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On the Laplacian spectral radius of weighted trees with fixed diameter andweight setShang-Wang Tana; Jing-Jing Jianga

a Department of Mathematics, China University of Petroleum, Dongying 257061, China

Online publication date: 18 February 2011

To cite this Article Tan, Shang-Wang and Jiang, Jing-Jing(2011) 'On the Laplacian spectral radius of weighted trees withfixed diameter and weight set', Linear and Multilinear Algebra, 59: 2, 173 — 192To link to this Article: DOI: 10.1080/03081080903279980URL: http://dx.doi.org/10.1080/03081080903279980

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Linear and Multilinear AlgebraVol. 59, No. 2, February 2011, 173–192

On the Laplacian spectral radius of weighted trees with

fixed diameter and weight set

Shang-Wang Tan* and Jing-Jing Jiang

Department of Mathematics, China University of Petroleum, Dongying 257061, China

Communicated by R. B. Bapat

(Received 28 January 2009; final version received 20 August 2009)

The spectrum of weighted graphs is often used to solve the problems in thedesign of networks and electronic circuits. We first give some perturba-tional results on the (signless) Laplacian spectral radius of weighted graphswhen some weights of edges are modified, then we determine the weightedtree with the largest Laplacian spectral radius in the set of all weighted treeswith fixed diameter and positive weight set.

Keywords: weighted graph; weighted tree; Laplacian spectral radius;Perron vector

AMS Subject Classification: 05C50

1. Introduction

In this article, we only consider simple weighted graphs with positive weight set. LetG be a weighted graph with vertex set {v1, v2, . . . , vn}, edge set E(G) 6¼ ; and weightset W(G)¼ {wj4 0 : j¼ 1, 2, . . . , jE(G)j}. The function wG :E(G)!W(G) is called aweight function of G. It is obvious that each weighted graph corresponds to a weightfunction. For convenience, define wG(uv)¼ 0 if uv =2 E(G). So G may be regarded as aweighted graph on a nonnegative weight set, where uv2E(G) if and only ifwG(uv)4 0. Thus the adjacency matrix of G is defined to be the n� n matrixA(G)¼ (wG(vivj)). The weight of vertex vi, denoted by wG(vi), is the sum of weights ofall edges incident to vi in G. Let W(G)¼ diag(wG(v1),wG(v2), . . . ,wG(vn)) be thediagonal matrix of vertex weights of G. Then L(G)¼W(G)�A(G) is called theLaplacian matrix of G. From this fact and Gersgorin’s theorem, it follows that itseigenvalues are nonnegative real numbers. Moreover, since its rows sum to 0, then 0is its smallest eigenvalue. Thus the eigenvalues of L(G) are denoted by

�1ðLðGÞÞ � �2ðLðGÞÞ � � � � � �n�1ðLðGÞÞ � �nðLðGÞÞ ¼ 0,

where �1(L(G)), denoted by �(G), is called the Laplacian spectral radius of G and�n�1(L(G)) is called the algebraic connectivity of G.

The matrix R(G)¼W(G)þA(G) is called the signless Laplacian matrix of G.Note that R(G) is a nonnegative symmetric matrix, its eigenvalues are all real

*Corresponding author. Email: [email protected]

ISSN 0308–1087 print/ISSN 1563–5139 online

� 2011 Taylor & Francis

DOI: 10.1080/03081080903279980

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numbers and its largest eigenvalue is a positive number. The largest eigenvalue

�1(R(G)) of R(G) is called the signless Laplacian spectral radius of G, denoted by

�(G). Since R(G) is nonnegative, there is a nonnegative eigenvector corresponding

to �(G). In particular, when G is connected, R(G) is irreducible and by the Perron–

Frobenius theorem (see [8], for example), �(G) is simple and there is a unique

positive unit eigenvector. We shall refer to such an eigenvector as the Perron vector

of R(G).Let H and G be two weighted graphs. G and H are called isomorphic, denoted by

G¼H, if there is a bijection f from V(G) to V(H ) such that ab2E(G) if and only

if f (a) f (b)2E(H ), and wG(ab)¼wH( f (a) f (b)) for each ab2E(G). H is called

a weighted subgraph of G if H is a subgraph of G and wH(e)¼wG(e) for each

e2E(H ).The Laplacian eigenvalues of weighted graphs have many important applications

in combinatorial optimization, the design of networks, the design of electronic

circuits and so on. On the other hand, unweighted graphs may be regarded as

weighted graphs whose edges have weight 1. Therefore, it is significant and necessary

to investigate the Laplacian eigenvalues of weighted graphs. Recently, this topic was

mostly investigated in the literature. Das and Bapat [4], Rojo [10] obtained two upper

bounds on the Laplacian spectral radius of weighted graphs, respectively. Rojo and

Robbiano [11] gave an upper bound on the Laplacian spectral radius of weighted

trees. Fernandes et al. [5] derived an upper bound on the Laplacian spectral radius of

a weighted graph defined by a weighted tree and a weighted triangle attached, by one

of its vertices, to a pendant vertex of the tree. Berman and Zhang [2] gave a lower

bound on the second smallest Laplacian eigenvalue of weighted graphs. Kirkland

and Neumann [9] researched the algebraic connectivity of weighted trees under

perturbation.The article studies the Laplacian spectral radius of weighted graphs. Throughout,

let NG(u) be the adjacent vertex set of a vertex u in G, dG(u) be the degree of u in G

and �(M )¼�(M,x)¼ det(xI�M ) be the characteristic polynomial of a matrix M.

In particular, let �(G)¼�(R(G)). All other notations and definitions not given in the

article are standard terminology of graph theory (see [3], for example).The article is organized as follows: in Section 2 we give some perturbational

results on the (signless) Laplacian spectral radius of weighted graphs when some

weights of edges are modified. Furthermore, we also present some properties of

characteristic polynomials of the (signless) Laplacian matrices of weighted graphs.

In Section 3, we determine the weighted tree with the largest (signless) Laplacian

spectral radius in the set of all weighted trees with fixed diameter and positive weight

set (see [13] for similar results on the spectral radius of adjacency matrices).

2. Some perturbational results

Any modification of a weighted graph gives rise to perturbations of eigenvalues of its

matrix. In the literature, this topic is extensively investigated for unweighted graphs.

In this section, we mainly present some perturbational results on the signless

Laplacian spectral radius of weighted graphs, which are useful and more ordinary

than those of unweighted graphs.

174 S.-W. Tan and J.-J. Jiang


LEMMA 2.1 [8] (Rayleigh–Ritz Theorem) Let M be a Hermitian matrix and let

�1(M ) be the largest eigenvalue of M. Then �1(M )¼maxkXk¼1,X2Rn XtMX, and

�1(M )¼XtMX if X is a unit eigenvector corresponding to �1(M ).

LEMMA 2.2 [7] Let M be a nonnegative symmetric matrix and let X be a unit vector

of Rn. If �1(M )¼XtMX, then MX¼ �1(M )X.

THEOREM 2.3 Let a, b, u, v be four vertices of a weighted graph G and let X¼ (x1,

x2, . . . , xn)t be a nonnegative unit eigenvector corresponding to �(G), where xi corres-

ponds to the vertex vi of G. For 05 ��wG(uv), let G1 be the weighted graph obtained

from G such that

wG1ðabÞ ¼ wGðabÞ þ �, wG1 ðuvÞ ¼ wGðuvÞ � �, wG1ðeÞ ¼ wGðeÞ, e2EðGÞnfab, uvg:

If xuþ xv� xaþ xb, then �(G)��(G1).

In addition, if xuþxv5 xaþxb or xuþ xv� xaþ xb and G is connected, then

�(G)5�(G1).

Proof From Lemma 2.1, we have that

�ðG1Þ � �ðGÞ ¼ maxkYk¼1

YtRðG1ÞY� XtRðGÞX � XtðRðG1Þ � RðGÞÞX

¼ �½ðxa þ xbÞ2� ðxu þ xvÞ

2� � 0: ð2:1Þ

Suppose that xuþxv5 xaþxb. By (2.1), it is obvious that �(G)5�(G1).Suppose that xuþxv� xaþxb and G is connected. From the Perron–Frobenius

theorem, X is a positive unit eigenvector. Assume �(G)¼�(G1). From (2.1), we have

that �(G1)¼XtR(G1)X. Again by Lemma 2.2, we get R(G1)X¼�(G1)X, i.e.

