inverse eigenvalue problem of distance matrix via orthogonal matrix

15
Linear Algebra and its Applications 450 (2014) 202–216 Contents lists available at ScienceDirect Linear Algebra and its Applications www.elsevier.com/locate/laa Inverse eigenvalue problem of distance matrix via orthogonal matrix A.M. Nazari , F. Mahdinasab Department of Mathematics, Faculty of Science, Arak University, Arak 38156-8-8349, Iran article info abstract Article history: Received 12 December 2012 Accepted 9 February 2014 Available online 27 March 2014 Submitted by R. Brualdi MSC: 15A18 15A48 15A51 Keywords: Distance matrix Inverse eigenvalue problem In this paper, for a given list σ of real numbers λ 1 2 ,...,λ n , with sum zero and some additional more technical conditions specified in the paper, we construct a Euclidean distance matrix (EDM) having σ as its list of eigenvalues, without using Hadamard matrices. © 2014 Elsevier Inc. All rights reserved. 1. Introduction and preliminaries A matrix D =(d ij ) M n×n is said to be a Euclidean distance matrix (EDM), if there are n points x 1 ,x 2 ,...,x n R r , such that d ij = x i x j 2 for all i, j =1, 2,...,n, where . denotes the Euclidean norm. By this definition of EDM the following properties immediately hold. 1.D is nonnegative matrix. * Corresponding author. E-mail addresses: [email protected] (A.M. Nazari), [email protected] (F. Mahdinasab). http://dx.doi.org/10.1016/j.laa.2014.02.017 0024-3795/© 2014 Elsevier Inc. All rights reserved.

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Page 1: Inverse eigenvalue problem of distance matrix via orthogonal matrix

Linear Algebra and its Applications 450 (2014) 202–216

Contents lists available at ScienceDirect

Linear Algebra and its Applications

www.elsevier.com/locate/laa

Inverse eigenvalue problem of distance matrix viaorthogonal matrix

A.M. Nazari ∗, F. MahdinasabDepartment of Mathematics, Faculty of Science, Arak University,Arak 38156-8-8349, Iran

a r t i c l e i n f o a b s t r a c t

Article history:Received 12 December 2012Accepted 9 February 2014Available online 27 March 2014Submitted by R. Brualdi

MSC:15A1815A4815A51

Keywords:Distance matrixInverse eigenvalue problem

In this paper, for a given list σ of real numbers λ1, λ2, . . . , λn,with sum zero and some additional more technical conditionsspecified in the paper, we construct a Euclidean distancematrix (EDM) having σ as its list of eigenvalues, without usingHadamard matrices.

© 2014 Elsevier Inc. All rights reserved.

1. Introduction and preliminaries

A matrix D = (dij) ∈ Mn×n is said to be a Euclidean distance matrix (EDM), if thereare n points x1, x2, . . . , xn ∈ R

r, such that dij = ‖xi − xj‖2 for all i, j = 1, 2, . . . , n, where‖ . ‖ denotes the Euclidean norm. By this definition of EDM the following propertiesimmediately hold.

1. D is nonnegative matrix.

* Corresponding author.E-mail addresses: [email protected] (A.M. Nazari), [email protected] (F. Mahdinasab).

http://dx.doi.org/10.1016/j.laa.2014.02.0170024-3795/© 2014 Elsevier Inc. All rights reserved.

