The trees for which maximum multiplicity implies the simplicity of other eigenvalues

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www.elsevier.com/locate/disc

The trees for which maximum multiplicity implies the

simplicity of other eigenvalues

夡

Charles R. Johnson

a

_{, Carlos M. Saiago}

b,∗

a_{Department of Mathematics, College of William and Mary, P.O. Box 8795, Williamsburg, VA 23187-8795, USA}

b_{Departamento de Matemática, Faculdade de Ciências e Tecnologia da Universidade Nova de Lisboa, 2829-516 Quinta da Torre, Portugal}

Received 15 November 2003; received in revised form 13 January 2005; accepted 1 April 2005 Available online 10 July 2006

Abstract

Among those real symmetric matrices whose graph is a given treeT, the maximum multiplicity is known to be the path cover number ofT. An explicit characterization is given for those trees for which whenever the maximum multiplicity is attained, all other multiplicities are 1.

Keywords:Real symmetric matrices; Eigenvalues; Multiplicities; Maximum multiplicity; Trees; NIM trees; Vertex degrees

LetGbe an undirected graph onnvertices. ByS_(G)_{we mean the set of all real symmetric}_n_-by-_n_{matrices whose}

graph is preciselyG. Note that no constraint, other than reality, is imposed upon the diagonal entries ofA∈S_(G)_.

An overriding problem is to understand the possible lists of multiplicities for the eigenvalues among the matrices in

S_(G)_{. As in prior work, we concentrate here upon the case in which}_G₌_T _{is a tree.}

For a treeT, there are several formulas for the maximum possible multiplicityM(T )for a single eigenvalue of a matrix inS_{(T )}_{. It is, for example,}_{P (T )}_{, the} _{path cover number}_of_T_[2]_{. For certain trees}_T_{, whenever a matrix} A∈S_{(T )}_{attains this maximum multiplicity, all other multiplicities are 1. This happens, for example, both for a path}

onnvertices and for a star onnvertices. Our purpose here is to characterize all such trees. We call such trees NIM trees (nointermediatemultiplicities). We note that not all trees are NIM. For example, the treeDP₃

has maximum multiplicity 2, but 2,2,1,1 is a multiplicity list. This is the smallest tree (fewest vertices) for which a non-NIM tree exists.

夡_{This work was done within the activities of Centro de Matemática e Aplicações da Universidade Nova de Lisboa.}

∗_{Corresponding author.}

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Our main result is a graph theoretic characterization of NIM trees. For a treeT, letu(T )= {v1, . . . , vk}be the set of

vertices ofTof degree at least 3 (“high” degree vertices) and letH=H (T )denote the subgraph ofTinduced byu(T ). For a given vertexv, we denote its degree in the graphGby deg_G(v). ByG−vwe mean the subgraph ofGinduced by the vertices ofGother thanv. For trees,T −vhas exactly deg_T(v)components, each one being a tree, which we

callbranchesofTatv.

Theorem 1. Let T be a tree on n vertices. Then T isNIMif and only if for eachv∈u(T ):

(i) at most two components ofT −vhave more than one vertex,and

(ii) deg_T(v)deg_H(v)+3.

In order to prove the claimed result, we need some auxiliary results and background. IfGis an undirected graph onnvertices andA∈ S_(G)_{, given}_{⊆ {}₁_{, . . . , n}}_{, we denote the principal submatrix of}_A_{resulting from retention}

(deletion) of the rows and columnsbyA[](A()). IfG′is the subgraph ofGinduced by, we may writeA[G′]

(A(G′)) instead ofA[](A()). We often refer to the “eigenvalues” ofG′ meaning the eigenvalues of the principal submatrixA[G′]ofA. We also denote by(A)the spectrum ofAand bymA()the multiplicity ofas an eigenvalue

ofA.

The following fact is easily verified, and we shall use its corollary to arrange a new common eigenvalue among the branches ofT, without changing the multiplicity of another eigenvalue.

