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Bounding the sum of the largest Laplacian eigenvalues of graphs
Discrete Applied Mathematics 170 (2014) 95–103
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Bounding the sum of the largest Laplacian eigenvalues of graphs I. Rocha ∗ , V. Trevisan Instituto de Matemática, UFRGS—Avenida Bento Gonçalves, 9500, 91501–970 Porto Alegre, RS, Brazil
article
abstract
info
Article history: Received 26 November 2012 Received in revised form 25 October 2013 Accepted 30 January 2014 Available online 20 February 2014 Keywords: Laplacian eigenvalues Brouwer’s conjecture
We prove that Brouwer’s conjecture holds for certain classes of graphs. We also give upper bounds for the sum of the largest Laplacian eigenvalues for graphs satisfying certain properties: those that contain a path or a cycle of a given size, graphs with a given matching number and graphs with a given maximum degree. Then we provide conditions for which these upper bounds are better than the previous known results. © 2014 Elsevier B.V. All rights reserved.
1. Introduction In this paper, we investigate the sum of the largest Laplacian eigenvalues of a graph, which is the subject of a conjecture proposed by A.E. Brouwer and which has recently, in [3,4], showed to be helpful for the study of the Laplacian energy of graphs. Given graph G on the vertex set V = {v1 , . . . , vn }, the Laplacian matrix of G is given by L = D − A, where D is the diagonal matrix whose entry (i, i) is equal to the degree of vi and A is the adjacency matrix of G. The Laplacian spectrum of G is defined as the set of eigenvalues of L, which we shall denote by µ1 ≥ µ2 ≥ · · · ≥ µn = 0. With this, the sum of the largest Laplacian eigenvalues of G is denoted by Sk (G) =
n
µi .
i =1
There are many results concerning Sk (G). In [8], there are upper bounds for Sk (G) in terms of the number of vertices and edges. Also, the author characterizes the extremal cases. Brouwer proposed the following conjecture. Conjecture 1. Let G = (V , E ) with n vertices. Then Sk (G) ≤ |E | +
k+1 2
,
for 1 ≤ k ≤ n.
∗
Corresponding author. Tel.: +55 51 3308 6174; fax: +55 5133087301. E-mail addresses: [email protected] (I. Rocha), [email protected] (V. Trevisan).
http://dx.doi.org/10.1016/j.dam.2014.01.023 0166-218X/© 2014 Elsevier B.V. All rights reserved.
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In [1], some advances on Conjecture 1 are mentioned: this inequality holds for split graphs, in particular for threshold graphs. It also holds for regular graphs. For k = 1, Conjecture 1 follows from the well-known inequality µ1 (G) ≤ n. In [5], Haemers, Mohammadian and Tayfeh-Rezaie showed that Conjecture 1 is true for all graphs when k = 2 and is also true for trees. We say that a connected graph is c-cyclic, or it has c cycles, if it has n − 1 + c edges and we say that the graph is unicyclic or bicyclic if it is 1-cyclic or 2-cyclic respectively. More recently, Du and Zhou [2] showed that the conjecture holds for unicyclic and bicyclic graphs. The authors of [2] have also proved the following lemma. Lemma 1. Let H be a subgraph of graph G, and |V (H )| = n1 ≥ 2. Then Sk (G) ≤ Sk (H ) + 2(e(G) − e(H )) for 1 ≤ k ≤ n1 . This lemma provides a useful tool to extend an upper bound on trees for the general class of connected graphs. Assume that we have a new upper bound for a tree, that is, Sk (T ) is bounded for any tree T . Now consider a connected graph G with n vertices. Then for any generating tree T of G using Lemma 1, we obtain the following result. Lemma 2. Let G be a connected graph on n vertices and let T be a generating tree of G. Then Sk (G) ≤ Sk (T ) + 2(e(G) − n + 1) for 1 ≤ k ≤ n. Therefore, for the purpose of bounding Sk for graphs, it may be interesting to start investigating this parameter for trees. In [3], the following upper bound is given for a tree T on n vertices. 2k − 2
. (2) n By applying this upper bound for trees, the authors of [6] proved the following upper bound for a connected graph G on n vertices. Sk (T ) ≤ n − 2 + 2k −
Sk (G) ≤ 2e(G) − n + 2k −
2k − 2
. (3) n They also settled Brouwer’s conjecture for unicyclic and bicyclic graphs, since for such graphs the upper bound (3) is tighter than (1). However, for c-cyclic graphs, whenever c ≥ 3 the upper bound (3) is not enough to settle Brouwer’s conjecture. For the general case, in [2] the following result was proved. Lemma 3. Let G be a graph with n1 are not isolated vertices. Then inequality (1) holds for 1 ≤ k ≤ n if n√vertices, of which 3+ 9−8(n1 −e(G)) ≤ k ≤ n if 9 − 8(n1 − e(G)) ≥ 0. 9 − 8(n1 − e(G)) < 0, and for 2 Lemma 3 will play a special role in our results. Also, in [6] the following result was proved. Lemma 4. Let G be a connected graph on n vertices. Then inequality (1) holds for 3n − 4 +
9n2 − 8n + 16 + 8e(G)n2 − 8n3 2n
≤ k ≤ n.
The bound (2) is tight when k = 1 and T is a star, and it cannot be improved by subtracting 1/n, even if we consider trees with diameter at least three [3]. However, in [4], the authors have shown that the upper bound (2) could be improved by 2/n for trees with diameter at least 4 and at least 6 vertices. Sk (T ) ≤ n − 2 + 2k −
2k
. (4) n This slight improvement bounding (2) was sufficient to obtain a sharper result than Lemma 4 and, consequently, new families of graphs satisfying (1), as showed in [4]. Naturally, Brouwer’s conjecture holds in case it is possible to show an upper bound tighter than (1). In an attempt to achieve such sharper bounds, we shall estimate Sk (G) for graphs depending on certain parameters, such as diameter, maximum degree, matching number and girth. These results allow us to settle Brouwer’s conjecture for such graphs or at least to prove that inequality (1) holds for some values of k. By obtaining a sharp bound for paths, we also give upper bounds for the sum of the largest Laplacian for graphs containing a path or a cycle of a given size. This paper is organized as follows. In Section 2 we study and bound Sk (G) for G satisfying certain conditions and in Section 3 we show new families of graphs for which Brouwer’s conjecture holds. We finish by comparing the bounds with existing bounds.
