On eigenvalue inequalities of a matrix whose graph is bipartite

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On eigenvalue inequalities of a matrix whose

graph is bipartite

Abdullah Alazemi


, Milica An ¯



and Slobodan K. Simi´c



alazmi95@gmail.com 1Department of Mathematics,

Kuwait University, Kuwait city, Kuwait

Full list of author information is available at the end of the article


We consider the set of real zero diagonal symmetric matrices whose underlying graph, if not told otherwise, is bipartite. Then we establish relations between the eigenvalues of such matrices and those arising from their bipartite complement. Some accounts on interval matrices are provided. We also provide a partial answer to the still open problem posed in (Zhan in SIAM J. Matrix Anal. Appl. 27:851–860,2006).

MSC: Primary 15A18; secondary 15A42; 05C50

Keywords: Bipartite complement of matrix; Eigenvalue bounds; Interlacing property; Interval matrices

1 Introduction

In this paper, we consider the eigenvalues of a real symmetric matrix of ordern≥2 with zero diagonal. The set of such matrices is denoted byMn, that is,


A= (aij)∈Rn×n:aij=aji,aii= 0


ByG(A) we denote theunderlying graphofAMn, that is, it is a simple graph (without

loops or multiple edges) whose vertices correspond to rows (or columns) ofAwith two verticesviandvjadjacent whenever the (i,j)th entry ofAis nonzero. Let


AMn:G(A) is bipartite


We denote byKnthe complete graph onnvertices and byKm,nthe complete bipartite

graph with corresponding color classes of sizesm,n. We also use the following notation:

Jndenotes the all-ones matrix of ordern, whereasJm,nis the all-ones matrix of orderm×n.

Note thatA(Kn) =JnIn, whereInis the unit matrix of ordern, whereas

A(Km,n) =

Om Jm,n

Jn,m On


Throughout the paper, we assume that the eigenvalues of a matrixA, that is, the roots of its characteristic polynomialϕA(λ) =det(λIA) (or, occasionally, its underlying graph

G=G(A)) are ordered in nonincreasing way (λi(A)≥λj(A) fori<j).


We will frequently exploit the well-known Courant–Weyl inequalities (see, e.g., [2, p. 239]).

Theorem 1.1 Let A and B be n×n Hermitian matrices.Then

λi(A+B)≤λj(A) +λij+1(B) (nij≥1), (1.1)

λi(A+B)≥λj(A) +λij+n(B) (1≤ijn), (1.2)

with equality in(1.1)for a pair i,j if and only if there is a nonzero vectorxsuch that Ax=

λj(A)x,Bx=λij+1(B)x,and(A+B)x =λi(A+B)xand in(1.2)if and only if there is a nonzero

vectorxsuch that Ax=λj(A)x,Bx=λij+n(B)x,and(A+B)x =λi(A+B)x.If A and B have

no common eigenvector,then every inequality in(1.1)and(1.2)is strict.

The rest of the paper is organized as follows. In Sect.2, we introduce the concept of a bipartite complement of a real symmetric matrix with zero diagonal whose underlying graph is bipartite, and then we show that the odd-indexed eigenvalues of the initial matrix interlace the even-indexed eigenvalues of its bipartite complement (and vice versa). In this section, we also show that the bounds on the second largest eigenvalue of matrices inMnwhose off-diagonal entries are in the (closed) interval [0, 1], as given in [3], can be

significantly improved in the case where the underlying graph of a matrix in question is bipartite. Also, we extend these results (i.e., bounds) in several directions. First, we prove that these bounds hold for all matrices inMn, not only for those withaij∈[0, 1]; second,

we prove that similar bounds hold for the other eigenvalues (not only for the second largest one). We also deduce, in the nonbipartite case, similar properties for a real symmetric matrix with zero diagonal with respect to its complement. In Sect.3, we focus on matrices whose underlying graphs are bipartite and all nonzero entries are in a given interval. We provide both upper and lower bounds on the largest eigenvalue of these matrices. If the corresponding matrices are nonnegative, then we also provide upper and lower bounds on the second largest eigenvalue. Here we give a partial answer to the problem posed in [4], that is, we determine extremal values for the second largest eigenvalue of a real symmetric matrix whose entries lie in a nonnegative interval.

