If we have a metric space $ (X,d) $, a finite set of distinct points $ a_1, ..., a_n $, do the distance functions $ d(a_1,x), ..., d(a_n,x) $ have to be linearly independent?

That is, if $ c_1d(a_1,x) + ... + c_nd(a_n,x) = 0$ for all $ x $, then $ c_1 = ... = c_n = 0 $? This can easily be seen to be true for $ n = 2 $ and also for $ n = 3 $, but what's the answer in general?

If we substitude all the $ a_i $ for x, we get an interesting system of n equations, where we can treat $ c_i $ as the unknowns, and then the matrix of the coefficients $ d(a_i, a_j) $ is symmetric with 0s on the diagonal(apparently this is called a "hollow matrix"). All this looks quite interesting and promising, but I don't really know where to take it.

  • $\begingroup$ By induction you can see that taking any finite ($n$) number of points of $X$ gives ($n$) linearly independent distance functions $\endgroup$ – Aram Nov 3 '14 at 19:29
  • $\begingroup$ Really? I'd like to see the details. Note that if there exists a metric space for which the answer is "no" for $n$ given points, then the finite metric space whose points are only $\{a_1,\dots,a_n\}$ is also such a metric space. Therefore if you can prove linear independence for any finite metric space, you can prove it for all metric spaces. $\endgroup$ – Greg Martin Nov 3 '14 at 19:35
  • $\begingroup$ I note that I found some symmetric $4\times4$ matrices with zeros on the diagonal and positive entries elsewhere that have $0$ as an eigenvalue; but I didn't find a way to make the off-diagonal entries satisfy the necessary triangle inequalities for it to have come from a metric (indeed they came right up to the borderline but didn't quite make it before one of the "distances" vanished). $\endgroup$ – Greg Martin Nov 3 '14 at 19:37

A counterexample is given by the four-point metric space with the distance matrix $$\begin{pmatrix} 0 & 1 & 2 & 1 \\ 1 & 0 & 1 & 2 \\ 2 & 1 & 0 & 1 \\ 1 & 2 & 1 & 0 \end{pmatrix}$$ Note that the sum of the $1$st and $3$rd rows is the same as the sum of the $2$nd and $4$th.

Geometrically, this metric space is realized by the vertices of the $4$-cycle $C_4$ with the path metric, i.e., the smallest number of edges to travel.


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