[W(G1)þA(G1)]X¼�(G1)X. Without loss of generality, assume a =2 {u, v}. Then

�ðG1Þxa ¼ wG1 ðaÞxa þ wG1 ðabÞxb þX

z2NG1ðaÞnfbg

wG1 ðzaÞxz

¼ �ðxa þ xbÞ þ wGðaÞxa þ wGðabÞxb þX

z2NGðaÞnfbg

wGðzaÞxz

¼ �ðxa þ xbÞ þ wGðaÞxa þX

z2NGðaÞ

wGðzaÞxz:

Also from R(G)X¼�(G)X, we have that

�ðGÞxa ¼ wGðaÞxa þX

z2NGðaÞ

wGðzaÞxz:

So we have that (�(G1)��(G))xa¼ �(xaþ xb)¼ 0. This implies that xaþ xb¼ 0,

a contradiction to xaþ xb4 0. Therefore, �(G)5�(G1). g

COROLLARY 2.4 Let G and G1 be the two weighted graphs defined in Theorem 2.3. Let

(x1, x2, . . . , xn)t and ð ~x1, ~x2, . . . , ~xnÞ

t be two nonnegative unit eigenvectors corresponding

to �(G) and �(G1), where xi and ~xi correspond to the vertex vi of G and G1, respectively.

If xuþ xv� xaþ xb, then ~xu þ ~xv � ~xa þ ~xb.In addition, if xuþxv5 xaþxb or xuþ xv� xaþ xb and G is connected, then

~xu þ ~xv 5 ~xa þ ~xb.

Linear and Multilinear Algebra 175


Proof It is easy to see that G can be obtained from G1 in the following way:

wGðuvÞ ¼ wG1ðuvÞ þ �,wGðabÞ ¼ wG1 ðabÞ � �,wGðeÞ ¼ wG1 ðeÞ, e2EðG1Þnfab, uvg:

We first prove that ~xu þ ~xv � ~xa þ ~xb. Assume that ~xu þ ~xv 4 ~xa þ ~xb. Then by

the additional claim of Theorem 2.3, we have �(G1)5�(G). On the other hand, since

xuþ xv� xaþ xb, again from Theorem 2.3, we get �(G)��(G1), a contradiction.

Therefore, ~xu þ ~xv � ~xa þ ~xb.We next prove the additional claim. Assume that ~xu þ ~xv � ~xa þ ~xb. Then by

Theorem 2.3, we get �(G1)��(G). On the other hand, since xuþxv5 xaþxb or

xuþ xv� xaþ xb and G is connected, from the additional claim of Theorem 2.3,

we have that �(G)5�(G1), a contradiction. Therefore, ~xu þ ~xv 5 ~xa þ ~xb. g

THEOREM 2.5 Let u, v be two distinct vertices of a connected weighted graph G and

u1, u2, . . . , us (ui 6¼ v, s 6¼ 0) be some vertices of NG(u)nNG(v). Let X¼ (x1,x2, . . . , xn)t

be the Perron vector of R(G), where xi corresponds to the vertex vi of G. Let G2 be the

weighted graph obtained from G by deleting edges uuj and adding edges vuj such that

wG2ðvujÞ ¼ wGðuujÞ, wG2ðeÞ ¼ wGðeÞ, e 6¼ uuj, j ¼ 1, 2, . . . , s:

If xu� xv, then �(G)5�(G2).

Proof Set H0¼G. For j¼ 1, 2, . . . , s, let Hj be the weighted graph obtained from

Hj�1 by deleting the edge uuj and adding the edge vuj such that

wHjðvujÞ ¼ wHj�1

ðuujÞ, wHjðeÞ ¼ wHj�1

ðeÞ, e2EðHj�1Þnfuujg:

Since wHj�1(uuj)4wHj�1

(vuj)¼ 0 for j¼ 1, 2, . . . , s, set �j¼wHj�1(uuj), then Hj can be

obtained from Hj�1 in the following way:

wHjðvujÞ ¼ wHj�1

ðvujÞ þ �j, wHjðuujÞ ¼ wHj�1

ðuujÞ � �j,

wHjðeÞ ¼ wHj�1

ðeÞ, e2EðHj�1Þnfvuj, uujg:

Let Xj ¼ ðx j1, x

j2, . . . , x j

nÞt be a nonnegative unit eigenvector corresponding to �(Hj),

where X 0¼X and x j

i corresponds to the vertex vi of Hj. Then x0u1 þ x0v � x0u1 þ x0u,

and by the additional claim of Corollary 2.4, we have that

x jujþ1þ x j

v 4 x jujþ1þ x j

u, j ¼ 1, 2, . . . , s� 1:

Since Hs¼G2, by the additional claim of Theorem 2.3, we get

�ðGÞ ¼ �ðH0Þ5�ðH1Þ5 � � � 5�ðHsÞ ¼ �ðG2Þ: g

THEOREM 2.6 Let a, b, u, v be four distinct vertices of a connected weighted graph G

and let X¼ (x1, x2, . . . , xn)t be the Perron vector of R(G), where xi corresponds to the

vertex vi of G. For 05 ��wG(ab), 05 ��wG(uv), let G3 be the weighted graph

obtained from G such that

wG3ðabÞ ¼ wGðabÞ � �, wG3ðauÞ ¼ wGðauÞ þ �, wG3ðuvÞ ¼ wGðuvÞ � �,

wG3ðvbÞ ¼ wGðvbÞ þ �, wG3 ðeÞ ¼ wGðeÞ, e2EðGÞnfab, uv, vb, aug:

If (xu�xb)[(2xaþ xbþ xu)�� (2xvþ xbþ xu)�]� 0, then �(G)��(G3).



In addition, �(G)¼�(G3) if and only if xu¼ xb and (2xaþ xbþ xu)�¼ (2xvþ

xbþ xu)�.

Proof From Lemma 2.1, we have

�ðG3Þ � �ðGÞ ¼ maxkYk¼1

YtRðG3ÞY� XtRðGÞX � XtðRðG3Þ � RðGÞÞX

¼ ðxu � xbÞ½ð2xa þ xb þ xuÞ� � ð2xv þ xb þ xuÞ�� 0: ð2:2Þ

Assume �(G)¼�(G3). By (2.2), we have �(G3)¼XtR(G3)X. Again from

Lemma 2.2, we get R(G3)X¼�(G3)X, i.e. [W(G3)þA(G3)]X¼�(G3)X. Thus

�ðG3Þxa ¼ wG3ðaÞxa þ wG3ðbaÞxb þ wG3ðuaÞxu þX

z2NG3ðaÞnfu, bg

wG3ðzaÞxz

¼ �ðxu � xbÞ þ wGðaÞxa þX

z2NGðaÞ

wGðzaÞxa

¼ �ðxu � xbÞ þ �ðGÞxa:

So we obtain xu¼ xb. In the similar way, we can get

�ðG3Þxb ¼ �ðxb þ xvÞ � �ðxa þ xbÞ þ �ðGÞxb:

Therefore, we have that �(xaþxb)¼ �(xbþxv). Combining xu¼xb, we can get

(2xaþxbþ xu)�¼ (2xvþxbþxu)�.Assume xu¼ xb, (2xaþ xbþ xu)�¼ (2xvþxbþxu)�, i.e. xu¼xb, �(xaþ xb)¼

�(xvþ xb), �(xaþ xu)¼ �(xvþ xu). Then we easily get

wG3ðaÞxa þX

z2NG3ðaÞ

wG3ðzaÞxz ¼ �ðxu � xbÞ þ �ðGÞxa ¼ �ðGÞxa,

wG3 ðvÞxv þX

z2NG3ðvÞ

wG3ðzvÞxz ¼ �ðxb � xuÞ þ �ðGÞxv ¼ �ðGÞxv,

wG3ðbÞxb þX

z2NG3ðbÞ

wG3ðzbÞxz ¼ �ðxb þ xvÞ � �ðxa þ xbÞ þ �ðGÞxb ¼ �ðGÞxb,

wG3ðuÞxu þX

z2NG3ðuÞ

wG3ðzuÞxz ¼ �ðxa þ xuÞ � �ðxv þ xuÞ þ �ðGÞxu ¼ �ðGÞxu:

It is obvious that, for p2V(G)� {a, b, u, v}, we have

wG3ð pÞxp þX

z2NG3ð pÞ

wG3 ðzpÞxz ¼ wGð pÞxp þX

z2NGð pÞ

wGðzpÞxz ¼ �ðGÞxp:

Thus [W(G3)þA(G3)]X¼�(G)X, i.e. R(G3)X¼�(G)X. Since X is the Perron vector

of R(G), by the Perron–Frobenius theorem, it follows that �(G)¼�(G3). g

For v2V(G), let Rv(G) be the principal submatrix of R(G) formed by deleting

the row and column corresponding to vertex v. If G¼ v, then suppose �(Rv(G))¼ 1.