2 Matrices with bipartite underlying graph

LetABnand assume thatG(A) has two color classes of sizespandq. We writeBp,q,

p+q=n,p,q> 0, ifpandqare relevant further on. ThenABp,q with suitable (and

simultaneous) permutation of rows and columns has the following form:


Op B



with B∈Rp×q. Thebipartite complementof A(with respect to the above form) is the




A(Kp,q) =

Op Jp,q

Jq,p Oq


that is,A+Ab is the adjacency matrix of the complete bipartite graphK

p,q. Note that,

for reducible symmetric matrices or, equivalently, for those with disconnected underlying graphs, such a representation (i.e., form) need not be unique. This stems from the fact that color classes only in disconnected bipartite graphs need not be uniquely determined. Note also that if we restrict ourselves to{0, 1}-matrices or just simple graphs, then the bipartite complement of a disconnected bipartite graph need be neither unique nor connected (ob-serve, for example, two copies ofK1,2). In the graph theory the bipartite complement may

be also considered as a Seidel switching regarding the color classes (cf. [1, p. 5]). Finally, note that (Ab)b=A, provided that color classes ofG(A) andG(Ab) are preserved in both


We first establish relations between the eigenvalues ofAandAb. Recall that the spectrum

of a symmetric matrix with bipartite underlying graph is symmetric with respect to the origin (see, e.g., [1, p. 56]).

Theorem 2.1 Let ABp,q(p+q≥3,p,q≥1),and let Abbe its bipartite complement.If



Ab, λn(A)≤λn–1

Ab, (2.3)




Ab (2≤jn– 1). (2.4)

Proof Since the eigenvalues ofA(Kp,q) are√pq, 0, . . . , 0 n–2

, –√pq, by Theorem1.1(case (1.2))

we obtain

0 =λ2



A+Abλj(A) +λ2–j+n

Ab (2≤jn). Hence

λj(A) +λ2–j+n

Ab≤0, (2.5)

that is,λj(A)≤–λ2–j+n(Ab). Since the underlying graph ofAbis bipartite, its spectrum is

symmetric with respect to the origin, which implies –λ2–j+n(Ab) =λj–1(Ab) for all 2≤jn.

So, we obtain


Ab (2jn).

Forj=n, we obtain the second inequality in (2.3). From (2.5), forj≥2, we can also obtain

λn+2–j(Ab)≤–λj(A) =λn+1–j(A), which in turn is equivalent toλj(Ab)≤λj–1(A) for 2≤j


This result tells that the even- and odd-indexed eigenvalues ofAinterlace the odd- and even-indexed eigenvalues ofAb, respectively. We now deduce when the first inequality in

(2.3) becomes equality under the constraint that all entries ofAbelong to [0, 1].

Theorem 2.2 If ABp,q(p+q≥3,p,q≥1)is a matrix with all entries in[0, 1],then

λ2(A) =λ1

Ab (2.6)

if and only if G(Ab)is disconnected,and either

A=A(Kp,q) or, equivalently, Ab=O,

or there exists a nonzero vectorx∈Rnsuch that Abx=λ


(0, . . . , 0


, 1, . . . , 1


)x = 0, (1, . . . , 1


, 0, . . . , 0


)x = 0. (2.7)

Proof IfA=A(Kp,q), then we are immediately done. So letA=A(Kp,q). Then

λ2(A) +λn




sinceλn(Ab) = –λ1(Ab) andλ2(A(Kp,q)) = 0. By Theorem1.1, case (1.2), the previous

equal-ity holds if and only if there exists a nonzero vector x such that

Ax=λ2(A)x, Abx=λn

Abx, A(Kp,q)x =λ2


x= 0.

The last condition is equivalent to (2.7).

We now prove thatG(Ab) is disconnected. Suppose on the contrary thatG(Ab) is

con-nected. ThenAbis irreducible. If x = (yt, zt)t is the eigenvector corresponding toλ n(Ab),

then (yt, –zt)t is the eigenvector corresponding toλ1(Ab). Therefore (yt, –zt)t is a scalar

multiple of the Perron eigenvector ofAb. (Note that bothAandAbare nonnegative.) This

implies that all entries of y must be either strictly positive or strictly negative. Hence their sum cannot be equal to 0 as expected (see (2.7)). A similar argument holds for z, and we arrive at a contradiction.

