# Are distance functions $d(a_1,x), …, d(a_n,x)$ for an arbitrary metric space linearly independent?

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.

• By induction you can see that taking any finite ($n$) number of points of $X$ gives ($n$) linearly independent distance functions – Aram Nov 3 '14 at 19:29
• 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. – Greg Martin Nov 3 '14 at 19:35
• 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). – Greg Martin Nov 3 '14 at 19:37

## 1 Answer

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.