Page 2: Inverse eigenvalue problem of distance matrix via orthogonal matrix

A.M. Nazari, F. Mahdinasab / Linear Algebra and its Applications 450 (2014) 202–216 203

2. D is symmetric.3. D has zero main diagonal and this means that the sum of its eigenvalues is zero.

(1.1)

If x1, x2, . . . , xn ∈ Rr are the constructive points of EDM D, then X = (x1 x2 . . . xn)T ∈

Mn×r is called its coordinate matrix, where the ith row of this matrix is the coordinateof xi. Since translation and rotation preserve the distance between two points, thencoordinate matrix associated with an EDM is not unique. The minimum rank of thecoordinate matrices associated with an EDM D is called an embedding dimension of Dand denoted by ed(D). If e is a vector with all elements 1, then D is a distance matrixif and only if D is negative semidefinite on e⊥ = {y ∈ R

n, yT e = 0} [10]. Therefore anEDM D has at most one positive eigenvalue on the set e⊥ with dimension n − 1. Byattention to property (3) of (1.1) we conclude that D has exactly one positive eigenvalue.Distance matrices are used in applications in geodesy, economics, genetics, psychology,biochemistry, engineering, etc.

Let SH denote the set of symmetric matrices of order n with zero diagonal and let SC

denote the set of symmetric matrices A of order n with Ae = 0. We define the followingmaps:

T : SH → SC and K : SC → SH

as

T (D) = −12

(I − eeT

n

)D

(I − eeT

n

),

K(B) = diag(B)eT + e(diag(B)

)T − 2B.

The linear maps T and K are mutually inverse, and D ∈ SH is an EDM of embeddingdimension r if and only if T (D) is positive semidefinite of rank r [1].

An EDM D is said to be spherical if the construction points of D lie on a hypersphere,otherwise, it said to be non-spherical. By [2] we know that a distance matrix D ofembedding dimension r is spherical if and only if its rank is r+1 and D is non-spherical ifand only if its rank is r+2. A spherical EDM D is called regular if the constructive pointsof D lie on a hypersphere whose center coincides with the centroid of those points. D isregular spherical if and only if e is the eigenvector of D corresponding to the eigenvalueeTDe

n [3]. If D is regular spherical of embedding dimension r and A = T (D), then the r

negative eigenvalues of D are exactly the eigenvalues of −2A [4].The characteristic polynomial of a spherical EDM D is given by

P (λ) = λn−r−1

[(λ− eTDe

n

) r∏(λ− ai) −

r∑bi

2r∏

(λ− aj)]

(1.2)

i=1 i=1 j=1,j �=i
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204 A.M. Nazari, F. Mahdinasab / Linear Algebra and its Applications 450 (2014) 202–216

where r is the embedding dimension of D, ai and bi are the ith coordinates of the vectors−2 diag(XTX), and 1√

n(XTX)− 1

2XTDe respectively [6] and if D is a regular sphericaldistance matrix, then the part of bi in formula (1.2) is deleted.

Distance matrices and its eigenvalues are studied in several papers such as [5,6,8].An n×n matrix H is called a Hadamard matrix if every entry is ±1 and HTH = nI.

If n > 2, then n is a multiple of 4, is a necessary condition for existence of these matrices,though it is conjectured to be also sufficient [9].

T.L. Hayden, R. Reams and J. Wells have solved the inverse eigenvalue problem fordistance matrices of order n = 3, 4, 5, 6, and any n for which there exist a Hadamardmatrix and also they solved this problem:

If for n ∈ N there exists a Hadamard matrix of order n, then there is an (n+1)×(n+1)and an (n + 2) × (n + 2) distance matrix with eigenvalues which hold under specialconditions for n � 16 [5].

In the paper [8], G. Jaklič and J. Modic offered a method for constructing a symmetricnonnegative matrix with zero diagonal and eigenvalues λi, where

∑ni=1 λi = 0 and λ1 >

0 > λ2 � · · · � λn, then they survey the inverse eigenvalue problem for Euclideandistance matrices, which are a subclass of such matrices. Although this method is usefulfor solving many problem of inverse eigenvalue problems of distance matrices, but it hassome constraints, for example large matrix dimension, and there is no solution for somespecial spectrum.

For a given set of eigenvalues which satisfy certain additional hypotheses specified inthe paper, we construct a distance matrix with those eigenvalues, using an orthogonalmatrix not derived from a Hadamard matrix.