Suppose now that there exists a nonzero vector x such thatAbx=λ

n(Ab)x and (2.7)

holds. Since A+Ab =A(Kp,q), we have (A+Ab)x =A(Kp,q)x and Ax+Abx= 0, since

A(Kp,q)x = 0 due to (2.7). ThereforeAx= –λn(Ab)x =λ1(Ab)x. This means thatλ1(Ab) is

the eigenvalue ofA, and we only need to prove that it is not equal toλ1(A). Since (2.7)

holds, x cannot be a Perron vector, because all entries of a Perron vector can be taken to be nonnegative.

This completes the proof.

Remark2.1 By applying Courant–Weyl inequalities to the largest eigenvalue ofA+Ab,

that is, toλ1(A(Kp,q)), we obtainλ1(A) +λ1(Ab)≥ √pq.


G=K5,6–e,λ2(AG) = 0.84076, andλ1(AGb) = 1. ForG=K5,6–e1–e2, wheree1ande2are

two nonadjacent edges inK5,6,λ2(AG) =λ1(AbG) = 1.

Next, we consider the Kronecker productA(K2)⊗Afor any matrixAMn. Then

0 1

1 0





which means that the underlying graph of A(K2)⊗Ais bipartite. The eigenvalues of

A(K2)⊗Aare±λi(A) for 1≤in(see [1, p. 44]). The bipartite complement ofA(K2)⊗A

has the following form:




O Ac+I n

Ac+I n O

, (2.8)


nInA(the underlying graph ofA+AcisKn). In view of Theorems2.1and

2.2and previous observations, together with he corresponding inequalities (2.3) and (2.4) applied to bothAandAc, we have the following:

Theorem 2.3 Let A be a real symmetric matrix with zero diagonal,and let Ac=J


be its complement.If n≥2,then


Ac+ 1, λn(A)≤λn–1

Ac– 1, (2.9)



Ac+ 1≤λj(A)≤λj–1

Ac– 1 (2≤jn– 1). (2.10)

If A is a matrix with all entries in[0, 1]and if G((A(K2)⊗A)b)is disconnected,then

λ2(A) =λ1

Ac– 1 (2.11)

if and only if either A=JnInor,equivalently,Ac=O,or there exists a nonzero vector

x∈Rnsuch that Acx=λ

n(Ac)xandjtx= 0.

Remark2.2 Theorem2.3generalizes Proposition 2.1 from [3] since it provides interlacing between all the eigenvalues ofA(i.e.,Ac), not only between the largest and second largest


We conclude this section with the following result.

Proposition 2.4 Let A=BO BtO

Bp,qbe an irreducible nonnegative matrix such that the

matrix B has the fixed row(column)sum equal to r(resp.,s).Thenσ(A),the spectrum of A,is

±√rs, ±μ2m2, . . . , ±μkmk,


σ(Ab) =±(pqrs),±μ

2m2, . . . ,±μkmk ifpq–√rs=λ2(A), • σ(Ab) =±(pqrs)m2+1,±μ

3m3, . . . ,±μkmkifpq–√rs=λ2(A).

Proof Since the matricesAandA(Kp,q) are symmetric and commute, they can be

simulta-neously diagonalized by, for example, matrixP, that is,P–1AP=D



1. The matrixA(Kp,q) has three distinct eigenvalues, namely

pq, 0, –√pqwith multiplicities 1,p+q– 2, 1, respectively. Also,λ1(A) = √

λ1(BBt) = √

rs, sinceBBthas the constant row sum equal tors. Now the result directly follows.

3 Interval matrices

By aninterval matrixwe assume a matrix whose all entries lie in some interval, say (closed) interval [a,b] (–∞<a<b< +∞). Its diagonal entries in general need not be equal to 0. LetSn[a,b] denote the set of all symmetric interval matrices over the interval [a,b]. In [4]

the range of extremal eigenvalues of a real symmetric interval matrix was considered. Here we consider the set of symmetric matrices whose all nonzero entries lie in the interval [a,b] and whose underlying graphs are bipartite with bipartitionUV, where

|U|=p,|V|=q,p+q=n, p,q> 0. Without loss of generality we may assume that this set, sayBp,q[a,b], consists of all matrices of the formBO BtO

, whereB∈[a,b]p×qfor some

p+q=n,p,q> 0.