We have the following theorem which is a generalization of Theorem 2.2 in [14] and

its proof is very similar to Lemma 8 in [6].



THEOREM 2.7 Let G be the weighted graph formed from two weighted graphs G1 and

G2 by joining the vertex u of G1 to the vertex v of G2 by an edge e¼ uv. Then

�ðGÞ ¼ �ðG1Þ�ðG2Þ � wGðeÞ½�ðG1Þ�ðRvðG2ÞÞ þ �ðG2Þ�ðRuðG1ÞÞ�: ð2:3Þ

Proof Let RðG�1ÞðRðG�2ÞÞ be the principal submatrix obtained by deleting the row

and column corresponding to the vertex v(u) from R(G1u : v)(R(G2v : u)), where G1u : v

is the weighted graph formed from G1 by joining a new pendant vertex v to u.

Without loss of generality, we may assume that

RðGÞ ¼RðG�1Þ E11

Et11 RðG�2Þ

� �,

where E11 is the jV(G1)j � jV(G2)j matrix whose unique nonzero entry is a wG(e) in

position (1, 1). By the Laplace theorem, we have

�ðGÞ ¼ �ðRðG�1ÞÞ�ðRðG�2ÞÞ � w2

GðeÞ�ðRuðG1ÞÞ�ðRvðG2ÞÞ: ð2:4Þ

Since

�ðRðG�1ÞÞ ¼ �ðG1Þ � wGðeÞ�ðRuðG1ÞÞ, �ðRðG�2ÞÞ ¼ �ðG2Þ � wGðeÞ�ðRvðG2ÞÞ,

by putting them into (2.4), it follows (2.3). g

COROLLARY 2.8

(1) Let uva be a path of a weighted graph G such that u is a vertex of degree 1 and

v is a vertex of degree 2 in G. Then

�ðGÞ ¼ ðx� 2wGðuvÞÞ�ðG� uÞ � wGðuvÞwGðvaÞ�ðG� u� vÞ: ð2:5Þ

(2) Let z be a vertex of a weighted graph G. Let ~Gt be the weighted graph obtained

from G by adding pendant edges zai with weights w(zai) (i¼ 1, 2, . . . , t). Then

�ð ~GtÞ ¼ �ðGÞ �Xti¼1

xwðzaiÞ�ðRzðGÞÞ

x� wðzaiÞ

" #Yti¼1

ðx� wðzaiÞÞ:

Proof

(1) By Theorem 2.7, we have

�ðGÞ ¼ ðx� wGðuvÞÞ�ðG� uÞ � wGðuvÞx�ðRvðG� uÞÞ, ð2:6Þ

�ðG� uÞ ¼ ðx� wGðvaÞÞ�ðG� u� vÞ � wGðvaÞx�ðRaðG� u� vÞÞ: ð2:7Þ

It is easy to see that

�ðRvðG� uÞÞ ¼ �ðG� u� vÞ � wGðvaÞ�ðRaðG� u� vÞÞ: ð2:8Þ

By removing �(Rv(G� u)) and �(Ra(G� u� v)) from (2.6) to (2.8), we

get (2.5).



(2) By induction on t we prove the result. By setting G1¼G, G2¼ {a1}, u¼ z,

v¼ a1 in Theorem 2.7, we get

�ð ~G1Þ ¼ x�ðGÞ � wðza1Þ½�ðGÞ þ x�ðRzðGÞÞ�

¼ �ðGÞ �X1i¼1


x� wðzaiÞ

" #Y1i¼1

ðx� wðzaiÞÞ,

i.e. the result holds for t¼ 1.

Assume that the result holds for t� 1(t� 2). By the definition of Rv(G), we have

�ðRzð ~Gt�1ÞÞ ¼ �ðRzðGÞÞYt�1i¼1

ðx� wðzaiÞÞ: ð2:9Þ

Taking G1 ¼ ~Gt�1,G2 ¼ fatg, u ¼ z, v ¼ at in Theorem 2.7, by (2.9) and the induction

hypothesis, we get

�ð ~GtÞ ¼ x�ð ~Gt�1Þ � wðzatÞ½�ð ~Gt�1Þ þ x�ðRzð ~Gt�1ÞÞ�

¼ ðx� wðzatÞÞ�ð ~Gt�1Þ � xwðzatÞ�ðRzð ~Gt�1ÞÞ

¼ ðx� wðzatÞÞ �ðGÞ �Xt�1i¼1


x� wðzaiÞ

" #Yt�1i¼1

ðx� wðzaiÞÞ

� xwðzatÞ�ðRzðGÞÞYt�1i¼1

ðx� wðzaiÞÞ

¼ �ðGÞ �Xti¼1


x� wðzaiÞ

" #Yti¼1

ðx� wðzaiÞÞ:

This completes the proof. g

THEOREM 2.9 Let H be a weighted proper spanning subgraph of a weighted tree T.

Then for x��(T ), we have �(H )4�(T ).

Proof Let E(T )nE(H )¼ {u1v1, u2v2, . . . , usvs}, where s� 1. Write T0¼T and

Ti ¼ Ti�1 � uivi, i ¼ 1, 2, . . . , s:

Then Ts¼H. Let Gi1 be the component containing ui in Ti and Gi

2 be the reminder of

Ti. Then for i¼ 1, 2, . . . , s, we have

�1ðRðTi�1ÞÞ � �1ðRðTiÞÞ � maxf�1ðRðGi1ÞÞ, �1ðRðG

i2ÞÞg:

In particular, since R(T0) is irreducible, we have �(T )¼ �1(R(T0))4 �1(R(T1)). Thus

when x��(T ), all of �ðGi1Þ,�ðRviðG

i2ÞÞ,�ðG

i2Þ and �ðRuiðG

i1ÞÞ are positive. Note that

�ðTiÞ ¼ �ðGi1Þ�ðG

i2Þ. So by (2.3), when x��(T ), for i¼ 1, 2, . . . , s, we have

�ðTi�1Þ ¼ �ðTiÞ � wTðuiviÞ½�ðGi1Þ�ðRviðG

i2ÞÞ þ �ðG

i2Þ�ðRuiðG

i1ÞÞ�5�ðTiÞ:

Therefore, the required result follows. g



3. Main results and proofs

In this section, we will apply the idea from [1,12] to determine the weighted tree with

the largest (signless) Laplacian spectral radius in the set of all weighted trees with

fixed diameter and positive weight set.Let �(d; m1, m2, . . . ,mn�1) be the set of all weighted trees with n vertices, diameter

d and positive weight set {m1, m2, . . . ,mn�1}. Let TM be a weighted tree in �(d; m1,

m2, . . . ,mn�1) with the largest signless Laplacian spectral radius. Without loss of

generality, we also assume d� 3 (otherwise, TM is a weighted star, so the problem is

trivial). Let X¼ (x1, x2, . . . , xn)t be the Perron vector of R(TM), where xi corresponds

to the vertex vi of TM. Next we will investigate the spectral and structural properties

of TM.

LEMMA 3.1 Let ab, uv be two distinct edges of TM.