2. A new method for construction of distance matrix

Let σ = {λ1, λ2, . . . , λn} be a set of n real numbers, such that∑n

i=1 λi = 0 andλ1 > 0 > λ2 � · · · � λn.

Step 1. Choose a matrix W ′ ∈ Mn×(n−1) such that W ′T e = 0 i.e. the sum of all columnsof W ′ is zero, and all columns of W ′ are linearly independent. In order that the distancematrix constructed is a regular spherical matrix and has eigenvalue set σ, it is sufficientthat the entries of each column of W ′ have the same absolute value and that the columnsbe orthogonal. So n must be even. (We explain this after algorithm.)

Step 2. By Gram–Schmidt process we construct orthonormal vectors of columns of W ′

and we denote this new matrix by W .Then Q = (W e√

n) is an orthogonal matrix of order n.

Step 3. We construct two diagonal matrices Λ and Λ′ of order n− 1 as follows

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A.M. Nazari, F. Mahdinasab / Linear Algebra and its Applications 450 (2014) 202–216 205

Λ =

⎛⎜⎜⎜⎜⎜⎜⎝

√−λ2

2 0√−λ3

2. . .

0√

−λn

2

⎞⎟⎟⎟⎟⎟⎟⎠

, Λ′ =

⎛⎜⎜⎜⎜⎝

λ2 0λ3

. . .0 λn

⎞⎟⎟⎟⎟⎠ ,

and obtain the matrices X = WΛ, A = XXT .

Step 4. Distance matrix D is constructed as

D = aeT + eaT − 2A,

where a is the main diagonal of A.

Now we know that Q is an orthogonal matrix, then D is similar to QTDQ, but wehave

QTDQ =(WT

eT√n

)D

(W e√

n

)=

(WTDW WTD e√

n

eT√nDW eTDe

n

)

=(

Λ′ WTD e√n

eT√nDW eTDe

n

),

since W = XΛ−1 and we have

WTDW = Λ−1XTDXΛ−1 = Λ−1XT(aeT + eaT − 2A

)XΛ−1. (2.1)

With regard to the first step, we have chosen the matrix W ′, such that W ′T e = 0, thenthis property holds for W and X, i.e. XT e = 0. The condition W ′T e = 0 implies thatXT e = 0, consequently we can write the relation (2.1) in the following form

WTDW = −2Λ−1XTAXΛ−1 = −2Λ−1XTXXTXΛ−1

= −2Λ−1(XTX)2Λ−1 = −2Λ−1(Λ2)2Λ−1 = −2Λ2 = Λ′,

because XTX = ΛWTWΛ = Λ2.Therefore QTDQ is a bordered diagonal matrix, and if D is regular spherical matrix,

this matrix reduces to the diagonal matrix and this occurs when at first we choose W ′

in orthogonal form, consequently W obtain from W ′ by normalization and the columnsof X are orthogonal.

Since the rows of W ′ have same entries, then main diagonal of matrix A = XXT

is fixed. Since the entries of the main diagonal of A showed the Euclidean norm ofconstructive points of distance matrix, and Ae = 0, then we have De = λe, for some

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206 A.M. Nazari, F. Mahdinasab / Linear Algebra and its Applications 450 (2014) 202–216

λ, this means that De is a multiple of the all-ones vector e. Therefore D is a regularspherical distance matrix.

If the main diagonal of A is a multiple of the all-ones vector e, then D is regularspherical matrix and QTDQ is a diagonal matrix such that its main diagonal is σ. Sothe eigenvalues of D are the elements of σ and we have an EDM matrix with set ofeigenvalues σ, thus, if we can construct a matrix W with mentioned conditions, then wecan find a distance matrix for given eigenvalues.