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2. Upper bounds on Sk Since Conjecture 1 holds for trees, it is natural to ask whether it is possible to improve that upper bound. For a path, we obtain a better upper bound, as the next result shows us. We will use that k−1
cos(nk) =
sin (nk) cos (n) + sin (nk) 2 sin (n)
i =0
1
−
2
cos (nk) +
1
(5)
2
which can be proved using exponential sum formulas. Proposition 1. If Ps is a path with s ≥ 5 vertices, then Sk (Ps ) < s − 1, for 1 ≤ k ≤ ⌊ 5s ⌋. Proof. Consider the sum of the k largest Laplacian eigenvalues of Ps Sk (Ps ) =
k−1
2 − 2 cos
π (s − 1 − i) s
i =0
.
It is easy to see that Sk (Ps ) is a nondecreasing function of k. Therefore, it is enough to show that S⌊ s ⌋ (Ps ) < s − 1. Using Eq. 5 (5), one can prove that Sk (Ps ) = 2k + cos
k s
cos πs sin ks π + sin ks π π − 1. π + sin
s
Since this expression is nondecreasing for k, we can write S⌊ s ⌋ (Ps ) ≤ 5
2 5
s + cos
1 5
π +
cos πs sin 15 π + sin 51 π
− 1.
sin πs
π3 We notice that, by the Taylor series expansion of sine at 0, we obtain sin πs > πs − 6s 3 . Now, we use the exact trigonometric constants cos
S⌊ s ⌋ (Ps ) ≤ 5
√
√
1 π = 5
1+ 5 4
and sin
1 π = 5
2(5− 5) , 4
and that cos(x) ≤ 1, to obtain
1 −48 π s3 + 8 π 3 s + 90 π s2 − 15 π 3 − 30
√
5 π s2 + 5
√
5π 3 − 60
√
10 − 2
5s3
π −6 s 2 + π 2
20
= p(s). It remains to prove that p(s) < s − 1. We can write p(s) − s + 1 =
1 72 π s3 − 12 π 3 s − 30 π s2 + 5 π 3 − 30
√
5π s2 + 5
√
5π 3 − 60
√ 10 − 2
π −6 s 2 + π 2
20
Now we notice that the polynomial 20 π −6 s2 + π 2 < 0
whenever s ≥ 2, and that 72 π s3 − 12 π 3 s − 30 π s2 + 5 π 3 − 30 whenever s ≥ 5. This completes the proof.
√
5π s2 + 5
√
5π 3 − 60
√
10 − 2
5s3 > 0
In order to estimate Sk (G) a tool that is often used is the classical inequality by Wielandt [7]. Lemma 5. Let A, B, C be Hermitian matrices of order n such that A = B + C . Then
i∈I
λi (A) ≤
|I |
λi (B) +
i =1
for any subset I ⊂ {1, 2, . . . , n}.
i∈I
λi ( C )
5s3
.
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Now, let F be a graph on n vertices and let G be an arbitrary subgraph. We define graph H from graph F by removing the edges of G from F . If G has m vertices, then we can define the graph G from G by adding n − m isolated vertices in G. Hence, both graphs F , G and H have n vertices. It is easy to see that we can decompose the Laplacian matrix of F such that L(F ) = L( G) + L(H )
(6)
using a suitable ordering of rows and columns. The spectrum of L( G) is the set of eigenvalues of L(G) together with n − m zeros. Therefore, for simplicity, we will write L(F ) = L(G) + L(H ) to refer to the decomposition (6). We will also use, for k = 1, 2, . . . , n, the notation Sk (G) instead of Sk ( G). Henceforth, we will use in our proofs a particular case of Lemma 5 with I = {1, 2, . . . , k}. The next result gives us an upper bound for a graph with a given diameter. Theorem 1. Let G be a graph on n vertices and diameter D − 1 and let PD be a path in G. Form the graph H by removing the edges of PD from G. Let p be the number of isolated vertices of H. We have that Sk (G) ≤ 2(e(G) − D) + 1 − n + 4k + p + cos
k D
π +
cos πD sin Dk π + sin Dk π sin πD
for k = 1, 2, . . . , D − 1. Proof. Denote by H the graph obtained from G by removing the D − 1 edges of path PD . Since G has a path with D − 1 edges, we can decompose the Laplacian matrix of G such that L(G) = L(PD ) + L(H ). By applying the Wielandt inequality for the sum of eigenvalues, we have Sk (G) ≤ Sk (PD ) + Sk (H ). Denote by H1 , H2 , . . . , Ht the connected components of H, where e(Hi ) ≥ 1 and ni denotes the number of vertices of Hi . Denote by p the number of isolated Assume that ki of the k largest Laplacian eigenvalues of H come from Hi . We t vertex of H. t t have that i=1 e(Hi ) = e(H ), i=1 ki = k and i=1 ni = n − p. For k = 1, 2, . . . , D − 1, path PD has eigenvalues such that Sk (PD ) =
k−1
2 − 2 cos
π (D − 1 − i) D
i=0
.
Using Eq. (5), one can prove that Sk (PD ) = 2k + cos
k D
π +
cos πD sin Dk π + sin Dk π sin πD
− 1.
By using (3) for each graph Hi , we obtain Sk (H ) ≤
t
2e(Hi ) − ni + 2ki = 2(e(G) − D + 1) − n + p + 2k.
i =1
Hence, we can write Sk (G) ≤ Sk (PD ) + 2(e(G) − D + 1) − n + 2k + p
≤ 2k + cos
k D
π +
cos πD sin Dk π + sin Dk π sin πD
= 2(e(G) − D) + 1 − n + 4k + p + cos This completes the proof.
k D
π +
− 1 + 2(e(G) − D + 1) − n + 2k + p
cos πD sin Dk π + sin Dk π sin πD
.
