Linear Algebra and its Applications 439 (2013) 2784–2789

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Linear Algebra and its Applications

www.elsevier.com/locate/laa

Edge-disjoint spanning trees and eigenvaluesof graphs

Guojun Li a,1, Lingsheng Shi b,∗,2

a Department of Mathematics, Beijing Jiaotong University, Beijing, 100044, Chinab Department of Mathematical Sciences, Tsinghua University, Beijing, 100084, China

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

Article history:Received 20 November 2012Accepted 30 August 2013Available online 18 September 2013Submitted by R. Brualdi

MSC:05C5015A1805C4015A42

Keywords:Edge-disjointSpanning treeEigenvalueConnectivity

It is proven that a graph of order n with minimum degree δ � 2khas k edge-disjoint spanning trees if its second largest eigenvalueis less than δ − 2k−1

δ+1 − O (1/n). This extends earlier results ofCioaba and Wong and improves results of Gu, Lai, Li and Yao, andalso confirms approximately their conjectures for sufficiently largegraphs with respect to the minor term. We also conjecture that theminor term can be removed.

© 2013 Elsevier Inc. All rights reserved.

1. Introduction

In this paper, we consider finite, undirected and simple graphs. For a graph G , κ ′(G) denotes theedge connectivity of G and τ (G) denotes the maximum number of edge-disjoint spanning trees in G .Let G be an undirected graph on n vertices with vertex set {v1, v2, . . . , vn}. The adjacency matrix of Gis an n by n matrix A(G) = (aij) where aij is the number of edges between vi and v j for 1 � i, j � n.

* Corresponding author.E-mail addresses: 11121752@bjtu.edu.cn (G. Li), lshi@math.tsinghua.edu.cn (L. Shi).

1 Supported by National Natural Science Foundation of China (No. 11101027) and Fundamental Research Funds for the CentralUniversities of China (No. 2011JBM136).

2 Supported by Tsinghua University Initiative Scientific Research Program.

0024-3795/$ – see front matter © 2013 Elsevier Inc. All rights reserved.http://dx.doi.org/10.1016/j.laa.2013.08.041

G. Li, L. Shi / Linear Algebra and its Applications 439 (2013) 2784–2789 2785

By the definition, if G is simple, then A(G) is a symmetric (0,1)-matrix. The eigenvalues of G are theeigenvalues of A(G). We use λi(G) to denote the i-th largest eigenvalue of G . With these notations,we always have λ1(G) � λ2(G) � · · · � λn(G).

Motivated by a question of Seymour, Cioaba and Wong [3] posed the following conjecture.

Conjecture 1. (See [3].) Let d � 4 and 2 � k � �d/2� be two integers. If G is a d-regular graph such thatλ2(G) < d − 2k−1

d+1 , then τ (G) � k.

Cioaba and Wong [3] proved the conjecture for k = 2,3 and showed by examples that the boundon λ2 is sharp in each of these cases. Their results were extended to general graphs by Gu, Lai, Li andYao [4].

Theorem 1. (See [4].) Let k be an integer with k � 2 and G be a graph with minimum degree δ � 2k andmaximum degree �. If λ2(G) < 2δ − � − 3k−1

δ+1 , then τ (G) � k.

In this paper, we obtain the following result which not only improves Theorem 1, but also impliesa stronger version of Conjecture 1 for graphs with large minimum degree, and an approximate versionof Conjecture 1 for large graphs with small minimum degree.

Theorem 2. Let k � 2 be an integer and let G be a graph of order n with minimum degree δ � 2k. Thenτ (G) � k if

δ − λ2(G) >

⎧⎪⎪⎨⎪⎪⎩

n(k−1)

n(δ+1)−(δ+1)2 for n � 3δ + 2,

8k−73(δ+1)

for 3(δ + 1) � n � 5(δ + 1),

2k−1δ+1 + 2(k−1)

n−2(δ+1)for n > 5(δ + 1).

In order to prove Theorem 2, we extend to a general version a theorem of Cioaba [2] in whichonly regular graphs are considered.

Theorem 3. Let k be an integer with k � 2 and let G be a graph of order n with minimum degree δ � k. Ifλ2(G) < δ − n(k−1)

n(δ+1)−(δ+1)2 , then κ ′(G) � k.

