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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
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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.
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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).
176 S.-W. Tan and J.-J. Jiang
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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].
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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).
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(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
xwðzaiÞ�ðRzðGÞÞ
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
xwðzaiÞ�ðRzðGÞÞ
x� wðzaiÞ
" #Yt�1i¼1
ðx� wðzaiÞÞ
� xwðzatÞ�ðRzðGÞÞYt�1i¼1
ðx� wðzaiÞÞ
¼ �ðGÞ �Xti¼1
xwðzaiÞ�ðRzðGÞÞ
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
Linear and Multilinear Algebra 179
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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:
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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.
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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.
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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 :
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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:
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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:
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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:
So by Lemma 3.1(1), we get
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:
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So by Lemma 3.1(3), we get
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:
So by Lemma 3.1(1), we get
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:
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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
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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
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� � � �
� 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.
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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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