Remark 2.1. The matrix W ′ is not necessarily a Hadamard matrix. We show this withan example. Let σ = {15,−1,−2,−3,−4,−5} and

W ′ =

⎡⎢⎢⎢⎢⎢⎢⎢⎣

1 −1 1 1 −5−1 −1 1 −1 51 1 1 1 5−1 1 −1 −1 −51 1 −1 −1 −5−1 −1 −1 1 5

⎤⎥⎥⎥⎥⎥⎥⎥⎦

then

W =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1/6√

6 −1/3√

3 0 0 −1/3√

3−1/6

√6 −1/6

√3 1/2 −1/2 1/6

√3

1/6√

6 1/6√

3 1/2 1/2 1/6√

3−1/6

√6 1/3

√3 0 0 −1/3

√3

1/6√

6 1/6√

3 −1/2 −1/2 1/6√

3−1/6

√6 −1/6

√3 −1/2 1/2 1/6

√3

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

and

D =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 196 7/2 5/3 7/2 19

6196 0 8/3 7/2 13

6 7/27/2 8/3 0 19

6 7/2 136

5/3 7/2 196 0 19

6 7/27/2 13

6 7/2 196 0 8/3

196 7/2 13

6 7/2 8/3 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦.

Then D is a regular spherical distance matrix, while we have not used the Hadamardmatrix.

Otherwise if e is not the eigenvector of D, then some bordered entries of QTDQ arenonzero and so some of eigenvalues of D are not exactly with some given elements of σ,

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A.M. Nazari, F. Mahdinasab / Linear Algebra and its Applications 450 (2014) 202–216 207

and the algorithm in this case fails. To eliminate this defect we can use Theorems 5.6,5.7 of [5], that are presented in the following.

Lemma 2.2. (See [5].) Let λ1 � 0 � λ2 � · · · � λn, where∑n

i=1 λi = 0, then (n−1)|λn| �λ1 and (n− 1)|λ2| � λ1.

Theorem 2.3. (See [5].) Let n be such that there exists a Hadamard matrix of order n.Let λ1 � 0 � λ2 � · · · � λn and

∑ni=1 λi = 0, then there is a distance matrix D with

eigenvalues λ1, λ2, . . . , λn.

Now we recall an important theorem due to Fiedler [7].

Theorem 2.4. Let α1 � α2 � · · · � αk be the eigenvalues of the symmetric nonnegativematrix A ∈ R

k×k and β1 � β2 � · · · � βl the eigenvalues of the symmetric nonnegativematrix B ∈ R

l×l, where α1 � β1. Moreover, Au = α1u, Bv = β1v, so that u and v arethe corresponding unit Perron eigenvectors. Then with ρ =

√σ(α1 − β1 + σ) the matrix

(A ρuvT

ρuT v B

)

has eigenvalues (α1 + σ), (β1 − σ), α2, . . . , αk, β2, . . . , βl for any σ � 0.

Next we also present two theorems from [5].

Theorem 2.5. Let n be such that there exists a Hadamard matrix of order n. Let λ1 � 0 �λ2 � · · · � λn+1 and

∑n+1i=1 λi = 0, then there is an (n + 1) × (n + 1) distance matrix D̂

with eigenvalues λ1, λ2, . . . , λn+1 in the following form

D̂ =(

D ρu

ρuT 0

).

The eigenvalues of D in the above theorem are λ1 + λn+1, λ2, . . . , λn, D ∈ Rn×n is a

distance matrix which is constructed by using Theorem 2.3, with ρ =√−λ1λn+1 and

u = 1√n(1 1 . . . 1)nT .