Lemma 2. Let G be an undirected graph on n vertices,A ∈ S_(G)_and _{⊆ {}₁_{, . . . , n}}_._If_, _∈ _R_, ₌ _0,_then

B=A+In∈S(G)andmB[](+)=mA[]().

Corollary 3. Let G be an undirected graph on n vertices,A∈S_(G)_and_{⊆ {}₁_{, . . . , n}}_._{Suppose that}_,_∈_(A[_])_,

= ,and let′,′ ∈ Rbe such that ′ = ′.Then,there is aB ∈ S_(G) _{such that}_m_B_[_]₍′₎₌_m_A_[_]₍₎_and

mB[](′)=mA[]().

LetTbe a tree andA∈S_{(T )}_{. When}_{is an eigenvalue of}_A_{of multiplicity}_m_{, by the interlacing inequalities (see,}

e.g.,[1]), we havemA(i)()∈ {m−1, m, m+1}. However, in[3]it was shown that if∈(A)∩(A(i))for a vertex

iofT, then there is a vertexvofTsuch thatmA(v)()=mA()+1. For historical reasons (see[6,7,3]) we call such a vertexv, aParter vertexofTforrelative toA(a Parter vertex, for short). Note that, whenmA()2, there is always a Parter vertex for. Moreover, there must exist a Parter vertexv′of degree at least 3 and such thatis an eigenvalue of at least 3 direct summands ofA(v′). We call such a vertexv′, astrong Parter vertexofTforrelative toA(a strong Parter vertex, for short). WhenmA()=1 andis an eigenvalue of a principal submatrixA(i)ofA, there exists a Parter vertexv′of degree at least 2 such thatis an eigenvalue of at least 2 direct summands ofA(v′).

WhenmA()1 and{v1, . . . , vk}is a set of vertices ofTsuch thatmA({v1,...,vk})()=mA()+k, we call{v1, . . . , vk}

aParter setof vertices ofTforrelative toA(a Parter set, for short). Each vertex in a Parter set of vertices must be individually Parter[4]. However, a collection of Parter vertices does not necessarily form a Parter set[4].

In[2], the authors show thatM(T )not only isP (T ), but is also max[p−q], such that there existqvertices ofT

whose removal fromTleavesppaths. We call such a set ofqvertices aresidual path maximizingset (an RPM set, for short). In general, an RPM set of vertices is not unique, not even in the value ofq.

If the removal ofqverticesv1, . . . , vqfromTleavesppaths such thatp−q=M(T ), i.e.,p=M(T )+q, a matrixA∈

S_{(T )}_having_∈_R_{as an eigenvalue of each summand corresponding to each of the}_p_{components (with multiplicity}

1, as a real symmetric matrix whose graph is a path has only simple eigenvalues) satisfiesmA({v1,...,vq})()=M(T )+q

and, therefore,mA()=M(T ). Since the removal of eachvi fromTmust have increased the multiplicity ofby 1, we may conclude that eachvi is Parter forand, of course,{v1, . . . , vq}is a Parter set for.

In[3], it was also shown that a Parter vertex for an eigenvaluerelative to a matrixA∈S_{(T )}_{always belongs to a}

Parter set whose removal fromTleaves components in which the corresponding summands ofAhaveas an eigenvalue of multiplicity 1. So, we may now state the following facts.

Lemma 4. Let T be a tree on n vertices and letbe an eigenvalue ofA∈S_{(T )}_{of multiplicity}_{M(T )}_._{If a vertex}_v_of

T is Parter forthendeg_T(v)2,mA(v)()=mA()+1andis an eigenvalue of each of thedegT(v)components

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Lemma 5. Let T be a tree on n vertices,and letbe an eigenvalue ofA∈S_{(T )}_{of multiplicity}_{M(T )}_._{A vertex}_v_is

Parter for the eigenvaluein A if and only if there is anRPMset Q containingvsuch thatmA(Q)()=M(T )+ |Q|.