We first determinemax{λ1(A) :ABp,q[a,b]} andmin{λ1(A) :ABp,q[a,b]}, and we

also identify matrices that attain these extremal values. LetA=BO BtO


p,q[a,b]. Without loss of generality, we assume thatpq. ThenA2=



andϕA2(λ) =λp2

BtB(λ). Henceλ1(A) = √

λ1(BtB), where

BtB∈ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

Sq[pa2,pb2], 0≤a<b;

Sq[pb2,pa2], a<b≤0;

Sq[pab,pb2], a≤0 <b,|a| ≤b;

Sq[pab,pa2], a< 0≤b,|a|>b.


Next, we apply the following results toBtB.

Theorem 3.1([4]) Let ASn[a,b]with n> 2and a<b.

(i) If|a|>b,then


⎧ ⎨ ⎩

n(ba)/2 ifnis even,

(nbb2+ (n2– 1)a2)/2 ifnis odd.

Ifnis even,then the equality holds if and only ifAis permutation similar to

bJn 2 aJn2

aJn 2 bJn2


Ifnis odd,then the equality holds if and only ifAis permutation similar to


2 aJn–12 ,n+12


2 ,n–12 bJn+12


(ii) If|a| ≤b,thenλ1(A)≤nb.If|a|<b,then the equality holds if and only ifA=bJn.

If|a|=b,then the equality holds if and only ifAis permutation similar to

bJk aJk,nk

aJnk,k bJnk

for somekwith1≤kn.

Theorem 3.2([4]) Let ASn[a,b]with n> 2.

If0 <a<b,thenλ1(A)≥nawith equality if and only ifA=aJn.

Ifa≤0 <b,thenλ1(A)≥awith equality if and only ifA=aIn.

According to the intervals where entries ofBtBbelong to, we distinguish the following


(1) If0 <a<b, thenλ1(A)≤ √pqb. The equality holds if and only ifB=bJpq. Also,

λ1(A)≥ √pqa. The equality holds if and only ifB=aJpq.

(2) Ifa<b< 0, thenλ1(A)≤ √pq|a|. The equality holds if and only ifB=aJpq. Here

λ1(A)≥ √pq|b|. The equality holds if and only ifB=bJpq.

(3) Ifa≤0 <b,|a| ≤b, thenλ1(A)≤ √pqb. If|a|=b, then the equality holds if and only ifBis permutation similar to

bJk aJk,pk

aJpk,k bJpk

for somekwith1≤kn. If|a|<b, then the holds if and only ifB=bJpq. In this

case,λ1(A)≥0. The equality holds if and only ifB=Opq, sinceBtBis positive


(4) Ifa< 0≤b,|a|>b, thenλ1(A)≤ √pq|a|. The equality holds if and only ifB=aJpq.

Also,λ1(A)≥0. The equality holds if and only ifB=Opq.

Next, we determine upper and lower bounds on the second largest eigenvalue of bipar-tite interval matrices with given cardinality of color classes in the case where 0≤a<bor


Recall first the following:

Theorem 3.3([2, p. 238]) Let A∈Rn×nbe a symmetric matrix,and letλ

1(A)≥ · · · ≥λn(A)

be its eigenvalues given in nonincreasing fashion.Let S be a subspace ofRn.Then λ2(A) = min


tAx. (3.2)

Next, we provide an upper bound for the second largest eigenvalue of a nonnegative matrix with entries in a given interval.

Theorem 3.4 Let ASn[a,b],n≥2, 0≤a<b,be an irreducible matrix.Then


⎧ ⎨ ⎩

n(ba)/2 if n is even,


Proof Let v1, v2be unit eigenvectors corresponding toλ1(A) andλ2(A), respectively. Then v2is orthogonal to v1. SinceAis a nonnegative matrix, all entries of v1can be taken to be

nonnegative (see [2, p. 529]). Due to orthogonality, v2has both nonnegative and negative

entries. By simultaneous row and column permutations ofA, if necessary, we may assume that the firstkentries of v2are nonnegative, whereas the remaining are negative for some

k<n. Without loss of generality, we may also assume thatknk. (In the opposite case, instead of v2, we would consider –v2.) Also,λ2(A) can be written in the form

