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This is actually a follow up question. I have to apologise for the length of it. I didn't anticipate to be that long. I hope that it will be proved interesting nevertheless. I used the name "vector turn map" due to lack of a better alternative.

Let $f_A$ be a map defined as following. Choose any $2\times 2$ real matrix $A$. Then take a normalised vector, apply the map on the vector and renormalise it. So $f_A$ takes a normal vector and gives a normal vector, so it can be thought as a map $f_A:S^1\to S^1$.

I thought of that because of an error propagation problem I am working on. My idea was to see the asymptotic behaviour of a random error vector. This lead to the Jacobian matrix applied to the error vector multiple times, so I want to see how this vector turns. My problem was over the complex numbers so I was searching for any known results. However after playing a bit with that, I figured out that it won't help me but I thought it is an interesting puzzle. So I thought I should write what I found out and ask for more information.

To compute the map, we do the following. Take $\phi\in[0,1)$ to be the coordinate on $S^1$. Construct the vector $[\cos 2\pi \phi,\sin 2\pi \phi]$. Apply the map $A\cdot[\cos 2\pi \phi,\sin 2\pi \phi]$. And finally take the angle of the new vector. This can be written as $$f_A(\phi)=\frac{1}{2\pi}\mbox{arctan2}(A\cdot[\cos 2\pi \phi,\sin 2\pi \phi]).$$

The multiplication and division by $2\pi$ is to ensure that $\phi\in[0,1)$ or rather $\phi\in[-\frac{1}{2},\frac{1}{2})$ depending on the definition on arctan2, it is not important at all.

A few things can be told about $f_A$ right away. If $Av=\lambda v$ with $\lambda>0$ then $\phi_\lambda=\frac{1}{2\pi}\mbox{arctan2}(v)$ is a fixed point of $f_A$. Moreover $\phi_\lambda+\frac{1}{2}$ is also a fixed point since $-v$ is also an eigenvector. Also through some calculations if $\lambda_1>\lambda_2>0$ are eigenvalues of $A$ then $\phi_{\lambda_1}$ and $\phi_{\lambda_1}+\frac{1}{2}$ are stable points and $\phi_{\lambda_2}$, $\phi_{\lambda_2}+\frac{1}{2}$ are unstable.

This is easy to see by taking the diagonal matrix $D=((1,0),(0,2))$. Then $D\cdot[x,y]=[x,2y]$, so we see that $D\cdot[x,y]$ turns a bit towards the vertical axis if $y\ne0$.

If $\lambda_1<\lambda_2<0$, then we see that $f_A(\phi_{\lambda_i})=\phi_{\lambda_i}+\frac{1}{2}$. The situation is similar, but now instead of a fixed point, we have a 2-periodic orbit. The one that corresponds to the eigenvalue with the largest absolute value is stable and the other is unstable. If one eigenvalue is positive and the other is negative, then there is a fixed point and a 2-periodic orbit, with their stability determined by the absolute value of the eigenvalues.

If $A$ has only one eigenvector (it is similar to a Jordan block) with positive eigenvalue. Then there are only two fixed points that are stable from one direction and unstable from the other. If the eigenvalue is negative then instead of a fixed point we have a 2-periodic orbit with similar properties.

An very useful observation is that similar matrices correspond to "similar" maps. Similar in the sense that they have the same number of fixed points (or 2-periodic orbits) with the same stability. I haven't really proven that but with a little thought it becomes almost obvious. I think it can be proven by writing down a couple of commutative diagrams.

Now with that said, it is natural to define an other map $g_A$ by taking $\phi\in[0,\frac{1}{2})$. This can be done by using arctan instead of arctan2 (and tweaking the details). Then fixed points and 2-periodic orbits of $f_A$ are fixed points of $g_A$ with the same stability. So in a sense one could say that $g_A$ is a simplification of $f_A$. However this is true only in the $2\times2$ case.

The real difference is that $f_A:S^1\to S^1$ but $g_A:\mathbb{P}^1\to \mathbb{P}^1$ ($\mathbb{P}^1$ is the projective line which here is practiacally the same as the circle).

By the way the $2\times 2$ case is nice because one can draw pictures. By plugging in the equation in a program like mathematica or matlab, one can get the cobweb diagram and see the behaviour of the map.

The real fun begins when the same is considered in higher dimensions. So let $A$ be an $n\times n$ real matrix. Then by the same procedure one can define $f_A:S^{n-1}\to S^{n-1}$. Again eigenvectors give fixed points or 2-periodic orbits. Based on intuition I would expect that that the stability is determined by the eigenvalue, so the largest eigenvalues (in absolute value) will correspond to a stable point, the smallest to an unstable and the intermediate to some kind of saddles.

Now we also have $g_A:\mathbb{P}^n\to \mathbb{P}^n$, so now the difference is rather significant. A conjugacy between $f_A$ and $g_A$ is expected, as in the $2\times 2$ case, but I would expect some complications.

Finally it would be also interesting to check complex cases of the map for their behaviour. One interesting question for example is whether the $4\times 4$ real case actually contains the $2\times 2$ complex one.

Anyway, I would hope that this is something that is known and that people have worked on it. If not I would be interested to anything that anyone finds. I just put it here as a mathematical curiosity.

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I have posted an answer to your last question which should be relevant. This is indeed well-known as has been studied extensively. (It is a major subject within group theory and dynamical systems.) – Jim Belk May 11 '13 at 2:58

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