It may happen that several sets ofq vertices achieve this maximum and that the maximum may be achieved for several values ofq. For example, if we consider again the treeDP3, we haveM(T )=2. In order to maximizep−q,

we may removeq=2 vertices (v1andv2) or onlyq=1 vertex (v1orv2). This means that there are matrices inS(T )

having an eigenvalue of multiplicity 2 for which, eitherv1andv2are Parter or only one of them is Parter. This is an

example of a tree in which each of the high degree vertices may be removed in order to maximizep−q. However there are trees in which some high degree vertices cannot be part of an RPM set. For example, the following treeT′

hasM(T′₎₌_{4. Observe that only}_{v

2, v3, v4}is an RPM set of vertices forT′. So, there is no matrix inS(T′)for

which the vertexv₁is Parter for an eigenvalue of multiplicity 4.

In[5], an algorithm was given to computeM(T )for a general treeT. The strategy was to determine an RPM set of vertices. For this purpose, (1) and (2) of the following lemma was shown.

Lemma 6. Let T be a tree and letv∈u(T ).Then we have the following:

(1) Ifdeg_T(v)deg_H(v)+3,thenvbelongs to everyRPMset of vertices.

(2) Ifdeg_T(v)=deg_H(v)+2and if Q is anRPMset of vertices,then eitherv∈Q,orQ∪ {v}is also anRPMset.

(3) Ifmaxv∈u(T )degH(v)=2and there existsv∈u(T )such thatdegT(v)=degH(v)+1,then noRPMset of vertices

contains all verticesvinu(T )such thatdeg_T(v)=deg_H(v)+1.

(4) Ifmaxv∈u(T )degH(v)=2,v ∈ u(T ), degT(v)=degH(v)+1andv does not belong to anRPMset Q,then at

least one vertex adjacent tovin H must belong to the set Q. Moreover,there is anRPMset containing both the

vertices adjacent tovin H.

Proof. We only need to prove (3) and (4). By hypothesis in (3) and (4), the maximum degree of a vertex inHis 2; thus, it follows that any vertex inu(T )satisfying deg_T(v)=deg_H(v)+1 has degree 3. We may also conclude that there are no vertices satisfying deg_T(v)=deg_H(v), so that deg_T(v)deg_H(v)+1 for all vertices inu(T ).

For (3), in order to obtain a contradiction, suppose that there is an RPM setQcontaining all verticesvofu(T )such that deg_T(v)=deg_H(v)+1. By (1),Qcontains all verticesvofu(T )such that deg_T(v)deg_H(v)+3 and, by (2), we may assume without loss of generality thatQcontains all verticesv ofu(T )such that deg_T(v)=deg_H(v)+2. Since all verticesvinu(T )satisfy deg_T(v)deg_H(v)+1,Qcontains all high degree vertices ofT. Now consider any particularvsuch deg_T(v)=deg_H(v)+1 andQ′=Q\{v}. Now, deg_T₋_Q′(v)=1 and removal ofvfromT−Q′could

not increase the number of components. This contradiction verifies claim (3).

For (4), letvbe a vertex guaranteed by the hypothesis. By (3),vdoes not belong to some RPM setQ. As there are RPM sets contained inu(T ), we may assume without loss of generality thatQ⊆u(T ). Because there are no vertices of degree greater than 2 among theppaths resulting from the removal ofQ, we may conclude that at least one of the vertices adjacent tovinHbelongs toQ(becausevhas one neighbor not inu(T )and thus not inQ). Suppose thatuis a vertex adjacent tovinH, andudoes not belong toQ. Thus,uandvbelong to one pathT′of the remainingppaths ofT−Q. Because deg_T(u)deg_H(u)+1, there is a neighbor ofuinTthat is not inu(T ). SinceT′is a path, we may conclude that deg_T′(u)=2. Thus, ifuis removed fromT′the number of paths remaining increases by 1 (as well the

number of removed vertices fromT). Therefore,Q∪ {u}is an RPM set.