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Using a similar proof, we can write the following result: Theorem 2. Let G be a graph on n vertices and containing a cycle Cs . Form the graph H by removing the edges of Cs from G. Let p be the number of isolated vertices of H. We have that Sk (G) ≤ 2(e(G) − s) + 1 − n + 4k + p + cos
k s
π +
cos πs sin ks π + sin ks π sin πs
for k = 1, 2, . . . , s − 1. Proof. We notice that Cs contains path Ps . Let H be the graph obtained from G by removing the s − 1 edges of path Ps . Therefore, we can decompose L(G) as follows: L(G) = L(Ps ) + L(H ). The rest of the proof follows as the proof of Theorem 1.
The next result gives us an upper bound for a graph with matching number m. Theorem 3. Let G be a graph on n vertices and let M be a maximum matching in G. Let H be the graph formed from G by removing the edges of M from G. Then Sk (G) ≤ 2e(G) − 2|M | − n + 4k + p, where p is the number of isolated vertices of H. Proof. Since G has a matching with m = |M | edges, we can see that G has a subgraph mK2 formed by m copies of K2 . Hence, we can decompose the Laplacian matrix of G such that L(G) = L(mK2 ) + L(H ). By applying the Wielandt inequality for the sum of eigenvalues, we have Sk (G) ≤ Sk (mK2 ) + Sk (H ). Denote by H1 , H2 , . . . , Ht the connected components of H, where e(Hi ) ≥ 1 and ni denotes the number of vertices of Hi . Denote by p the of H. Assume that ki of the k largest Laplacian eigenvalues of H come from Hi . tnumber of isolated vertices t t We have that i=1 e(Hi ) = e(H ), i=1 ki = k and i=1 ni = n − p. By using (3) for each graph Hi , we obtain Sk (H ) ≤
t
2e(Hi ) − ni + 2ki = 2(e(G) − m) − n + p + 2k.
i=1
Since the graph K2 has eigenvalues 0 and 2, we can write Sk (G) ≤ Sk (mK2 ) + 2(e(G) − m) − n + 2k + p
≤ 2k + 2(e(G) − m) − n + 2k + p = 2e(G) − 2m − n + 4k + p, which completes the proof.
Next, we give an upper bound for a graph with maximum degree ∆. Theorem 4. For a graph G on n vertices and maximum degree ∆ ≥ 2, let v be a vertex with maximum degree ∆ and p pendant vertices. Then we have Sk (G) ≤ 2e(G) + 3k − n − ∆ + p + 1. Proof. Since G has vertex v with degree ∆, the graph G has a subgraph S∆+1 , the star with ∆ + 1 vertices and center v . Denote by H the graph obtained from G by removing the ∆ edges adjacent to v . We can decompose the Laplacian matrix of G such that L(G) = L(S∆+1 ) + L(H ). The star S∆+1 has eigenvalues ∆ + 1 with multiplicity 1, 1 with multiplicity ∆ − 1 and 0. By applying the Wielandt inequality for the sum of eigenvalues, we have Sk (G) ≤ Sk (S∆+1 ) + Sk (H ). Denote by H1 , H2 , . . . , Ht the connected components of H, where e(Hi ) ≥ 1 and ni is the number of vertices in Hi . Since v is an isolated vertex in H and v has p pendant vertices we have p + 1 isolated vertices in H. Assume that ki of the k largest
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Laplacian eigenvalues of H come from Hi . We have that (3) for each graph Hi , we obtain
t
i=1
e(Hi ) = e(H ),
t
i =1
ki = k and
t
i =1
ni = n − p − 1. By using
Sk (G) ≤ Sk (S∆+1 ) + 2(e(G) − ∆) − n + 2k + p + 1
≤ ∆ + 1 + (k − 1) + 2(e(G) − ∆) − n + 2k + p + 1 = 2e(G) + 3k − n − ∆ + p + 1, which completes the proof.
3. On Brouwer’s conjecture for special graphs Proposition 2. Let G be a graph and let H1 , H2 , . . . , Ht be its connected components. If Conjecture 1 holds for each Hi , then Conjecture 1 holds for G. Proof. Let ni denote the number of of Hi . Assume that ki of the k largest Laplacian eigenvalues of G come from Hi . t vertices t t We have that i=1 e(Hi ) = e(G), i=1 ki = k and i=1 ni = n. Therefore, we can write Sk (G) =
t
Ski (Hi )
i=1
≤
t
e(Hi ) +
ki + 1 2
i=1
= e(G) +
t k2 + ki i
i=1
≤ e(G) + as required.
2
k +k 2
2
,
We recall that the girth of a graph is the size of the smallest cycle in the graph. Theorem 5. Let G be a graph with girth g ≥ 5. Then inequality (1) is true for 1 ≤ k ≤
g 5
.
Proof. First, as proved in [5,2], we observe that Conjecture 1 is true for trees, unicyclic and bicyclic graphs. Hence, we can suppose that G is a c-cyclic graph with c ≥ 3. Suppose, by contradiction, that the following inequality holds e(G) +
k+1
2
< Sk (G),
g
for some 1 ≤ k ≤ 5 . Since G has girth g, it contains subgraph Pg . We define G1 , as the graph obtained from G by removing the edges of Pg . Therefore, we can decompose the Laplacian matrix of G as follows: L(G) = L(Pg ) + L(G1 ). Then, by Proposition 1 and the Wielandt inequality for the sum of eigenvalues, we can write e(G) +
k+1
2
< Sk (G) ≤ Sk (Pg ) + Sk (G1 ) ≤ e(Pg ) + Sk (G1 ).
Since e(G) − e(Pg ) = e(G1 ), we have e(G1 ) +
k+1 2
g
< Sk (G1 ),
for some 1 ≤ k ≤ 5 . Let c1 be the maximum number of cycles at each connected component of G1 . By the construction of G1 , we can see that c1 < c. If 0 ≤ c1 ≤ 2, then each connected component is a tree, a unicyclic or a bicyclic graph. Then, by Proposition 2, Conjecture 1 holds for G1 , which is a contradiction. In this case, the proof is done. Otherwise, suppose that c1 > 2. By the construction of G1 , it has girth g1 ≥ g. Hence, the graph G1 contains path Pg1 . We can define graph G2 as the graph obtained from G1 by removing the edges of path Pg1 . The graph G2 has girth g2 ≥ g. Let c2
I. Rocha, V. Trevisan / Discrete Applied Mathematics 170 (2014) 95–103
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be the maximum number of cycles at each connected component of G2 . By the construction of G2 , we can see that c2 < c1 . Using the same argument as above, we obtain that e(G2 ) +
k+1 2
< Sk (G2 ),
g
for some 1 ≤ k ≤ 5 . Therefore, by successive applications of the argument above, we can define graph Gi as the graph obtained from Gi−1 by removing the edges of path Pgi−1 . The graph Gi has girth gi ≥ g. Let ci be the maximum number of cycles at each connected component of Gi . By the construction of Gi , we can see that ci < ci−1 . Therefore, since G is finite, we eventually end up with graph Gl such that each of its connected component is a tree, a unicyclic or a bicyclic graph. Then, by Proposition 2, Conjecture 1 holds for Gl , which is a contradiction. Hence, graph G satisfies Sk (G) ≤ e(G) + for 1 ≤ k ≤
k+1
2
g 5
, as required.