Theorem 2.6. Let n = 4, 8, 12 or 16 be such that there exists a Hadamard matrix oforder n. Let λ1 � 0 � λ2 � · · · � λn+1 � λn+2 and

∑n+2i=1 λi = 0, then there is an

(n + 2) × (n + 2) distance matrix D̂ with eigenvalues λ1, λ2, . . . , λn+1, λn+2 in the fol-lowing form

D̂ =

⎛⎝ D1 ρ e1√

ne2

T√

2

ρ e2√ e1T

√ D2

⎞⎠ ,

2 n

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208 A.M. Nazari, F. Mahdinasab / Linear Algebra and its Applications 450 (2014) 202–216

where the eigenvalues of D1 are λ1 + λn+1 + λn+2, λ2, . . . , λn, and

D2 =(

0 −λn+1−λn+1 0

), ρ =

√(−λn+1 − λn+2)(λ1 + λn+1),

e1 = ( 1 1 . . . 1 )nT, e2 = ( 1 1 )T .

Theorem 2.7. Let n be such that there exists a Hadamard matrix of order n. Letλ1 � 0 � λ2 � · · · � λn+1 � λn+2 � λn+3, where

∑n+3i=1 λi = 0. Then there is a sym-

metric nonnegative matrix with zero diagonal with eigenvalues λ1, λ2, . . . , λn+2, λn+3. Ifthe following conditions hold, then that matrix is a distance matrix.

R2 − λn+3

4 � k,−λn+3

4 � l,

where

R2 = λ1 + λn+1 + λn+2 + λn+3

2n ,

k = 1√2n

√−λn+1(λ1 + λn+2 + λn+3)(−λn+2 − λn+3)(λ1 + λn+3)√

(−λn+1)2 − λn+1(λ1 + λn+2 + λn+3),

l = 1√2

(−λn+1)√

(−λn+2 − λn+3)(λ1 + λn+3)√(−λn+1)2 − λn+1(λ1 + λn+2 + λn+3)

.

Proof. Let λ1′ = λ1 + λn+1 + λn+2 + λn+3. By Lemma 2.2 we have

λ′1 � 0 � λ2 � · · · � λn, λ′

1 + λ2 + · · · + λn = 0.

Let D1 ∈ Mn×n be a distance matrix constructed using a Hadamard matrix, as inTheorem 2.3, with eigenvalues λ′

1, λ2, . . . , λn.Let λ1

′′ = λ1+λn+2+λn+3, and D2 ∈ M(n+1)×(n+1) be a distance matrix constructedusing Theorem 2.5, with eigenvalues λ′′

1, λ2, . . . , λn, λn+1 in the following form:

D2 =(

D1 ρ1u1

ρ1u1T 0

),

where

ρ1 =√

−λn+1λ′′1 =√−λn+1(λ1 + λn+2 + λn+3),

u1 = 1√n

( 1 1 . . . 1 )nT.

Now by Theorem 2.6, we construct the matrix D of order n + 3 as:

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A.M. Nazari, F. Mahdinasab / Linear Algebra and its Applications 450 (2014) 202–216 209

D =(

D2 ρ2u2vT

ρ2vu2T B

),

where

B =(

0 −λn+3

−λn+3 0

),

ρ2 =√

(−λn+2 − λn+3)(λ1 + λn+3), u2 is the unit Perron eigenvalue of D2, and v =1√2(1 1)T .The normalized Perron eigenvector of D2 associated with the Perron eigenvalue λ′′

1has the following form:

u2 = ( f f . . . f g )T ∈ M(n+1)×1,

where

f = 1√n

ρ1√1 + ( ρ1

−λn+1)2 × (−λn+1)

, g = 1√1 + ( ρ1

−λn+1)2.

With these values of f, g for i = 1, . . . , n we have

[D2u2]i = f.[D1e]i + g.[ρ1u1]i = f.λ1′ + g.

ρ1√n

= f.(λ1

′ + (−λn+1))

= f.λ1′′ =

[λ′′

1u2]i.

Moreover

[D2u2]n+1 =n∑

i=1

[ρ1u1

T]i.f =nρ1√

n.f

= nρ1√n.

ρ1√n(−λn+1)

.g = ρ12

(−λn+1).g = λ1

′′.g

=[λ′′

1u2]n+1,

so D2u2 = λ′′1u2.