(1) If xaþ xb� xuþxv, then wTM(ab)�wTM

(uv).(2) If wTM

(ab)4wTM(uv), then xaþ xb4 xuþ xv.

(3) If xaþ xb¼ xuþ xv, then wTM(ab)¼wTM

(uv).

Proof It is obvious that (2) and (3) can be immediately deduced from (1) and (2),

respectively. Hence we only give the proof of (1). Assume that wTM(ab)5wTM

(uv).

Put �¼wTM(uv)�wTM

(ab) and let T 0 be the weighted tree obtained from TM such

that

wT 0 ðabÞ ¼ wTMðabÞ þ �, wT 0 ðuvÞ ¼ wTM

ðuvÞ � �, wT 0 ðeÞ ¼ wTMðeÞ, e 6¼ ab, uv,

i.e. T 0 is the weighted tree obtained from TM by exchanging the weights of edges ab

and uv while making the weights of other edges fixed. Then T 0 2�(d;m1,m2, . . . ,

mn�1), and by the additional claim of Theorem 2.3, we have that �(T 0)4�(TM),

a contradiction to the choice of TM. g

LEMMA 3.2 Let n4 dþ 1 and let P¼ u1u2� � � ududþ1 be a longest path of TM.

(1) TM has a unique vertex z such that dTM(z)� 3.

(2) z2V(P) and each vertex not in P is a pendant vertex adjacent to z.(3) For each v2V(TM)n{z}, we have that xz4 xv. In addition, for each vertex

ui(2� i� d ), we have that xui4max{xu1, xudþ1}.

Proof TM can be obtained from P by attaching a proper weighted tree to the vertex

ui for each i¼ 2, 3, . . . , d.

(1) It is obvious that TM has a vertex z such that dTM(z)� 3. Apart from z,

suppose that TM has another vertex with degree greater than 2. We will

distinguish the two following cases.

Case 1 TM is a caterpillar.

It is obvious that there are two distinct vertices uk and ul of P such that TM has

two pendant edges auk and bul not in P. Without loss of generality, assume xuk� xul.

Let T 0 be the weighted tree obtained from TM by deleting the edge bul and adding the

new edge buk such that

wT 0 ðbukÞ ¼ wTMðbulÞ, wT 0 ðeÞ ¼ wTM

ðeÞ, e2EðTMÞnfbulg:



Then T 0 2�(d; m1, m2, . . . ,mn�1), and by Theorem 2.5, we have �(T 0)4�(TM),

a contradiction to the choice of TM.

Case 2 TM is not a caterpillar.It is clear that TM has a nonpendant edge uv not in P. Without loss of generality,

assume xu� xv. Let NTM(v)¼ {v1, v2, . . . , vk, u} and let T 0 be the weighted tree

obtained from TM by deleting edges vvi and adding new edges uvi such that

wT 0 ðuviÞ ¼ wTMðvviÞ, wT 0 ðeÞ ¼ wTM

ðeÞ, e 6¼ vvi, i ¼ 1, 2, . . . , k:

Then T 0 2�(d;m1,m2, . . . ,mn�1), and by Theorem 2.5, we have �(T 0)4�(TM),

a contradiction to the choice of TM.

(2) It is obvious that z is in P, i.e. there is a vertex ur(2� r� d ) such that z¼ ur.

The remainder can be proved in the similar way to Case 2 of Claim (1).(3) Write NTM

(z)¼ {z1, z2, . . . , zs, zsþ1, zsþ2}, where zsþ1¼ urþ1, zsþ2¼ ur�1.

First by the Perron–Frobenius theorem, we easily see that �(TM)4 2wTM(zzi).

So for i¼ 1, 2, . . . , s, from R(TM)X¼�(TM)X, we get

xzi ¼wTMðzziÞ

�ðTMÞ � wTMðziÞ� xz ¼

wTMðzziÞ

�ðTMÞ � wTMðzziÞ� xz 5 xz:

Next suppose that there is a vertex uj 6¼ z such that xuj� xz. If uj =2 {u1, udþ1}, then

let T 0 be the weighted tree obtained from TM by deleting the edge z1z and adding the

new edge z1uj such that

wT 0 ðz1ujÞ ¼ wTMðz1zÞ, wT 0 ðeÞ ¼ wTM

ðeÞ, e2EðTMÞnfz1zg:

If uj2 {u1, udþ1}, without loss of generality, assume uj¼ u1, then let T 0 be the

weighted tree obtained from TM by deleting edges ziz and adding edges ziujsuch that

wT 0 ðziujÞ ¼ wTMðzizÞ, wT 0 ðeÞ ¼ wTM

ðeÞ, e 6¼ ziz, i ¼ 1, 2, . . . , sþ 1:

Then T 0 2�(d; m1, m2, . . . ,mn�1), and by Theorem 2.5, we have that �(T 0)4�(TM),

a contradiction to the choice of TM. The additional claim also follows in the similar

method to the case uj2 {u1, udþ1}. g

By Lemma 3.2(1), (2), if we consider no weights of edges, then TM is the tree

shown in Figure 1. In the following we always suppose that

P ¼ 1 2 3 � � � d ðdþ 1Þ

is the path of TM and ei¼ i(iþ 1)2E(TM), i¼ 1, 2, . . . , d. For convenience, write

a1¼ dþ 2, a2¼ dþ 3, . . . , at¼ n (t¼ n� d� 1) when n4 dþ 1.

Figure 1. The tree TM without considering weights.



LEMMA 3.3 Let a¼ k, b¼ kþ 1, u¼ q, v¼ qþ 1 be distinct vertices of P. Then

(1) (xu� xb)[(2xaþ xbþ xu)wTM(ab)� (2xvþ xbþ xu)wTM

(uv)]� 0. In addition,xu¼ xb if and only if (2xaþxbþxu)wTM

(ab)¼ (2xvþ xbþ xu)wTM(uv).

(2) (xv� xa)[(2xbþ xaþ xv)wTM(ab)� (2xuþ xaþxv)wTM

(uv)]� 0. In addition,xv¼ xa if and only if (2xbþ xaþ xv)wTM

(ab)¼ (2xuþ xaþ xv)wTM(uv).

Proof We only give the proof of (1). Assume the contrary, that is

ðxu � xbÞ½ð2xa þ xb þ xuÞwTMðabÞ � ð2xv þ xb þ xuÞwTM

ðuvÞ�4 0,

or only one between xu¼xb and (2xaþxbþxu)wTM(ab)¼ (2xvþ xbþ xu)wTM

(uv)holds. Put �¼wTM

(ab), �¼wTM(uv). Let T 0 be the weighted tree obtained from TM

such that

wT 0 ðabÞ ¼ wTMðabÞ � �, wT 0 ðauÞ ¼ wTM

ðauÞ þ �, wT 0 ðuvÞ ¼ wTMðuvÞ � �,

wT 0 ðvbÞ ¼ wTMðvbÞ þ �, wT 0 ðeÞ ¼ wTM

ðeÞ, e2EðTMÞnfab, uvg,

i.e. T 0 is the weighted tree obtained from TM by deleting the edges ab, uv andadding the new edges au, vb such that wT 0(au)¼wTM

(ab), wT 0(vb)¼wTM(uv). Then

T 0 2�(d;m1,m2, . . . ,mn�1), and from Theorem 2.6, we have that �(T 0)4�(TM),a contradiction to the choice of TM. g

LEMMA 3.4 Let r, l (r5 l ) be two distinct vertices of P such that xr¼ xl. Then

(1) rþ l¼ dþ 2.(2) xi ¼ xd�iþ2, i ¼ 1, 2, . . . , bdþ12 c.

Proof

(1) Assume the contrary, namely rþ l 6¼ dþ 2. Suppose that r� 2 and l� d.We first prove xr�1¼ xlþ1. Assume that xr�1 6¼ xlþ1, then, without loss ofgenerality, assume that xr�14 xlþ1. Let a¼ r� 1, b¼ r, u¼ l, v¼ lþ 1.Then xaþ xb4 xuþxv. By Lemma 3.1(1), we have wTM

(ab)�wTM(uv).