,
The following result shows that Brouwer’s conjecture holds whenever the graph has a large girth compared with the number of cycles. Theorem 6. Let G be a connected graph with girth g ≥ 5. If
√ 9 + 8e − 8n
3+
2
≤
g 5
,
then Conjecture 1 holds. Proof. Since G is connected, we have 9 + 8e − 8n ≥ 0. Also, we notice that Lemma 3 implies that inequality (1) holds for
√ 3+
9 + 8e − 8n 2
≤ k ≤ n.
We also recall that Theorem 5 implies that inequality (1) holds for 1 ≤ k ≤
g 5
. Hence, the hypothesis
√ 9 + 8e − 8n
3+
2 settles the result.
≤
g 5
Next, we show that inequality (1) is true for graphs with certain conditions on the maximum degree. Theorem 7. Let G be a graph and let v be a vertex with maximum degree ∆ and p pendant vertices. Then inequality (1) is true for
√
25 + 8(e(G) − n − ∆ + p + 1)
5+
2
≤ k ≤ n.
Proof. Since Theorem 4 gives us Sk (G) ≤ 2e(G) + 3k − n − ∆ + p + 1 it is enough to check that e(G) + 3k − n − ∆ + p + 1 ≤ Since
k+1 2
=
k(k + 1) 2
k(k+1) 2
k+1 2
.
we observe that we just need to check that
− (e(G) + 3k − n − ∆ + p + 1)
(7)
is positive for 1 ≤ k ≤ n Rewriting Eq. (7), we obtain k2 − 5k − 2e(G) + 2n + 2∆ − 2 2
.
(8)
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This way, Eq. (8) can be seen as a polynomial in k. We need that k2 − 5k − 2e(G) + 2n + 2∆ − 2p − 2 2
≥ 0.
(9)
The largest root of the polynomial (8) equals to
√ 5+
25 + 8(e(G) − n − ∆ + p + 1) 2
.
And since the polynomial in k (8) is positive for
√ 5+
25 + 8(e(G) − n − ∆ + p + 1) 2
the result follows.
≤k≤n
Now, given the proof above, it is easy to settle Conjecture 1 for the graphs described. Theorem 8. Let G be a connected graph that has vertex v with maximum degree ∆ and p pendant vertices. If G has c cycles and
∆ ≥ c + p + 4, then Conjecture 1 is true.
Proof. It is enough to see that Eq. (9) is positive for 1 ≤ k ≤ n. This holds, whenever the discriminant of the polynomial (9) is negative. We notice that ∆ ≥ c + p + 4 implies that ∆ ≥ e(G) − n + 1 + p + 4. Hence, e(G) − n − ∆ + p + 1 ≤ −4. Therefore, the discriminant of the polynomial (9) satisfies 25 + 8(e(G) − n − ∆ + p + 1) < 0 and we conclude that Eq. (9) is positive for 1 ≤ k ≤ n, as required. This completes the proof. Thus, to see that Conjecture 1 holds for a graph with c cycles, it is enough to see that the graph has ∆ ≥ c + 4 and a vertex with degree ∆ with no pendant vertices. 4. Final remarks We now compare the upper bounds on Sk (G) obtained in Section 2 with the bound of Eq. (3), providing conditions for which these results represent an improvement. For instance, if we consider the upper bound (3), we can see that the upper bound of Theorem 3 is sharper whenever 2 m ≥ 2k + p + 1, where m is the number of edges on maximum matching of G. Hence, our result presents an improvement for graphs having large matchings. Performing the same analysis, it is easy to see that the upper bound of Theorem 4 is sharper than (3) whenever ∆ ≥ k + 2 + p, meaning that our result is better for graphs having large maximum degree. To simplify the results given by Theorems 1 and 2 we notice that
cos
kπ D
+
cos πD sin kDπ + sin kDπ sin πD
≤ 1 + 2k.
Therefore Theorem 1 provides a sharper bound than the bound (3) for graphs having large diameter d, more precisely if 2d ≥ 4k + p + 1, whereas Theorem 2 provides a sharper bound than the bound (3) for graphs having a large cycle Cs , that is if 2s ≥ 4k + p + 3. To better understand for what graphs Theorem 6 proves Conjecture 1, we can start by fixing the number of cycles of the graph. We recall that a connected graph with n vertices and c cycles has n − 1 + c edges. Hence, we can rewrite the hypothesis as
√ 3+
1 + 8c 2
≤
g 5
.
With this, we can see that the more cycles a graph has, the larger the girth has to be for Theorem 6 to be true. For instance, on the set of graphs with 3 cycles, Theorem 6 settles Conjecture 1 for those graphs with girth at least 20; on the set of graphs with 4 cycles, it settles Conjecture 1 for those graphs with girth at least 22 and so forth. Acknowledgments V. Trevisan was partially supported by CNPq (Proc. 309531/2009-8 and 481551/2012-3) and FAPERGS (Proc. 11/1619-2).