In view of Theorem 2, we extend Conjecture 1 to general graphs.

Conjecture 2. Let k be an integer with k � 2 and let G be a graph with minimum degree δ � 2k. If λ2(G) <

δ − 2k−1δ+1 , then τ (G) � k.

Note that every graph of order n with minimum degree δ � �n/2� is δ-edge connected and so has�δ/2� edge-disjoint spanning trees by Nash-Williams and Tutte’s theorem [5,6]. Then Conjecture 2 istrue for δ � (n − 2)/3 by Theorem 2, and also true for k = 2,3 by applying Lemma 2 in the proofof [4, Theorem 1.7]. In Section 2, we display some preliminaries and mechanisms, including conse-quences of the Cauchy interlace theorem and quotient matrices. These will be applied in the proofsof the main results, to be presented in the last section.

2. Preliminaries

In this section, we present some of the preliminaries and former result to be used in our argu-ments. Throughout this section, G always denotes a simple graph.

Lemma 1. (See [4].) Let G be a graph with minimum degree δ and U be a non-empty proper subset of V (G). Ife(U , V \ U ) � δ − 1, then |U | � δ + 1.

2786 G. Li, L. Shi / Linear Algebra and its Applications 439 (2013) 2784–2789

The following lemma follows easily from the Perron–Frobenius theorem, see [1] for instance.

Lemma 2. Let B = (bij) be a real nonnegative n × n matrix whose underlying directed graph is stronglyconnected. Then

min1�i�n

n∑j=1

bij � λ1(B) � max1�i�n

n∑j=1

bij .

Given two real sequences θ1 � θ2 � · · · � θn and η1 � η2 � · · · � ηm with n > m, the second se-quence is said to interlace the first one if θi � ηi � θn−m+i , for i = 1,2, . . . ,m. When we say theeigenvalues of a matrix B interlace the eigenvalues of a matrix A, it means the non-increasing eigen-value sequence of B interlaces that of A.

Suppose that we partition V (G) into s non-empty subsets V 1, V 2, . . . , V s . The quotient matrix As =A(V 1, V 2, . . . , V s) of G with respect to this partition, is an s by s matrix (bij) such that bij is theaverage number of neighbors in V j of the vertices of V i for 1 � i, j � s. As As is an s by s square realmatrix, the following equality is well known from linear algebra,

λ1(As) + λ2(As) + · · · + λs(As) = tr(As).

Lemma 3. (See [1].) The eigenvalues of a quotient matrix of a graph G interlace the eigenvalues of G.

Let S and T be disjoint subsets of V (G). We denote by E(S, T ) the set of edges each of whichhas one vertex in S and the other in T . Let e(S, T ) = |E(S, T )|. We denote by G[S] the subgraph of Ginduced by S , and by d(G) the average degree of G . The next useful lemma follows immediately fromthe Cauchy interlace theorem [1].

Lemma 4. (See [3].) Let S and T be disjoint subsets of V (G) and e(S, T ) = 0. Then

λ2(G) � λ2(G[S ∪ T ]) � min

{λ1

(G[S]), λ1

(G[T ])} � min

{d(G[S]),d

(G[T ])}.

The following two lemmas are easy exercises of calculus.

Lemma 5. Let p � 6 and f (x) := p−3xx2−1

. Then − 92 (p + √

p2 − 9 )−1 � f (x) � p/3 − 2 for x � 2.

Lemma 6. Let 0 < a � b � c, 0 < d1 � d2 and n � 2d1 + d2 . Define f (x, y, z) := ax + b

y + cz . Then for all

(x, y, z) ∈ {(x, y, z) ∈ R3 | x + y + z = n, x, y � d1, z � d2},

(√

a + √b + √

c )2/n � f � b

d1+ max

{a

d1+ c

n − 2d1,

a

n − d1 − d2+ c

d2

}.

In particular, if d1 = d2 then f � an−2d1

+ bd1

+ cd1

.

