TITLE: Are distance functions $ d(a_1,x), ..., d(a_n,x) $ for an arbitrary metric space linearly independent?
QUESTION [8 upvotes]: 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.

REPLY [5 votes]: 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.