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On the sum of the two largest Laplacian eigenvalues of trees

Journal of Inequalities and Applications20142014:242

https://doi.org/10.1186/1029-242X-2014-242

  • Received: 23 December 2013
  • Accepted: 5 June 2014
  • Published:

Abstract

For S ( T ) , the sum of the two largest Laplacian eigenvalues of a tree T, an upper bound is obtained. Moreover, among all trees with n 4 vertices, the unique tree which attains the maximal value of S ( T ) is determined.

MSC:05C50.

Keywords

  • upper bound
  • tree
  • Laplacian matrix

1 Introduction

Let V ( G ) be the vertex set and E ( G ) be the edge set of a graph G. The numbers of vertices and edges of G are denoted by n ( G ) and m ( G ) , respectively. For a vertex v V ( G ) , let N G ( v ) be the set of vertices adjacent to v and d G ( v ) = | N G ( v ) | be the degree of v. Particularly, denote by Δ ( G ) the maximum degree of G. The diameter of a connected graph G, denoted by d ( G ) , is the maximum distance among all pairs of vertices in G. Let A ( G ) be the adjacency matrix of G and D ( G ) be the diagonal matrix of vertex degrees. The matrix D ( G ) A ( G ) is called the Laplacian matrix of G and its eigenvalues are called the Laplacian eigenvalues of G. Let μ 1 μ 2 μ n be the Laplacian eigenvalues of a graph G with n vertices. It is well known that μ n = 0 and i = 1 n 1 μ i = 2 m ( G ) . In particular, μ n 1 is called the algebraic connectivity of G and it is denoted by α ( G ) .

The Laplacian matrix is an important topic in the theory of graph spectra. Particularly, much literature has paid attention to μ 1 , μ 2 , μ n 1 or μ 1 μ n 1 for trees (see, for example, [16]). Let S n be the star of order n, S a , b k be the tree obtained from two stars S a + 1 , S b + 1 by joining a path of length k between their central vertices (see Figure 1). As is well known, among all trees of order n, S n has the largest value of μ 1 (see [7]) and S n 3 , 1 1 has the second largest value of μ 1 (see [6]). On the other hand, Guo [4] proved that these two trees also attain the first two smallest values of μ 2 , respectively. This implies that μ 1 , μ 2 cannot attain simultaneously the maximal (or minimal) value and even the relation between them seems like a seesaw. Therefore, it is interesting to investigate the value of μ 1 + μ 2 . Moreover, Zhang [6] showed that the S k 1 , k 1 1 , S k 1 , k 2 2 , S k 2 , k 2 3 attain simultaneously the largest value of μ 2 among all trees with 2k vertices. Then Shao et al.[5] showed that S k 1 , k 1 2 attains the largest value of μ 2 among all trees with 2 k + 1 vertices.
Figure 1
Figure 1

S a , b k : a tree of order a + b + k + 1 .

Another motivation to study the value of μ 1 + μ 2 came from a result of Haemers et al.[8], who showed that μ 1 + μ 2 m ( G ) + 3 for any graph G. This result implies that Brouwer’s conjecture [9],
μ 1 + μ 2 + + μ k m ( G ) + ( k + 1 2 ) ,

is true for k = 2 . Considering a tree T, we have μ 1 + μ 2 n ( T ) + 2 . Recently, Fritscher et al.[10] improved this bound by giving μ 1 + μ 2 < n ( T ) + 2 2 n ( T ) . This paper determines the extremal tree that attains the bound of μ 1 + μ 2 . Moreover, for general connected graphs, we also give a conjecture on the extremal graphs for μ 1 + μ 2 .

2 A sharp upper bound of μ 1 + μ 2

Let S k ( G ) be the sum of the largest k Laplacian eigenvalues of a graph G. When k = 2 , we shall write S ( G ) instead of S k ( G ) for simplicity. For graphs G and H, we denote by G H the graph with vertex set V ( G ) V ( H ) and edge set E ( G ) E ( H ) . The following lemmas come from an important result as regards a real symmetric matrix.

