# Generalized Euclidean distance matrices

R. Balaji a , R.B. Bapatb and Shivani Goelc CONTACT: Shivani Goel. Email:
<EMAIL_ADDRESS>aDepartment of Mathematics, IIT Madras, Chennai,
India; bTheoretical Statistics and Mathematics Unit, Indian Statistical
Institute, Delhi, India;cDepartment of Mathematics, IISc Bangalore,
Bangalore,India

###### Abstract

Euclidean distance matrices ($\mathop{\rm EDM}$) are symmetric nonnegative
matrices with several interesting properties. In this article, we introduce a
wider class of matrices called generalized Euclidean distance matrices
($\mathop{\rm GEDM}$s) that include $\mathop{\rm EDM}$s. Each $\mathop{\rm
GEDM}$ is an entry-wise nonnegative matrix. A $\mathop{\rm GEDM}$ is not
symmetric unless it is an $\mathop{\rm EDM}$. By some new techniques, we show
that many significant results on Euclidean distance matrices can be extended
to generalized Euclidean distance matrices. These contain results about
eigenvalues, inverse, determinant, spectral radius, Moore-Penrose inverse and
some majorization inequalities. We finally give an application by constructing
infinitely divisible matrices using generalized Euclidean distance matrices.

###### keywords:

Euclidean distance matrix, Haynsworth inertia formula, Laplacian matrices,
Majorization, Infinitely divisible matrices.

15A57

##  1 Introduction

An $n\times n$ real matrix $D=[d_{ij}]$ is a Euclidean distance matrix (EDM)
if there exist vectors $x^{1},x^{2},\dotsc,x^{n}$ in a Euclidean space
$(V,\langle\cdot,\cdot\rangle)$ such that

$d_{ij}=\langle x^{i}-x^{j},x^{i}-x^{j}\rangle~{}~{}~{}\mbox{for
all}~{}i,j=1,2,\dotsc,n;$ (1)

or equivalently,

$d_{ij}=\|x^{i}-x^{j}\|^{2}~{}~{}\mbox{for all}~{}i,j=1,2,\dotsc,n.$

Euclidean distance matrices appear in several fields. For instance, in
approximation theory, Micchelli [9] proved the striking result: If
$x^{1},\dotsc,x^{n}$ are $n$ distinct points in the plane, then

$(-1)^{n-1}\det[\sqrt{1+\|x^{i}-x^{j}\|^{2}}]>0.$

As a consequence, there exists a unique function

$f(x)=\sum_{k=1}^{n}c_{k}\sqrt{1+\|x-x^{k}\|^{2}}$

interpolating a given data $y_{1},\dotsc,y_{n}$ at $x^{1},\dotsc,x^{n}$.
Euclidean distance matrices have several interesting properties. For example,
we have the following basic result.

######  Theorem 1.1 (Menger, Schoenberg. [7]).

Let $D=[d_{ij}]$ be an $n\times n$ symmetric matrix with zero diagonal. Then
the following are equivalent.

  1. 1.

$D$ is an $\mathop{\rm EDM}$.

  2. 2.

If $(x_{1},\dotsc,x_{n})^{\prime}$ is such that $\sum_{i=1}^{n}x_{i}=0$, then
$\sum_{i,j}d_{ij}x_{i}x_{j}\leq 0$.

  3. 3.

Let $\mathbf{1}=(1,1,\dotsc,1)^{\prime}\in\mathbb{R}^{n}$. Then the bordered
matrix $\left[\begin{matrix}D&\mathbf{1}\\\
\mathbf{1}^{\prime}&0\end{matrix}\right]$ has exactly one (simple) positive
eigenvalue.

Symmetric matrices with exactly one simple positive eigenvalue are called
elliptic matrices. Elliptic matrices play an important role in Alexandrov
inequalities for mixed volumes [3, chapter 5]. Elliptic matrices are useful in
obtaining infinitely divisible matrices. If $[a_{ij}]$ is an elliptic matrix
with all entries positive, then $[\displaystyle\frac{1}{a_{ij}^{r}}]$ is
positive semidefinite for all $r>0$, i.e. $[\frac{1}{a_{ij}}]$ is a infinitely
divisible matrix. The main theme of [4] is to obtain infinitely divisible
matrices via elliptic matrices.

###  1.1 Objective of the paper

In this paper, we investigate the so-called generalized Euclidean distance
matrices ($\mathop{\rm GEDM}$). All Euclidean distance matrices ($\mathop{\rm
EDM}$) are generalized Euclidean distance matrices. If a generalized Euclidean
distance matrix is not an $\mathop{\rm EDM}$, then it is not symmetric.
Despite this fact, we extend many properties of $\mathop{\rm EDM}$s to
$\mathop{\rm GEDM}$s. For example, we show that all eigenvalues of a
$\mathop{\rm GEDM}$ are real and has at most one positive eigenvalue, null
space of a $\mathop{\rm GEDM}$ is a subspace of $\mathbf{1}^{\perp}$, the
Moore-Penrose inverse of a $\mathop{\rm GEDM}$ is negative semidefinite on
$\mathbf{1}^{\perp}$ and so on. These results are obtained by using new
techniques that circumvent the standard arguments on symmetric matrices.

###  1.2 Definition of generalized Euclidean distance matrices

We begin with the following observation. Let $D=[d_{ij}]$ be an $n\times n$
Euclidean distance matrix. In view of (1), if $D=[d_{ij}]$ is a EDM, then

$d_{ij}=\|x^{i}\|^{2}+\|x^{j}\|^{2}-2\langle
x^{i},x^{j}\rangle~{}~{}~{}\forall i,j=1,2,\dotsc,n.$

Recall that the Gram matrix of an ordered system of vectors
$\\{v^{1},\dotsc,v^{n}\\}$ in a Euclidean space is the matrix $[\langle
v^{i},v^{j}\rangle]$. As every positive semidefinite matrix is a Gram matrix
of some system of vectors in some Euclidean space, it follows that
$D=[d_{ij}]$ is an EDM if and only if there exists an $n\times n$ positive
semidefinite matrix $F=[f_{ij}]$ such that

$d_{ij}=f_{ii}+f_{jj}-2f_{ij}.$ (2)

Let $\mathbf{1}$ be the vector of all ones in $\mathbb{R}^{n}$. Define
$J:=\mathbf{1}\mathbf{1}^{\prime}$. Then (2) can be rewritten as

$D=\mathop{\rm diag}(F)J+J\mathop{\rm diag}(F)-2F.$ (3)

Define

$P:=I-\frac{1}{n}J~{}~{}\mbox{and}~{}~{}Y=[y_{ij}]:=-\frac{1}{2}PDP,$ (4)

where $I$ is the $n\times n$ identity matrix. As the diagonal entries of $D$
are zero, from equation $(\ref{df11})$, it can be verified that

$d_{ij}=y_{ii}+y_{jj}-2y_{ij},$

or equivalently,

$D=\mathop{\rm diag}(Y)J+J\mathop{\rm diag}(Y)-2Y.$

By an easy verification, it follows that $Y$ is positive semidefinite and
$Y\mathbf{1}=0$. Thus, a symmetric matrix $D=[d_{ij}]$ is an EDM if and only
if there exists a positive semidefinite matrix $G=[g_{ij}]$ such that
$G\mathbf{1}=0$ (equivalently, row sums and column sums of $G$ are zero) and

$d_{ij}=g_{ii}+g_{jj}-2g_{ij}.$

Several important properties of Euclidean distance matrices depend on this
characterization.

######  Definition 1.2.

We say that an $n\times n$ real symmetric matrix $L$ is a generalized
Laplacian matrix if $L$ is positive semidefinite and $L\mathbf{1}=0$.

It is easy to note that $L$ is a generalized Laplacian if and only if there
exists a positive semidefinite matrix $F$ such that $L=PFP$, where $P$ is the
matrix in $(\ref{df1})$. Laplacian matrices of connected graphs are examples
of generalized Laplacian matrices.

