How can I prove this fact using Linear Algebra:
I know that in Linear Algebra a matrix $R$ is a rotation matrix $\iff R^T=R^{-1}$ and $\det R=1$
How to show that $R$ is obtained as a reflection of two reflection?
Any help
On
Look at the image below:
Let we want to reflect the green point on the right side of the red line $y=mx$ or $y=x\cdot\tan\theta_1$ to the other green point on the left side of the red line. Let the polar coordinates of the primary point be $(r,\theta)$; then within the reflection the angle of the points increases as much as $2(\theta_1-\theta)$ and the secondary angle of the reflected point would be $\theta+2(\theta_1-\theta)=2\theta_1-\theta$ therefore the point $(r,\theta)$ would be mapped to $(r,2\theta_1-\theta)$. If another reflection happens along $y=x\cdot \tan\theta_2$ the final position of the point becomes $$(r,2\theta_2-(2\theta_1-\theta))=(r,2\theta_2-2\theta_1+\theta)$$which means that two reflections in a row lead to a rotation as much as $2\theta_2-2\theta_1$
If we have two lines $y=kx$ and $y=lx$ (lines through the origin) then reflections across those two are respectively represented with matrix
$$R_1=\begin{bmatrix} {2-k^2\over k^2+2} & {4k\over k^2+2} \\ {2k\over k^2+2}& {k^2-2\over k^2+2} \\ \end{bmatrix}$$ and
$$R_2=\begin{bmatrix} {2-l^2\over l^2+2} & {4l\over l^2+2} \\ {2l\over l^2+2}& {l^2-2\over l^2+2} \\ \end{bmatrix}$$
where $k=\tan \alpha$ and $l=\tan \beta$
While rotation matrix around origin for angle $\varphi$ is $$R=\begin{bmatrix} \cos \varphi & -\sin \varphi \\ \sin \varphi & \cos \varphi \\ \end{bmatrix}$$
So you have to calculate the $${2-k^2\over k^2+2} {2-l^2\over l^2+2}+ {4k\over k^2+2} {2l\over l^2+2}$$ and the result should be $\cos 2(\beta -\alpha)$