3. Proof of theorems

Proof of Theorem 3. Assume to the contrary that κ ′(G) � k −1. Then there exists a non-empty propersubset U ⊂ V (G) such that e(U , V \ U ) � k − 1. Let W = V \ U and r = e(U , W ). By Lemma 1, |U | �δ + 1 and |W | � δ + 1. The quotient matrix of G with respect to the partition (U , W ) is

Q :=(

d − r|U |

r|U |

r|W | d′ − r

|W |

),

where d denotes the average degree of the vertices in U and d′ denotes the average degree of thevertices in W respectively. We emphasize that d is not the average degree of the subgraph of Ginduced by U , and so d � δ and similarly d′ � δ. By Lemma 2, λ1(Q ) � max{d,d′}, and so by Lemma 3,

G. Li, L. Shi / Linear Algebra and its Applications 439 (2013) 2784–2789 2787

λ2(G) � λ2(Q ) = tr(Q ) − λ1(Q ) � tr(Q ) − max{

d,d′} = min{

d,d′} − r

|U | − r

|W |� δ − k − 1

δ + 1− k − 1

n − δ − 1= δ − n(k − 1)

n(δ + 1) − (δ + 1)2,

which is a contradiction. This completes the proof. �Proof of Theorem 2. Since Conjecture 2 is true for k = 2 and 3, we may assume that k � 4. Assume tothe contrary that τ (G) � k−1. By Nash-Williams and Tutte’s theorem, there exists an edge subset X ⊂E(G) such that |X | � k(t − 1) − 1 where t is the number of components of G − X . Let G1, G2, . . . , Gtbe the components of G − X . For 1 � i � t , let V i = V (Gi) and ri = e(V i, V \ V i). Without loss ofgenerality, we always assume that

r1 � r2 � · · ·� rt .

With these notations and by |X |� k(t − 1) − 1, we have∑1�i�t

e(V i, V j) � k(t − 1) − 1 = kt − k − 1.

Note that

n(k − 1)

n(δ + 1) − (δ + 1)2<

2k − 1

δ + 1+ 2(k − 1)

n − 2(δ + 1), and

n(k − 1)

n(δ + 1) − (δ + 1)2<

3k − 1

δ + 1for n � 3(δ + 1).

Theorem 3 implies that r1 � k. Let xl denote the multiplicity of l in {r1, r2, . . . , rt} for 1 � l � 2k − 1.Then x j = 0 for j = 1,2, . . . ,k − 1. It follows that

r1xr1 + (r1 + 1)xr1+1 + · · · + (2k − 1)x2k−1 + 2k[t − (xr1 + xr1+1 + · · · + x2k−1)

]

�t∑

i=1

ri � 2kt − 2(k + 1),

which implies that

2(k + 1)� (2k − r1)xr1 + · · · + 2x2k−2 + x2k−1 � (2k − r1)

2k−1∑i=r1

xi . (1)

Thus∑2k−1

i=k xi � 2 + 2/k� = 3 and r1 � r2 � r3 � 2k − 1. By Lemma 1, |V i| � δ + 1 for 1 � i � 3 whichcontradicts n � 3δ + 2. This completes the proof for n � 3δ + 2.

Now we consider n � 3(δ + 1). Let U = V \ (V 1 ∪ V 2). Then |U | � |V 3| � δ + 1. The following claimmakes Lemma 2 applicable to the quotient matrix with respect to the partition (V 1, V 2, U ).

Claim. There exist no indices p and q with 1 � p �= q � t such that e(V p, Vq) = 0 and rp, rq � 2k − 1.

Assume to the contrary that there exists such a pair of indices. By Lemma 1, |V p | � δ + 1 and|Vq| � δ+1. It follows that d(G p) � δ− 2k−1

|V p | � δ− 2k−1δ+1 and d(Gq) � δ− 2k−1

|Vq | � δ− 2k−1δ+1 . By Lemma 4,

λ2(G) � min{d(G p),d(Gq)} � δ − 2k−1δ+1 , contrary to all assumptions. Thus the claim is proven.