######  Definition 1.3.

Let $a$ and $b$ be any two positive numbers and $L$ be an $n\times n$
generalized Laplacian matrix. Define

$d_{ij}=a^{2}l_{ii}+b^{2}l_{jj}-2abl_{ij}~{}~{}~{}i,j=1,2,\dotsc,n.$

We now say that $[d_{ij}]$ is a generalized Euclidean distance matrix
($\mathop{\rm GEDM}$).

An easy computation shows that

$d_{ji}=b^{2}l_{ii}+a^{2}l_{jj}-2abl_{ij}.$

Thus, $D$ is symmetric if and only if $a=b$ and in this case, all the diagonal
entries of $D$ are zero; hence $D$ will be a Euclidean distance matrix. To
illustrate the definition, consider the following example.

######  Example 1.4.

Let

$L:=[l_{ij}]=\left[\begin{array}[]{cccc }~{}~{}3&-1&-2\\\ -1&~{}~{}3&-2\\\
-2&-2&~{}~{}4\\\ \end{array}\right].$

Then $L$ is a generalized Laplacian matrix. Set $a=1$ and $b=3$. For
$i,j=1,2,3$, define

$\displaystyle d_{ij}:$ $\displaystyle=a^{2}l_{ii}+b^{2}l_{jj}-2abl_{ij}$
$\displaystyle=l_{ii}+9l_{jj}-6l_{ij}.$

Now,

$D:=[d_{ij}]=\left[\begin{array}[]{cccc}12&36&51\\\ 36&12&51\\\
43&43&16\end{array}\right]$

is a $\mathop{\rm GEDM}$.

We note the following proposition.

######  Proposition 1.5.

An $n\times n$ matrix $D$ is a $\mathop{\rm GEDM}$ if and only if there exist
$x^{1},\dotsc,x^{n}$ in some Euclidean space $(V,\langle\cdot,\cdot\rangle)$
such that

$d_{ij}=\langle
ax^{i}-bx^{j},ax^{i}-bx^{j}\rangle~{}~{}\mbox{and}~{}~{}\sum_{j=1}^{n}x^{j}=0.$

###### Proof.

Let $D=[d_{ij}]$ be a $\mathop{\rm GEDM}$. Then,

$d_{ij}=a^{2}l_{ii}+b^{2}l_{jj}-2abl_{ij},$

where $L=[l_{ij}]$ is a generalized Laplacian matrix and $a,b$ are some
positive numbers. Because $L$ is positive semidefinite and $L\mathbf{1}=0$,
there exist $x^{1},\dotsc,x^{n}$ in a Euclidean space
$(V,\langle\cdot,\cdot\rangle)$ such that

$l_{ij}=\langle
x^{i},x^{j}\rangle~{}~{}\mbox{and}~{}~{}\sum_{j=1}^{n}x^{j}=0.$

Thus,

$\displaystyle d_{ij}=a^{2}\langle x^{i},x^{i}\rangle+b^{2}\langle
x^{j},x^{j}\rangle-2ab\langle x^{i},x^{j}\rangle=\langle
ax^{i}-bx^{j},ax^{i}-bx^{j}\rangle.$

The converse is immediate. ∎

###  1.3 Preliminaries

We fix the notation and mention a few results/definitions that are needed in
the sequel.

  1. (N1)

All matrices considered are real. The transpose of a matrix $A$ is written
$A^{\prime}$. The notation $\mathbf{1}$ will denote the vector of all ones in
$\mathbb{R}^{n}$, $J$ will be the matrix $\mathbf{1}\mathbf{1}^{\prime}$ and
$\mathbf{1}^{\perp}$ will be the subspace containing all vectors that are
orthogonal to $\mathbf{1}$. As usual $e_{1},\dotsc,e_{n}$ will denote the
standard orthonormal basis vectors in $\mathbb{R}^{n}$; hence the first column
of an $n\times n$ matrix $A$ will be $Ae_{1}$ and so on. We use $P$ to denote
the orthogonal projection $I-\frac{1}{n}J$ onto $\mathbf{1}^{\perp}$.

  2. (N2)

The null space of $A$ is denoted by $\mathop{\rm null}(A)$ and the column
space (range) of $A$ by $\mathop{\rm col}(A)$. As usual, $\rho(A)$ will be the
spectral radius of $A$ and the Moore-Penorse inverse of $A$ will be written
$A^{\dagger}$.

  3. (N3)

Let $x:=(x_{1},\dotsc,x_{n})^{\prime}$ and $y:=(y_{1},\dotsc,y_{n})^{\prime}$
be any two vectors. Let $\sigma$ and $\pi$ be permutations on
$\\{1,\dotsc,n\\}$ such that

$x_{\sigma(1)}\geq\cdots\geq
x_{\sigma_{n}}~{}~{}\mbox{and}~{}~{}y_{\pi_{1}}\geq\cdots\geq y_{\pi_{n}}.$

We say that $x$ is majorized by $y$ if

$\sum_{i=1}^{n}x_{i}=\sum_{i=1}^{n}y_{i}~{}~{}~{}\mbox{and}~{}~{}\sum_{j=1}^{k}x_{\sigma(j)}\leq\sum_{j=1}^{k}y_{\pi(j)}~{}~{}~{}(k=1,\dotsc,n-1).$

To say that $x$ is majorized by $y$, we use the notation $x\prec y$.

  4. (N4)

Given an $n\times n$ matrix $A=[a_{ij}]$, we use $\mathop{\rm diag}(A)$ to
denote the vector $(a_{11},\dotsc,a_{nn})^{\prime}$ in $\mathbb{R}^{n}$. If
$(p_{1},\dotsc,p_{n})^{\prime}$ is a vector in $\mathbb{R}^{n}$, we use
$\mathop{\rm Diag}(p_{1},\dotsc,p_{n})$ to denote the diagonal matrix with
diagonal entries $p_{1},\dotsc,p_{n}$.

  5. (N5)

Let $H$ be an $n\times n$ symmetric matrix and the eigenvalues are
$\lambda_{1}\dotsc,\lambda_{n}$ such that

$\lambda_{1}\geq\cdots\geq\lambda_{n}.$

We now define $\lambda(H)$ by

$\lambda(H)=(\lambda_{1},\dotsc,\lambda_{n})^{\prime}.$

  6. (N6)

By a theorem of Schur, if $H$ is an $n\times n$ symmetric matrix, then

$\mathop{\rm diag}(H)\prec\lambda(H).$

If $A$ and $B$ are $n\times n$ symmetric matrices, then a result of Ky Fan
states that

$\lambda(A+B)\prec\lambda(A)+\lambda(B).$

(see e.g., [11, Theorem 7.14 and 7.15])

  7. (N7)

The inertia of a symmetric matrix $A$ is denoted by $\mathop{\rm
In}(A)=(\nu,\delta,\mu)$ where $\nu$ is the number of negative eigenvalues of
$A$, $\delta$ is the nullity of $A$ and $\mu$ is the number of positive
eigenvalues of $A$. If $A$ is $n\times n$, $p\in\mathbb{R}^{n}$, and if

$\widetilde{A}=\left[\begin{array}[]{ccc}A&p\\\
p^{\prime}&0\end{array}\right],$

then the generalized Schur complement $\widetilde{A}/A$ is
$-p^{\prime}A^{{\dagger}}p$. If $p\in\mathop{\rm col}(A)$, then

$\mathop{\rm In}(\widetilde{A}/A)+\mathop{\rm In}(A)=\mathop{\rm
In}(\widetilde{A}).$

(see [5, Theorem 2])

  8. (N8)

An $n\times n$ matrix $A$ is called an ${\bf M}$-matrix, if $A=\rho I-S$,
where $S$ is a nonnegative matrix and $\rho\geq\rho(S)$.

  9. (N9)

Let $A$ be an $n\times n$ symmetric matrix and $p\in\mathbb{R}^{n}$. If the
bordered matrix

$\left[\begin{array}[]{ccccccc}A&p\\\ p^{\prime}&0\end{array}\right]$

has exactly one positive eigenvalue, then

$y\in p^{\perp}\implies y^{\prime}Ay\leq 0.$

(see [6, Theorem 2.9])

##  2 Results

In the sequel, we assume that $D$ is a non-zero $\mathop{\rm GEDM}$ and $L$ is
the generalized Laplacian such that

$D=a^{2}\widetilde{L}J+b^{2}J\widetilde{L}-2abL,$ (5)

where $a$ and $b$ are fixed positive numbers and $\widetilde{L}:=\mathop{\rm
Diag}(L)$.

