Let $D$ be an $n\times n$ Euclidean distance matrix i.e. a real symmetric positive semi definite matrix where $D_{ij} = || x_i - x_j||$ for some set of points $\{X_n\}$.
If I define Periodic Boundary Conditions over $\{X_n\}$, we have $x_i^\alpha \in [0,L^\alpha)\forall i$ where $i$ indexes an element of $\{X_n\}$ and $\alpha \in [0,1,...,d]$ indexes the spatial dimensions. Let's consider for simplicity that $L^\alpha = L$ for every dimension, which is equivalent to bound the points in a periodic hypercube in $d$ dimensions.
Then we have $D_{ij} \le \sqrt{d}L$.
Let's say now that we want to reconstruct the set of points back from the distance matrix D. The way to go is the following :
- Build the matrix $$M_{ij} = \frac{D_{1i}^2 + D_{1j}^2 - D_{ij}^2}{2}$$
- Perform its eigendecomposition $M = USU^T$
- Up to an orthogonal transformation, the set of points is $$\tilde{X} = U\sqrt{S}$$ (taking the non-zero columns of this $n\times n$ matrix)
Question :
Starting from a distance matrix $D$ whose entries satisfy $D_{ij} \le \sqrt{d}L$ , will the points $\tilde{X}$ always lie inside the hypercube $[-L,L]^d$ ?
If not, how can I ensure that ?
Edit
I found an interesting paper on the topic, but it treats existence of a circumhypersphere, whereas I'm looking for a bounding hypercube.
Edit 2
I found a partial solution :
Starting from a set a points $X$ inside the hypercube $[-L,L]^d$, its distance matrix $D$ satisfies $D_{ij} \leq \sqrt{d}L$.
Then, the reconstructed points $\tilde{X}$ will be, up to an orthogonal transformation, inside the hypercube $[-\sqrt{d}L, \sqrt{d}L]^d$ since the maximum distance between two points will be $\sqrt{d}L$ and this distance is the length of the hypercube's diagonal.
The question now is : will an orthogonal transformation always bring back $\tilde{X}$ inside the hypercube $[-L,L]^d$ ?
