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We call an $n\times n$-matrix ${\bf A}\in \text{Mat}(n\times n, \mathbb{R})$ a metric matrix if

  1. ${\bf A}_{ii} = 0$ for all $i\in \{1,\ldots,n\}$,
  2. ${\bf A}_{ij} = {\bf A}_{ji}$ for all $i,j \in \{1,\ldots,n\}$ (that is, ${\bf A}$ is symmetric), and
  3. ${\bf A}_{ik} \leq {\bf A}_{ij} + {\bf A}_{jk}$ for all $i,j,k \in \{1,\ldots, n\}$ (that is, the triangle inequality holds).

We say that ${\bf A}\in \text{Mat}(n\times n, \mathbb{R})$ is $\mathbb{R}^k$-realizable for some positive integer $k$ if there is a subset $S\subseteq \mathbb{R}^k$ having $n$ elements, and a bijection $\varphi:\{1,\ldots,n\}\to S$ such that $$\|\varphi(i) - \varphi(j)\| = {\bf A}_{ij} \text{ for all } i,j\in \{1,\ldots n\}.$$

(Note that $\|\cdot\|$ denotes the Euclidean norm.) So for instance, the following metric matrix ${\bf A}\in \text{Mat}(4\times 4, \mathbb{R})$ is not $\mathbb{R}^2$-realizable, but it is $\mathbb{R}^3$-realizable:

$${\bf A} = \begin{pmatrix} 0 & 1 & 1 & 1\\ 1 & 0 & 1 & 1\\ 1 & 1 & 0 & 1\\ 1 & 1 & 1 & 0\\ \end{pmatrix}$$

Questions. (Only the first question needs to be answered for acceptance.)

  1. Given an integer $n>1$, is every metric $n\times n$-matrix $\mathbb{R}^{n-1}$-realisable?
  2. For every metric matrix ${\bf A}\in \text{Mat}(n\times n, \mathbb{N})$ let its metric dimension $\text{mdim}({\bf A})$ denote the smallest positive integer $k$ such that ${\bf A}$ is $\mathbb{R}^k$-realizable. Is the problem of finding $\text{mdim}({\bf A})$ given ${\bf A}\in \text{Mat}(n\times n, \mathbb{N})$ a polynomial-time problem with respect to $n$?
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    $\begingroup$ I think for (1) the answer is no and a counterexample is: \begin{pmatrix}0&1&1&0.5\\1&0&1&0.5\\1&1&0&0.5\\0.5&0.5&0.5&0\end{pmatrix} The first three points must form a regular triangle and then it is impossible to place the forth point. $\endgroup$ Commented Aug 17, 2022 at 12:11
  • $\begingroup$ Great example - thanks Henrik! $\endgroup$ Commented Aug 17, 2022 at 14:55

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In other words, you ask whether every finite metric space may be isometrically embedded to Euclidean space $\mathbb{R}^n$. Not every. A necessary and sufficient condition is given by the non-negativity of the so called Cayley--Menger determinants.

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