Consequently the matrix [ρ2u2vT ](n+1)×2 has the following form

⎛⎜⎜⎜⎜⎜⎜⎝

k k

k k...

...k k

l l

⎞⎟⎟⎟⎟⎟⎟⎠

,

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210 A.M. Nazari, F. Mahdinasab / Linear Algebra and its Applications 450 (2014) 202–216

where

k = 1√2n

ρ2ρ1√(−λn+1)2 + ρ12

, l = 1√2

ρ2(−λn+1)√(−λn+1)2 + ρ12

.

By applying Theorem 2.4 where A = D2, α1 = λ1 + λn+2 + λn+3, αi = λi for i =2, 3, . . . , n+1, β1 = −λn+3, β2 = λn+3 and σ = −λn+2 −λn+3, the eigenvalues of D are:

α1 + σ = λ1, β1 − σ = λn+2, β2 = λn+3, λ2, . . . , λn, λn+1.

Now we show that D can be a distance matrix. We know that D1en = λ′

1n e, so by [12],

D1 is a circum-Euclidean distance matrix and the points which generate it lie on ahypersphere with radius R, where R2 = ‖xi‖2 = λ1

2n .Let x1, x2, . . . , xn ∈ R

n be the n points which constructed D1, and the vectors thatcorrespond to the n + 1 vertices of D2 be x1

′ = (x1, 0), . . . , xn′ = (xn, 0), xn+1

′ =(0n, t) ∈ R

n+1. Now assume that the following n + 3 points construct the matrix D:

x1′′ =

(x1

′, 0, 0), . . . , xn

′′ =(xn

′, 0, 0), xn+1

′′ =(xn+1

′, 0, 0)

= (0n, t, 0, 0),

xn+2′′ =

(0n+1, s,

√−λn+3

2

), xn+3

′′ =(

0n+1, s,−√

−λn+3

2

)∈ R

n+3.

Therefore we have

dij =∥∥xi

′′ − xj′′∥∥2 =

∥∥xi′ − xj

′∥∥2,

for i, j = 1, 2, . . . , n + 1, and

dij =∥∥xi

′′ − xj′′∥∥2 = R2 + s2 − λn+3

4 ,

for j = n + 2, n + 3, i = 1, 2, . . . , n, and

dij =∥∥xi

′′ − xj′′∥∥2 = (t− s)2 − λn+3

4 ,

for j = n + 2, n + 3, i = n + 1.To prove that D is a distance matrix we must show that the entries of matrix ρ2u2v

T

and the corresponding entries of the matrix D that follows from the above relations areequal.

So it is sufficient to have:

∃s ∈ R; R2 + s2 − λn+3

4 = k or R2 − λn+3

4 � k

∃s, t ∈ R; (t− s)2 − λn+3 = l or −λn+3 � l.

4 4
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A.M. Nazari, F. Mahdinasab / Linear Algebra and its Applications 450 (2014) 202–216 211

So that if the above inequalities hold, then D is a distance matrix. �Since there are Hadamard matrices for all n, a multiple of 4 such that 4 � n < 668

[11], Theorem 2.7 solves the inverse eigenvalue problem in these special cases.

Step 1: Let λ1′ = λ1 + λn+1 + λn+2 + λn+3. So we have

λ′1 � 0 � λ2 � · · · � λn, λ′

1 + λ2 + · · · + λn = 0.

Let D1 ∈ Mn×n be a distance matrix constructed using the previous algorithm, witheigenvalues λ′

1, λ2, . . . , λn.

Step 2: Let λ1′′ = λ1 + λn+2 + λn+3, and D2 ∈ M(n+1)×(n+1) be a distance matrix con-

structed using Theorem 5.4 of [5], with eigenvalues λ′′1, λ2, . . . , λn, λn+1 in the following

form:

D2 =(

D1 ρ1u1

ρ1u1T 0

),

where

ρ1 =√

−λn+1λ′′1 =√

−λn+1(λ1 + λn+2 + λn+3),

u1 = 1√n

( 1 1 . . . 1 )nT.