Let e(V 1, V 2) = y with 0 < y < r1. The quotient matrix of G with respect to the partition(V 1, V 2, U ) is

Q :=⎛⎜⎝

d1 − r1|V 1|y

|V 1|r1−y|V 1|

y|V 2| d2 − r2|V 2|

r2−y|V 2|

r1−y r2−y d′ − r1+r2−2y

⎞⎟⎠ ,

|U | |U | |U |

2788 G. Li, L. Shi / Linear Algebra and its Applications 439 (2013) 2784–2789

where di denotes the average degree of V i in G for i = 1,2 and d′ denotes the average degree of Uin G respectively. Let λ2 := λ2(G). By Lemma 2, we obtain that

λ2 � λ2(Q ) = 1

2

[tr(Q ) − λ1(Q )

]� 1

2

[tr(Q ) − max

{d1,d2,d′}]

� δ − 1

2

(r1

|V 1| + r2

|V 2| + r1 + r2 − 2y

|U |)

� δ − 1

2

(r1

|V 1| + r2

|V 2| + r1 + r2 − 2

|U |)

(2)

as y � 1. Since r2 � r1 � k, we have that r1+22k−r2

� 1 + 2k which implies c := r1+2

2k−r2�� 2. Then r1 + 2 �

c(2k − r2) and thus

(c + 1)r1 � r1 + cr2 � 2ck − 2. (3)

By (1), we have

(2k − r2)xr2 + (2k − r2 − 1)xr2+1 + · · · + 2x2k−2 + x2k−1 � 2(k + 1) − (2k − r1) = r1 + 2.

It follows that xr2 + xr2+1 +· · ·+ x2k−2 + x2k−1 � c and r1 � r2 � · · · � rc � rc+1 � 2k − 1. By Lemma 1,|V i | � δ + 1 for i = 1,2, . . . , c + 1. Then |U | � |V 3| + · · · + |V c+1| � (c − 1)(δ + 1). So by (2), (3) andLemma 5, we have

δ − λ2 �1

2

(r1

|V 1| + r2

|V 2| + r1 + r2 − 2

|U |)

� 1

2

(r1

δ + 1+ r2

δ + 1+ r1 + r2 − 2

(c − 1)(δ + 1)

)

= cr1 + cr2 − 2

2(c − 1)(δ + 1)�

( c−1c+1 + 1)(ck − 1) − 1

(c − 1)(δ + 1)

= 2k

δ + 1+ 2k − 1 − 3c

(c2 − 1)(δ + 1)� 8k − 7

3(δ + 1)

as c � 2 and k � 4. This contradicts the condition for 3(δ + 1) � n � 5(δ + 1). So next we assume thatn > 5(δ + 1). Now the proof splits into two cases according to the value of c.

Case 1. c = 2.

By (3), we have 3r1 � r1 + 2r2 � 4k − 2, and so by (2) and Lemma 6,

δ − λ2 �1

2

(r1

|V 1| + r2

|V 2| + r1 + r2 − 2

|U |)

� 1

2

(r1

n − 2(δ + 1)+ r2

δ + 1+ r1 + r2 − 2

(δ + 1)

)

� 2(k − 1)

(δ + 1)+ 2k − 1

3n − 6(δ + 1)

<2k − 1

δ + 1+ 2(k − 1)

n − 2(δ + 1).

Case 2. c � 3.

Note that r1 � r2 � 2k − 1 and so by (2) and Lemma 6,

G. Li, L. Shi / Linear Algebra and its Applications 439 (2013) 2784–2789 2789

δ − λ2 �1

2

(r1

|V 1| + r2

|V 2| + r1 + r2 − 2

|U |)

� 1

2

(r2

δ + 1+ max

{r1

δ + 1+ r1 + r2 − 2

n − 2(δ + 1),

r1

n − 3(δ + 1)+ r1 + r2 − 2

2(δ + 1)

})

� max

{2k − 1

δ + 1+ 2(k − 1)

n − 2(δ + 1),

4k − 3

2(δ + 1)+ 2k − 1

2n − 6(δ + 1)

}

= 2k − 1

δ + 1+ 2(k − 1)

n − 2(δ + 1)

as n > 5(δ + 1). In either case we reach a contradiction. This completes the proof. �Acknowledgement

The authors thank an anonymous referee for pointing out an error in Lemma 6 in an earlier versionof the paper and many useful suggestions.

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