Lemma 2.1 ([8])

Let G 1 , G 2 , , G r be some edge-disjoint graphs. Then S k ( i = 1 r G i ) i = 1 r S k ( G i ) for any k.

Lemma 2.2 ([8])

For any graph G, S ( G ) m ( G ) + 3 .

Lemma 2.3 Let G be a connected graph, d i = d G ( v i ) and m i = Σ v j N G ( v i ) d j / d i . Then
  1. (i)

    [11] μ 1 ( G ) Δ ( G ) + 1 , with equality if and only if Δ ( G ) = n ( G ) 1 .

     
  2. (ii)

    [7] μ 1 ( G ) n ( G ) , with equality if and only if the complement of G is disconnected.

     
  3. (iii)

    [12] μ 1 ( G ) max { d i + m i | v i V ( G ) } .

     

Lemma 2.4 ([6])

Let T be a tree of order n. If T S n , then μ 1 ( T ) μ 1 ( S n 3 , 1 1 ) , with equality if and only if T S n 3 , 1 1 .

Corollary 2.5 Let T be a tree with n vertices and diameter d 3 . Then μ 1 ( T ) < n 0.5 .

Proof Note that any tree T has diameter d 3 if T S n . According to Lemma 2.4, μ 1 ( T ) μ 1 ( S n 3 , 1 1 ) . Further, by Lemma 2.3,
μ 1 ( S n 3 , 1 1 ) max { d i + m i } = n 2 + n 1 n 2 = n 1 + 1 n 2 < n 0.5

for n 5 . For n = 4 , a straightforward calculation shows that μ 1 ( S 1 , 1 1 ) = 2 + 2 < 3.5 . □

Lemma 2.6 ([2])

Let T be a tree of order n and diameter d 3 . Then α ( T ) α ( S n d + 1 2 , n d + 1 2 d 2 ) , with equality if and only if T S n d + 1 2 , n d + 1 2 d 2 .

Lemma 2.7 ([11])

Let G be a graph with a vertex u of degree one. Then α ( G ) α ( G u ) .

Lemma 2.7 implies that the algebraic connectivity of a tree is not greater than that of its subtree.

Lemma 2.8 ([4])

Let T n k ( n 2 k + 1 ) be a tree obtained from a star S n k by replacing its k edges with k paths of length two, respectively. If k 2 , then μ 2 ( T n k ) = 3 + 5 2 .

The following lemma can be found in [13] and is known as the Interlacing Theorem of Laplacian eigenvalues.

Lemma 2.9 Let G be a graph of order n and H be a graph obtained from G by deleting an edge. Then
μ 1 ( G ) μ 1 ( H ) μ n ( G ) μ n ( H ) = 0 .

Next we give the main theorem of this section. Its proof is divided into several sequent claims.

Theorem 2.10 For any tree T with order n 4 , S ( T ) S ( S n 2 2 , n 2 2 1 ) . The equality holds if and only if T S n 2 2 , n 2 2 1 .

Claim 2.11 For any tree T with order n 4 and diameter d 3 , S ( T ) < S ( S n 2 2 , n 2 2 1 ) except that T S n 2 2 , n 2 2 1 .

Proof If d ( T ) = 3 , then T S a , b 1 for some positive integers a, b with a + b = n 2 . It is well known that the Laplacian characteristic polynomial of S a , b 1 is μ ( μ 1 ) n 4 f a , b ( μ ) , where
f a , b ( μ ) = μ 3 ( n + 2 ) μ 2 + ( a b + 2 n + 1 ) μ n .
(1)
Note that S a , b 1 contains S 1 , 1 1 as a subtree. By Lemma 2.9, μ 2 ( S a , b 1 ) μ 2 ( S 1 , 1 1 ) = 2 . Moreover, we know that for any tree T, α ( T ) 1 , with equality if and only if T is a star. These imply that μ 1 ( S a , b 1 ) , μ 2 ( S a , b 1 ) , and α ( S a , b 1 ) consist of the three roots of f a , b ( μ ) . As follows from (1), we have
μ 1 ( S a , b 1 ) + μ 2 ( S a , b 1 ) + α ( S a , b 1 ) = n + 2 .
(2)

