Bounds for the singless Laplacian energy - Universidade de Aveiro

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Bounds for the singless Laplacianenergy

Nair Abreu∗

Universidade Federal do Rio de Janeiro, [email protected]

Domingos M. Cardoso†

Universidade de Aveiro, [email protected]

Ivan GutmanUniversity of Kragujevac, Serbia

[email protected]

Enide Andrade Martins†

Universidade de Aveiro, [email protected]

Marıa Robbiano‡

Universidad Catolica del Norte, [email protected]

May 5, 2010

∗The author is grateful to CNPq, Projeto PQ- 305016/2006-2007†Research supported by the Center for Research and Development in Mathematics and

Applications from the Fundacao para a Ciencia e a Tecnologia, cofinanced by the EuropeanCommunity Fund FEDER/POCI 2010.‡Research partially supported by Proyecto Mecesup 2 UCN 0605, Chile y Fondecyt-IC

Project 11090211, Chile.


Abstract

The energy of a graph G is the sum of the absolute values of theeigenvalues of the adjacency matrix of G. The Laplacian (respectively,the signless Laplacian) energy of G is the sum of the absolute values ofthe differences between the eigenvalues of the Laplacian (respectively,signless Laplacian) matrix and the arithmetic mean of the vertex de-grees of the graph. In this paper, among some results which relatethese energies, we point out some bounds to them using the energy ofthe line graph of G. Most of these bounds are valid for both energies,Laplacian and singless Laplacian. However, we present two new upperbounds on the signless Laplacian which are not upper bounds for theLaplacian energy.

AMS classification: 5C50, 15A48Graph spectrum, Laplacian graph spectrum, signless Laplacina spectrum,

Laplacian energy, signless Laplacian energy.

1 Preliminaries

In this paper we consider simple unordered graphs G, herein just calledgraphs, with vertex set V(G) and edge set E(G). We say that G is an(n,m)−graph if G has n vertices, vi, 1 ≤ i ≤ n and m edges ek = vivj =vjvi, 1 ≤ k ≤ m, with i, j ∈ {1, . . . , n}. The complement of a graph G isdenoted G. The complete graph on n vertices, Kn, is such that each pair ofvertices is connected by an edge. Then, the null graph is Kn. The line graphL(G) of G is the graph whose vertex set is in one-to-one correspondence withthe edge set of the graph G and where two vertices of L(G) are adjacent ifand only if the corresponding edges in G have a vertex in common [14].

The adjacency matrix of G, A(G), is a binary matrix of order n suchthat aij = 1 if the vertex vi is adjacent to the vertex vj and 0 otherwise.Let D(G) be the diagonal matrix of order n whose (i, i)−entry is the degreedi of the vertex vi ∈ V(G). Then the matrices L(G) = D(G) − A(G) andL+(G) = D(G)+A(G) are the Laplacian and the signless Laplacian matrices,respectively. For details on their spectral properties see [3, 4].

The eigenvalues λ1(G), λ2(G), . . . , λn(G) of the adjacency matrix A(G)of the graph G are also called the eigenvalues of G. For details on spectralgraph theory see [2]. The energy of the graph G is defined as


E (G) =n∑

j=1

|λj(G)|.

Several properties of graph energy can be founded in [11, 12, 13, 15].Let µ1, µ2, . . . , µn and µ+

1 , µ+2 , . . . , µ

+n be the eigenvalues of the matrices

L(G) and L+(G), respectively. Then, the Laplacian energy of G is defined,as in [9],

LE(G) =n∑

i=1

∣∣∣∣µi −2m

n

∣∣∣∣and, in analogy to LE(G), the signless Laplacian energy is defined

LE+(G) =n∑

i=1

∣∣∣∣µ+i −

2m

n

∣∣∣∣ .The concept of matrix energy, put forward by Nikiforov [19], represents

a far-reaching generalization of graph energy. Let C be a real matrix oforder s × t, with singular values s1(C), s2(C), . . . , sq(C). Its energy, E (C),is defined as s1(C) + s2(C) + · · ·+ sq(C), where q ≤ min{s, t}. Therefore, ifC is real and symmetric of order n, then si(C), i = 1, · · · , n are equal to theabsolute values of their eigenvalues.

According to the above definitions, E(G) = E(A(G)), LE(G) = E(L(G)−2mnIn) and LE+(G) = E(L+(G)− 2m

nIn), where In stands for the identity ma-

trix of order n.

2 Energy and signless Laplacian energy

This section begins with the following two well known results which will beused along the paper.

