Apply Cayley transformation on vector x If I have $Q = (I + S)(I - S)^{-1}$ ($Q$ is the Cayley transformation of skew-symmetric matrix $S$) then how do I construct a rank-2 $S$ such that $Qx$ has all zeros except the first component?
 A: Let your $Q$ be $n\times n$ and let $\{e_1,\ldots,e_n\}$ be the standard basis of $\mathbb R^n$. Let also $K=\pmatrix{0&-1\\ 1&0}$.
If the desired construction is possible, then $S=U\left((bK)\oplus0\right)U^T$ for some real number $b\ne0$ and some orthogonal matrix $U$. Therefore,
$$
Q = U\pmatrix{(I_2+bK)(I_2-bK)^{-1}\\ &I_{n-2}}U^T.\tag{1}
$$
Let $R$ be the Cayley transform of $bK$, i.e. $R=(I_2+bK)(I_2-bK)^{-1}$. By construction, $R\ne-I$. Also, as $b$ is nonzero, $R\ne I$. Now, if $Qx=te_1$ for some $t\ne0$, obviously $t=\pm\|x\|$. So, without loss of generality, we may assume that $x$ is a unit vector, and the equation $Qx=\pm e_1$ can be rewritten as
$$
\pmatrix{R\\ &I_{n-2}}U^Tx = \pm U^Te_1.\tag{2}
$$
So, the question boils down to this:

Given that $x$ is a unit vector, can we find a rotation matrix $R\ne\pm I$ and an orthogonal matrix $U$ such that $(2)$ is satisfied?

If so, we can recover $bK$ by applying Cayley transform on $R$ and hence we can calculate $S=U\left((bK)\oplus0\right)U^T$.
Unfortunately, $(2)$ is not always solvable. If $n=2$ and $x=\pm e_1$, then equation $(2)$ becomes $URU^Te_1=\pm e_1$. Since $URU^T$ is a $2\times2$ rotation matrix, we must have $URU^T=\pm I$ and in turn $R=\pm I$.
Fortunately, in all other cases, $(2)$ is solvable (recall the assumption that $x$ is a unit vector):


*

*If $n>2$ and $x=\pm e_1$, let $U$ be any orthogonal matrix such that $U^Te_1=e_n$. Then $(2)$ is satisfied by every rotation matrix $R$.

*If $x\ne\pm e_1$, they are linearly independent. Let $U$ be any orthogonal matrix such that the linear span of $U^Tx$ and $U^Te_1$ is equal to the linear span of $e_1$ and $e_2$. So $(2)$ boils down to solving $Rv=\pm w$ for two given linearly independent vectors $v,w\in\mathbb R^2$. Obviously this can be solved by some $R\ne\pm I$.