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References [1] [2] [3] [4] [5] [6] [7] [8]
A.E. Brouwer, W.H. Haemers, Spectra of Graphs, Springer, New York, 2012. Z. Du, B. Zhou, Upper bounds for the sum of Laplacian eigenvalues of graphs, Linear Algebra Appl. 436 (2012) 3672–3683. E. Fritscher, C. Hoppen, I. Rocha, V. Trevisan, On the sum of the Laplacian eigenvalues of a tree, Linear Algebra Appl. 435 (2011) 371–399. E. Fritscher, C. Hoppen, I. Rocha, V. Trevisan, Characterizing trees with large Laplacian energy, Linear Algebra Appl. 442 (2014) 20–49. W.H. Haemers, A. Mohammadian, B. Tayfeh-Rezaie, On the sum of Laplacian eigenvalues of graphs, Linear Algebra Appl. 432 (2010) 2214–2221. S. Wang, Y. Huang, B. Liu, On a conjecture for the sum of Laplacian eigenvalues, Math. Comput. Modelling 56 (2012) 60–68. H. Wielandt, An extremum property of sums of eigenvalues, Proc. Amer. Math. Soc. 6 (1955) 106–110. B. Zhou, On Laplacian eigenvalues of a graph, Z. Naturforsch. A 59 (2004) 181–184.
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Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam
Bounding the sum of the largest Laplacian eigenvalues of graphs I. Rocha ∗ , V. Trevisan Instituto de Matemática, UFRGS—Avenida Bento Gonçalves, 9500, 91501–970 Porto Alegre, RS, Brazil
article
abstract
info
Article history: Received 26 November 2012 Received in revised form 25 October 2013 Accepted 30 January 2014 Available online 20 February 2014 Keywords: Laplacian eigenvalues Brouwer’s conjecture
We prove that Brouwer’s conjecture holds for certain classes of graphs. We also give upper bounds for the sum of the largest Laplacian eigenvalues for graphs satisfying certain properties: those that contain a path or a cycle of a given size, graphs with a given matching number and graphs with a given maximum degree. Then we provide conditions for which these upper bounds are better than the previous known results. © 2014 Elsevier B.V. All rights reserved.
1. Introduction In this paper, we investigate the sum of the largest Laplacian eigenvalues of a graph, which is the subject of a conjecture proposed by A.E. Brouwer and which has recently, in [3,4], showed to be helpful for the study of the Laplacian energy of graphs. Given graph G on the vertex set V = {v1 , . . . , vn }, the Laplacian matrix of G is given by L = D − A, where D is the diagonal matrix whose entry (i, i) is equal to the degree of vi and A is the adjacency matrix of G. The Laplacian spectrum of G is defined as the set of eigenvalues of L, which we shall denote by µ1 ≥ µ2 ≥ · · · ≥ µn = 0. With this, the sum of the largest Laplacian eigenvalues of G is denoted by Sk (G) =
n
µi .
i =1
There are many results concerning Sk (G). In [8], there are upper bounds for Sk (G) in terms of the number of vertices and edges. Also, the author characterizes the extremal cases. Brouwer proposed the following conjecture. Conjecture 1. Let G = (V , E ) with n vertices. Then Sk (G) ≤ |E | +
k+1 2
,
for 1 ≤ k ≤ n.
∗
Corresponding author. Tel.: +55 51 3308 6174; fax: +55 5133087301. E-mail addresses: [email protected] (I. Rocha), [email protected] (V. Trevisan).
http://dx.doi.org/10.1016/j.dam.2014.01.023 0166-218X/© 2014 Elsevier B.V. All rights reserved.
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In [1], some advances on Conjecture 1 are mentioned: this inequality holds for split graphs, in particular for threshold graphs. It also holds for regular graphs. For k = 1, Conjecture 1 follows from the well-known inequality µ1 (G) ≤ n. In [5], Haemers, Mohammadian and Tayfeh-Rezaie showed that Conjecture 1 is true for all graphs when k = 2 and is also true for trees. We say that a connected graph is c-cyclic, or it has c cycles, if it has n − 1 + c edges and we say that the graph is unicyclic or bicyclic if it is 1-cyclic or 2-cyclic respectively. More recently, Du and Zhou [2] showed that the conjecture holds for unicyclic and bicyclic graphs. The authors of [2] have also proved the following lemma. Lemma 1. Let H be a subgraph of graph G, and |V (H )| = n1 ≥ 2. Then Sk (G) ≤ Sk (H ) + 2(e(G) − e(H )) for 1 ≤ k ≤ n1 . This lemma provides a useful tool to extend an upper bound on trees for the general class of connected graphs. Assume that we have a new upper bound for a tree, that is, Sk (T ) is bounded for any tree T . Now consider a connected graph G with n vertices. Then for any generating tree T of G using Lemma 1, we obtain the following result. Lemma 2. Let G be a connected graph on n vertices and let T be a generating tree of G. Then Sk (G) ≤ Sk (T ) + 2(e(G) − n + 1) for 1 ≤ k ≤ n. Therefore, for the purpose of bounding Sk for graphs, it may be interesting to start investigating this parameter for trees. In [3], the following upper bound is given for a tree T on n vertices. 2k − 2
. (2) n By applying this upper bound for trees, the authors of [6] proved the following upper bound for a connected graph G on n vertices. Sk (T ) ≤ n − 2 + 2k −
Sk (G) ≤ 2e(G) − n + 2k −
2k − 2
. (3) n They also settled Brouwer’s conjecture for unicyclic and bicyclic graphs, since for such graphs the upper bound (3) is tighter than (1). However, for c-cyclic graphs, whenever c ≥ 3 the upper bound (3) is not enough to settle Brouwer’s conjecture. For the general case, in [2] the following result was proved. Lemma 3. Let G be a graph with n1 are not isolated vertices. Then inequality (1) holds for 1 ≤ k ≤ n if n√vertices, of which 3+ 9−8(n1 −e(G)) ≤ k ≤ n if 9 − 8(n1 − e(G)) ≥ 0. 9 − 8(n1 − e(G)) < 0, and for 2 Lemma 3 will play a special role in our results. Also, in [6] the following result was proved. Lemma 4. Let G be a connected graph on n vertices. Then inequality (1) holds for 3n − 4 +
9n2 − 8n + 16 + 8e(G)n2 − 8n3 2n
≤ k ≤ n.
