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When implementing a clustering algorithm earlier today I investigated (in Matlab syntax):

$$\text{det}(\text{abs}({\bf d - d}^T)), {\bf d }\in \mathbb R^{N}$$

edit: In non-Matlab syntax: $$\text{det}(\text{abs}({\bf d1}^T-{\bf 1d}^T))$$

In other words the determinant of a matrix consisting of absolute value of pairwise differences.

Now since determinants are usually terrible to calculate, this is mostly of a curiosity of mine, but it seems that if any two elements of $\bf d$ are the same, then the above determinant is 0, and if no two elements of $\bf d$ are the same, then determinant $\neq 0$.

Can we prove this?

For example the vector ${\bf d} = [1,2,3]^T$:

$$\text{det}(\text{abs}({\bf d - d}^T)) = \begin{bmatrix}|1-1|&|1-2|&|1-3|\\|2-1|&|2-2|&|2-3|\\|3-1|&|3-2|&|3-3|\end{bmatrix} = \left\|\begin{bmatrix}0&1&2\\1&0&1\\2&1&0\end{bmatrix}\right\|\neq 0$$ And the vector : ${\bf d} = [3,2,3]^T$: $$\text{det}(\text{abs}({\bf d - d}^T)) = \begin{bmatrix}|3-3|&|3-2|&|3-3|\\|2-3|&|2-2|&|3-3|\\|3-3|&|3-2|&|3-3|\end{bmatrix} = \left\|\begin{bmatrix}0&1&0\\1&0&1\\0&1&0\end{bmatrix}\right\|= 0$$

mathreadler
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    I think your notation is confusing here, because if $\mathbf{d}$ is a vector, then $\mathbf{d}-\mathbf{d}^T$ isn't well-defined; one is an $n\times1$ matrix while the other is a $1\times n$ matrix. – Steven Stadnicki Dec 05 '19 at 16:46
  • Yes, it is true. Write $d=[a,b,c]$ and consider your matrix. It has two parallel columns if two elements are the same. – A.Γ. Dec 05 '19 at 16:49
  • @StevenStadnicki yes it is a bit confusing, which is why I wrote "Matlab syntax". I will try and find a better way to write it. ${\bf 1 d}^T - {\bf d 1}^T$ should be more understandable, I suppose? – mathreadler Dec 05 '19 at 16:53
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    @mathreadler That works; another would be to use $\otimes$ for the (implicit) tensor product there: $\mathbf{1\otimes d}-\mathbf{d\otimes 1}$. – Steven Stadnicki Dec 05 '19 at 16:55
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    Even with the products cleared up, I suggest restraint about using $\left|\cdot\right|$ for the entrywise absolute value of a matrix... Also, I'd say explicitly that the $d_i$ are real numbers, if you mean them to be. – darij grinberg Dec 05 '19 at 18:40