λ2(A) = v2tAv2= jt



where◦denotes the Hadamard entrywise product of matrices, and j is the all-ones vector. Note thatAv2vt2has the sign pattern

+kk,n–kn–k,k +n–k

. Let


A=J(k,a,b) =

bJk aJk,nk

aJnk,k bJnk



Rn k,nk=

(x1, . . . ,xn)t∈Rn:x1, . . . ,xk≥0,xk+1, . . . ,xn< 0


For any x∈Rn

k,nk, xtAxxtAˆx, and thus



xtAx≤ max

{x:xS∩Rnk,n–k,x=1},dimS=n–1 xtAˆx,


{S≤Rn:dimS=n–1}{x:xSRmaxn k,n–k,x=1}

xtAx≤ min

{S≤Rn:dimS=n–1}{x:xSRmaxn k,n–k,x=1}


Since bothAandAˆ have at least one eigenvector corresponding toλ2(A), that is,λ2(Aˆ)

belonging toRn

k,nk, by (3.2) it follows

λ2(A) = min

{S≤Rn:dimS=n–1}xSRmaxn k,n–k,x=1


≤ min

{S≤Rn:dimS=n–1}{x:xSRmaxn k,n–k,x=1}


that is,


J(k,a,b). (3.5)

Therefore the maximal value forλ2(A) is attained at some matrix of the form (3.4). The

eigenvalues ofAˆ are 0 with multiplicityn– 2, andλ1,2(Aˆ) =


2 . Since


λ2(Aˆ) =

nb–(n– 2k)b2+ 4k(nk)a2

2 > 0. (3.6)

Ifnis even, then the right-hand side of (3.6) attains its maximum atk=n2, and thenλ2(A)≤


Ifnis odd, then the maximal value in (3.6) is attained atk=n–12 ork=n+12 . Hence



n– 1

2 ,a,b


b2+ (n2– 1)a2

2 .

This completes the proof.

Theorem 3.5 Let ASn[a,b],n≥2, 0≤a<b,be an irreducible matrix.If n is even,then

λ2(A) =n2(ba)if and only if A is permutation similar to


2 aJn2

aJn 2 bJn2


If n is odd,thenλ2(A) =12(nb

b2+ (n2– 1)a2)if and only if A is permutation similar to


2 aJn–12 ,n+12


2 ,n–12 bJn+12


Proof For evenn, let ASn[a,b] be a matrix satisfying λ2(A) = n2(ba), and let x =

(x1, . . . ,xn)tbe a corresponding unit eigenvector. First, suppose that x has no zero entries

and thatkof them are positive andnknegative. Ifk=n2, then by (3.5) we know that

λ2(A)≤λ2(J(k,a,b)) <λ2(J(2n,a,b)) = n(b2a), a contradiction. Thereforek=n2. By

simulta-neous row and column permutations ofA, if necessary, we may assume thatxi> 0 forin2

andxj< 0 forj>n2. Then

λ2(A) = xtAxxtJ




implies thatA=J(n

2,a,b). Otherwise, the first inequality would be strict, a contradiction.

Therefore, the initial matrixAis permutation similar toJ(n2,a,b).

The equality condition for oddnwhen x has no zero components can be proved simi-larly, and therefore we omit it. Note that


2 aJn+12 ,n–12


2 ,n+12 bJn–12



2 aJn–12 ,n+12


2 ,n–12 bJn+12

are permutation similar.

Otherwise, ifnis even and x has zero entries, then without loss of generality it can be represented in the form x = (u, 0, . . . , 0


), where u∈Rk,k<n, and u has no zero entries.

Consequently,Acan be partitioned in the formA B˜ BtC

, whereA˜ ∈Rk×k,B∈Rk×(nk), andC∈R(nk)×(nk), provided that


Au=λ2(A)u, (3.7)


From (3.7) it follows thatλ2(A)∈Spec(A˜), and thereforeλ2(A) =λ1(A˜) orλ2(A) =λ2(A˜). If

the second equality holds, then u is the corresponding eigenvector forλ2(A˜) with no zero

entries, and hence from the first case it follows thatλ2(A) =λ2(A˜)≤k

2(ba) <


2(ba) if

kis even and similarly ifkis odd, a contradiction.

