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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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    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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    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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    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