###  2.1 Eigenvalues of a $\mathop{\rm GEDM}$

We shall now show that all eigenvalues of $D$ are real and in fact $D$ is
similar to a symmetric matrix.

######  Theorem 2.1.

$D$ is similar to a symmetric matrix and has exactly one positive eigenvalue.

###### Proof.

If all the diagonal entries of $D$ are zero, then $D$ is a Euclidean distance
matrix and in this case, the result is known. Now assume that $D$ has at least
one positive diagonal entry. Let the eigenvalues of $L$ be arranged

$\alpha_{1}\leq\alpha_{2}\leq\cdots\leq\alpha_{n}.$

Since $L$ is positive semidefinite and $L\mathbf{1}=0$, $\alpha_{1}=0$. Let
$U$ be an orthogonal matrix with first column equal to the unit vector
$\frac{1}{\sqrt{n}}\mathbf{1}$ and such that

$U^{\prime}LU=\mathop{\rm Diag}(0,\alpha_{2},\dotsc,\alpha_{n}).$

Define

$(s_{1},s_{2},\dotsc,s_{n})^{\prime}:=U^{\prime}\widetilde{L}JUe_{1}.$

By an easy verification,

$s_{1}={\rm trace}(L).$

Furthermore,

$U^{\prime}\widetilde{L}JUe_{i}=0~{}~{}~{}~{}i=2,\dotsc,n.$

Thus,

$\displaystyle U^{\prime}DU$
$\displaystyle=a^{2}U^{\prime}\widetilde{L}JU+b^{2}U^{\prime}J\widetilde{L}U-2abU^{\prime}LU$
(6)
$\displaystyle=\left[\begin{array}[]{ccccc}(a^{2}+b^{2})s_{1}&b^{2}s_{2}&\ldots&b^{2}s_{n}\\\
a^{2}s_{2}&-2ab\alpha_{2}&\ldots&0\\\ \vdots&\vdots&\ddots&\vdots\\\
a^{2}s_{n}&0&\ldots&-2ab\alpha_{n}\end{array}\right].$

Define

$W:=\mbox{Diag}(\frac{\displaystyle b}{\displaystyle a},1,\dotsc,1).$

Now,

$W^{-1}U^{\prime}DUW=\begin{array}[]{ccccc}\left[\begin{array}[]{ccccc}(a^{2}+b^{2})s_{1}&abs_{2}&\ldots&abs_{n}\\\
abs_{2}&-2ab\alpha_{2}&\ldots&0\\\ \vdots&\vdots&\ddots&\vdots\\\
abs_{n}&0&\ldots&-2ab\alpha_{n}\end{array}\right];\end{array}$ (7)

hence $D$ is similar to a symmetric matrix. By interlacing theorem,
$W^{-1}U^{\prime}DUW$ has $n-1$ non-positive eigenvalues. Since ${\rm
trace}(D)>0$, we see that

$(W^{-1}U^{\prime}DUW)_{11}>0.$

Hence, $W^{-1}U^{\prime}DUW$ has at least one positive eigenvalue. So, $D$ has
exactly one positive eigenvalue. The proof is now complete. ∎

In the rest of the paper, we shall use $W$ and $U$ to denote the matrices
defined in Theorem 2.1. We have the following corollary now.

######  Corollary 2.2.

All the eigenvalues of the bordered matrix

$\widetilde{D}:=\left[\begin{array}[]{cccc}D&\mathbf{1}\\\
\mathbf{1}^{\prime}&0\end{array}\right]$

are real and $\widetilde{D}$ has exactly one positive eigenvalue.

###### Proof.

Setting

$K:=\left[\begin{array}[]{cccc}U&0\\\ 0&1\\\
\end{array}\right]\left[\begin{array}[]{cccc}W&0\\\ 0&1\\\
\end{array}\right],$

we find that the bordered matrix

$\widetilde{D}:=\left[\begin{array}[]{cccccc}D&\mathbf{1}\\\
\mathbf{1}^{\prime}&0\end{array}\right]$

is similar to

$Q:=K^{-1}\widetilde{D}K=\left[\begin{array}[]{cccccc}W^{-1}U^{\prime}DUW&W^{-1}U^{\prime}\mathbf{1}\\\
\mathbf{1}^{\prime}UW&0\end{array}\right].$ (8)

Since $U$ is an orthogonal matrix with first column equal to
$\frac{1}{\displaystyle\sqrt{n}}\mathbf{1}$,

$\displaystyle W^{-1}U^{\prime}\mathbf{1}$ $\displaystyle=\mathop{\rm
Diag}(\frac{a}{b},1,\dotsc,1)(\sqrt{n}e_{1})$ (9)
$\displaystyle=(\sqrt{n}\frac{a}{b},0,\dotsc,0)^{\prime}$
$\displaystyle=\sqrt{n}\frac{a}{b}e_{1}.$

Put

$\delta:=\sqrt{n}\dfrac{a}{b}.$

By $(\ref{m4})$ and $(\ref{m5})$,

$\displaystyle Q$
$\displaystyle=\left[\begin{array}[]{ccccc}W^{-1}U^{\prime}DUW&\delta e_{1}\\\
\frac{1}{\delta}e_{1}^{\prime}&0\\\ \end{array}\right]$ (10)
$\displaystyle=\left[\begin{array}[]{ccccc}(a^{2}+b^{2})s_{1}&abs_{2}&\ldots&abs_{n}&\delta\\\
abs_{2}&-2ab\alpha_{2}&\ldots&0&0\\\ \vdots&\vdots&\ddots&\vdots&\vdots\\\
abs_{n}&0&\ldots&-2ab\alpha_{n}&0\\\
\frac{1}{\displaystyle\delta}&0&0\ldots&0&0\end{array}\right]$

Define

$G:=\mbox{Diag}(I,\frac{1}{\delta}),$

where $I$ is the $n\times n$ identity matrix. Now,

$\displaystyle G^{-1}QG$ $\displaystyle=\left[\begin{array}[]{ccccc}I&0\\\
0&\delta\\\
\end{array}\right]\left[\begin{array}[]{ccccc}W^{-1}U^{\prime}DUW&\delta
e_{1}\\\ \frac{1}{\delta}e_{1}^{\prime}&0\\\
\end{array}\right]\left[\begin{array}[]{ccccc}I&0\\\ 0&\frac{1}{\delta}\\\
\end{array}\right]$ (11)
$\displaystyle=\left[\begin{array}[]{ccccc}W^{-1}U^{\prime}DUW&e_{1}\\\
e_{1}^{\prime}&0\\\ \end{array}\right].$

As $W^{-1}U^{\prime}DUW$ is symmetric (see $(\ref{m4})$), $G^{-1}QG$ is
symmetric. Thus, $\widetilde{D}$ is similar to a symmetric matrix. Therefore,
$\widetilde{D}$ has only real eigenvalues.