By virtue of Lemma 2.6, we have α ( S a , b 1 ) > α ( S n 2 2 , n 2 2 1 ) except that ( a , b ) = ( n 2 2 , n 2 2 ) . Equivalently, S ( S a , b 1 ) < S ( S n 2 2 , n 2 2 1 ) except that ( a , b ) = ( n 2 2 , n 2 2 ) .

If d ( T ) = 2 , then T S n . We first give a lower bound of S ( S n 2 2 , n 2 2 1 ) for n 6 :
S ( S n 2 2 , n 2 2 1 ) > n + 1.5 .
(3)

Indeed, by (2) it suffices to show α ( S n 2 2 , n 2 2 1 ) < 0.5 . Note that for n 6 , S n 2 2 , n 2 2 1 contains S 2 , 2 1 as a subtree. By Lemma 2.7, α ( S n 2 2 , n 2 2 1 ) α ( S 2 , 2 1 ) = 5 17 2 < 0.5 .

Note that S ( S n ) = n + 1 for n 3 . According to (3), we have S ( S n ) < S ( S n 2 2 , n 2 2 1 ) for n 6 . As for n { 4 , 5 } , a straightforward calculation shows that
S ( S 1 , 1 1 ) 5.4142 , S ( S 2 , 1 1 ) 6.4811 .
(4)

Also we have S ( S n ) < S ( S n 2 2 , n 2 2 1 ) . □

Claim 2.12 For any tree T with order n and diameter d 5 , S ( T ) < S ( S n 2 2 , n 2 2 1 ) .

Proof Since d ( T ) 5 , then n 6 and there is a path of length 5 in T. By inequality (3), it suffices to show S ( T ) n + 1.5 . First suppose that there is a path v 0 v 1 v 5 in T such that either max { d T ( v 0 ) , d T ( v 5 ) } 2 or max { d T ( v 2 ) , d T ( v 3 ) } 3 . Let T 1 , T 2 be the two components of T v 2 v 3 . Clearly, both T 1 and T 2 have at least two edges.

If μ 1 , μ 2 of T 1 T 2 attain at the same component, say T 1 , then by Lemma 2.2,
S ( T 1 T 2 ) = S ( T 1 ) m ( T 1 ) + 3 m ( T 1 T 2 ) + 1 .
(5)
Note that S ( v 2 v 3 ) = S ( S 2 ) = 2 . By Lemma 2.1,
S ( T ) S ( T 1 T 2 ) + S ( v 2 v 3 ) m ( T 1 T 2 ) + 3 = m ( T ) + 2 = n + 1 .
(6)
Otherwise, S ( T 1 T 2 ) = μ 1 ( T 1 ) + μ 1 ( T 2 ) . Whether max { d T ( v 0 ) , d T ( v 5 ) } 2 or max { d T ( v 2 ) , d T ( v 3 ) } 3 , we can observe that max { d ( T 1 ) , d ( T 2 ) } 3 . Say d ( T 2 ) 3 , then by Corollary 2.5, μ 1 ( T 2 ) < n ( T 2 ) 0.5 . By Lemma 2.3(ii), μ 1 ( T 1 ) n ( T 1 ) . Hence,
S ( T ) S ( T 1 T 2 ) + S ( v 2 v 3 ) = n ( T 1 ) + n ( T 2 ) 0.5 + 2 = n + 1.5 .
Next, we may assume that each path v 0 v 1 v 5 of length 5 in T has d T ( v 0 ) = d T ( v 5 ) = 1 and d T ( v 2 ) = d T ( v 3 ) = 2 . This implies that d ( T ) = 5 and T S a , b 3 for some integers a, b with a + b = n 4 . If a = b = 1 , then T is isomorphic to a path of order 6 and a straightforward calculation shows that S ( T ) = 5 + 3 < n + 1.5 , as claimed. Otherwise, assume without loss of generality that a 2 . Then d T ( v 1 ) 3 . Let T 3 , T 4 be the two components of T v 1 v 2 with v 0 v 1 E ( T 3 ) . Then both T 3 and T 4 have at least two edges. If μ 1 , μ 2 of T 3 T 4 attain at the same component, say T 3 , then by Lemmas 2.1 and 2.2,
S ( T ) S ( T 3 T 4 ) + S ( v 1 v 2 ) = S ( T 3 ) + 2 m ( T 3 ) + 5 m ( T ) + 2 = n + 1 .
Otherwise, S ( T 3 T 4 ) = μ 1 ( T 3 ) + μ 1 ( T 4 ) . Note that μ 1 ( T 3 ) n ( T 3 ) . Since d ( T 4 ) = 3 , by Corollary 2.5, μ 1 ( T 4 ) < n ( T 4 ) 0.5 . So
S ( T ) S ( T 3 T 4 ) + S ( v 1 v 2 ) n ( T 3 ) + n ( T 4 ) 0.5 + 2 = n + 1.5 .