If wTM(ab)4wTM

(uv), then

xu ¼ xb, ð2xa þ xb þ xuÞwTMðabÞ4 ð2xv þ xb þ xuÞwTM

ðuvÞ: ð3:1Þ

If wTM(ab)¼wTM

(uv), then

xa 4 xv, ð2xb þ xa þ xvÞwTMðabÞ ¼ ð2xu þ xa þ xvÞwTM

ðuvÞ: ð3:2Þ

(3.1) and (3.2) contradict the additional claims of Lemma 3.3. Therefore,xr�1¼ xlþ1. If r� 3 and l� d� 1, then replacing the vertices r and l above byr� 1 and lþ 1 respectively, in the similar way above we can get xr�2¼xlþ2.By proceeding in this way, we can get

xr�k ¼ xlþk, k ¼ 0, 1, . . . , minfr� 1, dþ 1� l g: ð3:3Þ

Clearly, (3.3) also holds for r¼ 1 or l¼ dþ 1. If rþ l� dþ 1, then min{r� 1,dþ 1� l}¼ r� 1. Take k¼ r� 1 in (3.3) and set v¼ rþ l� 1. Then x1¼ xv.It is easy to see that 2� v� d, so by Lemma 3.2(3), we get xv4 x1,a contradiction. If rþ l� dþ 3, then min{r� 1, dþ 1� l }¼ dþ 1� l.



Take k¼ dþ 1� l in (3.3) and set v¼ rþ l� d� 1. Then xdþ1¼ xv. It is easy

to see that 2� v� d, so by Lemma 3.2(3), we get xv4 xdþ1, a contradiction.

Therefore, rþ l¼ dþ 2.(2) From rþ l¼ dþ 2 and (3.3), it follows that xi¼ xd�iþ2, i¼ 1, 2, . . . , r.

If r ¼ bdþ12 c, then the results have followed. Hence now assume that r5 bdþ12 cand prove that xrþ1¼ xdþ1�r. Assume that xrþ1 6¼xdþ1�r, then, without loss

of generality, assume that xrþ14 xdþ1�r. Let a¼ rþ 1, b¼ r, u¼ dþ 2� r,

v¼ dþ 1� r. In the similar way with the proof of (1), we will get (3.1) and

(3.2), contradictions to the additional claims of Lemma 3.3. Therefore,

xrþ1¼ xdþ1�r. Repeat the above steps by replacing r with rþ 1, we can get

xi¼ xd�iþ2, i ¼ rþ 1, rþ 2, . . . , bdþ12 c. g

From now on, we always suppose that the vertices of P are relabelled by v1,

v2, . . . , vdþ1 such that xv1� xv2� � � � � xvdþ1.

LEMMA 3.5 For the path P, {i, d� iþ 2}¼ {vd�2iþ2, vd�2iþ3}, i ¼ 1, 2, . . . , bdþ12 c.

Proof For 2� i� d, by the additional claim of Lemma 3.2(3) we have that xi4 x1and xi4 xdþ1. It follows that {x1, xdþ1}¼ {xvd,xvdþ1}. Therefore, {1, dþ 1}¼ {vd, vdþ1},

i.e. the results hold for i¼ 1.

Case 1 Let d be an even number.Suppose that the results hold for 1, 2, . . . , i� 1ð2 � i � bdþ12 cÞ, i.e.

f j, d� jþ 2g ¼ fvd�2jþ2, vd�2jþ3g, j ¼ 1, 2, . . . , i� 1: ð3:4Þ

Next we will show that {i, d� iþ 2}¼ {vd�2iþ2, vd�2iþ3}. By (3.4), we get

f1, 2, . . . , i� 1, d� iþ 3, . . . , d, dþ 1g ¼ fvdþ1, vd, vd�1, vd�2, . . . , vd�2iþ5, vd�2iþ4g:

ð3:5Þ

Assume {i, d� iþ 2} 6¼ {vd�2iþ2, vd�2iþ3}, then we have

fi, d� iþ 2g 6 fvd�2iþ2, vd�2iþ3g, fvd�2iþ2, vd�2iþ3g 6 fi, d� iþ 2g:

So by combining (3.5), we have

fi, d� iþ 2g 6 fvdþ1, vd, vd�1, vd�2, . . . , vd�2iþ5, vd�2iþ4, vd�2iþ3, vd�2iþ2g, ð3:6Þ

fvd�2iþ2, vd�2iþ3g 6 f1, 2, . . . , i� 1, i, d� iþ 2, d� iþ 3, . . . , d, dþ 1g: ð3:7Þ

By (3.6), it is easy to see that there is a k(1� k� d� 2iþ 1) such that either i¼ vk or

d� iþ 2¼ vk. By (3.7), it is easy to see that there is a l(iþ 1� l� d� iþ 1) such that

either l¼ vd�2iþ2 or l¼ vd�2iþ3.First assume that i¼ vk. Write a¼ i� 1, b¼ i, u¼ l, v¼ lþ 1. Then by Lemma 3.3,

we have

ðxu � xbÞ½ð2xa þ xb þ xuÞwTMðabÞ � ð2xv þ xb þ xuÞwTM

ðuvÞ� � 0, ð3:8Þ

ðxv � xaÞ½ð2xb þ xa þ xvÞwTMðabÞ � ð2xu þ xa þ xvÞwTM

ðuvÞ� � 0: ð3:9Þ

Note that k� d� 2iþ 15 d� 2iþ 25 d� 2iþ 3, so we have

xb ¼ xvk � xvd�2iþ1 � xvd�2iþ2 � xvd�2iþ3 :



Therefore, by u¼ l2 {vd�2iþ2, vd�2iþ3}, we get

xb � xu: ð3:10Þ

Since lþ 12 {iþ 2, iþ 3, . . . , d� iþ 1, d� iþ 2}, by (3.5), we have

v ¼ lþ 12 fv1, v2, . . . , vd�2iþ2, vd�2iþ3g:

By taking j¼ i� 1 in (3.4), we have

a ¼ i� 12 fvd�2iþ4, vd�2iþ5g:

Therefore, by the above two equations, we get

xv � xa: ð3:11Þ

If xb4 xu and xv4 xa, then by (3.8) and (3.9), we get, respectively,

wTMðabÞ �

2xv þ xb þ xu2xa þ xb þ xu

� wTMðuvÞ4wTM

ðuvÞ, ð3:12Þ

and

wTMðabÞ �

2xu þ xa þ xv2xb þ xa þ xv

� wTMðuvÞ5wTM

ðuvÞ, ð3:13Þ

a contradiction. Thus we must have that xb� xu or xv� xa. Again by (3.10) and

(3.11), we have that xb¼ xu or xv¼ xa. By Lemma 3.4(1), we get bþ u¼ dþ 2 or

aþ v¼ dþ 2. These contradict bþ u¼ aþ v¼ iþ l� dþ 1.Next assume that d� iþ 2¼ vk. Write a¼ d� iþ 3, b¼ d� iþ 2, u¼ l, v¼ l� 1.

Then (3.8)–(3.10) hold. Since l� 12 {i, iþ 1, . . . , d� i� 1, d� i}, by (3.5), we have

v ¼ l� 12 fv1, v2, . . . , vd�2iþ2, vd�2iþ3g:

By taking j¼ i� 1 in (3.4), we have

a ¼ d� iþ 32 fvd�2iþ4, vd�2iþ5g:

Therefore, by the above two equations, (3.11) still holds. Note that

bþ u ¼ aþ v ¼ dþ ðl� iÞ þ 2 � dþ 3,

so in the similar way to the case i¼ vk, we will also get contradictions.