The bound (2) is tight when k = 1 and T is a star, and it cannot be improved by subtracting 1/n, even if we consider trees with diameter at least three [3]. However, in [4], the authors have shown that the upper bound (2) could be improved by 2/n for trees with diameter at least 4 and at least 6 vertices. Sk (T ) ≤ n − 2 + 2k −
2k
. (4) n This slight improvement bounding (2) was sufficient to obtain a sharper result than Lemma 4 and, consequently, new families of graphs satisfying (1), as showed in [4]. Naturally, Brouwer’s conjecture holds in case it is possible to show an upper bound tighter than (1). In an attempt to achieve such sharper bounds, we shall estimate Sk (G) for graphs depending on certain parameters, such as diameter, maximum degree, matching number and girth. These results allow us to settle Brouwer’s conjecture for such graphs or at least to prove that inequality (1) holds for some values of k. By obtaining a sharp bound for paths, we also give upper bounds for the sum of the largest Laplacian for graphs containing a path or a cycle of a given size. This paper is organized as follows. In Section 2 we study and bound Sk (G) for G satisfying certain conditions and in Section 3 we show new families of graphs for which Brouwer’s conjecture holds. We finish by comparing the bounds with existing bounds.
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2. Upper bounds on Sk Since Conjecture 1 holds for trees, it is natural to ask whether it is possible to improve that upper bound. For a path, we obtain a better upper bound, as the next result shows us. We will use that k−1
cos(nk) =
sin (nk) cos (n) + sin (nk) 2 sin (n)
i =0
1
−
2
cos (nk) +
1
(5)
2
which can be proved using exponential sum formulas. Proposition 1. If Ps is a path with s ≥ 5 vertices, then Sk (Ps ) < s − 1, for 1 ≤ k ≤ ⌊ 5s ⌋. Proof. Consider the sum of the k largest Laplacian eigenvalues of Ps Sk (Ps ) =
k−1
2 − 2 cos
π (s − 1 − i) s
i =0
.
It is easy to see that Sk (Ps ) is a nondecreasing function of k. Therefore, it is enough to show that S⌊ s ⌋ (Ps ) < s − 1. Using Eq. 5 (5), one can prove that Sk (Ps ) = 2k + cos
k s
cos πs sin ks π + sin ks π π − 1. π + sin
s
Since this expression is nondecreasing for k, we can write S⌊ s ⌋ (Ps ) ≤ 5
2 5
s + cos
1 5
π +
cos πs sin 15 π + sin 51 π
− 1.
sin πs
π3 We notice that, by the Taylor series expansion of sine at 0, we obtain sin πs > πs − 6s 3 . Now, we use the exact trigonometric constants cos
S⌊ s ⌋ (Ps ) ≤ 5
√
√
1 π = 5
1+ 5 4
and sin
1 π = 5
2(5− 5) , 4
and that cos(x) ≤ 1, to obtain
1 −48 π s3 + 8 π 3 s + 90 π s2 − 15 π 3 − 30
√
5 π s2 + 5
√
5π 3 − 60
√
10 − 2
5s3
π −6 s 2 + π 2
20
= p(s). It remains to prove that p(s) < s − 1. We can write p(s) − s + 1 =
1 72 π s3 − 12 π 3 s − 30 π s2 + 5 π 3 − 30
√
5π s2 + 5
√
5π 3 − 60
√ 10 − 2
π −6 s 2 + π 2
20
Now we notice that the polynomial 20 π −6 s2 + π 2 < 0
whenever s ≥ 2, and that 72 π s3 − 12 π 3 s − 30 π s2 + 5 π 3 − 30 whenever s ≥ 5. This completes the proof.
√
5π s2 + 5
√
5π 3 − 60
√
10 − 2
5s3 > 0
In order to estimate Sk (G) a tool that is often used is the classical inequality by Wielandt [7]. Lemma 5. Let A, B, C be Hermitian matrices of order n such that A = B + C . Then
i∈I
λi (A) ≤
|I |
λi (B) +
i =1
for any subset I ⊂ {1, 2, . . . , n}.
i∈I
λi ( C )
5s3
.
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Now, let F be a graph on n vertices and let G be an arbitrary subgraph. We define graph H from graph F by removing the edges of G from F . If G has m vertices, then we can define the graph G from G by adding n − m isolated vertices in G. Hence, both graphs F , G and H have n vertices. It is easy to see that we can decompose the Laplacian matrix of F such that L(F ) = L( G) + L(H )
(6)
using a suitable ordering of rows and columns. The spectrum of L( G) is the set of eigenvalues of L(G) together with n − m zeros. Therefore, for simplicity, we will write L(F ) = L(G) + L(H ) to refer to the decomposition (6). We will also use, for k = 1, 2, . . . , n, the notation Sk (G) instead of Sk ( G). Henceforth, we will use in our proofs a particular case of Lemma 5 with I = {1, 2, . . . , k}. The next result gives us an upper bound for a graph with a given diameter. Theorem 1. Let G be a graph on n vertices and diameter D − 1 and let PD be a path in G. Form the graph H by removing the edges of PD from G. Let p be the number of isolated vertices of H. We have that Sk (G) ≤ 2(e(G) − D) + 1 − n + 4k + p + cos
k D
π +
cos πD sin Dk π + sin Dk π sin πD
for k = 1, 2, . . . , D − 1. Proof. Denote by H the graph obtained from G by removing the D − 1 edges of path PD . Since G has a path with D − 1 edges, we can decompose the Laplacian matrix of G such that L(G) = L(PD ) + L(H ). By applying the Wielandt inequality for the sum of eigenvalues, we have Sk (G) ≤ Sk (PD ) + Sk (H ). Denote by H1 , H2 , . . . , Ht the connected components of H, where e(Hi ) ≥ 1 and ni denotes the number of vertices of Hi . Denote by p the number of isolated Assume that ki of the k largest Laplacian eigenvalues of H come from Hi . We t vertex of H. t t have that i=1 e(Hi ) = e(H ), i=1 ki = k and i=1 ni = n − p. For k = 1, 2, . . . , D − 1, path PD has eigenvalues such that Sk (PD ) =
k−1
2 − 2 cos
π (D − 1 − i) D
i=0
.
Using Eq. (5), one can prove that Sk (PD ) = 2k + cos
k D
π +
cos πD sin Dk π + sin Dk π sin πD
− 1.