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Proof of Theorem 1. We start by showing that conditions (i) and (ii) are together sufficient. Suppose thatTis a tree onnvertices satisfying conditions (i) and (ii) and letA∈S_{(T )}_having_{as an eigenvalue of multiplicity}_{M(T )}_{. By}

(1) of Lemmas 6 and 5, any vertexvofTsatisfying (ii) must be a Parter vertex for, which implies that all vertices in

u(T )are Parter for. Because of Lemma 4, for any vertexv∈u(T ),is an eigenvalue of deg_T(v)direct summands ofA(v), and, sinceT satisfies (i), each vertex inu(T )may be a strong Parter for at most one multiple eigenvalue. Therefore, (i) and (ii) together imply that each vertex inu(T )must be Parter for exactly one multiple eigenvalue, the eigenvalue of multiplicityM(T ), which proves the sufficiency of the stated conditions.

For the necessity of the stated conditions, our strategy is to show that if either (i) or (ii) does not hold for a treeT, then a matrix inS_{(T )}_{may be constructed with an eigenvalue of maximum multiplicity}_{M(T )}_{and another multiple}

eigenvalue. We first show that not (i) implies not NIM, and then, when we show that not (ii) implies not NIM, we may and do assume that (i) holds.

First suppose that (i) is not satisfied. Then, there is a vertexvof degree at least 3 such that T −v has at least 3 components of more than 1 vertex; we use only 3. We consider two cases: (a)v can be Parter for, the maximum multiplicity eigenvalue (in someA∈S_{(T )}_{); or (b)}_v_{is never Parter for}_.

In case (a), letA∈ S_{(T )}_{be such that}_m_A₍₎₌_{M(T )}_{, the maximum possible, and}_v_{is Parter for}_in_A_{. Let}_T₁_,

T2andT3be 3 components ofT −vwith at least 2 vertices and letA1,A2andA3be the corresponding principal

submatrices ofA. Choose∈Rsuch that=, andi,_i,i=1,2,3, according to Lemma 2, so that, by Corollary 3,Bi=iAi+iIn∈S(Ti),mBi()=mAi()andmBi()1,i=1,2,3. Now, definingBby replacingA1,A2and

A3inAbyB1,B2andB3, respectively (and no other changes),B ∈S(T ),mB()=M(T )(since, by construction,

we havemB(v)()=mA(v)()=M(T )+1) andmB()2 (by the interlacing inequalities for the eigenvalues of a

symmetric matrix), so thatTis not NIM.

In case (b), letA∈S_{(T )}_satisfy_m_A₍₎₌_{M(T )}_{. Then, there is an RPM set}_Q_,_{|Q| =}_q_{, of vertices whose removal}

fromTleavep=M(T )+qpaths, in each of whichoccurs as an eigenvalue of multiplicity 1. By Lemma 5, vertexv

is not in any RPM set, so that vertexvmust remain as part of one of thesep(possibly degenerate) paths. The path that containsvmust havevas an endpoint or else it is possible to removev(increasingqby 1) and increasepby 1, so that

Q∪ {v}would be an RPM set, contradicting Lemma 5. Now decomposeTinto the branches atv, and either makev

a separate part of the decomposition (if it is a single vertex among theppaths) or include it in the unique branch that its path (among theppaths) intersects (otherwise). Now, each of theppaths lies fully within one of the parts of this decomposition: call themT1, . . . , Tk. Each part corresponds to a principal submatrixAi ofA,i=1, . . . , k. We may

now apply Lemma 2 to eachAi, producingBi, and then replace eachAiinAbyBito produceB∈S(T ). Choose, for

eachisuch thatTihas more than 1 vertex (there are at least 3)i =0 andi so thati+i=and∈(Bi), except

that ifvis a vertex ofTi, choosei,i so as to attain∈(Bi(v)), while applying the linear transformation toAi to

obtainBi. Now inB,mB()=mA()=M(T ), asis still an eigenvalue of the principal submatrix corresponding to

each of theppaths. But alsomB()2, asmB(v)()3. SinceB ∈S(T ),Tis not NIM, completing this portion of

the proof.

