Give a Skew-symmetric matrix such that $Qx=\alpha e_1$

Question

Let $S$ is a Skew-symmetric matrix, then $Q=(I+S)(I-S)^{-1}$ is an orthogonal matrix. If $x\in\mathbb{R}^2$, please give a Skew-symmetric matrix such that $Qx=\alpha e_1$, where $\alpha$ is a constant, $e_1 = (1,0)^{\mathsf{T}}$. After the following attempts, I did not get the desired result $$ Q^{\mathsf{T}}Qx=\alpha Q^{\mathsf{T}}e_1 \\ \Rightarrow x=\alpha Q^{\mathsf{T}}e_1 \\ \Rightarrow x=\alpha \left( I+S \right) ^{-1}\left( I-S \right) e_1 \\ \Rightarrow \left( I+S \right) x=\alpha \left( I-S \right) e_1 \\ \Rightarrow x+Sx=\alpha e_1-\alpha Se_1 \\ \Rightarrow Sx+\alpha Se_1=\alpha e_1-x \\ \Rightarrow S\left( x+\alpha e_1 \right) =\alpha e_1-x \\ $$ Then, I can't find a Skew-symmetric $S$ matrix that satisfies this.

Answer 1

If $S=\left[\begin{matrix} 0 & a \\ -a & 0 \\\end{matrix}\right]$ then we can show that $Q=\left[\begin{matrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\\end{matrix}\right]$ where $\cos\theta=\frac{1-a^2}{1+a^2}$ and $\sin\theta=\frac{-2a}{1+a^2}.$

Let the given $x$ be $x=\alpha e^{i\gamma}$ then since $Q$ is the rotation matrix which rotates $\theta$ angle counter-clockwise, $Qx=\alpha e^{i(\gamma+\theta)}$ and if this is real then $\theta=-\gamma$.

Then we have the equations $\cos\gamma=\frac{1-a^2}{1+a^2}$ and $\sin\gamma=\frac{2a}{1+a^2}.$ Hence, $a=\tan\frac{\gamma}{2}$ and $S=\left[\begin{matrix} 0 & \tan\frac{\gamma}{2} \\ -\tan\frac{\gamma}{2} & 0 \\\end{matrix}\right]$. We can check that $$\left[\begin{matrix} 0 & \tan\frac{\gamma}{2} \\ -\tan\frac{\gamma}{2} & 0 \\\end{matrix}\right]\left[\begin{matrix} \cos\gamma +1 \\ \sin\gamma\\ \end{matrix}\right]=\left[\begin{matrix} 1-\cos\gamma \\ -\sin\gamma\\ \end{matrix}\right]$$ so that the equation $S(x+\alpha e_1)=\alpha e_1 -x$ is satisfied.