finding a suitable orthogonal matrix Assume that $x,y \in \text{R}^n$ are two arbitrary vectors. Assuming that $x^\top y >0$, I want to prove that there exist an orthogonal matrix $U \in \text{O}(n)$ such that all elements of the vectors $U x$ and $U y$ are positive.
I intuitively feel that the above claim should be correct. At least, when $n=1$, $n=2$, $n=3$ I know that the above claim is correct. For example for $n=2$, for any $x,y \in \text{R}^2$ with $x^\top y >0$, there exist a rotation matrix so that rotates the vectors $x$ and $y$ such that the resulting vectors fall in the first quarter of the plane (where all the elements of resulting vectors are positive). That rotation matrix can then be chosen as $U$. Same ratiocination can be done for the case that $n=3$. However, I do not know how to prove the above claim for $n >3$. Any suggestions?
Thanks in advanced :-)
 A: Here is a brute-force approach. It's not elegant but it's a no-brainer. WLOG we may assume that both $x$ and $y$ are unit vectors. We will first map $x$ and $y$ to the first quadrant of $\mathbb{R}^2$. More specifically, let $\theta=\cos^{-1}(x^\top y)\in[0,\frac{\pi}2)$ and $z=\frac{y-(y^\top x)x}{\|y-(y^\top x)x\|}$ (if $y=x$, just take $z$ to be any unit vector orthogonal to $x$), so that $x\perp z$ and $y=\cos(\theta)x+\sin(\theta)z$. Let $Q$ be a real orthogonal matrix whose first two columns are $x$ and $z$. Then $Q^\top x=(1,0,\ldots,0)^\top$ and $Q^\top y=(\cos\theta,\sin\theta,0,\ldots,0)^\top$.
It remains to steal some positive amounts from the first elements of $Q^\top x$ and $Q^\top y$ and transfer them to the other entries of the two vectors. Let $G(1,j,\varepsilon)$ denotes the Givens rotation matrix such that its $2\times2$ principal submatrix for indices $\{1,j\}$ is the $2\times2$ rotation matrix for angle $\varepsilon$. Now, for sufficiently small $\varepsilon>0$, the matrix
$$U=G(1,n,\varepsilon)\cdots G(1,3,\varepsilon)G(1,2,\varepsilon)Q^\top$$
will satisfy your requirement.
