How do we show that $\text{Fix}(\beta)=\frac{1}{2}v+\mathbb{R}e_{\alpha}$?

Question

Let $a\in \mathbb{R}$, $v\in \mathbb{R}^2$ amd $\beta=\tau_v\circ\sigma_{\alpha}$.

I have shown that $\text{Kern}(s_a-I_2)=\mathbb{R}e_{\alpha}$ and $\text{Im}(s_a-I_2)=\mathbb{R}f_{\alpha}$, where $s_{\alpha}=\begin{pmatrix} \cos (\alpha )& \sin (\alpha ) \\ \sin (\alpha) & -\cos(\alpha)\end{pmatrix}$, $e_{\alpha}=\begin{pmatrix}\cos \left (\frac{\alpha}{2}\right ) \\ \sin \left (\frac{\alpha}{2}\right )\end{pmatrix}$ and $f_{\alpha}=\begin{pmatrix}-\sin \left (\frac{\alpha}{2}\right ) \\ \cos \left (\frac{\alpha}{2}\right )\end{pmatrix}$.

I want to show that $\text{Fix}(\beta)\neq \emptyset$ iff $v\in \mathbb{R}f_{\alpha}$.

In other words we have to show that $v$ is an element of $\text{Im}(s_a-I_2)$, right?

We have that \begin{equation*}\beta (p)=p \Rightarrow \left (\tau_v\circ \sigma_{\alpha}\right )(p)=p \Rightarrow \tau_v\left (\sigma_{\alpha}(p)\right )=p\Rightarrow \sigma_{\alpha}p+v=p\Rightarrow \sigma_{\alpha}p-p=-v\Rightarrow (\sigma_{\alpha}-I_2)p=-v\end{equation*}

That means that \begin{equation*}p\in \text{Fix}(\beta)\iff v\in \text{Bild}(s_a-u_2)\iff v\in \mathbb{R}f_{\alpha}\end{equation*} right?

Now I want to show that if $\text{Fix}(\beta)\neq \emptyset$ then $\text{Fix}(\beta)=\frac{1}{2}v+\mathbb{R}e_{\alpha}$ and $\beta=\tau_{\frac{1}{2}v}\circ\sigma_{\alpha}\circ\tau_{-\frac{1}{2}v}$.

Could you give me a hint for that?

Do we have to solve the system $(\sigma_{\alpha}-I_2)p=-v$ ?

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EDIT:

I have also shown that $$\tau_w\circ\beta(x)=\beta\circ\tau_w(x)\iff w \in \mathbb{R}e_{\alpha}$$ Let $p=\frac{1}{2}(v\cdot f_{\alpha})f_{\alpha}\in \mathbb{R}^2$, $\sigma=\tau_p\circ\sigma_{\alpha}\circ\tau_p^{-1}$ and $q=(v\cdot e_{\alpha})e_{\alpha}$.

Then how can we show that $\beta=\tau_q\circ\sigma=\sigma\circ\tau_q$ ?

Do we assume that this relatin is true and then we find something that is true, or how can we show that?

Answer 1

What you have done is correct. The following is a complete answer to the remaining problem.

Note that $e_\alpha,f_\alpha$ form an orthogonal basis of $\Bbb R^2$. Suppose $p=k_1e_\alpha+k_2f_\alpha$, then $(s_\alpha-I_2)p=k_2(s-I_2)f_\alpha=-v$. But $(s-I_2)f_\alpha=-2f_\alpha$ (verify!) so we get $v=k_2(2f_\alpha)$ or $p=k_1e_\alpha+v/2$.

Since $v\in\Bbb Rf_\alpha,v=kf_\alpha$ and $(s_\alpha-I_2)v=k(-2f_\alpha)=-2v$. Thus $\sigma_\alpha v=-v$.

Hence$$\begin{align*}\beta(p)&=\left(\sigma_\alpha p+\frac v2\right)+\frac v2\\&=\left(\sigma_\alpha p-\sigma_\alpha\left(\frac v2\right)\right)+\frac v2\\&=\sigma_\alpha(p-v/2)+v/2\\&=\tau_{v/2}\circ\sigma_\alpha\circ\tau_{\color{red}{-v/2}}\end{align*}$$