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