Case 2 Let d be an odd number.Set d¼ 2r� 1 and suppose the results hold for 1, 2, . . . , i� 1(2� i� r� 1), i.e.

f j, 2r� jþ 1g ¼ fv2r�2jþ1, v2r�2jþ2g, j ¼ 1, 2, . . . , i� 1:

In the similar way to Case 1 we can show that {i, 2r� iþ 1}¼ {v2r�2iþ1, v2r�2iþ2}. So

f j, 2r� jþ 1g ¼ fv2r�2jþ1, v2r�2jþ2g, j ¼ 1, 2, . . . , r� 1:

Again from {1, 2, . . . , 2r}¼ {v1, v2, . . . , v2r}, we get {r, rþ 1}¼ {v1, v2}. g

Let T be a weighted tree shown Figure 1, and without loss of generality,

assume thatwTðza1Þ � wTðza2Þ � � � � � wTðzatÞ:



Then T (any weighted tree isomorphic to it) is called an alternating weighted tree in

edge weights if d¼ 2r, z¼ rþ 1 (if t 6¼ 0) and the weights of all edges of T satisfy

wTðerþ1Þ � wTðerÞ � wTðza1Þ � wTðza2Þ � � � � � wTðzatÞ

� wTðerþ2Þ � wTðer�1Þ � � � � � wTðe2rÞ � wTðe1Þ, ð3:14Þ

or d¼ 2r� 1, z¼ rþ 1 (if t 6¼ 0) and the weights of all edges of P satisfy

wTðerÞ � wTðerþ1Þ � wTðza1Þ � wTðza2Þ � � � � � wTðzatÞ

� wTðer�1Þ � wTðerþ2Þ � wTðer�2Þ � � � � � wTðe2r�1Þ � wTðe1Þ: ð3:15Þ

In particular, when n¼ dþ 1 (i.e. T is a weighted path), T is called an alternating

weighted path in edge weights.By Lemma 3.5, we have {1, dþ 1}¼ {vdþ1, vd}. Without loss of generality, now

assume that (for the other case the proof is quite analogous and the resulting

weighted tree is isomorphic)

vdþ1 ¼ 1, vd ¼ dþ 1: ð3:16Þ

LEMMA 3.6 P is an alternating weighted path in edge weights and v1 ¼ bdþ32 c.

Proof We will distinguish the two following cases.

Case 1 Assume that x1, x2, . . . , xdþ1 are distinct, i.e. xv14 xv24 � � �4 xvdþ1.We first prove the following claim:

vd�2iþ3 ¼ i, vd�2iþ2 ¼ d� iþ 2, i ¼ 1, 2, . . . ,dþ 1

2

� �: ð3:17Þ

By (3.16), the results hold for i¼ 1. For 2 � i � bdþ12 c, suppose that

vd�2jþ3 ¼ j, vd�2jþ2 ¼ d� jþ 2, j ¼ 1, 2, . . . , i� 1:

Next we prove that vd�2iþ3¼ i, vd�2iþ2¼ d� iþ 2. Assume the contrary, i.e. either

vd�2iþ3 6¼ i or vd�2iþ2 6¼ d� iþ 2. By Lemma 3.5, {i, d� iþ 2}¼ {vd�2iþ3, vd�2iþ2}.

So wemust have that vd�2iþ3¼ d� iþ 2, vd�2iþ2¼ i. Write a¼ i� 1, b¼ i, u¼ d� iþ 2,

v¼ d� iþ 3. Then (3.8) and (3.9) hold. It is easy to see that

xb ¼ xi ¼ xvd�2iþ2 4 xvd�2iþ3 ¼ xd�iþ2 ¼ xu:

By the assumptions of induction (take j¼ i� 1), we have

xv ¼ xd�iþ3 ¼ xvd�2iþ4 4 xvd�2iþ5 ¼ xi�1 ¼ xa:

So by (3.8) and (3.9), we get (3.12) and (3.13), a contradiction. Therefore, vd�2iþ3¼ i,

vd�2iþ2¼ d� iþ 2. By induction principle, the claim (3.17) holds.First assume that d¼ 2r. From (3.17), we get v1 ¼ rþ 1 ¼ bdþ32 c. For con-

venience, set v0¼ v1. Then by (3.17), for i¼ r, r� 1, . . . , 2, 1, we have

ei ¼ iðiþ 1Þ ¼ v2r�2iþ3v2r�2iþ1, e2r�iþ1 ¼ ð2r� iþ 1Þð2r� iþ 2Þ ¼ v2r�2iv2r�2iþ2:

It is easy to see that x2rþ x2rþ14 x1þ x2, and for i¼ r, r� 1, . . . , 3, 2,

x2r�iþ1 þ x2r�iþ2 4 xi þ xiþ1 4 x2r�iþ2 þ x2r�iþ3:



So by Lemma 3.1(1), we get wTM(e2r)�wTM

(e1), and for i¼ r, r� 1, . . . , 3, 2,

wTMðe2r�iþ1Þ � wTM

ðeiÞ � wTMðe2r�iþ2Þ:

Hence it is easy to see that the weights of all edges of P satisfies (3.14) (by taking

t¼ 0), i.e. P is an alternating weighted path in edge weights.Next assume that d¼ 2r� 1. By (3.17), we get v1 ¼ rþ 1 ¼ bdþ32 c. For con-

venience, set v0¼ v1, v�1¼ v2. Then by (3.17), for i¼ r, r� 1, . . . , 2, 1, we have that

ei ¼ iðiþ 1Þ ¼ v2r�2iþ2v2r�2i, e2r�i ¼ ð2r� iÞð2r� iþ 1Þ ¼ v2r�2i�1v2r�2iþ1:

It is easy to see that x2r�1þ x2r4 x1þ x2, and for i¼ r, r� 1, . . . , 3, 2,

x2r�i þ x2r�iþ1 4 xi þ xiþ1 4 x2r�iþ1 þ x2r�iþ2:

So by Lemma 3.1(1), we get wTM(e2r�1)�wTM

(e1), and for i¼ r, r� 1, . . . , 3, 2,

wTMðe2r�iÞ � wTM

ðeiÞ � wTMðe2r�iþ1Þ:

Hence it is easy to see that the weights of all edges of P satisfies (3.15) (by taking

t¼ 0), i.e. P is an alternating weighted path in edge weights.

Case 2 Assume that at least two of x1, x2, . . . , xdþ1 are equal.By Lemmas 3.4(2) and 3.5, we get, respectively,

xi ¼ xd�iþ2, i ¼ 1, 2, . . . ,dþ 1

2

� �, ð3:18Þ

fi, d� iþ 2g ¼ fvd�2iþ2, vd�2iþ3g, i ¼ 1, 2, . . . ,dþ 1

2

� �: ð3:19Þ

First assume that d¼ 2r. By (3.19), we get v1 ¼ rþ 1 ¼ bdþ32 c. By (3.18), we get

xi þ xiþ1 ¼ x2r�iþ1 þ x2r�iþ2, i ¼ 1, 2, . . . , r:

So by Lemma 3.1(3), we get

wTMðeiÞ ¼ wTM

ðe2r�iþ1Þ, i ¼ 1, 2, . . . , r: ð3:20Þ

Again by (3.18) and (3.19), we get xi¼xv2r�2iþ3, i¼ 1, 2, . . . , rþ 1. Therefore, for

i¼ 1, 2, . . . , r� 1, we have

xi þ xiþ1 ¼ xv2r�2iþ3 þ xv2r�2iþ1 � xv2r�2iþ1 þ xv2r�2i�1 ¼ xiþ1 þ xiþ2:


wTMðeiÞ � wTM

ðeiþ1Þ, i ¼ 1, 2, . . . , r� 1: ð3:21Þ

By (3.20) and (3.21), it is easy to see that the weights of all edges of P satisfies (3.14)

(by taking t¼ 0), i.e. P is an alternating weighted path in edge weights.Next assume that d¼ 2r� 1. From (3.18) and (3.19), we have xr¼ xrþ1,

{r, rþ 1}¼ {v1, v2}. It follows xv1¼ xv2. So, without loss of generality, we may take

v1 ¼ rþ 1 ¼ bdþ32 c. Again by (3.18), we have

xi þ xiþ1 ¼ x2r�i þ x2r�iþ1, i ¼ 1, 2, . . . , r� 1:




wTMðeiÞ ¼ wTM

ðe2r�iÞ, i ¼ 1, 2, . . . , r� 1: ð3:22Þ

For convenience, set v0¼ v1. Again by (3.18) and (3.19), we get xi¼ xv2r�2iþ2,i¼ 1, 2, . . . , rþ 1. Therefore, for i¼ 1, 2, . . . , r� 1, we have

xi þ xiþ1 ¼ xv2r�2iþ2 þ xv2r�2i � xv2r�2i þ xv2r�2i�2 ¼ xiþ1 þ xiþ2:


wTMðeiÞ � wTM

ðeiþ1Þ, i ¼ 1, 2, . . . , r� 1: ð3:23Þ

By (3.22) and (3.23), it is easy to see that the weights of all edges of P satisfies (3.15)(by taking t¼ 0), i.e. P is an alternating weighted path in edge weights. g

LEMMA 3.7 Suppose that n4 dþ 1 and wTM(za1)�wTM

(za2)� � � � �wTM(zat). Then

wTM(ez�1)�wTM

(za1) and wTM(ez)�wTM

(za1).