We now claim that $\widetilde{D}$ has exactly one positive eigenvalue. Let
$x=(x_{1},\dotsc,x_{n})^{\prime}$ be orthogonal to $e_{1}$. Because $x_{1}=0$,
from $(\ref{m4})$, we have

$x^{\prime}W^{-1}U^{\prime}DUWx=-2ab\sum_{i=2}^{n}x_{i}^{2}\alpha_{i}\leq 0.$
(12)

The subspace

$\nabla:=\\{(x,r):x^{\prime}e_{1}=0,~{}r\in\mathbb{R}\\}$

of $\mathbb{R}^{n+1}$ has dimension $n$. Consider a vector
$w:=(x,r)^{\prime}\in\nabla$. By $(\ref{GQG})$ and $(\ref{eperp})$,

$\displaystyle w^{\prime}G^{-1}QGw$
$\displaystyle=x^{\prime}W^{-1}U^{\prime}DUWx+re_{1}^{\prime}x$
$\displaystyle=x^{\prime}W^{-1}U^{\prime}DUWx\leq 0.$

Hence, $G^{-1}QG$ has at least $n$ non-positive eigenvalues. Because $D$ has
exactly one positive eigenvalue, $G^{-1}QG$ has at least one positive
eigenvalue. So, $G^{-1}QG$ has exactly one positive eigenvalue. Because
$\widetilde{D}$ and $G^{-1}QG$ are similar, $\widetilde{D}$ has exactly one
positive eigenvalue. The result is proved. ∎

###  2.2 Sign pattern of $(\rho(D)I-D)^{r}$

Let the eigenvalues of $D$ be $\delta_{1},\dotsc,\delta_{n}$. Since $D$ is a
nonnegative matrix and has exactly one positive eigenvalue, $\rho(D)$ is the
only positive eigenvalue of $D$. Therefore,

$\gamma_{i}:=\rho(D)-\delta_{i}\geq 0~{}~{}~{}i=1,\dotsc,n.$

Now, let $A$ be an invertible matrix such that

$A^{-1}DA=\mathop{\rm Diag}(\delta_{1},\dotsc,\delta_{n}).$

Define

$S:=\rho(D)I-D.$

Then, for any $r>0$

$S^{r}:=A\mathop{\rm Diag}(\gamma_{1}^{r},\dotsc,\gamma_{n}^{r})A^{-1}.$

As a consequence of Theorem 2.1, we now have the following result.

######  Corollary 2.3.

If $0<r<1$, then $S^{r}$ is an ${\bf M}$-matrix.

###### Proof.

Fix $0<r<1$. Since $S$ is an ${\bf M}$-matrix, by Theorem $3.1$ in [1],
$S^{r}$ is an ${\bf M}$-matrix. The proof is complete. ∎

To illustrate, we give the following example.

######  Example 2.4.

Consider the matrix $D$ given in Example 1.4. The eigenvalues of $D$ are
approximately

$-24,-36.1322~{}\mbox{and}~{}100.1322.$

Now the matrix $S$ is

$S=\rho(D)I-D=\left[\begin{array}[]{ccc}88.1322&-36&-51\\\ -36&88.1322&-51\\\
-43&-43&84.1322\end{array}\right].$

Then

$S^{\frac{1}{2}}=\left[\begin{array}[]{ccc}7.8037&-3.3378&-4.3690\\\
-3.3378&7.8037&-4.3690\\\ -3.6836&-3.6836&7.2073\end{array}\right].$

It is easy to see that $S^{\frac{1}{2}}$ is an ${\bf M}$-matrix.

###  2.3 Null space of a $\mathop{\rm GEDM}$

The main result about the null space of a $\mathop{\rm GEDM}$ is $\mathop{\rm
null}(D)\subseteq\mathbf{1}^{\perp}$; so $\mathbf{1}\in\mathop{\rm col}(D)$
and this in turn will be useful to investigate the Moore-Penorse inverse of
$D$.

######  Theorem 2.5.

If $D$ is a $\mathop{\rm GEDM}$, then $\mathbf{1}\in\mathop{\rm null}(D)$.

###### Proof.

Following same notation as in Theorem 2.1, we have by equation $(\ref{pdp})$,

$\Delta:=U^{\prime}DU=\left[\begin{array}[]{ccccc}(a^{2}+b^{2})s_{1}&b^{2}s_{2}&\ldots&b^{2}s_{n}\\\
a^{2}s_{2}&-2ab\alpha_{2}&\ldots&0\\\ \vdots&\vdots&\ddots&\vdots\\\
a^{2}s_{n}&0&\ldots&-2ab\alpha_{n}\end{array}\right].$

As $\Delta_{11}=(a^{2}+b^{2})s_{1}$ and $s_{1}={\rm trace}(L)$,
$\Delta_{11}>0$. Let $Dx=0$. We claim that $\mathbf{1}^{\prime}x=0$. Since
$U\Delta U^{\prime}=D$ and $Dx=0$, we have

$\Delta U^{\prime}x=0.$ (13)

If $v:=(s_{2},s_{3},\dotsc,s_{n})^{\prime}$ and $S:=\mathop{\rm
Diag}(2ab\lambda_{2},\dotsc,2ab\lambda_{n})$, then

$\Delta=\left[\begin{array}[]{cccc}\Delta_{11}&b^{2}v^{\prime}\\\
a^{2}v&-S\end{array}\right].$

Put $y:=U^{\prime}x$. By writing $y=(y_{1},\bar{y})^{\prime}$, where
$y_{1}\in\mathbb{R}$ and $\bar{y}^{\prime}\in\mathbb{R}^{n-1}$, from
$(\ref{Delta})$, we have

$y_{1}\Delta_{11}+b^{2}v^{\prime}\bar{y}=0$ (14) $y_{1}a^{2}v=S\bar{y}.$ (15)

If possible, let $y_{1}\neq 0$. Then from $(\ref{e22})$ we have,

$v=\frac{1}{a^{2}y_{1}}S\bar{y}.$

Thus by equation $(\ref{e11})$,

$y_{1}\Delta_{11}+\frac{b^{2}}{a^{2}y_{1}}\bar{y}^{\prime}S\bar{y}=0,$

or equivalently,

$y_{1}^{2}\Delta_{11}+\frac{b^{2}}{a^{2}}\bar{y}^{\prime}S\bar{y}=0.$ (16)

As $S$ is positive semidefinite, $\bar{y}^{\prime}S\bar{y}$ is nonnegative.
Because $\Delta_{11}>0$,

$y_{1}^{2}\Delta_{11}+\frac{b^{2}}{a^{2}}\bar{y}^{\prime}S\bar{y}>0,$

contradicting $(\ref{ss1})$. Hence, $y_{1}=0$. Since $Uy=x$ and
$y=(0,y_{2},\dotsc,y_{n})^{\prime}$,

$x\in\mathop{\rm span}\\{Ue_{2},\dotsc,Ue_{n}\\}=\mathbf{1}^{\perp}.$

So, $\mathbf{1}^{\prime}x=0$. The proof is complete. ∎

######  Corollary 2.6.

$\mathbf{1}\in\mathop{\rm null}(D^{\prime})$.

###### Proof.

As $D^{\prime}$ is also a $\mathop{\rm GEDM}$, $\mathbf{1}\in\mathop{\rm
null}(D^{\prime})$. ∎

######  Corollary 2.7.

$\mathbf{1}\in\mathop{\rm col}(D)\cap\mathop{\rm col}(D^{\prime})$.

###### Proof.

This follows easily since $\mathop{\rm null}(D)=\mathop{\rm
col}(D^{\prime})^{\perp}$. ∎

######  Corollary 2.8.

$\mathbf{1}^{\prime}D^{{\dagger}}\mathbf{1}=0$ if and only if there exists
$f\in\mathbf{1}^{\perp}$ such that $Df=\mathbf{1}$.

###### Proof.

Suppose $\mathbf{1}^{\prime}D^{{\dagger}}\mathbf{1}=0$. Let $Dx=\mathbf{1}$.
Since $DD^{{\dagger}}\mathbf{1}=\mathbf{1}$, we have
$x-D^{{\dagger}}\mathbf{1}\in\mathop{\rm null}(D)$ and because $\mathop{\rm
null}(D)\subseteq\mathbf{1}^{\perp}$,
$\mathbf{1}^{\prime}(x-D^{{\dagger}})\mathbf{1}=0$. So,
$(\mathbf{1}^{\prime}x)\mathbf{1}=0$; hence $\mathbf{1}^{\prime}x=0$.
Conversely, let $Df=\mathbf{1}$ and $\mathbf{1}^{\prime}f=0$. Since,
$DD^{{\dagger}}\mathbf{1}=\mathbf{1}$ and $\mathop{\rm
null}(D)\subseteq\mathbf{1}^{\perp}$,
$f-D^{\dagger}\mathbf{1}\in\mathbf{1}^{\perp}$. Thus,
$\mathbf{1}^{\prime}D^{\dagger}\mathbf{1}=0$. ∎

######  Corollary 2.9.

The following are equivalent.

