduce the complexity and allow me to determine these functions.

EDIT: I began by asking a simpler question than I ultimately want to know, which is how to replace g by some function of f. Suppose I have a nonlinear system of equations which involve f and g, how can I eliminate g from these equations, given that I essentially "know" what g is due to it's symmetry with respect to f?

Answer 1

If you want the output rotated by a matrix $R$, then $g(\vec{x}) = f(R^{-1}\vec{x})$ where $R^{-1}$ is just the inverse rotation.

Answer 2

You wish to find

$$ g(x, y) = f(x, y)\ \text{rotated by} \ 2\pi/3 $$

To do this, let us assume for the moment that there is some matrix $R$ that rotates the axis by $2 \pi / 3$ for us. In that case, $g(x, y)$ will be given be $f(R^{-1} \cdot (x, y))$. This is because, the value at $g$ is $2\pi/3$ from the value of $f$. Hence, we need to rotate backwards by the same amount (which is why we take $R^{-1}$).

Let the rotation matrix be $R$. Then

$$ g(x, y) = f(R^{-1} \cdot(x, y)^T) $$ There is a $(x, y)^T$ since I'm treating $(x, y)$ as a column vector rather than a row vector.

It is well known that a rotation matrix which rotates a column vector by an angle $\theta$ is

$$ R = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{pmatrix} $$

In your case, substitute $\theta = 2\pi/3$ for the right answer.