Proof By Lemmas 3.2(3) and 3.6, we have z ¼ v1 ¼ bdþ32 c � 3.

First prove wTM(ez�1)�wTM

(za1). Assume the contrary. By Lemma 3.1(2), itfollows that xz�1þ xz5 xzþ xa1, i.e. xz�15 xa1. Let T

0 be the weighted tree obtainedfrom TM by deleting the edge ez�2 and adding the edge (z� 2)a1 such that

wT 0 ððz� 2Þa1Þ ¼ wTMðez�2Þ, wT 0 ðeÞ ¼ wTM

ðeÞ, e2EðTMÞnfez�2g:

Then T 0 2�(d;m1,m2, . . . ,mn�1), and by Theorem 2.5, we have �(T 0)4�(TM),a contradiction to the choice of TM.

Next prove wTM(ez)�wTM

(za1). If d� 4, in the similar way above, we can showwTM

(ez)�wTM(za1). If d¼ 3, then ez, za1, . . . , zat are symmetric pendant edges. Thus

without loss of generality, we may assume wTM(ez)�wTM

(za1). g

LEMMA 3.8 Suppose that n4 dþ 1 and wTM(za1)�wTM

(za2)� � � � �wTM(zat). Then

wTM(zat)�wTM

(ez�2) for d� 3 and wTM(zat)�wTM

(ezþ1) for d� 4.

Proof Assume d� 3. Now we prove that wTM(zat)�wTM

(ez�2). By Lemmas 3.2(3)and 3.6, z ¼ v1 ¼ b

dþ32 c � 3. Let P(k, l ) be the weighted subpath between the vertex

k and the vertex l in P, including k and l. In particular, P(i, i ) is an isolated vertex P1.Let H be the weighted tree containing the vertex z in TM� ez�2. For convenience,denote wTM

(e) by w(e). Set H1¼H� at, q¼ dþ 1, �¼�(Rz(P(z, q))),

gðkÞ ¼Yt�1i¼k

ðx� wðzaiÞÞ, hðkÞ ¼Xt�1i¼k

x � � � wðzaiÞ

x� wðzaiÞ,

where for k� t, set g(k)¼ 1 and h(k)¼ 0; while for s¼ 0, set �(P(1, s))¼ 1 andw(es)¼�(Rs(P(1, s)))¼ 0. Write u¼ z� 2, v¼ z� 1¼ a0. By Theorem 2.7, we get

�ðTMÞ ¼ �ðPð1, uÞÞ�ðH Þ � wðeuÞ½�ðPð1, uÞÞ�ðRvðH ÞÞ þ �ðH Þ�ðRuðPð1, uÞÞÞ�,

�ðH Þ ¼ x�ðH1Þ � wðzatÞ½�ðH1Þ þ x�gð0Þ�:

By the definition of Rv(G), we get

�ðRvðH ÞÞ ¼ �ðH� a0Þ � wðza0Þ�ðRzðH� a0ÞÞ ðuse Theorem 2.7 to �ðH� a0ÞÞ

¼ ðx� wðzatÞÞ�ðH1 � a0Þ � ½wðzatÞxþ wðza0Þðx� wðzatÞÞ��gð1Þ,

�ðRuðPð1, uÞÞÞ ¼ �ðPð1, z� 3ÞÞ � wðez�3Þ�ðRz�3ðPð1, z� 3ÞÞÞ:



Let ¼w(eu)w(zat). Then by the above four equations, we get

�ðTMÞ ¼ F1ðwðeuÞ,wðzatÞÞ þ F2ðwðeuÞ,wðzatÞÞ, ð3:24Þ

where

F1ðwðeuÞ,wðzatÞÞ ¼ x�ðPð1, uÞÞ�ðH1Þ

þ �ðPð1, uÞÞ½�ðH1 � a0Þ þ x�gð1Þ � wðza0Þ�gð1Þ�

þ ½�ðPð1, z� 3ÞÞ � wðez�3Þ�ðRz�3ðPð1, z� 3ÞÞÞ�½�ðH1Þ þ x�gð0Þ�,

F2ðwðeuÞ,wðzatÞÞ ¼ �wðzatÞ�ðPð1, uÞÞ�ðH1Þ � xwðeuÞ�ðPð1, uÞÞ�ðH1 � a0Þ

þ x��ðPð1, uÞÞ½�wðzatÞ gð0Þ þ wðeuÞwðza0Þ gð1Þ�

� xwðeuÞ�ðH1Þ½�ðPð1, z� 3ÞÞ � wðez�3Þ�ðRz�3ðPð1, z� 3ÞÞÞ�:

ð3:25Þ

Let ¼w(eu)þw(zat). By the definition of g(k), we have

�wðzatÞ gð0Þ þ wðeuÞwðza0Þ gð1Þ ¼ ½�xwðzatÞ þ wðza0Þ� gð1Þ: ð3:26Þ

By Theorem 2.7, we have

�ðPð1, uÞÞ ¼ x�ðPð1, z� 3ÞÞ � wðez�3Þ½�ðPð1, z� 3ÞÞ þ x�ðRz�3ðPð1, z� 3ÞÞÞ�,

i.e.

x½�ðPð1, z� 3ÞÞ � wðez�3Þ�ðRz�3ðPð1, z� 3ÞÞÞ� ¼ �ðPð1, uÞÞ þ wðez�3Þ�ðPð1, z� 3ÞÞ:

ð3:27Þ

By placing (3.26) and (3.27) into (3.25), it follows that

F2ðwðeuÞ,wðzatÞÞ ¼ ½��ðH1Þ þ x�wðza0Þ gð1Þ��ðPð1, uÞÞ

� xwðeuÞ�ðPð1, uÞÞ�ðH1 � a0Þ � x2�wðzatÞ�ðPð1, uÞÞ gð1Þ

� wðeuÞwðez�3Þ�ðPð1, z� 3ÞÞ�ðH1Þ: ð3:28Þ

Since F1(w(eu),w(zat)) contains no w(eu) and w(zat) apart from , it follows that

F1ðwðeuÞ,wðzatÞÞ ¼ F1ðwðzatÞ,wðeuÞÞ: ð3:29Þ

Assume w(ez�2)4w(zat), i.e. w(eu)4w(zat). Let T0 be the weighted graph formed

from TM by exchanging the weights of eu and zat while keeping the weights of other

edges not changed. By (3.24), (3.28) and (3.29), we get

�ðTMÞ � �ðT0Þ ¼ F2ðwðeuÞ,wðzatÞÞ � F2ðwðzatÞ,wðeuÞÞ ¼ ðwðeuÞ � wðzatÞÞ4,

where

4 ¼ x�ðPð1, z� 2ÞÞ½��ðH1 � a0Þ þ x�gð1Þ� � wðez�3Þ�ðPð1, z� 3ÞÞ�ðH1Þ: ð3:30Þ

By setting G¼P(z, q) in Corollary 2.8(2), we get

�ðH1Þ ¼ ½�ðPðz, qÞÞ � hð0Þ� gð0Þ: ð3:31Þ

�ðH1 � a0Þ ¼ ½�ðPðz, qÞÞ � hð1Þ� gð1Þ: ð3:32Þ



By placing (3.31) and (3.32) into (3.30), we get i¼ g(1)�, where

� ¼x�ðPð1, z� 2ÞÞ

wðza0Þ½�wðza0Þ�ðPðz, qÞÞ þ x�wðza0Þ�

� wðez�3Þ�ðPð1, z� 3ÞÞ½ðx� wðza0ÞÞ�ðPðz, qÞÞ � x�wðza0Þ�

þ wðez�3Þðx� wðza0ÞÞ�ðPð1, z� 3ÞÞhð1Þ þ xhð1Þ�ðPð1, z� 2ÞÞ:

