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- Apr 14, 2013

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Let $1\leq n\in \mathbb{N}$ and for $x=\begin{pmatrix}x_1\\ x_2\\ \vdots \\ x_n\end{pmatrix}, \ x=\begin{pmatrix}x_1\\ x_2\\ \vdots \\ x_n\end{pmatrix}\in \mathbb{R}^n$ and let $x\cdot y=\sum_{i=1}^nx_iy_i$ the dot product of $x$ and $y$.

Let $S=\{v\in \mathbb{R}^n\mid v\cdot v=1\}$ and for $v\in S$ let $\sigma_v$ be a map defined by $\sigma_v:\mathbb{R}^n\rightarrow \mathbb{R}^n, \ x\mapsto x-2(x\cdot v)v$.

I have shown that it holds for $v\in S$ and $x,y\in \mathbb{R}^n$ that $\sigma_v(x)\cdot \sigma_v(y)=x\cdot y$.

Let $v\in S$. I have shown that $\sigma_v^2=\text{id}_{\mathbb{R}^n}$. To show that $\sigma_v$ is diagonalizable do we have to calculate the matrix of that transformation?

Let $n=4$ and $v,w\in S$. I want to show that there is $0\leq \alpha\leq 2\pi$ and an orthogonal basis $B$ of $\mathbb{R}^4$ such that the matrix $\sigma_v\circ\sigma_w$ as for the basis $B$ is of the form $$D=\begin{pmatrix}\cos \alpha & -\sin \alpha & 0 & 0 \\ \sin \alpha & \cos \alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}$$ Could you give me a hint for that? Do we have to find the matrix of $\sigma_v\circ\sigma_w$ and calculate the images of the elements of the basis $B$ and write then write the result as a linear combination of the elements of $B$ and the result should be the matrix $D$?