2 Answers2

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This is true. By permuting the rows and columns of the matrix, we may assume that the entries of $\mathbf d=(d_1,d_2,\ldots,d_n)^T$ are arranged in descending order. Moreover, since the rows and columns are simultaneously permuted, the determinant is unchanged. Now let $\Delta_i=d_i-d_{i+1}$, $$ L=\pmatrix{1\\ &\ddots\\ &&\ddots\\ &&&1\\ \frac12&\frac12&\cdots&\frac12&1} \ \text{ and } \ U=\pmatrix{1&-1\\ &\ddots&\ddots\\ &&\ddots&\ddots\\ &&&\ddots&-1\\ &&&&1}. $$ Then \begin{aligned} A:=|\mathbf d\mathbf 1^T-\mathbf 1\mathbf d^T| &=\pmatrix{0&d_1-d_2&d_1-d_3&\cdots&d_1-d_n\\ d_1-d_2&0&d_2-d_3&\cdots&d_2-d_n\\ d_1-d_3&d_2-d_3&\ddots&\ddots&\vdots\\ \vdots&\vdots&\ddots&0&d_{n-1}-d_n\\ d_1-d_n&d_2-d_n&\cdots&d_{n-1}-d_n&0},\\ B:=UAU^T &=\pmatrix{-2\Delta_1&0&\cdots&0&\Delta_1\\ 0&-2\Delta_2&\ddots&\vdots&\Delta_2\\ \vdots&\ddots&\ddots&0&\vdots\\ 0&\cdots&0&-2\Delta_{n-1}&\Delta_{n-1}\\ \Delta_1&\Delta_2&\cdots&\Delta_{n-1}&0},\\ C:=LBL^T &=\pmatrix{-2\Delta_1&0&\cdots&0&0\\ 0&-2\Delta_2&\ddots&\vdots&0\\ \vdots&\ddots&\ddots&0&\vdots\\ 0&\cdots&0&-2\Delta_{n-1}&0\\ 0&0&\cdots&0&\frac12(d_1-d_n)}. \end{aligned} Since $U$ and $L$ have determinants $1$, \begin{aligned} \det A=\det C &=\left[\prod_{i=1}^{n-1}(-2\Delta_i)\right]\left[\frac12(d_1-d_n)\right]\\ &=\left[\prod_{i=1}^{n-1}(-2\Delta_i)\right]\left[-\frac12(d_n-d_1)\right]\\ &=(-2)^{n-2}\prod_{cyc}(d_i-d_{i+1}). \end{aligned} (If the $d_i$s are arranged in ascending order, the formula becomes $\det A=2^{n-2}\prod_{cyc}(d_i-d_{i+1})$ instead.) Hence $A$ is nonsingular if and only if all entries of $\mathbf d$ are distinct.

Remark. As pointed out by darij grinberg, there is actually a more general result, namely, if $\mathbf x_1,\ldots,\mathbf x_n\in\mathbb R^m$ and $A$ is the distance matrix defined by $a_{ij}=\|\mathbf x_i-\mathbf x_j\|_2$, then:
  1. (cf. A matrix involving distances of $n$ points in $\mathbb{R}^3$ ) $A$ is negative semidefinite on $\mathbf1^\perp$ and if all $\mathbf x_i$s are distinct, it is negative definite on $\mathbf1^\perp$;
  2. (cf. Nonsingularity of Euclidean distance matrix ) when all $\mathbf x_i$s are distinct, since we also have $\mathbf1^TA\mathbf1>0$, $A$ must be nonsingular.

My argument above only proves the non-singularity of $A$ when $m=1$. It cannot deal with the higher-dimensional case and it doesn't concern about the negative semidefiniteness of $A$ on $\mathbf1^\perp$.

user1551
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  • Cool! Thank you. – mathreadler Dec 05 '19 at 19:58
  • Oh nice! I figured a cyclic product might be the result but I'd missed the notion of sorting $\mathbf{d}$ in increasing order as a way of ensuring specific signs on the absolute values, and so my result didn't satisfy the requirement of being zero if any two $d_i$ were equal. Very slick! – Steven Stadnicki Dec 05 '19 at 20:09
  • Now try the case when $d_1, d_2, \ldots, d_n$ are complex numbers, or more generally, vectors in a Euclidean space :) (No guarantee the claim is true in this case, but your argument definitely doesn't work.) – darij grinberg Dec 05 '19 at 20:29
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    Ah: The claim is indeed true when $d_1, d_2, \ldots, d_n$ are vectors in any Euclidean space. This is proved in https://math.stackexchange.com/questions/186830/nonsingularity-of-euclidean-distance-matrix/3010694#3010694 . – darij grinberg Dec 05 '19 at 21:04
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The second half — the assertion that if two elements are the same, then the determinant is zero — is trivial; if two elements are the same then their vector of (absolute values of) pairwise differences with all elements will be the same, and so your matrix will have two identical columns.

Steven Stadnicki
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