Theorem 1 [5] Let A,B ∈ Rn×n and let C = A+B. Then

E (C) ≤ E (A) + E (B) . (1)

Moreover, equality holds if and only if there exists an orthogonal matrix Psuch that PA and PB are both positive semidefinite matrices.


Consider the matrix B =(ATA

)1/2such that BTB = ATA and denote

|A| ,(ATA

)1/2. Then, we have the following version of the polar decompo-

sition theorem.

Theorem 2 [17] If A ∈ Rn×n, then there exist positive semidefinite matricesX, Y ∈ Rn×n and orthogonal matrices P,Q ∈ Rn×n such that A = PX = Y Q.

Moreover, the matrices X = |A| , Y =(AAT

)1/2are unique that satisfy those

equalities. Also, the matrices P and Q are uniquely determined if and onlyif A is nonsingular.

The inequality LE (G) ≤ E(G) +∑n

i=1

∣∣di − 2mn

∣∣ was proven in [20] andthe equality case was studied in [21]. In analogous manner it can be shownthat

LE+ (G) ≤ E(G) +n∑

i=1

∣∣∣∣di − 2m

n

∣∣∣∣ . (2)

If the graph G is connected then the equality holds if and only if G is regular.

The Zagreb index of a graph G is defined as

Zg(G) =n∑

i=1

d2i . (3)

So, using this invariant, we can obtain:

Theorem 3 If G is an (n,m)−connected graph, then

LE+(G) ≤ E(G) +√nZg(G)− 4m2. (4)

This inequality holds as equality if and only if G is regular.

Proof. By Cauchy-Schwarz inequality, we get

n∑j=1

|dj − 2mn| ≤

√√√√n

n∑j=1

[dj − 2m

n

]2=√nZg(G)− 4m2.

The inequality holds as equality if and only if∣∣dj − 2m

n

∣∣ = c, ∀j = 1, . . . , n.Thus, from inequality (2), the result follows.


LetG be an (n,m)−graph with edge set {e1, e2, . . . , em} and letG(ek), k ∈{1, 2, . . . ,m} be a spanning subgraph of G with only one edge ek connect-ing the vertices vi and vj for some i, j ∈ {1, 2, . . . , n}. Then, the signlessLaplacian matrix of G(ek) is

L+(G(ek))rs =

{1, if r, s ∈ {i, j};0, otherwise.

If α ∈ R, then

E(L+ (G(ek))− αIn

)= (n− 1) |α|+ |2− α| . (5)

Therefore, for k = 1, . . . ,m, the signless Laplacian matrix of G can beexpressed in terms of L+ (G(ek)) as

L+ (G) =m∑k=1

L+ (G(ek)) . (6)

In order to conclude that the upper bound for the Laplacian energy,obtained in [[21], Theorem 7], is also valid for the signless Laplacian energyof an (n,m)−graph, we need the following lemma.

Lemma 4 [21] Let 0 < a < 1 and consider the following matrices.

A =

[a 11 a

], S =

1√2

[1 −11 1

], Q =

[0 11 0

]and D =

[a+ 1 0

0 a− 1

].

Then, A = SDS−1, |A| = S|D|S−1 and A = Q|A|.

Using the Lemma 4 and adapting the proof given in [21], it is straight-forward to conclude the next upper bound on LE+(G).

Theorem 5 Let G be an (n,m)− graph. Then

LE+(G) ≤ 4m(1− 1

n

). (7)

The equality holds if and only if either G is a null graph (that is a graph withn vertices and without edges) or G is a graph with only one edge plus n− 2isolated vertices.


3 Relations between Laplacian and signless

Laplacian energy

It is well known that the spectra of L(G) and L+(G) coincide if and onlyif the graph G is bipartite (see [7] and [8]). Then, of course, LE(G) =LE+(G). It is also elementary to see that if the graph G is regular, thenLE(G) = LE+(G) = E(G). Therefore, the concept of signless Laplacianenergy could be of interest only for non-bipartite, non-regular graphs, inwhich case LE+(G) would differ from LE(G).

There exist non-bipartite non-regular graphs for which the inequality

LE+(G) < LE(G) (8)

holds, and other such graphs for which

LE+(G) > LE(G) (9)

is satisfied. The graphs G1 and G2, depicted if Figure 1 are examples for thevalidity of (8) and (9), respectively.

The following result provides arbitrarily many examples for the equality

LE+(G) = LE(G) (10)

for disconnected non-bipartite, non-regular graphs. Whether there are con-nected graphs of this kind, satisfying (10) remains an open problem.

