Existence of rotations between two points Let $x,y\in\mathbb R^n$ ($n\in\mathbb N$) be two given points with the same Euclidean norm: $\|x\|=\|y\|$. Does there, in this case, exist an orthogonal matrix $U\in\mathbb R^{n\times n}$ such that $$Ux=y\quad\text{ and }\quad U^{\mathsf T}U=I?$$ I don’t need an explicit construction, just mere existence. Would the Gram–Schmidt procedure help construct an appropriate orthonormal basis as the columns of $U$?
 A: Let $G$ be an $n\times n$ matrix whose first column is $x/\|x\|$ and whose columns are orthonormal.
Let $H$ be an $n\times n$ matrix whose first column is $y/\|y\|$ and whose columns are orthonormal.
Let $e$ be the $n\times 1$ vector whose first entry is $1$ and whose other entries are $0$.
The $G^T = G^{-1}$ and $Ge = x/\|x\|$, and $H^T = H^{-1}$ and $He = y/\|y\|$.
Consequently $U= HG^T$ satisfies $U^T = U^{-1}$ and $Ux=y$.
A: If you are just looking for existence, consider the span $S = \operatorname{span} \{x,y\}$, take two orthogonal vectors $u_1,u_2\in S$, and using Gram-Schmidt extend this set to an orthogonal basis $u_1,u_2,\ldots,u_n$ of $\mathbb R^n$. Now Consider the matrix
$$U=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\\&&1\\&&&\ddots\\&&&&1\end{pmatrix},$$
Then clearly $U^TU=I$, and all you need to do is find the coordinates of $x$ given by $(u^1(x),u^2(x),0,\ldots,0)$ where $u^i(u_j)=\delta_{ij}$. Then for the right $\theta$, $Ux=y$.
A: I suggest the following link http://mathforum.org/library/drmath/view/74532.html. One can construct the hyperplane that bisects the two vectors, then the matrix $U$ is the reflection with respect to this hyperplane. For the given problem, $x-y$ is perpendicular to this hyperplane. One can show this by $(x-y)\cdot(x+y)=0$. Then for $v=(x-y)/||x-y||$, the reflection operator is given by the Housholder matrix $I-2v v^T$
