4

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 $U$?

triple_sec
  • 23,935
  • 1
    Most certainly there exists such a map. Try a rotation in the plane spanned by $x$ and $y$ (if they're parallel, plus or minus the identity fits the bill). In some nice orthonormal basis, I'd say that the matrix of the operator will have block form $$\begin{bmatrix} R_\theta & 0 \ 0 & {\rm Id}{n-2}\end{bmatrix}$$ for some convenient $\theta$, where $R\theta$ is the usual rotation matrix in $\Bbb R^2$. If you can't manage to do it, give me a holler and I'll try to write it for you. – Ivo Terek Jun 16 '16 at 03:19
  • 1
    @IvoTerek Thank you very much for your comment. I prefer to go with another answer that relies more on the algebraic, as opposed to the geometric, structure of $\mathbb R^n$. – triple_sec Jun 16 '16 at 04:34

3 Answers3

4

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 $Ux=y$.

Michael Hardy
  • 1
2

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 $\theta$, $Ux=y$.

Michael Hardy
  • 1
Couchy
  • 2,796
  • Thank you very much. I accepted Michael’s answer because I prefer the elegance of the algebraic approach to the geometric point of view of trigonometry. – triple_sec Jun 16 '16 at 04:36
2

I suggest the following link Link. 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 $I-2v v^T$

Glorfindel
  • 4,000
Andrei
  • 39,869
  • 3
    A summarization of what's in the link would be nice, lest the remote resource becomes suddenly inaccessible. – J. M. ain't a mathematician Jun 16 '16 at 03:33
  • Of course. 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 $I-2v v^T$ – Andrei Jun 16 '16 at 03:43
  • Please add that into your answer, then. – J. M. ain't a mathematician Jun 16 '16 at 03:44
  • Thank you, @Andrei. This is a very nice explicit construction. – triple_sec Jun 16 '16 at 05:05