Suppose now that (ii) is not satisfied and assume that (i) holds. Letv∈u(T )be such that deg_T(v)deg_H(v)+2. Thus, deg_T(v)=deg_H(v)+2 or deg_T(v)=deg_H(v)+1 or deg_T(v)=deg_H(v). By (i), we have deg_H(v)2, so that deg_T(v)=deg_H(v) cannot occur. We consider the remaining two cases:(a′) there exists v ∈ u(T ) with deg_T(v)=deg_H(v)+1; or(b′)there existsv∈u(T )with deg_T(v)=deg_H(v)+2.

In case(a′), letv∈u(T )with deg_T(v)=deg_H(v)+1. Observe that, since deg_H(v)2 we have deg_T(v)=3 and degH(v)=2, so that there are exactly two high degree vertices adjacent tov. Because of (i), we conclude that there

is a vertex pendant atv. By part (3) of Lemma 6, we may assume thatvis a vertex such that deg_T(v)=deg_H(v)+1 and thatvdoes not belong to an RPM set of vertices. By part (4) of Lemma 6, there is an RPM setQofqvertices containing both high degree neighbors ofvand one of thep=M(T )+q components resulting from deletion ofQ

is a pathT1on 2 vertices including the vertexv. By Lemma 5, we may conclude that there is a matrixA ∈ S(T ),

havingas an eigenvalue of multiplicityM(T ), such thatvis not Parter for. Now letT2andT3be the two branches

ofTatv that do not include the vertex pendant atv. Consider the decomposition ofTinto the componentsT1,T2

andT3. Using this decomposition and following the procedure used to prove case (b) above, we may obtain a matrix

B∈S_{(T )}_{such that}_m_B₍₎₌_m_A₍₎₌_{M(T )}_{but with an additional multiple eigenvalue. Note that either}_A₁₌_A[T₁_]

need not be transformed (and then is the single entry of A1(v), which cannot be ) or may be transformed as

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In case(b′₎_{, we may now assume that deg}

T(v)degH(v)+2 for all vertices inu(T ). Suppose that there is a particular

vertexv ∈u(T )such that deg_T(v)=deg_H(v)+2. Becausev ∈u(T ), (i) implies that deg_H(v)∈ {1,2}and there is at least one pendant vertexv′atv. Consider the set of all verticesuinu(T )such that deg_T(u)deg_H(u)+2, except vertexv. By part (1) and (2) of Lemma 6, the removal fromTof such a set ofqvertices leavesppaths and maximizes

p−q; i.e., this is an RPM setQ. Since deg_T(v)=deg_H(v)+2, one of thepcomponents is a pathT1havingv as

an interior vertex. By Lemma 5, we may conclude that there is a matrixA ∈ S_{(T )}_{, having} _{as an eigenvalue of}

multiplicityM(T )and such thatmA[T1]()=1. SinceA[T1]may be chosen so thatdoes not occur as an eigenvalue

ofA[T₁−v], we assume thatvis not Parter forinA[T₁].

Suppose first that deg_H(v)=1. Recall that one of the endpoints ofT₁isv′and denote byv′′ the other endpoint of T1. Observe that, when degH(v)=1 it may occur that v′′ is adjacent, inT, to a vertex v′′′ of the RPM set Q.