By using (3) for each graph Hi , we obtain Sk (H ) ≤
t
2e(Hi ) − ni + 2ki = 2(e(G) − D + 1) − n + p + 2k.
i =1
Hence, we can write Sk (G) ≤ Sk (PD ) + 2(e(G) − D + 1) − n + 2k + p
≤ 2k + cos
k D
π +
cos πD sin Dk π + sin Dk π sin πD
= 2(e(G) − D) + 1 − n + 4k + p + cos This completes the proof.
k D
π +
− 1 + 2(e(G) − D + 1) − n + 2k + p
cos πD sin Dk π + sin Dk π sin πD
.
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Using a similar proof, we can write the following result: Theorem 2. Let G be a graph on n vertices and containing a cycle Cs . Form the graph H by removing the edges of Cs from G. Let p be the number of isolated vertices of H. We have that Sk (G) ≤ 2(e(G) − s) + 1 − n + 4k + p + cos
k s
π +
cos πs sin ks π + sin ks π sin πs
for k = 1, 2, . . . , s − 1. Proof. We notice that Cs contains path Ps . Let H be the graph obtained from G by removing the s − 1 edges of path Ps . Therefore, we can decompose L(G) as follows: L(G) = L(Ps ) + L(H ). The rest of the proof follows as the proof of Theorem 1.
The next result gives us an upper bound for a graph with matching number m. Theorem 3. Let G be a graph on n vertices and let M be a maximum matching in G. Let H be the graph formed from G by removing the edges of M from G. Then Sk (G) ≤ 2e(G) − 2|M | − n + 4k + p, where p is the number of isolated vertices of H. Proof. Since G has a matching with m = |M | edges, we can see that G has a subgraph mK2 formed by m copies of K2 . Hence, we can decompose the Laplacian matrix of G such that L(G) = L(mK2 ) + L(H ). By applying the Wielandt inequality for the sum of eigenvalues, we have Sk (G) ≤ Sk (mK2 ) + Sk (H ). Denote by H1 , H2 , . . . , Ht the connected components of H, where e(Hi ) ≥ 1 and ni denotes the number of vertices of Hi . Denote by p the of H. Assume that ki of the k largest Laplacian eigenvalues of H come from Hi . tnumber of isolated vertices t t We have that i=1 e(Hi ) = e(H ), i=1 ki = k and i=1 ni = n − p. By using (3) for each graph Hi , we obtain Sk (H ) ≤
t
2e(Hi ) − ni + 2ki = 2(e(G) − m) − n + p + 2k.
i=1
Since the graph K2 has eigenvalues 0 and 2, we can write Sk (G) ≤ Sk (mK2 ) + 2(e(G) − m) − n + 2k + p
≤ 2k + 2(e(G) − m) − n + 2k + p = 2e(G) − 2m − n + 4k + p, which completes the proof.
Next, we give an upper bound for a graph with maximum degree ∆. Theorem 4. For a graph G on n vertices and maximum degree ∆ ≥ 2, let v be a vertex with maximum degree ∆ and p pendant vertices. Then we have Sk (G) ≤ 2e(G) + 3k − n − ∆ + p + 1. Proof. Since G has vertex v with degree ∆, the graph G has a subgraph S∆+1 , the star with ∆ + 1 vertices and center v . Denote by H the graph obtained from G by removing the ∆ edges adjacent to v . We can decompose the Laplacian matrix of G such that L(G) = L(S∆+1 ) + L(H ). The star S∆+1 has eigenvalues ∆ + 1 with multiplicity 1, 1 with multiplicity ∆ − 1 and 0. By applying the Wielandt inequality for the sum of eigenvalues, we have Sk (G) ≤ Sk (S∆+1 ) + Sk (H ). Denote by H1 , H2 , . . . , Ht the connected components of H, where e(Hi ) ≥ 1 and ni is the number of vertices in Hi . Since v is an isolated vertex in H and v has p pendant vertices we have p + 1 isolated vertices in H. Assume that ki of the k largest
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Laplacian eigenvalues of H come from Hi . We have that (3) for each graph Hi , we obtain
t
i=1
e(Hi ) = e(H ),
t
i =1
ki = k and
t
i =1
ni = n − p − 1. By using
Sk (G) ≤ Sk (S∆+1 ) + 2(e(G) − ∆) − n + 2k + p + 1
≤ ∆ + 1 + (k − 1) + 2(e(G) − ∆) − n + 2k + p + 1 = 2e(G) + 3k − n − ∆ + p + 1, which completes the proof.
3. On Brouwer’s conjecture for special graphs Proposition 2. Let G be a graph and let H1 , H2 , . . . , Ht be its connected components. If Conjecture 1 holds for each Hi , then Conjecture 1 holds for G. Proof. Let ni denote the number of of Hi . Assume that ki of the k largest Laplacian eigenvalues of G come from Hi . t vertices t t We have that i=1 e(Hi ) = e(G), i=1 ki = k and i=1 ni = n. Therefore, we can write Sk (G) =
t
Ski (Hi )
i=1
≤
t
e(Hi ) +
ki + 1 2
i=1
= e(G) +
t k2 + ki i
i=1
≤ e(G) + as required.
2
k +k 2
2
,
We recall that the girth of a graph is the size of the smallest cycle in the graph. Theorem 5. Let G be a graph with girth g ≥ 5. Then inequality (1) is true for 1 ≤ k ≤
g 5
.
Proof. First, as proved in [5,2], we observe that Conjecture 1 is true for trees, unicyclic and bicyclic graphs. Hence, we can suppose that G is a c-cyclic graph with c ≥ 3. Suppose, by contradiction, that the following inequality holds e(G) +
k+1
2
< Sk (G),
g
for some 1 ≤ k ≤ 5 . Since G has girth g, it contains subgraph Pg . We define G1 , as the graph obtained from G by removing the edges of Pg . Therefore, we can decompose the Laplacian matrix of G as follows: L(G) = L(Pg ) + L(G1 ). Then, by Proposition 1 and the Wielandt inequality for the sum of eigenvalues, we can write e(G) +
k+1
2
< Sk (G) ≤ Sk (Pg ) + Sk (G1 ) ≤ e(Pg ) + Sk (G1 ).