Step 3: Now by Theorem 5.7 of [5] (this theorem holds for any n for which a Hadamardmatrix of order n exists) and Theorem 2.3 of [7], we construct the matrix D of ordern + 3 by Theorem 2.7.

D =(

D2 ρ2u2vT

ρ2vu2T B

),

where

B =(

0 −λn+3

−λn+3 0

),

ρ2 =√

(−λn+2 − λn+3)(λ1 + λn+3), u2 is the unit Perron eigenvalue of D2, and v =1√2(1 1)T . u2 has the following form:

u2 = ( f f . . . f g )T ∈ M(n+1)×1,

where

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212 A.M. Nazari, F. Mahdinasab / Linear Algebra and its Applications 450 (2014) 202–216

f = 1√n

ρ1√1 + ( ρ1

−λn+1)2 × (−λn+1)

, g = 1√1 + ( ρ1

−λn+1)2.

Consequently D is a distance matrix if the following conditions hold

R2 − λn+3

4 � k,−λn+3

4 � l,

where

R2 = λ1 + λn+1 + λn+2 + λn+3

2n ,

k = 1√2n

√−λn+1(λ1 + λn+2 + λn+3)(−λn+2 − λn+3)(λ1 + λn+3)√

(−λn+1)2 − λn+1(λ1 + λn+2 + λn+3),

l = 1√2

(−λn+1)√

(−λn+2 − λn+3)(λ1 + λn+3)√(−λn+1)2 − λn+1(λ1 + λn+2 + λn+3)

.

Example 2.8. Let n = 8, σ = {28,−1,−2,−3,−4,−5,−6,−7}. If we choose

W ′ =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 2 3 4 5 6 7−1 −2 3 4 −5 −6 71 −2 −3 4 5 −6 −7−1 2 −3 4 −5 6 −71 2 3 −4 −5 −6 −7−1 −2 3 −4 5 6 −71 −2 −3 −4 −5 6 7−1 2 −3 −4 5 −6 7

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

,

then

A =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

74 0 −1

2−14 −1 0 0 0

0 74

−14

−12 0 −1 0 0

−12

−14

74 0 0 0 −1 0

−14

−12 0 7

4 0 0 0 −1−1 0 0 0 7

4 0 −12

−14

0 −1 0 0 0 74

−14

−12

0 0 −1 0 −12

−14

74 0

0 0 0 −1 −14

−12 0 7

4

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

,

and

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A.M. Nazari, F. Mahdinasab / Linear Algebra and its Applications 450 (2014) 202–216 213

D =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

0 72

92 4 11

272

72

72

72 0 4 9

272

112

72

72

92 4 0 7

272

72

112

72

4 92

72 0 7

272

72

112

112

72

72

72 0 7

292 4

72

112

72

72

72 0 4 9

272

72

112

72

92 4 0 7

272

72

72

112 4 9

272 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

is a regular spherical distance matrix with eigenvalues σ.

Example 2.9. Let n = 4, σ = {6,−1,−2,−3}. By the choice

W ′ =

⎛⎜⎜⎝

4 −2 5−4 2 54 2 −5−4 −2 −5

⎞⎟⎟⎠ , D =

⎛⎜⎜⎜⎜⎝

0 32

52 2

32 0 2 5

252 2 0 3

2

2 52

32 0

⎞⎟⎟⎟⎟⎠

is a spherical distance matrix.

