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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.

j4GGy
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  • I am not sure if I understand you correctly. Suppose $X={x_1,x_2,\ldots,x_n}$ and $y=x_n$. What do you mean by $M(y)$? Do you mean an $(n-1)\times(n-1)$ matrix $A$ such that $a_{ij}=d(x_i,y)+d(y,x_j)-d(x_i,x_j)$? – user1551 Jan 13 '15 at 23:31
  • @user1551 That is correct. – j4GGy Jan 14 '15 at 06:37

2 Answers2

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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.

Dejan Govc
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  • Maybe it would be interesting to investigate the case $X={v_1,\ldots,v_n}\subseteq\mathbb R^m$? – Dejan Govc Jan 13 '15 at 23:14
  • Very nice, indeed. Since I might be interested in characterizing metric spaces for which $M(y)$ is invertible for all $y\in X$, may I ask if you just stumbled across this example or did you 'construct' it with some special property in mind? – j4GGy Jan 14 '15 at 07:53
  • @GenericNickname: I was looking for spaces where a large number of triangle inequalities are equalities (which I'm not sure is helpful), and stumbled upon this example in the process. Anyway, I experimented with cycles a bit more. It seems to me that the resulting matrix $M(1)$ is singular for cycles of even length and non-singular for odd cycles. – Dejan Govc Jan 15 '15 at 10:54
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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

mathmandan
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    No wait, that's not correct, since $x$ could equal $z$ in which case the error term is $2$. Sorry. – mathmandan Jan 13 '15 at 20:57
  • I was thinking along the same lines :) – j4GGy Jan 13 '15 at 21:26
  • Indeed, for the discrete metric, the error matrix has 1's everywhere, except for 2's down the main diagonal. Such an $n\times n$ matrix has determinant exactly equal to $n+1$, by taking $a = 2$ and $b = 1$ at: http://math.stackexchange.com/questions/86644/determinant-of-a-specially-structured-matrix – mathmandan Jan 13 '15 at 21:41
  • I was just thinking about this! So for the discrete metric it is always invertible. More general if $d(x,y) = \delta_{xy} \cdot D$ for some $D > 0$, then the determinant will be $D \cdot (n+1)$. – j4GGy Jan 13 '15 at 21:44
  • I think you'll actually get $(n+1)$ times a power of $D$ for the determinant, since the determinant gains a factor of $D$ for each row of the matrix. Still an invertible matrix, though. – mathmandan Jan 13 '15 at 22:10
  • Oops, you're right. It's $D^n(n+1)$. – j4GGy Jan 13 '15 at 22:27