We shall use the following decomposition ofT: if v′′ is a pendant vertex inT we consider a decomposition ofT

into componentsT1andT2, in whichT2is the branch ofTatvnot containing vertices ofT1; ifv′′ is not a pendant

vertex in T (i.e., there is a vertexv′′′ of the RPM set Q, adjacent tov′′ inT) we consider a decomposition of T

into components T1,T2 andT3, in which T2 is the branch ofT at v not containing vertices of T1, and T3 is the

branch of Tatv′′ not containing vertices ofT1 (i.e., containing the vertexv′′′). Now, making the above described

decomposition into 2 or 3 components, each of theppaths lies fully within one of the parts of this decomposition. Suppose that such a decomposition has partsTi, each part corresponding to a principal submatrixAi ofA, in which

i=1,2,3 ori=1,2, depending on the number of components of the decomposition ofT. AsT1is a path, we have

mA1()=1. Choose ∈ R, = , and replace (inA) A1 by a matrixB1 ∈ S(T1)having as an eigenvalue

and such that mB1(v)()=2. (SinceT1 is a path, every eigenvalue of a matrix inS(T1)has multiplicity one. As

v is an interior vertex of T1, any matrix A′ ∈ S(T1)such thatmA′_(v)()=2 for a given ∈ R, we necessarily

havemA′()=1. By choosing an eigenvalueofA′,= , we may use Lemma 2, and by a linear transformation

toA′ we obtain such a matrixB₁.) We may now apply Lemma 2 to A₂ producing B₂, and then replaceA₂ inA

by B₂ to produce B ∈ S_{(T )}_{. Choose}₂ ₌ _{0 and}

2 so that 2+2= and ∈ (B2), while applying the

linear transformation toA2 to obtainB2. If the above described decomposition has only 2 components T1andT2,

we get a matrixB ∈S_{(T )}_{such that}_m_B₍₎₌_m_A₍₎₌_{M(T )}_{, as}_{is still an eigenvalue of the principal submatrix}

corresponding to each of theppaths. But also with an additional multiple eigenvalue because, by construction, we havemB(v)()3 and, by the interlacing inequalities, we have thatmB()2, which proves thatT is not NIM. If

the above described decomposition has 3 components, we also choose 3 = 0 and3 so that 3+3= and

∈ (B3(v′′′)), while applying the linear transformation to A3 to obtain B3 (recall that v′′′ is the vertex of T3

adjacent tov′′ inT and that belongs to the RPM setQ). As in the case in which we have a decomposition with 2 components, we may conclude that we get a matrixB ∈ S_{(T )}_{such that}_m_B₍₎₌_m_A₍₎₌_{M(T )}_{. By construction}

we havemB({v,v′′′_}₎()4 and, by the interlacing inequalities, we conclude thatm_B()2, which proves thatTis not

NIM.

To finish the proof, we suppose now that deg_H(v)=2. Since deg_T(v)=deg_H(v)+2, by (i), we conclude that

v has exactly 2 pendant vertices. In this case, we shall use the following decomposition of T: let T1 be the path

on 3 vertices having v as the interior vertex and the 2 endpoints are the pendant vertices at v in T, and let Ti,

i=2,3, be the 2 branches ofTatv that do not contain vertices ofT₁. This decomposition has 3 partsT₁,T₂ and

T₃, each part corresponding to a principal submatrix Ai of A, i=1,2,3. AsT1 is a path, we have mA1()=1.

Choose ∈ R, = , and replace (in A) A1 by a matrix B1 ∈ S(T1) having as an eigenvalue and such

that mB1(v)()=2. We may now apply Lemma 2 to Ai,i =2,3, producing Bi, and then replace each Ai inA

by Bi to produce B ∈ S(T ). For eachi, i=2,3, choose i = 0 andi so that i+i = and ∈ (Bi),

while applying the linear transformation to Ai to obtainBi. Now in B,mB()=mA()=M(T ), as is still an

eigenvalue of the principal submatrix corresponding to each of the p paths. But also mB()3, because by

con-struction we havemB(v)()4 and, by the interlacing inequalities, we have thatmB()3, which proves thatTis

not NIM.

We conclude by noting that a topological description of NIM trees may be deduced from the Theorem 1, and a list of “minimal” NIM trees (no pendant vertex may be removed from a high degree vertex and no NIM tree in the list is homeomorphic to a prior one in the list) could be given.

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