  1. (a)

$\mathbf{1}^{\prime}D^{{\dagger}}\mathbf{1}\neq 0$.

  2. (b)

$\mathop{\rm null}(D)=\mathop{\rm null}(D^{\prime})=\mathop{\rm
null}(L)\cap\mathbf{1}^{\perp}$.

###### Proof.

Assume (a). To prove (b), it suffices to show that

$\mathop{\rm null}(D)=\mathop{\rm null}(L)\cap\mathbf{1}^{\perp}.$

Let $x\in\mathop{\rm null}(D)$. By Theorem 2.5, $x\in\mathbf{1}^{\perp}$.
Since

$\displaystyle x^{\prime}Dx$
$\displaystyle=a^{2}x^{\prime}\widetilde{L}Jx+b^{2}x^{\prime}J\widetilde{L}x-(2ab)x^{\prime}Lx$
$\displaystyle=-2abx^{\prime}Lx=0.$

Because $x^{\prime}Lx=0$ and $L$ is positive semidefinite, $Lx=0$. So,
$x\in\mathop{\rm null}(L)$. Thus,

$\mathop{\rm null}(D)\subseteq\mathop{\rm null}(L)\cap\mathbf{1}^{\perp}.$

Now, let $f\in\mathop{\rm null}(L)\cap\mathbf{1}^{\perp}$. Then,

$\displaystyle Df$
$\displaystyle=a^{2}\widetilde{L}Jf+b^{2}J\widetilde{L}f-2abLf$ (17)
$\displaystyle=b^{2}J\widetilde{L}f=b^{2}\mathbf{1}1^{\prime}\widetilde{L}f=b^{2}(\mathbf{1}^{\prime}\widetilde{L}f)\mathbf{1}.$

From Corollary 2.8, we find that $\mathbf{1}^{\prime}\widetilde{L}f=0$. Thus,
$f\in\mathop{\rm null}(D)$. This proves (a) $\Rightarrow$ (b).

Assume (b). If $\mathbf{1}^{\prime}D^{{\dagger}}\mathbf{1}=0$, then by
Corollary 2.8, there exists $f\in\mathbf{1}^{\perp}$ such that
$Df=\mathbf{1}$. Therefore, $f^{\prime}Df=0$ and this gives $f^{\prime}Lf=0$.
Since $L$ is positive semidefinite, $Lf=0$ and therefore by our assumption,
$f\in\mathop{\rm null}(D)$ contradicting $Df=\mathbf{1}$. Thus (b)
$\Rightarrow$ (a). The proof is complete. ∎

######  Corollary 2.10.

$\mathop{\rm null}(D)=\mathop{\rm null}(\widetilde{L}J+J\widetilde{L}-2L)$.

###### Proof.

Put

$E:=\widetilde{L}J+J\widetilde{L}-2L.$

Let $x\in\mathop{\rm null}(D)$. Then by Theorem 2.5, $x\in\mathbf{1}^{\perp}$.
So,

$\displaystyle 0=Dx$
$\displaystyle=a^{2}\widetilde{L}Jx+b^{2}J\widetilde{L}x-2abLx$ (18)
$\displaystyle=b^{2}J\widetilde{L}x-2abLx$
$\displaystyle=b^{2}\mathbf{1}\mathbf{1}^{\prime}\widetilde{L}x-2abLx.$

Because $Lx$ is an element in $\mathbf{1}^{\perp}$, we get by $(\ref{dx})$,

$\mathbf{1}^{\prime}\widetilde{L}x=0~{}~{}\mbox{and}~{}~{}Lx=0.$ (19)

On the other hand, we have

$\displaystyle Ex=\widetilde{L}Jx+J\widetilde{L}x-2Lx$
$\displaystyle=J\widetilde{L}x-2Lx$
$\displaystyle=\mathbf{1}\mathbf{1}^{\prime}\widetilde{L}x-2Lx.$

By $(\ref{1'Lx})$, it now follows that $Ex=0$. Thus, $x\in\mathop{\rm
null}(E)$. So,

$\mathop{\rm null}(D)\subseteq\mathop{\rm null}(E).$

Similar argument leads to $\mathop{\rm null}(E)\subseteq\mathop{\rm null}(D)$.
The proof is complete. ∎

###  2.4 Moore-Penrose inverse of a $\mathop{\rm GEDM}$

We now obtain some properties of the Moore-Penrose inverse of $D$. The
following result says that $\mathbf{1}^{\prime}D^{-}\mathbf{1}$ is invariant
for any choice of $g$-inverse of $D$.

######  Theorem 2.11.

If $D$ is a $\mathop{\rm GEDM}$, and if $D^{-}$ is a generalized inverse of
$D$, then
$\mathbf{1}^{\prime}D^{-}\mathbf{1}=\mathbf{1}D^{\dagger}\mathbf{1}$.

###### Proof.

As $\mathbf{1}$ is an element in the column space of $D$ and $D^{\prime}$, we
have

$\mathbf{1}^{\prime}DD^{\dagger}=\mathbf{1}^{\prime}~{}~{}\mbox{and}~{}~{}DD^{{\dagger}}\mathbf{1}=\mathbf{1}.$

So,

$\mathbf{1}^{\prime}D^{-}\mathbf{1}=\mathbf{1}^{\prime}D^{\dagger}DD^{-}DD^{{\dagger}}\mathbf{1}.$

As $DD^{-}D=D,$ we get

$\mathbf{1}^{\prime}D^{-}\mathbf{1}=\mathbf{1}^{\prime}D^{\dagger}DD^{\dagger}\mathbf{1}=\mathbf{1}^{\prime}D^{\dagger}\mathbf{1}.$

∎

######  Theorem 2.12.

$1^{\prime}D^{{\dagger}}\mathbf{1}\geq 0$.

###### Proof.

In view of equation (11), $\widetilde{D}$ is similar to

$S:=\left[\begin{array}[]{cccccc}W^{-1}U^{\prime}DUW&e_{1}\\\
e_{1}^{\prime}&0\end{array}\right],$ (20)

which is symmetric. We claim the following.
Claim: There exists $x\in\mathbb{R}^{n}$ such that
$W^{-1}U^{\prime}DUWx=e_{1}$.

Since $UWe_{1}\in\mathop{\rm span}\\{\mathbf{1}\\}$ and $\mathbf{1}$ is in the
column space of $D$, there exists $y\in\mathbb{R}^{n}$ such that $Dy=UWe_{1}$.
As $U$ and $W$ are non-singular, $y=UWx$ for some $x\in\mathbb{R}^{n}$. Thus,
$DUWx=UWe_{1}$ and the claim is true.

Applying inertia formula to $S$ (see (N7)), we have

$\mathop{\rm In}(S)=\mathop{\rm In}(W^{-1}U^{\prime}DUW)+\mathop{\rm
In}(-e_{1}^{\prime}W^{-1}U^{\prime}D^{{\dagger}}UWe_{1}).$

By corollary 2.2, $\widetilde{D}$ has exactly one positive eigenvalue. Since
$W^{-1}U^{\prime}DUW$ has exactly one positive eigenvalue, by the above
formula,

$-e_{1}^{\prime}W^{-1}U^{\prime}D^{{\dagger}}UWe_{1}\leq 0.$

From the definition of $U$ and $W$, we see that

$-e_{1}^{\prime}W^{-1}U^{\prime}D^{\dagger}UWe_{1}=-\mathbf{1}^{\prime}D^{{\dagger}}\mathbf{1},$

and hence $\mathbf{1}^{\prime}D^{\dagger}\mathbf{1}\geq 0$. This proves the
result. ∎

######  Theorem 2.13.

$-PD^{\dagger}P$ is positive semidefinite.

