Random rotation of a set of distinct points in $R^n$

Question

Consider a set $\{\mathbf{X}_1,\cdots , \mathbf{X}_M\}$ of distinct points in $\mathbb{R}^n$ with $M$ finite. The $M$ values of the $i$-th coordinate do not all have to be dinstinct. For example, in $\mathbb{R}^2$ the points could be placed on a regular grid so that several points have the same x or y coordinates.

Let's now consider a rotation, that maps the original set into a new set $\{\mathbf{X}_1', \cdots, \mathbf{X}_M'\}$.

I would like to show that if we select the rotation randomly, i.e. if we project on a random orthogonal basis, the $M$ values of the $i$-th coordinates of the new set will be distinct for all $i=1, \cdots, N$ with probability almost one.

Intuitively it sounds like that should be the case. But I have no experience with random matrices and I wonder where I could start from.

Answer 1

Consider

$y = Ux$

where $U$ is some random matrix (orthogonal or otherwise). Since you are interested in only the $m^{th}$ coordinate, only the $m^{th}$ row of $U$ is relevant to your work. Therefore, you are basically only interested in the properties of

$z_{i} = u^{T}x_{i}$,

where $u$ is the $m^{th}$ row of the matrix. The orthogonal matrix part does not seem to be particularly relevant; for your case, $u$ is essentially a random unit vector.

Suppose the $i^{th}$ and $j^{th}$ rotation are the same. Then we have

$u^{T}(x_{i} -x_{j}) = 0$.

This implies that

$u^{T}(x_{i} - x_{k}) = - u^{T}(x_{j} - x_{k})$

for all $k$. Maybe you can use something like this if your points have some structure?

Generally, nothing can be said. If $x_{i} = x_{j}$ for any $(i,j)$ pair the result you are thinking of doesnt hold

Answer 2

Restatement of the question. Let $X_1,\ldots,X_M\in \mathbb R^n$ be deterministic vectors such that no two vectors $X_j$ and $X_k$ are identical in all components, and let $A$ be a random unitary matrix drawn from the circular real ensemble CRE$(n)$. Your question asks us to show that for all $1\leq i\leq n$, the random numbers $\{(AX_j)_i\colon j=1,\ldots,M\}$ have the following property: $$ \mathbb P\bigl(\exists j\not= k\in 1,\ldots, M\colon (AX_j)_i=(AX_k)_i\bigr)=0. $$ Proof of the property. By the union bound, $$ \mathbb P\bigl( \exists j\not= k\in 1,\ldots, M\colon (AX_j)_i=(AX_k)_i\bigr)\leq \sum_{j\not=k}\mathbb P\bigl((AX_j)_i=(AX_k)_i\bigr). $$ Since the random vector $A(X_j-X_k)$ is uniformly distributed on the sphere of radius $\|X_j-X_k\|$ (and $\|X_j-X_k\|>0$ by hypothesis), it follows that the event $$\Bigl\{\bigl(A(X_j-X_k)\bigr)_i=0\Bigr\}$$ has probability $0$. Indeed, this event describes a uniformly random element of the sphere belonging to a certain great circle, which has codimension $1$. Thus the probability in question is zero as well.