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I have a rotation of the form:

$$(z(s),w(s))=B(s)(u(s),v(s))$$

where $z(s),w(s),u(s),v(s)$ are in $\mathbb{R}$ and $s$ is a complex number and $B(s)$ is a $2\times 2$ matrix defined by $$ B(s) = \begin{bmatrix} \cos(\theta(s)) & -\sin(\theta(s)) \\ \sin(\theta(s)) & \cos(\theta(s)) \end{bmatrix} $$

The fixed point of the rotation must satisfies $(I_2-B(s))(u(s),v(s))=0$ where $I_2$ is the $2\times 2$ unit matrix. The determinant of the matrix $(I_2-B(s))$ is $-2(\cos\theta(s)-1)$ and it is not zero if $\theta (s)\ne 0\,\pmod{2\pi}$. This means that $s$ is a solution of $(u(s),v(s))=0$.

My question is what happen if $\theta(s)=0\,\pmod{2\pi}$? The reason is that this is the trivial rotation corresponding to the identity matrix, in which no rotation takes place. Does there exist zeros of $(u(s),v(s))=0$ in this case or not?

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    I feel like you have an explicit example in mind. Is that so? This would make things clearer.2012-12-22
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    Yes, we need only to know $B(s)$. Take $B(s)$ by lines as follow: line 1: $cosθ(s)$, $-sinθ(s)$, line 2: $sinθ(s)$, $cosθ(s)$. See the edited question.2012-12-22
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    Is the edit I did okay? I think this is better.2012-12-22
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    Yes, thank you very much.2012-12-22

3 Answers 3

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It seems to me that you're just confused by the terms that you introduced, so I'm going to lay out clearly what you have done.

Let $B_\theta$ be the matrix of anti clockwise rotation of $\theta$ about the origin. The matrix form of $B_\theta$ is $$ B_\theta = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\\ \end{bmatrix} $$

You want the fixed points of $B_\theta$, i.e. $B_\theta (u, v) = (u,v)$, which is equivalent to $(I_2 - B_\theta) (u,v) = 0$, where $I_2$ is the 2 by 2 Identity matrix. You further show that $\det(I_2 - B_\theta) = -2(\cos \theta-1) $

Now, from Linear Algebra, we know that this equation has the trivial solution $(0,0)$ if the determinant is non-zero. This corresponds to the case where $\cos \theta \neq 1$, or $\theta \neq 0 \pmod{2 \pi}$. This agrees with our understand of rotations - The only fixed point of $B_\theta$ is $(0,0)$ itself.

The 'interesting' case will be when $\det(I_2 - B_\theta) = 0$, and we can then ask what the null space of this $(I-B)$ matrix is. As mentioned, this happens when $\theta \equiv 0 \pmod{2 \pi}$. Furthermore, the null space will be the entire space, since we can actually show that $I_2 - B_{2 k \pi} = 0$. Hence, in this case, every single point will be a fixed point of $B_{2k\pi}$.

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    @ Calvin Lin : Thank you very much for your answer. But $(u,v)$ depend on the variable $s$. When $θ(s)≠0 (mod2π)$ you get $(u(s),v(s))=0$ and I know that $(u,v)=0$ have infinitely many roots with respect to $s$. So what happen to the unicity of the menstioned fixed point (0,0).2012-12-30
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    So, it sounds like what you're talking about, is that you want to look at the sets $F_\theta$, which are the fixed points of the rotation by $\theta$, and you're asking why does the fixed points jump from 1, to the entire plane, at $\theta \equiv 0 \mod{2 \pi}$, and why this jump is so abrupt / discontinuous. Is that so?2012-12-30
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    @ Calvin Lin : Yes, that is the case.2012-12-30
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    There is no reason why the set of fixed points has to have that behavior of 'continuity'. Just because it is currently a fixed point, doesn't mean that it will still be a fixed point (or that there will be one nearby), when the transformation is perturbed slightly. For example, consider translations of the form (a, b). There are no fixed points, except when (a,b)=(0,0), and in this case, the entire plane is a fixed point. A similar scenario to what you described, is to look at the family of graphs $y= Tx$. There only fixed point is $(0,0)$, except when $T=0$.2012-12-30
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    Also, if this is what you're interested in, then your question needs to be rephrased to reflect this. Currently, it talks more about how to find fixed points, then why the jump is so abrupt.2012-12-30
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If $\theta(s) \equiv 0 \pmod{2 \pi}$, you have $$ B(s) = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} $$ which is just the identity matrix. You know how to solve such a system of equations. The only point that will be mapped to $(0,0)$ by $B(s)$ is the point $(0,0)$.

Hope that helps,

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    But for the fixed point we have $0.(u(s),v(s))=0$.2012-12-22
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    @user53124 : I don't see your point? Is there a problem? If I read everything right, you asked to solve $(z,w) = B(u,v)$ (where everything is a function of $s$, but to be honest that doesn't seem to add a lot to the question) and that is what I believe I did.2012-12-22
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    I am asking about the existence of fixed points of the rotation when $θ(s)=0 (mod2π)$ and hence: Is there exists zeros of $(u(s),v(s))=0$ in this case or not2012-12-22
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For each $s$ the map $$B(s):\quad {\mathbb R}^2\to{\mathbb R}^2,\qquad \left[\matrix{u\cr v\cr}\right]\mapsto \left[\matrix{z\cr w\cr}\right]:=\bigl[B(s)\bigr]\left[\matrix{u\cr v\cr}\right]$$ is a counterclockwise rotation of the plane by the angle $\theta(s)$. From elementary geometry we know: If $\theta(s)\ne 0$ modulo $2\pi$ then the map $B(s)$ has a unique fixed point in ${\mathbb R}^2$, namely the origin $(0,0)$. If, on the other hand, $\theta(s)=0$ modulo $2\pi$ then $B(s)$ is the identity map of ${\mathbb R}^2$, and any point $(u,v)\in{\mathbb R}^2$ is a fixed point of $B(s)$.

In your setup the point ${\bf x}=(u,v)$ is not an independent variable, but together with $\theta(s)$ depends on the (complex) variable $s$. I understand your question as follows: "For which values of the variable $s$ is the point ${\bf x}(s)=\bigl(u(s),v(s)\bigr)$ a fixed point of the map $B(s)\>$?" On account of our preliminary considerations we now can answer this question as follows:

The point ${\bf x}(s)$ is a fixed point of $B(s)$ iff at least one of the following two conditions is fulfilled: $${\bf x}(s)=(0,0)\>,\qquad \theta(s)=0\ {\rm mod}\ 2\pi\ .$$

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    : The first case of the last equivalence corresponding to the case: $\theta(s)\ne 0$ modulo $2\pi$. No.2012-12-29