Given a finite metric space, are the matrices of triangle inequality errors invertible?

Question

I have been working on some problems regarding finite metric spaces and have already proven/positively answered the following statement/question if the underlying metric has additional properties. Now I'm wondering if the statement is true in general.

Suppose we have a finite metric space $(X,d)$ and we fix $y \in X$. Now consider the matrix $$M(y) := ( d (x, y) + d (y, z) - d (x, z))_{x, z \in X\setminus \{y\}}$$ of all possible errors that arise in the triangle inequality where the 'middle point' is fixed.

Is $M(y)$ invertible?

Any ideas are greatly appreciated.

Answer 1

Here's a counterexample. Let $X=\{1,2,3,4\}$ with the metric given by the following matrix D of distances, i.e. the $(i,j)$-th entry is $d(i,j)$: $$D=\begin{bmatrix}0&1&2&1\\1&0&1&2\\2&1&0&1\\1&2&1&0\end{bmatrix}$$

In other words, this is a graph metric on the $4$-cycle graph, whose vertices are $1,2,3,4$ respectively. (We can also verify that this is a metric by definition, in which case the triangle inequality follows easily from the fact that $2\leq 1+1$.)

Then, we have $$M(1)=\begin{bmatrix}2&2&0\\2&4&2\\0&2&2\end{bmatrix},$$ which is singular, since the middle row is the sum of the other two.

Answer 2

Edit: This answer is not correct (see my comment directly following this answer). I'll leave my incorrect answer here in case it is of interest.

Original:

Under the discrete metric, where $d(x, y) = 1$ whenever $x \neq y$, the errors $d(x, y) + d(y, z) - d(x, z)$ are all exactly $1$. Thus for any $y$, the matrix $M(y)$ will have all entries equal to $1$, and will not be invertible (assuming the space has at least three points so that $M(y)$ has size at least $2 \times 2$).

http://en.wikipedia.org/wiki/Discrete_space