Example 2.10. For n = 6, σ = {15,−1,−2,−3,−4,−5}, if we choose

W ′ =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

1 −2 3 4 −5−1 −2 3 −4 51 2 3 −4 5−1 2 −3 4 51 −2 −3 4 −5−1 2 −3 −4 −5

⎞⎟⎟⎟⎟⎟⎟⎟⎠,

then

D =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

0 4915

185

353120

15940

7730

4915 0 5

312140

5524

175

185

53 0 323

120218

4615

353120

12140

323120 0 77

30358

15940

5524

218

7730 0 353

1207730

175

4615

358

353120 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

is an EDM which is not spherical with eigenvalues

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214 A.M. Nazari, F. Mahdinasab / Linear Algebra and its Applications 450 (2014) 202–216

⎛⎜⎜⎜⎜⎜⎜⎜⎝

15.06857044−1.002439569−2.001141510−3.007884565−4.010967084−5.046137716

⎞⎟⎟⎟⎟⎟⎟⎟⎠

,

and for

W ′ =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 −2 3 4 −5−1 −2 3 −4 51 2 3 4 5−1 2 −3 4 −51 2 −3 −4 −5−1 −2 −3 −4 5

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

, D =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

0 6724

72

53

258

196

6724 0 73

24258

136

318

72

7324 0 19

6318

136

53

258

196 0 67

2472

258

136

318

6724 0 73

24196

318

136

72

7324 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

is a distance matrix with eigenvalues

⎛⎜⎜⎜⎜⎜⎜⎜⎝

−3.−2.−1.

15.01897846−5.014117791−4.004860665

⎞⎟⎟⎟⎟⎟⎟⎟⎠

,

and D is similar to the matrix

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

−1 0 0 0 0 00 −2 0 0 0 00 0 −3 0 0 00 0 0 −4 0 1

8√

60 0 0 0 −5 −1

8√

6√

30 0 0 1

8√

6 −18√

6√

3 15

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

But by using Theorem 2.6, and Example 2.9, we have ρ =√

(−λn+1 − λn+2)(λ1 + λn+1)=

√(−λ5 − λ6)(λ1 + λ5) = 3

√11, the spectrum of D1 is {6,−1,−2,−3}, and

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A.M. Nazari, F. Mahdinasab / Linear Algebra and its Applications 450 (2014) 202–216 215

D =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

0 32

52 2 3

4√

22 34√

2232 0 2 5

234√

22 34√

2252 2 0 3

234√

22 34√

222 5

232 0 3

4√

22 34√

2234√

22 34√

22 34√

22 34√

22 0 434√

22 34√

22 34√

22 34√

22 4 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

is an EDM with spectrum σ.

Example 2.11. We construct a matrix of order 7, by using the second algorithm andExample 2.10 with the following spectrum:

σ = {21,−1,−2,−3,−4,−5,−6}.

D1 =

⎛⎜⎜⎜⎜⎝

0 52 2 3

252 0 3

2 22 3

2 0 52

32 2 5

2 0

⎞⎟⎟⎟⎟⎠ ,

with the spectrum σ1 = {6,−1,−2,−3}.

D2 =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

0 52 2 3

2√

1052 0 3

2 2√

102 3

2 0 52

√10

32 2 5

2 0√

10√

10√

10√

10√

10 0

⎞⎟⎟⎟⎟⎟⎟⎟⎠

,

with the spectrum σ2 = {10,−1,−2,−3,−4}, and B = ( 0 66 0 ).

D =(

D2 X

XT B

), where X =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

528√

462 528√

462528√

462 528√

462528√

462 528√

462528√

462 528√

46217√

1155 17√

1155

⎞⎟⎟⎟⎟⎟⎟⎟⎠

. (2.2)

Moreover we have: R2 − λn+34 = 9

4 � 528√

462 and −λn+34 = 6

4 � 17√

1155. Therefore D

is a distance matrix.

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216 A.M. Nazari, F. Mahdinasab / Linear Algebra and its Applications 450 (2014) 202–216

Acknowledgements

The authors would like to thank the Arak University for financial support and ananonymous referee, whose sensible suggestions led to a significant rewrite which greatlysimplified the overall presentation of the results.

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