###### Proof.

We first claim that $PD^{\dagger}P$ is symmetric. Let $U$ be the orthogonal
matrix given in Theorem 2.1. Define

$f_{i}:=Ue_{i}~{}~{}~{}i=2,\dotsc,n.$ (21)

Each $f_{i}\in\mathbf{1}^{\perp}$. To complete the proof of the claim, it
suffices to show that

$f_{i}^{\prime}D^{\dagger}f_{j}=f_{j}^{\prime}D^{\dagger}f_{i}~{}~{}~{}i,j=2,\dotsc,n.$

Fix $i\neq j$ and $i,j\geq 2$. We know from Theorem 2.1 that
$W^{-1}U^{\prime}D^{\dagger}UW$ is symmetric. Hence,

$e_{i}^{\prime}W^{-1}U^{\prime}D^{\dagger}UWe_{j}=e_{j}^{\prime}W^{-1}U^{\prime}D^{\dagger}UWe_{i}.$
(22)

Since

$e_{i}^{\prime}W^{-1}=e_{i}^{\prime}~{}~{}\mbox{and}~{}~{}We_{j}=e_{j}~{}~{}~{}\mbox{for
all}~{}i,j\geq 2,$

we have

$e_{i}^{\prime}W^{-1}U^{\prime}D^{\dagger}UWe_{j}=e_{i}^{\prime}U^{\prime}D^{\dagger}Ue_{j};$

hence by $(\ref{f})$,

$e_{i}^{\prime}W^{-1}U^{\prime}D^{\dagger}UWe_{j}=f_{i}^{\prime}D^{\dagger}f_{j}.$

By a similar reasoning,

$e_{j}^{\prime}W^{-1}U^{\prime}D^{\dagger}UWe_{i}=f_{j}^{\prime}D^{\dagger}f_{i}.$

In view of $(\ref{equal})$,

$f_{i}^{\prime}D^{\dagger}f_{j}=f_{j}^{\prime}D^{\dagger}f_{i}.$

Thus, $PD^{\dagger}P$ is symmetric.

Consider the bordered matrix

$\widetilde{S}:=\left[\begin{array}[]{cccccc}W^{-1}U^{\prime}D^{{\dagger}}UW&e_{1}\\\
e_{1}^{\prime}&0\end{array}\right].$

Because $\mathbf{1}\in\mathop{\rm col}(D)$, $e_{1}\in\mathop{\rm
col}(W^{-1}U^{\prime}D^{{\dagger}}UW)$. Since $W^{-1}U^{\prime}DUW$ is
symmetric, $(W^{-1}U^{\prime}DUW)^{\dagger}$ is symmetric as well. Because

$(W^{-1}U^{\prime}DUW)^{\dagger}=W^{-1}U^{\prime}D^{\dagger}UW,$

$\widetilde{S}$ is symmetric. By inertia formula in (N7), we have

$\mathop{\rm In}(-e_{1}^{\prime}W^{-1}U^{\prime}DUWe_{1})+\mathop{\rm
In}(W^{-1}U^{\prime}D^{\dagger}UW)=\mathop{\rm In}(\widetilde{S}).$ (23)

By an easy computation, we see that

$e_{1}^{\prime}W^{-1}U^{\prime}DUWe_{1}=\frac{1}{n}\mathbf{1}^{\prime}D\mathbf{1}.$

So,

$\mathop{\rm In}(-e_{1}^{\prime}W^{-1}U^{\prime}DUWe_{1})=(1,0,0).$

Because $W^{-1}U^{\prime}D^{\dagger}UW$ has exactly one positive eigenvalue,
it follows from $(\ref{inertiafo})$ that $\widetilde{S}$ has exactly one
positive eigenvalue. By (N9),

$x\in e_{1}^{\perp}\implies x^{\prime}W^{-1}U^{\prime}D^{{\dagger}}UWx\leq 0.$

Specializing

$x=e_{i}~{}~{}~{}i=2,\dotsc,n$

in the above inequality leads to

$f_{i}^{\prime}D^{{\dagger}}f_{i}\leq 0~{}~{}i=2,\dotsc,n.$

This proves $-PD^{\dagger}P$ is positive semidefinite. The proof is complete.
∎

######  Corollary 2.14.

$D^{\dagger}$ is negative semidefinite on $\mathbf{1}^{\perp}$.

###### Proof.

Let $x\in\mathbf{1}^{\perp}$. Then,
$x^{\prime}D^{\dagger}x=x^{\prime}PD^{\dagger}Px$; hence by Theorem 2.13,
$x^{\prime}D^{\dagger}x\leq 0$. ∎

######  Theorem 2.15.

Let $E=\widetilde{L}J+J\widetilde{L}-2L$. Then,

$\mathbf{1}^{\prime}D^{{\dagger}}\mathbf{1}>0~{}~{}\mbox{if and only
if}~{}~{}\mathbf{1}^{\prime}E^{{\dagger}}\mathbf{1}>0.$

###### Proof.

Suppose $\mathbf{1}^{\prime}D^{\dagger}\mathbf{1}>0$. If possible, let
$\mathbf{1}^{\prime}E^{\dagger}\mathbf{1}=0$. We shall now get a
contradiction. By Corollary 2.8, there exists $f\in\mathbf{1}^{\perp}$ such
that

$Ef=\mathbf{1}.$ (24)

As $f^{\prime}Ef=0$, we have $f^{\prime}Lf=0$; so $Lf=0$. We now have

$\displaystyle Df$
$\displaystyle=a^{2}\widetilde{L}Jf+b^{2}J\widetilde{L}f-2abLf$ (25)
$\displaystyle=b^{2}\mathbf{1}\mathbf{1}^{\prime}\widetilde{L}f=b^{2}(\mathbf{1}^{\prime}\widetilde{L}f)\mathbf{1}.$

Since $\mathbf{1}^{\prime}D^{\dagger}\mathbf{1}>0$, from (25) and Corollary
2.8, we find that $\mathbf{1}^{\prime}\widetilde{L}f=0$. Thus,
$f\in\mathop{\rm null}(D)$. By Corollary 2.10, $f\in\mathop{\rm null}(E)$.
This contradicts $(\ref{ealpha})$. So,
$\mathbf{1}^{\prime}E^{\dagger}\mathbf{1}>0$.

By using a similar argument, we get the reverse implication. ∎

###  2.5 Generalized Circum Euclidean distance matrix

Suppose $p^{1},\dotsc,p^{n}$ are some vectors in a Euclidean space
$(V,\langle\cdot,\cdot\rangle)$. If there exists a vector $v\in V$ and $r>0$
such that

$\langle p^{i}-v,p^{i}-v\rangle=\|p^{i}-v\|^{2}=r~{}~{}\mbox{for
all}~{}i=1,\dotsc,n,$

then we say that $p^{1},\dotsc,p^{n}$ lie on the surface of a hypersphere. Now
the Euclidean distance matrix $[\|p^{i}-p^{j}\|^{2}]$ is called a circum
$\mathop{\rm EDM}$. A well-known result ([10, Theorem $3.4$]) characterizes
all circum $\mathop{\rm EDM}$s. This says that $E$ is a circum $\mathop{\rm
EDM}$ if and if there exists a vector $s$ and a scalar $\beta$ such that

$Es=\beta\mathbf{1}~{}~{}\mbox{and}~{}~{}s^{\prime}\mathbf{1}=1.$

This is equivalent to saying that $E$ is a circum $\mathop{\rm EDM}$ if and
only if $\mathbf{1}^{\prime}E^{\dagger}\mathbf{1}>0$. We now introduce the
following definition.

######  Definition 2.16.

We say that $D=[d_{ij}]$ is a circum $\mathop{\rm GEDM}$ if there exist
$a,b>0$ and vectors $x^{1},\dotsc,x^{n}$ on the surface of a hypersphere such
that

$d_{ij}=\|ax^{i}-bx^{j}\|^{2},$

where $\sum_{i=1}^{n}x^{i}=0$.

We now have the following result for $\mathop{\rm GEDM}$s.

######  Theorem 2.17.

The following are equivalent.

  1. (1)

$D$ is a circum $\mathop{\rm GEDM}$.

  2. (2)

$\mathbf{1}^{\prime}D^{{\dagger}}\mathbf{1}>0$.