 □

Claim 2.13 For any tree T with order n and diameter 4, S ( T ) < S ( S n 2 2 , n 2 2 1 ) .

Proof First suppose that T contains a path v 0 v 1 v 4 such that max { d T ( v 1 ) , d T ( v 3 ) } 3 . Now n 6 and it suffices to show S ( T ) n + 1.5 . Without loss of generality assume that d T ( v 1 ) 3 . Let T 1 , T 2 be the two components of T v 1 v 2 with v 0 v 1 E ( T 1 ) . Then both T 1 and T 2 have at least two edges.

If μ 1 , μ 2 of T 1 T 2 attain at the same component, say T 1 , then similarly to inequalities (5) and (6), we can observe that S ( T ) n + 1 .

Now let S ( T 1 T 2 ) = μ 1 ( T 1 ) + μ 1 ( T 2 ) . If d T ( v 2 ) 3 , then d ( T 2 ) 3 and hence μ 1 ( T 2 ) < n ( T 2 ) 0.5 . So
S ( T ) S ( T 1 T 2 ) + S ( v 1 v 2 ) < n ( T 1 ) + n ( T 2 ) 0.5 + 2 = n + 1.5 .
If d T ( v 2 ) = 2 , then T S a , b 2 for some positive integers a, b with 3 a + b = n 3 . Moreover, since d T ( v 1 ) 3 , then a 2 . If ( a , b ) { ( 2 , 1 ) , ( 3 , 1 ) } , a straightforward calculations show that S ( S a , b 2 ) < n + 1.5 . Otherwise, S a , b 2 contains either S 4 , 1 2 or S 2 , 2 2 as a subtree. Since
μ 3 ( S 4 , 1 2 ) 1.5068 , μ 3 ( S 2 , 2 2 ) 1.5858 ,
it follows from Lemma 2.9 that μ 3 ( S a , b 2 ) > 1.5 . Since S a , b 2 is not a star, μ n 1 ( S a , b 2 ) < 1 . On the other hand, note that the matrix 1 I n [ D ( S a , b 2 ) A ( S a , b 2 ) ] has a identical rows and b different identical rows, so the multiplicity of eigenvalue 1 is at least a + b 2 and else five eigenvalues are μ 1 , μ 2 , μ 3 , μ n 1 and μ n = 0 . Since i = 1 n μ i ( S a , b 2 ) = 2 ( n 1 ) , we have
i = 1 3 μ i ( S a , b 2 ) + μ n 1 ( S a , b 2 ) + μ n ( S a , b 2 ) = 2 ( n 1 ) ( a + b 2 ) = n + 3 .