Thereforeλ2(A) =λ1(A˜), and u is the corresponding positive eigenvector (since we

as-sumed that u has no zero entries). From (3.8) it follows that any row ofBtis orthogonal to

u, and consequentlyBt=OsinceBtis a nonnegative matrix (note thata= 0 in this case).

Assume that v1= [vt, wt], v∈Rk

, w∈Rnk. Having in mind thatB=O, it easily follows

thatCw=λ1(A)w, that is,λ1(A)∈Spec(C), which impliesλ1(A) =λ1(C).

According to Theorem 3.1, ifnis even and k< n2, thenλ2(A) =λ1(A˜)≤kb< n2b, a

contradiction. Hencek= n2. Moreover, again by Theorem3.1,λ1(A˜) = n2b if and only if ˜


2. Consequently,




2b, that is,λ1(C) =


2b, which


2. HenceA=J(


2, 0,b).

Ifnis odd and x contains zeros, in a similar way, we can prove that the equality holds if and only ifA=J(n–12 , 0,b).

This completes the proof.

For the remaining cases, where not bothaandb are nonnegative, we conjecture the following:

Conjecture 1 Let


A(r,s,t) =

⎡ ⎢ ⎣

bJr aJrs aJrt

aJsr bJs aJst

aJtr aJts bJt

⎤ ⎥

⎦, B˜(k,l) =

bJk Okl

Olk bJl


and let ASn[a,b],n≥2,be an irreducible matrix.

(i) Ifa< 0≤b,then

λ2(A)≤ max






˜ B(k,l).

(ii) Ifa<b≤0,then

λ2(A)≤ max


ˆ A(r,s,t).


, whereB∈Rp×q[a,b],BtBis nonnegative if 0a<bora<b0. Since

λ2(A) = √

λ2(BtB), it follows that


⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩


2 ifqis even and 0≤a<bora<b≤0,


2 ifqis odd and 0≤a<b,


2 ifqis odd anda<b≤0.

We next consider the lower bounds for second largest eigenvalue of the matrix inSn[a,b]


Recall thatλ2(A)≥λ2(C), whereCis any principal submatrix ofAof orderk. In



aii aij

aji ajj



(aiiajj)2+ 4a2ij

2 ≥ab. (3.9)

Thusλ2(A)≥ab. Ifλ2(A) =ab, then all the inequalities in (3.9) must be equalities.

This forcesaii=ajj=aandaij=aji=b. As this should be true for alli<j,A=bJ+ (ab)I.

Therefore we proved the following:

Theorem 3.6 Let ASn[a,b],n≥2,|a|<b,then


The equality holds if and only if A=bJ+ (ab)I.

Next, we focus only on bipartite interval matrices belonging to the setBp,q[0, 1].

Proposition 3.7 Let ABp,q[0, 1],n≥3.Thenλ2(A)≤q/2.

Proof Letdi(A) be the row sum ofith row ofA,D=diag(d1(A), . . . ,dn(A)), andQ=D+A=

(DA) + 2A. The matrixDAis positive semidefinite, and thereforeλ2(A)≤λ2(Q)/2

(see [2, p. 495]). On the other hand,qIJqpppIJpqq=Q+JqIqpppIJpqqQ, andqIJqpppIJpqqQis pos-itive semidefinite. Consequently, QqIJqpppIJpqqandλ2(Q)≤q, and thusλ2(A)≤q/2, as


Proposition 3.8 Let ABp,q[0, 1].Then λ2(A)≤max


where r(B)and s(B)are the minimal row and column sums in B,respectively.

Proof The bound is a direct consequence of Theorem2.1and the Frobenius upper bound:


Ab≤ max




The authors acknowledge the support of the Research Sector, Kuwait University. The authors would like to thank the anonymous referees for careful reading and helpful comments and suggestions that lead to the improvement of the original manuscript.

Funding Not applicable.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All three authors contributed equally in writing this manuscript. They all read and approved the final version.

Author details

1Department of Mathematics, Kuwait University, Kuwait city, Kuwait.2ENEL, D.O.O., Belgrade, Serbia.3Mathematical


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Received: 30 October 2018 Accepted: 18 February 2019


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