###### Proof.

As $D=[d_{ij}]$ is a $\mathop{\rm GEDM}$, there exist vectors
$x^{1},\dotsc,x^{n}$ such that

$\sum_{j=1}^{n}x^{i}=0~{}~{}\mbox{and}~{}~{}d_{ij}=\|ax^{i}-bx^{j}\|^{2}.$

Define

$e_{ij}:=[\|x^{i}-x^{j}\|^{2}]~{}~{}\mbox{and}~{}~{}E:=[e_{ij}].$

Assume (1). Then, $x^{1},\dotsc,x^{n}$ lie on the surface of a hypersphere.
So, $E$ is a circum $\mathop{\rm EDM}$ and hence
$\mathbf{1}^{\prime}E^{\dagger}\mathbf{1}>0$. By Theorem 2.15,
$\mathbf{1}^{\prime}D^{\dagger}\mathbf{1}>0$. This proves (2).

Assume (2). Then by Theorem 2.15,
$\mathbf{1}^{\prime}E^{\dagger}\mathbf{1}>0$. So, $x^{1},\dotsc,x^{n}$ lie on
the surface of a hypersphere. Hence $D$ is a circum $\mathop{\rm GEDM}$. ∎

######  Corollary 2.18.

If $D$ is a circum $\mathop{\rm GEDM}$, then $\mathop{\rm rank}(D)=\mathop{\rm
rank}(L)+1$; otherwise $\mathop{\rm rank}(D)=\mathop{\rm rank}(L)+2$.

###### Proof.

Let $E=\widetilde{L}J+J\widetilde{L}-2L$. By Corollary 2.15, $\mathop{\rm
rank}(D)=\mathop{\rm rank}(E)$. By the previous result $D$ is a circum
$\mathop{\rm GEDM}$ if and only if
$\mathbf{1}^{\prime}D^{\dagger}\mathbf{1}>0$. By Theorem 2.15,
$\mathbf{1}^{\prime}E^{\dagger}\mathbf{1}>0$ if and only if
$\mathbf{1}^{\prime}D^{\dagger}\mathbf{1}>0$. In view of Proposition $1$ in
[8], we get

$\mathop{\rm rank}(E)=\begin{cases}\mathop{\rm
rank}(L)+1&\mbox{if}~{}\mathbf{1}^{\prime}E^{\dagger}\mathbf{1}>0\\\
\mathop{\rm rank}(L)+2&\mbox{else}.\end{cases}$

The proof now follows easily. ∎

We now obtain a formula to compute the Moore-Penrose inverse of a circum
$\mathop{\rm GEDM}$.

######  Theorem 2.19.

If $D$ is a circum $\mathop{\rm GEDM}$, then

$D^{{\dagger}}=-\frac{1}{2ab}L^{{\dagger}}+\frac{1}{\mathbf{1}^{\prime}D^{{\dagger}}\mathbf{1}}(D^{{\dagger}}\mathbf{1})(\mathbf{1}^{\prime}D^{{\dagger}}).$

###### Proof.

Let

$\alpha:=\mathbf{1}^{\prime}D^{{\dagger}}\mathbf{1},~{}S:=D^{{\dagger}}-\frac{\displaystyle
1}{\displaystyle\alpha}(D^{{\dagger}}\mathbf{1})(\mathbf{1}^{\prime}D^{{\dagger}})~{}~{}\mbox{and}~{}K:=PDP.$

We now prove that $SKS=S$. Let $x\in\mathbb{R}^{n}$. Then,

$x=c_{1}\mathbf{1}+c_{2}f,$

for some $c_{1},c_{2}\in\mathbb{R}$ and $f\in\\{\mathbf{1}\\}^{\perp}$. Since
$S\mathbf{1}=0$, we see that

$Sx=c_{2}Sf.$

As $P\mathbf{1}=0$ and $Pf=f$,

$SPx=c_{2}Sf.$

Therefore, $SP=S$. In a similar manner, by using $\mathbf{1}^{\prime}S=0$ and
$\mathbf{1}^{\prime}P=0$, we get $PS=S$. Thus, to prove $SKS=S$, it suffices
to show that $SDS=S$. As $\mathbf{1}^{\prime}S=0$, we note that

$\displaystyle SDS$
$\displaystyle=(D^{{\dagger}}-\frac{1}{\alpha}(D^{{\dagger}}\mathbf{1})(\mathbf{1}^{\prime}D^{{\dagger}}))DS$
(26) $\displaystyle=D^{{\dagger}}DS$
$\displaystyle=D^{{\dagger}}D(D^{{\dagger}}-\frac{1}{\alpha}(D^{{\dagger}}\mathbf{1})(\mathbf{1}^{\prime}D^{{\dagger}}))$
$\displaystyle=S.$

Hence, $SKS=S$. We claim that $KSK=K.$ Again by using $PS=SP=S$ and
$P\mathbf{1}=0$, we see that

$\displaystyle KSK$ $\displaystyle=(PDP)S(PDP)$ (27)
$\displaystyle=(PDS)(PDP)$ $\displaystyle=PDSDP$
$\displaystyle=PD(D^{{\dagger}}-\frac{1}{\alpha}D^{{\dagger}}\mathbf{1}\mathbf{1}^{\prime}D^{{\dagger}})DP$
$\displaystyle=PDP$ $\displaystyle=K.$

Since $SK=KS=P$, we conclude that $K$ is the Moore-Penrose inverse of $S$.
Thus,

$(PDP)^{{\dagger}}=S.$

Since $L\mathbf{1}=0$ and $L$ is symmetric, $PL=LP=L$ and hence

$\displaystyle PDP$ $\displaystyle=-2abPLP$ (28) $\displaystyle=-2abL.$

So,

$(PDP)^{{\dagger}}=-\frac{1}{2ab}L^{{\dagger}}.$

Thus,

$S=(PDP)^{{\dagger}}=D^{{\dagger}}-\frac{1}{\alpha}(D^{{\dagger}}\mathbf{1})(\mathbf{1}^{\prime}D^{{\dagger}}).$

This completes the proof of the formula

$D^{{\dagger}}=-\frac{1}{2ab}L^{{\dagger}}+\frac{1}{\mathbf{1}^{\prime}D^{{\dagger}}\mathbf{1}}(D^{{\dagger}}\mathbf{1})(\mathbf{1}^{\prime}D^{{\dagger}}).$

∎

###  2.6 Some majorization results

Suppose all the eigenvalues of an $n\times n$ matrix $A$ are real. Let the
eigenvalues of $A$ be $\lambda_{1}(A),\dotsc,\lambda_{n}(A)$, where

$\lambda_{1}(A)\geq\cdots\geq\lambda_{n}(A).$

We now use $\lambda(A)$ to denote the vector
$(\lambda_{1}(A),\dotsc,\lambda_{n}(A))^{\prime}$. In the following, we obtain
a Schur-type majorization result for $\mathop{\rm GEDM}$s.

######  Theorem 2.20.

$\mathop{\rm diag}(D)$ is majorized by $\lambda(D)$.

###### Proof.

In view of (6),

$U^{\prime}DU=\left[\begin{array}[]{cccc}(a^{2}+b^{2}){\rm trace}(L)&b^{2}s\\\
a^{2}s&-2ab\mbox{Diag}(\alpha_{2},\dotsc,\alpha_{n})\end{array}\right].$ (29)

Define

$x:=((a^{2}+b^{2}){\rm
trace}(L),-2ab\alpha_{2},\dotsc,-2ab\alpha_{n})^{\prime}.$

Without loss of generality, we can assume that

$-2ab\alpha_{2}\geq\cdots\geq-2ab\alpha_{n}~{}~{}\mbox{and}~{}~{}-l_{11}\geq\cdots\geq-
l_{nn}.$

We now prove the following.
Claim: $(a-b)^{2}\mathop{\rm diag}(L)\prec x.$
Set $\alpha_{1}=0$. By the majorization result of Schur,

$-2ab\mathop{\rm diag}(L)\prec-2ab\lambda(L).$

So, for each $1\leq k\leq n-1$,

$-2ab\sum_{i=1}^{k}l_{ii}\leq-2ab\sum_{i=1}^{k}\alpha_{k}.$ (30)

As $L$ is positive semidefinite, for each $1\leq k\leq n-1$,

$(a^{2}+b^{2})\sum_{i=1}^{k}l_{ii}\leq(a^{2}+b^{2}){\rm trace}(L).$ (31)

Using $(\ref{lii})$ and (31), for each $1\leq k\leq n-1$, we find that

$(a-b)^{2}\sum_{i=1}^{k}l_{ii}\leq(a^{2}+b^{2}){\rm
trace}(L)-2ab\sum_{i=1}^{k}\alpha_{i}.$

Furthermore,

$(a-b)^{2}\sum_{i=1}^{n}l_{ii}=(a^{2}+b^{2}){\rm
trace}(L)-2ab\sum_{i=1}^{n}\alpha_{i}.$

Hence,

$(a-b)^{2}\mathop{\rm diag}(L)\prec x.$

This proves the claim.

