Is there a simple formula for the rotation with similar properties as the Householder reflection? A Householder reflection maps a vector $v$ to $||v||\ e_i$, and leaves all vectors orthogonal to both $v$ and $e_i$ unchanged. There is a simple formula for it, namely $x\to x - 2\frac{(x,w)}{(w,w)}w$ with $w:=v-||v||\ e_i$.
Instead of a reflection, I would like to have a rotation with similar properties, namely $v$ should be mapped to $||v||\ e_i$, and all vectors orthogonal to both $v$ and $e_i$ should be left unchanged. It is clear in principle that there is such a rotation, but it would be nice if there were also a simple formula for it.
 A: WLOG, let $||e_i||=||v||=1$. We are looking for an orthogonal map $R$ with determinant 1, $Re_i=v$, $(R-I)x=\alpha_xe_i+\beta_xv$, and ??? Any formula for $R$ must be more complicated than that for Householder reflections, since $R-I$ has rank 2 in our case, but rank 1 for Householder reflections.
One simple formula has been pointed out by amd:
$$Rx = x - \frac{(e_i+v) \cdot x}{1+e_i\cdot v} (e_i+v) + 2 (e_i \cdot x) v$$
It is easy to check that $Re_i=e_i - \frac{1+v\cdot e_i}{1+e_i\cdot v} (e_i+v) + 2 v = v$ and $R^Tv=e_i$. It is an orthogonal map, because $R=(I-2bb^T)(I-2e_ie_i^T)$ with $b:=\frac{e_i+v}{||e_i+v||}$, as also pointed out by amd.
To see that the formula above qualifies as easy, let me also show the formula which I had implemented before:
$$Rx = x -  \frac{w \cdot x}{||w||^2} ((1-v\cdot e_i)w+(v\cdot w)e_i) +  (e_i \cdot x) (v-e_i)$$ with $w:=v-(v\cdot e_i)e_i$. It is easy to verify $Re_i=v$, because $w\cdot e_i=0$. But verifying that it is an orthogonal map, i.e. that $R^TRx=x$ seem ugly. But because $R$ is unique, those two expressions for $R$, which look completely different, must be identical.
Maybe I will try to check it some day.