Theorem 6 Let G be a connected, non-bipartite (n,m)−graph with the Lapla-cian eigenvalues µ1 ≥ · · · ≥ µn−1 > µn = 0 and signless Laplacian eigenval-ues µ+

1 ≥ · · · ≥ µ+n−1 ≥ µ+

n > 0. If we choose an integer p, such that

2m

n+ p< min{µn−1, µ

+n },

then the graph G = G ∪Kp is such that LE(G) = LE+(G).

Proof. Recall that the Laplacian spectrum of G consistes of positive numbers


µ1 ≥ · · · ≥ µn−1 and p+ 1 zeros. Therefore,

LE(G) =n−1∑i=1

∣∣∣∣µi −2m

n+ p

∣∣∣∣+ (p+ 1)2m

n+ p

=n−1∑i=1

(µi −

2m

n+ p

)+ (p+ 1)

2m

n+ p

= 2m+ (p− n)2m

n+ p.

The signless Laplacian spectrum of G consistes of positive numbers µ+1 ≥

· · · ≥ µ+n and p zeros. Therefore,

LE+(G) =n∑

i=1

∣∣∣∣µ+i −

2m

n+ p

∣∣∣∣+ p2m

n+ p

=n∑

i=1

(µ+i −

2m

n+ p

)+ p

2m

n+ p

= 2m+ (p− n)2m

n+ p.

By using Theorem 1 we may prove the following result.

Theorem 7 Let G be an (n,m)− graph. Then∣∣LE+(G)− LE(G)∣∣ ≤ 2E(G). (11)

The equality holds if G is the null graph.

Proof. From the equalities

L(G)− 2mnIn = D(G)− 2m

nIn − A(G)

L+(G)− 2mnIn = D(G)− 2m

nIn + A(G),

it follows that(L+(G)− 2m

nIn)−(L(G)− 2m

nIn)

= 2A(G). Then,

L(G)− 2mnIn = −2A(G) +

(L+(G)− 2m

nIn)

and

L+(G)− 2mnIn = 2A(G) +

(L(G)− 2m

nIn).


By application of Theorem 1, we obtain

LE(G) = E(L(G)− 2m

nIn)≤ E (−2A(G)) + E

(L+(G)− 2m

nIn)

= 2E(G) + LE+(G) (12)

LE+(G) = E(L+(G)− 2m

nIn)≤ E (2A(G)) + E

(L(G)− 2m

nIn)

= 2E(G) + LE(G) (13)

and thus the inequality in (11) follows. Finally, if G is the null graph, thenit is immediate that the inequality (11) holds as equality.

Theorem 8 Let G be an (n,m)−graph, A = A(G) its adjacency matrix andD = D(G)− 2m

nIn. Then,

max{2E(G), 2E(D)} ≤ LE+(G) + LE(G) ≤ 2E(G) + 2E(D). (14)

All the inequalities hold as equalities if and only if G is regular.

Notice that E(D) =∑n

i=1

∣∣di − 2mn

∣∣ , cf. Eq. (2).

Proof. Consider the 2n×2nmatricesX =

(D AA D

)and J = 1√

2

(In InIn −In

).

It is not difficult to prove that J2 = I2n and

JXJ =

(D + A 0

0 D − A

)= Y.

Therefore, E(X) = E(Y ) = LE+(G) + LE(G). We observe that X =(D 00 D

)+

(0 AA 0

)and by applying Theorem 1, the right inequality is

obtained. This result may also be obtained from Theorems 1 and 4 in[21]. According to [21], the right inequality holds as equality if and onlyif G is regular. Regarding the left inequality, since the matrices X and

Z =

(D −A−A D

)are similar, by applying Theorem 1, we obtain

4E(D) = E(X + Z) ≤ E(X) + E(Z) = 2E(X)⇒ 2E(D) ≤ E(X)

and

4E(A) = E(X − Z) ≤ E(X) + E(Z) = 2E(X)⇒ 2E(A) ≤ E(X).


Thus, the left inequality follows. From Theorem 1, the left inequality holdsas equality if and only if there exist orthogonal matrices P and Q such thatPX = M1, PZ = M2, QX = N1, Q(−Z) = N2 are positive semidefinite

matrices. From Theorem 2, M1 = |X| = |Z| = M2, and N1 = |X| = |Z| =

N2, which implies X = Z or X = −Z. In both cases G is a regular graph.

As direct consequence of Theorems 7 and 8, we may conclude that if Gis an (n,m)−graph, then

|LE+(G)− LE(G)|2

≤ E(G) ≤ LE+(G) + LE(G)

2.

Based on the results obtained in Section 2 and also in [10], it seems thatthe Laplacian and signless Laplacian energies satisfy bounds of equal form.However, here we point out some differences between both graph invariants.In particular, we introduce some bounds for the signless Laplacian energywhich, for certain kind of graphs, are not valid for the Laplacian energy.