By an easy verification,

$\mathop{\rm diag}(D)=(a-b)^{2}\mathop{\rm diag}(L).$

Therefore,

$\mathop{\rm diag}(D)\prec x.$ (32)

We now recall equation (7):

$W^{-1}U^{\prime}DUW=\begin{array}[]{ccccc}\left[\begin{array}[]{ccccc}(a^{2}+b^{2})s_{1}&abs_{2}&\ldots&abs_{n}\\\
abs_{2}&-2ab\alpha_{2}&\ldots&0\\\ \vdots&\vdots&\ddots&\vdots\\\
abs_{n}&0&\ldots&-2ab\alpha_{n}\end{array}\right].\end{array}$ (33)

Since $W^{-1}U^{\prime}DUW$ is symmetric and $\mathop{\rm
diag}(W^{-1}U^{\prime}DUW)=x$, by Schur majorization result,

$x\prec\lambda(W^{-1}U^{\prime}DUW)=\lambda(D).$ (34)

By (32) and $(\ref{maj2})$,

$\mathop{\rm diag}(D)\prec\lambda(D).$

The proof is complete. ∎

We now prove another result.

######  Theorem 2.21.

$\lambda(D)\prec\lambda(\frac{D+D^{\prime}}{2})$.

###### Proof.

By (6),

$U^{\prime}DU=\left[\begin{array}[]{cccc}(a^{2}+b^{2}){\rm trace}(L)&b^{2}s\\\
a^{2}s&-2ab\mbox{Diag}(\alpha_{2},\dotsc,\alpha_{n})\end{array}\right].$

Hence,

$U^{\prime}(D+D^{\prime})U=\left[\begin{array}[]{ccccc}2(a^{2}+b^{2}){\rm
trace}(L)&(a^{2}+b^{2})s\\\
(a^{2}+b^{2})s&-4ab\mbox{Diag}(\alpha_{2},\dotsc,\alpha_{n})\\\
\end{array}\right].$

By $(\ref{m4})$,

$W^{-1}U^{\prime}DUW=\begin{array}[]{ccccc}\left[\begin{array}[]{ccccc}(a^{2}+b^{2}){\rm
trace}(L)&abs\\\ abs&-2ab\mathop{\rm Diag}(\alpha_{2},\dotsc,\alpha_{n})\\\
\end{array}\right].\end{array}$

Put

$F:=W^{-1}U^{\prime}DUW.$

Since

$ab\leq\frac{a^{2}+b^{2}}{2},$

we can write

$U^{\prime}(D+D^{\prime})U=2F+\left[\begin{array}[]{cccc}0&\alpha s\\\ \alpha
s&0\\\ \end{array}\right],$

for some $\alpha>0$. Put

$Q:=\left[\begin{array}[]{cccc}0&s\\\ s&0\\\ \end{array}\right].$

We now have

$U^{\prime}(D+D^{\prime})U-\alpha Q=2F.$

By Ky Fan majorization theorem,

$\lambda(U^{\prime}(D+D^{\prime})U-\alpha
Q)\prec\lambda(D+D^{\prime})-\alpha\lambda(Q).$

As,

$\lambda(Q)=(\lambda_{1}(Q),0,\dotsc,0,-\lambda_{1}(Q))~{}~{}\mbox{and}~{}~{}\lambda_{1}(Q)>0,$

we have

$\lambda(D+D^{\prime})-\alpha\lambda(Q)\prec\lambda(D+D^{\prime}).$

Thus,

$\lambda(U^{\prime}(D+D^{\prime})U-\alpha Q)\prec\lambda(D+D^{\prime}),$

i.e.

$\lambda(2F)\prec\lambda(D+D^{\prime}).$

Since

$\lambda(2F)=\lambda(2D),$

we get

$\lambda(D)\prec\lambda(\frac{D+D^{\prime}}{2}).$

This completes the proof. ∎

As an immediate corollary of the above result, we have the following.

######  Corollary 2.22.

$\rho(D)\leq\rho(\frac{D+D^{\prime}}{2})$.

##  3 Application

We end this paper with an application.

###  3.1 Constructing infinitely divisible matrices

Generalized distance matrices can be used to construct infinitely divisible
matrices. Recall that a symmetric matrix $E=[e_{ij}]$ is infinitely divisible
if $E$ is an entry-wise nonnegative and $[e_{ij}^{r}]$ is a positive
semidefinite matrix for all $r\geq 0$.

######  Theorem 3.1.

Let $S=[s_{ij}]$ be an $n\times n$ generalized Laplacian matrix. If $a,b>0$,
define

$d_{ij}:=a^{2}s_{ii}+b^{2}s_{jj}-2abs_{ij},~{}~{}f_{ij}:=\max{(d_{ij},d_{ji})};$
$D:=[d_{ij}]~{}~{}\mbox{and}~{}~{}F:=[f_{ij}].$

If $\mathop{\rm rank}(S)=n-1$, then each $f_{ij}>0$ and $[\frac{1}{f_{ij}}]$
is an infinitely divisible matrix.

###### Proof.

We claim that $d_{ij}>0$ for all $i,j$. Since $S\mathbf{1}=0$, all cofactors
of $S$ are equal. Hence all principal minors of $S$ are positive. In
particular, every $2\times 2$ principal submatrix of $S$ is positive definite.
So, $d_{ij}>0$ and thus $f_{ij}>0$ for all $i,j$.

Put $G:=[f_{ij}]:=[\max(d_{ij},d_{ji})]$. By Theorem 4.2.9 in Bapat [2], it
suffices to show that $G$ has exactly one (simple) positive eigenvalue. We
shall use the following identity: If $\alpha$ and $\beta$ are positive, then

$\max{(\alpha,\beta)}=\frac{1}{2}[(\alpha+\beta)+|\alpha-\beta|].$

Hence,

$2G=[d_{ij}+d_{ji}]+[|d_{ij}-d_{ji}|]$

has exactly one simple positive eigenvalue. Put

$A:=[d_{ij}+d_{ji}]~{}~{}\mbox{and}~{}~{}B:=[|d_{ij}-d_{ji}|].$

If $D=[d_{ij}]$, then for any $x\in\mathbf{1}^{\perp}$,

$x^{\prime}Dx=-2ab(x^{\prime}Sx)\leq 0,$

and hence $x^{\prime}Ax\leq 0$. We now claim that $B$ is negative semidefinite
on $\mathbf{1}^{\perp}$ as well. It can be noted easily that

$|d_{ij}-d_{ji}|=|a^{2}-b^{2}||s_{ii}-s_{jj}|.$

If $\alpha,\beta\geq 0$, then

$|\alpha-\beta|=\alpha+\beta-2{\rm min}(\alpha,\beta),$

and therefore,

$|d_{ij}-d_{ji}|=|a^{2}-b^{2}|(s_{ii}+s_{jj}-2{\rm min}(s_{ii},s_{jj})).$

It is well known that ${\rm min}(s_{ii},s_{jj})$ is positive semidefinite. So,
$B$ is negative semidefinite on $\mathbf{1}^{\perp}$, and consequently,
$2G=A+B$ is negative semidefinite on $\mathbf{1}^{\perp}$. Thus, $G$ has at
least $n-1$ non-positive eigenvalues. Since diagonal entries of $G$ are
positive, $G$ has at least one positive eigenvalue. Thus, $G$ has exactly one
simple positive eigenvalue. This completes the proof. ∎

## Funding

The second author acknowledges the support of the Indian National Science
Academy under the INSA Senior Scientist scheme.

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