Now assume x��(TM). Then x4 2w(za0) and �(P(k, l ))h(1)� 0, so we get

� �x�ðPð1, z� 2ÞÞ

wðza0Þ½�wðza0Þ�ðPðz, qÞÞ þ x�wðza0Þ�

� wðez�3Þ�ðPð1, z� 3ÞÞ½ðx� wðza0ÞÞ�ðPðz, qÞÞ � x�wðza0Þ�: ð3:33Þ

By using Theorem 2.7 to �(P(z� 1, q)), we get

�ðPðz� 1, qÞÞ ¼ ðx� wðza0ÞÞ�ðPðz, qÞÞ � x�wðza0Þ: ð3:34Þ

By using Corollary 2.8(1) to �(P(z� 1, q)) in (3.34), we get

wðza0ÞwðezÞ�ðPðzþ 1, qÞÞ ¼ �wðza0Þ�ðPðz, qÞÞ þ x�wðza0Þ: ð3:35Þ

So by (3.33) to (3.35), it follows that

� � xwðezÞ�ðPð1, z� 2ÞÞ�ðPðzþ 1, qÞÞ � wðez�3Þ�ðPð1, z� 3ÞÞ�ðPðz� 1, qÞÞ:

Since P1 [ Pðzþ 1, qÞ is a proper spanning subgraph of P(z, q), from Theorem 2.9,

x�ðPðzþ 1, qÞÞ ¼ � P1

[Pðzþ 1, qÞ

� �4�ðPðz, qÞÞ:

Therefore, by combining w(ez)�w(ez�3), we get

�4wðezÞ�ðPð1, z� 2ÞÞ�ðPðz, qÞÞ � wðez�3Þ�ðPð1, z� 3ÞÞ�ðPðz� 1, qÞÞ

� wðez�3Þ�,

where

� ¼ �ðPð1, z� 2ÞÞ�ðPðz, qÞÞ � �ðPð1, z� 3ÞÞ�ðPðz� 1, qÞÞ:

DefineQ0

i¼1 w2ðeiÞ ¼ 1. Since w(e1)�w(e2)� � � � �w(ez�2)�w(ez�1), by repeating

using Corollary 2.8(1), we get

� ¼ ½ðx� 2wðez�3ÞÞ�ðPð1, z� 3ÞÞ � wðez�3Þwðez�4Þ�ðPð1, z� 4ÞÞ��ðPðz, qÞÞ

� �ðPð1, z� 3ÞÞ½ðx� 2wðez�1ÞÞ�ðPðz, qÞÞ � wðee�1ÞwðezÞ�ðPðzþ 1, qÞÞ�

¼ 2½wðez�1Þ � wðez�3Þ��ðPð1, z� 3ÞÞ�ðPðz, qÞÞ

þ wðez�1ÞwðezÞ�ðPð1, z� 3ÞÞ�ðPðzþ 1, qÞÞ

� wðez�3Þwðez�4Þ�ðPð1, z� 4ÞÞ�ðPðz, qÞÞ

� wðez�1ÞwðezÞ�ðPð1, z� 3ÞÞ�ðPðzþ 1, qÞÞ

� wðez�3Þwðez�4Þ�ðPð1, z� 4ÞÞ�ðPðz, qÞÞ

� ½�ðPð1, z� 3ÞÞ�ðPðzþ 1, qÞÞ � �ðPð1, z� 4ÞÞ�ðPðz, qÞÞ�w2ðez�3Þ



� � � �

� ½�ðPð1, 1ÞÞ�ðPð2z� 3, qÞÞ � �ðPð1, 0ÞÞ�ðPð2z� 4, qÞÞ�Yz�3i¼1

w2ðeiÞ

¼ ½�ðP1

[Pð2z� 3, qÞÞ � �ðPð2z� 4, qÞÞ�

Yz�3i¼1

w2ðeiÞ:

Note that P1 [ Pð2z� 3, qÞ is a weighted proper subgraph of P(2z� 4, q), by Theorem

2.9, for x��(P(2z� 4, q)), we have �ðP1 [ Pð2z� 3, qÞÞ � �ðPð2z� 4, qÞÞ4 0: But�(TM)4�(P(2z� 4, q)), so for x��(TM), �4 0, i.e. D¼ g(1)�4 0. Thus when

x��(TM), �(TM)4�(T 0). This implies �(TM)5�(T 0), a contradiction to the choice

of TM. Therefore, wTM(zat)�wTM

(ez�2). In the similar way, we can prove that

wTM(zat)�wTM

(ezþ1) for d� 4. g

Combining Lemmas 3.6–3.8, it is easy to see that the weights of all edges of TM

satisfy (3.14) or (3.15), i.e. TM is an alternating weighted tree in edge weights. Note

that, for given n, d and positive weight set {m1,m2, . . . ,mn�1}, the alternating

weighted tree in edge weights is uniquely determined. Therefore, by the choice of TM,

we immediately get the following result.

THEOREM 3.9 The alternating weighted tree in edge weights is the unique graph in

�(d;m1,m2, . . . ,mn�1) having the largest signless Laplacian spectral radius.

Remark Let G¼ (V1,V2,E ) be a connected bipartite weighted graph with n vertices

and suppose that V1¼ {v1, v2, . . . , vk}, V2¼ {vkþ1, vkþ2, . . . , vn}. Let U¼ (uij) be the

n� n diagonal matrix with uii¼ 1 if 1� i� k, and uii¼�1 if kþ 1� i� n. It is easy

to show that U�1 L(G)U¼R(G). This indicates that L(G) and R(G) have the same

spectrum. Hence by Theorem 3.9, we get the main result in the article.

THEOREM 3.10 The alternating weighted tree in edge weights is the unique graph in

�(d;m1,m2, . . . ,mn�1) having the largest Laplacian spectral radius.

Example In Figure 2, two alternating weighted trees are displayed, where the

numbers denote the weights corresponding to the edges. The first has the largest

Laplacian spectral radius in �(8; 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2), while the second one

has the largest Laplacian spectral radius in �(7; 11, 10, 9, 8, 7, 6, 5, 4, 3, 2).

Figure 2. Examples of two alternating weighted trees.



THEOREM 3.11 Let TM¼Tn,d. Then �(Tn,d) is strictly decreasing in d(2� d� n� 1).

Proof Assume d� 3. By Lemma 3.6, we have v1 ¼ bdþ32 c � 3. Let T 0 be the weighted

tree formed from Tn,d by deleting the edge e1¼ 12 and adding the edge v11 such that

wT 0 ðv11Þ ¼ wTn,dðe1Þ, wT 0 ðeÞ ¼ wTn,d

ðeÞ, e2EðTn,dÞnfe1g:

Then T 0 2�(d� 1;m1,m2, . . . ,mn�1). Since xv1� x2, by Remark, Theorems 2.5 and3.10, we get

�ðTn,dÞ ¼ �ðTn,dÞ5�ðT 0Þ � �ðTn,d�1Þ ¼ �ðTn,d�1Þ: g

By Theorems 3.10 and 3.11, we immediately get the following results.

COROLLARY 3.12 Let T be a weighted tree with n vertices, diameter at least d andpositive weight set. Then �(T )� �(Tn,d), with equality if and only if T¼Tn,d.

COROLLARY 3.13 Let T be a weighted tree with n� 3 vertices and positive weight set.Then �(T )� �(Tn,2), with equality if and only if T¼Tn,2 (a weighted star).

COROLLARY 3.14 Let T 6¼Tn,2 be a weighted tree with n� 4 vertices and positiveweight set. Then �(T )� �(Tn,3), with equality if and only if T¼Tn,3.

Acknowledgements

The author would like to thank the referees for giving many valuable comments andsuggestions towards improving this article. This article was supported by the National NaturalScience Foundation of China (10871204).

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