1. Show that the matrix $A$ of a reflection with respect to a line $L$ through the origin has the form $$A=\begin{bmatrix}\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{bmatrix}$$ for some $\theta \in [0,2\pi)$

This I can prove by seeing what happens with $e_1$ and $e_2$. I have trouble with the logic proving the next one:

  1. Show that every matrix with this form is a reflection by the line spanned by the vector $(\cos(\theta/2),\sin(\theta/2))^T$

What should I do? All I know about reflections is that they are given by $2P_{L}(X)-X$ where $P$ is the projection. This is a very basic course so we don't know about orthogonal matrices.

  • 1
    $\begingroup$ Hint: the line of reflection is unchanged by the reflection. $\endgroup$ – amd Oct 3 '17 at 21:09

Since$$A.\begin{bmatrix}\cos\left(\frac\theta2\right)\\\sin\left(\frac\theta2\right)\end{bmatrix}=\begin{bmatrix}\cos\left(\frac\theta2\right)\\\sin\left(\frac\theta2\right)\end{bmatrix}\text{ and }A.\begin{bmatrix}-\sin\left(\frac\theta2\right)\\\cos\left(\frac\theta2\right)\end{bmatrix}=-\begin{bmatrix}-\sin\left(\frac\theta2\right)\\\cos\left(\frac\theta2\right)\end{bmatrix},$$then $A$ maps each point of the line $\mathbb{R}\left[\begin{smallmatrix}\cos\left(\frac\theta2\right)\\\sin\left(\frac\theta2\right)\end{smallmatrix}\right]$ into itself and each point of the line $\mathbb{R}\left[\begin{smallmatrix}-\sin\left(\frac\theta2\right)\\\cos\left(\frac\theta2\right)\end{smallmatrix}\right]$ into its symmetric. Since the lines are orthogonal, $A$ is the reflection on the first of these lines.

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