Since e(G) − e(Pg ) = e(G1 ), we have e(G1 ) +
k+1 2
g
< Sk (G1 ),
for some 1 ≤ k ≤ 5 . Let c1 be the maximum number of cycles at each connected component of G1 . By the construction of G1 , we can see that c1 < c. If 0 ≤ c1 ≤ 2, then each connected component is a tree, a unicyclic or a bicyclic graph. Then, by Proposition 2, Conjecture 1 holds for G1 , which is a contradiction. In this case, the proof is done. Otherwise, suppose that c1 > 2. By the construction of G1 , it has girth g1 ≥ g. Hence, the graph G1 contains path Pg1 . We can define graph G2 as the graph obtained from G1 by removing the edges of path Pg1 . The graph G2 has girth g2 ≥ g. Let c2
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be the maximum number of cycles at each connected component of G2 . By the construction of G2 , we can see that c2 < c1 . Using the same argument as above, we obtain that e(G2 ) +
k+1 2
< Sk (G2 ),
g
for some 1 ≤ k ≤ 5 . Therefore, by successive applications of the argument above, we can define graph Gi as the graph obtained from Gi−1 by removing the edges of path Pgi−1 . The graph Gi has girth gi ≥ g. Let ci be the maximum number of cycles at each connected component of Gi . By the construction of Gi , we can see that ci < ci−1 . Therefore, since G is finite, we eventually end up with graph Gl such that each of its connected component is a tree, a unicyclic or a bicyclic graph. Then, by Proposition 2, Conjecture 1 holds for Gl , which is a contradiction. Hence, graph G satisfies Sk (G) ≤ e(G) + for 1 ≤ k ≤
k+1
2
g 5
, as required.
,
The following result shows that Brouwer’s conjecture holds whenever the graph has a large girth compared with the number of cycles. Theorem 6. Let G be a connected graph with girth g ≥ 5. If
√ 9 + 8e − 8n
3+
2
≤
g 5
,
then Conjecture 1 holds. Proof. Since G is connected, we have 9 + 8e − 8n ≥ 0. Also, we notice that Lemma 3 implies that inequality (1) holds for
√ 3+
9 + 8e − 8n 2
≤ k ≤ n.
We also recall that Theorem 5 implies that inequality (1) holds for 1 ≤ k ≤
g 5
. Hence, the hypothesis
√ 9 + 8e − 8n
3+
2 settles the result.
≤
g 5
Next, we show that inequality (1) is true for graphs with certain conditions on the maximum degree. Theorem 7. Let G be a graph and let v be a vertex with maximum degree ∆ and p pendant vertices. Then inequality (1) is true for
√
25 + 8(e(G) − n − ∆ + p + 1)
5+
2
≤ k ≤ n.
Proof. Since Theorem 4 gives us Sk (G) ≤ 2e(G) + 3k − n − ∆ + p + 1 it is enough to check that e(G) + 3k − n − ∆ + p + 1 ≤ Since
k+1 2
=
k(k + 1) 2
k(k+1) 2
k+1 2
.
we observe that we just need to check that
− (e(G) + 3k − n − ∆ + p + 1)
(7)
is positive for 1 ≤ k ≤ n Rewriting Eq. (7), we obtain k2 − 5k − 2e(G) + 2n + 2∆ − 2 2
.
(8)
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This way, Eq. (8) can be seen as a polynomial in k. We need that k2 − 5k − 2e(G) + 2n + 2∆ − 2p − 2 2
≥ 0.
(9)
The largest root of the polynomial (8) equals to
√ 5+
25 + 8(e(G) − n − ∆ + p + 1) 2
.
And since the polynomial in k (8) is positive for
√ 5+
25 + 8(e(G) − n − ∆ + p + 1) 2
the result follows.
≤k≤n
Now, given the proof above, it is easy to settle Conjecture 1 for the graphs described. Theorem 8. Let G be a connected graph that has vertex v with maximum degree ∆ and p pendant vertices. If G has c cycles and
∆ ≥ c + p + 4, then Conjecture 1 is true.
Proof. It is enough to see that Eq. (9) is positive for 1 ≤ k ≤ n. This holds, whenever the discriminant of the polynomial (9) is negative. We notice that ∆ ≥ c + p + 4 implies that ∆ ≥ e(G) − n + 1 + p + 4. Hence, e(G) − n − ∆ + p + 1 ≤ −4. Therefore, the discriminant of the polynomial (9) satisfies 25 + 8(e(G) − n − ∆ + p + 1) < 0 and we conclude that Eq. (9) is positive for 1 ≤ k ≤ n, as required. This completes the proof. Thus, to see that Conjecture 1 holds for a graph with c cycles, it is enough to see that the graph has ∆ ≥ c + 4 and a vertex with degree ∆ with no pendant vertices. 4. Final remarks We now compare the upper bounds on Sk (G) obtained in Section 2 with the bound of Eq. (3), providing conditions for which these results represent an improvement. For instance, if we consider the upper bound (3), we can see that the upper bound of Theorem 3 is sharper whenever 2 m ≥ 2k + p + 1, where m is the number of edges on maximum matching of G. Hence, our result presents an improvement for graphs having large matchings. Performing the same analysis, it is easy to see that the upper bound of Theorem 4 is sharper than (3) whenever ∆ ≥ k + 2 + p, meaning that our result is better for graphs having large maximum degree. To simplify the results given by Theorems 1 and 2 we notice that
cos
kπ D
+
cos πD sin kDπ + sin kDπ sin πD
≤ 1 + 2k.
Therefore Theorem 1 provides a sharper bound than the bound (3) for graphs having large diameter d, more precisely if 2d ≥ 4k + p + 1, whereas Theorem 2 provides a sharper bound than the bound (3) for graphs having a large cycle Cs , that is if 2s ≥ 4k + p + 3. To better understand for what graphs Theorem 6 proves Conjecture 1, we can start by fixing the number of cycles of the graph. We recall that a connected graph with n vertices and c cycles has n − 1 + c edges. Hence, we can rewrite the hypothesis as
√ 3+
1 + 8c 2
≤
g 5
.
With this, we can see that the more cycles a graph has, the larger the girth has to be for Theorem 6 to be true. For instance, on the set of graphs with 3 cycles, Theorem 6 settles Conjecture 1 for those graphs with girth at least 20; on the set of graphs with 4 cycles, it settles Conjecture 1 for those graphs with girth at least 22 and so forth. Acknowledgments V. Trevisan was partially supported by CNPq (Proc. 309531/2009-8 and 481551/2012-3) and FAPERGS (Proc. 11/1619-2).
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