This implies that S ( S a , b 2 ) < n + 3 μ 3 ( S a , b 2 ) < n + 1.5 .

Next, it suffices to consider the case that each path v 0 v 1 v 4 of T has d T ( v 1 ) = d T ( v 3 ) = 2 . This implies that T T n k for some k 2 and n 2 k + 1 , since d ( T ) = 4 . According to Lemma 2.8, μ 2 ( T n k ) = 3 + 5 2 . Moreover, by Lemma 2.3,
μ 1 ( T n k ) max { d i + m i } = n k 1 + n 1 n k 1 n 2 + 2 n 3 .
Thus for n 6 ,
S ( T n k ) n 2 + 2 n 3 + 3 + 5 2 < n + 1.5 < S ( S n 2 2 , n 2 2 1 ) .

When n = 5 , T n k is a path. Comparing with (4), S ( T n k ) = 4 + 5 < S ( S 2 , 1 1 ) . This completes the proof. □

Following from Claims 2.11-2.13, Theorem 2.10 holds and the unique tree with maximal S ( T ) is S n 2 2 , n 2 2 1 . According to (2),
μ ( S n 2 2 , n 2 2 1 ) < n + 2 = m ( S n 2 2 , n 2 2 1 ) + 3 .

Theorem 2.14 Let m, n be two positive integers with n m 2 n 3 and G m , n be a graph of order n and size m obtained from a given edge uv by joining m n + 1 independent vertices with u and v, respectively, and another 2 n m 3 independent vertices with u. Then S ( G m , n ) = m + 3 .

Proof Let H s , t be a graph obtained by joining a vertex to s vertices of a given complete graph of order s + t and H s , t c be its complement graph. Then H s , t c is isomorphic to the union of S t + 1 and s isolated vertices. Clearly, the Laplacian eigenvalues of H s , t c consist of t + 1 , 1 with multiplicity t 1 and 0 with multiplicity s + 1 . Recall that for any graph G with n vertices, μ i ( G ) = n μ n i ( G c ) for 1 i n 1 and μ n ( G ) = 0 . So the Laplacian eigenvalues of H s , t consist of s + t + 1 with multiplicity s, s + t with multiplicity t 1 , s and 0.

Now G m , n c is isomorphic to the union of H 2 n m 3 , m n + 1 and an isolated vertex. So the Laplacian eigenvalues of G m , n c consist of n 1 with multiplicity 2 n m 3 , n 2 with multiplicity m n , 2 n m 3 , and 0 with multiplicity 2. Therefore, the Laplacian eigenvalues of G m , n consist of n, m n + 3 , 2 with multiplicity m n , 1 with multiplicity 2 n m 3 and 0. So S ( G m , n ) = n + ( m n + 3 ) = m + 3 . □

Recall that μ 1 ( G ) n ( G ) for any graph G. When m ( G ) > 2 n ( G ) 3 , Haemers’ bound is clearly not attainable. Theorem 2.14 implies that if m ( G ) 2 n ( G ) 3 , Haemers’ bound is always sharp for connected graphs other than trees. Ending the paper, we present a conjecture on the uniqueness of the extremal graph.

Conjecture 2.15 Among all connected graphs with n vertices and n m 2 n 3 edges, G m , n is the unique graph with maximal value of μ 1 + μ 2 .

Declarations

Acknowledgements

The authors are grateful to the referees for carefully reading the manuscript and for providing some comments and suggestions, which led to improvements in the paper. The research was supported by the National Natural Science Foundation of China (11101057, 11201432), the Foundation for Young Talents in College of Anhui Province (2012SQRL170) and the Natural Science Foundation of Anhui Province (1308085MA03).

Authors’ Affiliations

(1)
Department of Mathematics and Physics, Hefei University, Hefei, 230601, China
(2)
School of Mathematical Science, Chuzhou University, Chuzhou, 239012, China
(3)
College of Mathematics and Computer Science, Tongling University, Tongling, 244000, China

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