Let G be an (n,m)− graph whose vertex degrees are d1, . . . , dn and eigen-values λ1 ≥ · · · ≥ λn. Then, the following is known:

1. The line graph L(G) has m vertices and q edges, where

q = −m+1

2

n∑i=1

d2i . (15)

2. According to [16],

E(G) ≤ λ1 +√n− 1

√2m− λ2

1. (16)

Furthermore, if G is an (n,m)−graph with m ≥ n, using (15), (16) andTheorem 10-(b) and (c), we may conclude the inequalities

LE+(G) ≤ E(L(G)) ≤ λ1(L(G)) +√m− 1

√2q − λ2

1(L(G)), (17)

where λ1(L(G)) denotes the index of L(G).


In some cases, we can see that while the above upper bound is true forthe signless Laplacian energy, it is not valid for the Laplacian energy. Forinstance, considering the graph G1 depicted in Figure 1, we obtain

5.123 = LE+(G1) ≤ λ1(L(G1)) +√m− 1

√2q − λ2

1(L(G1))

= 5.773 < 6 = LE(G1).

HHHHH

��

��t

tt

tG1

HHHHH

��

��

��

HHHHH

tt

tt

tt

G2

Figure 1: The graphs G1 and G2 are such that LE+(G1) < LE(G1) andLE+(G2) = 8.666 > 8.456 = LE(G2).

Taking into account the definition of Zagreb index (3) and the equality(15), it is straightforward to conclude that

q = −m+1

2Zg(G). (18)

Let us consider an (n,m)−graph G. Then, it follows:

1. As proven in [1] and [18], the inequalities

2√m ≤ E(G) ≤

√2mn (19)

hold and the upper bound is attained if and only if G is either the nullgraph or m copies of K2.

2. As proven in [16], the inequality

E(G) ≤ 2m

n+

√√√√(n− 1)

[2m−

(2m

n

)2]

(20)

holds and the upper bound is attained if and only if G is either thenull graph, m copies of K2, the complete graph Kn−1 or a stronglyregular connected graph with two non-trivial eigenvalues, both having

the absolute value equal to

√(2m−

(2mn

)2)/(n− 1).


Therefore, using the lower bound in (19) and the upper bound (20) (since,for every graph, the upper bound in (19) is worse than the (20)), it follows√

2Zg(G)− 4m ≤ E(L(G)) ≤ Υ+(G), (21)

where

Υ+(G) =Zg(G)−2m+

√(m−1)(Zg(G)−2m)(m2−Zg(G)+2m)

m.

Therefore, applying (21) and Theorem 10, beyond the inequality√2(Zg(G)− 2m) ≤ LE+(G), (22)

we may conclude the inequalities:

LE+(G) ≤

{Υ+(G), if m ≥ n

Υ+(G) + 4mn (n−m), if m < n.

(23)

The upper bound (23) (for m ≥ n) also shows us that LE+(G) and LE(G)behave differently. In fact, using the same graph displayed in Figure 1, wehave

5.123 = LE+(G) ≤Zg(G)− 2m+

√(m− 1)(Zg(G)− 2m)(m2 − Zg(G) + 2m)

m= 5.854

< 6 = LE(G).

Now, let us compare Υ+(G) with the lower bound in (22). Since,

Zg(G) =n∑

i=1

d2i =

∑ij∈E(G)

(di + dj) ≤ 2m∆(G)⇒ Zg(G)− 2m ≤ 2m(∆(G)− 1),

where ∆(G) denotes the maximum degree of the vertices of G, and

4m2 = (∑n

i=1 di)2 ≤ n

∑ni=1 d

2i ⇒ 2m(2m

n− 1) ≤ Zg(G)− 2m,


denoting d(G) = 2mn

, it follows

Υ+(G) =

√Zg(G)−2m+

√(m−1)(m2−Zg(G)+2m)

m

√Zg(G)− 2m

≤√

2m(∆(G)−1)+√

2m(m−1)(m2−(d−1))

m

√Zg(G)− 2m

=

(√∆(G)−1

m+

√(m−1)(m

2−(d(G)−1))

m

)√2(Zg(G)− 2m)

≤(√

m−1m

+

√(m−1)(m

2−(d(G)−1))

m

)√2(Zg(G)− 2m)

≤(

1 +√

m2− (d(G)− 1)

)√2(Zg(G)− 2m).

Therefore,

√2(Zg(G)− 2m) ≤ LE+(G) ≤ Υ+(G) ≤

(1 +

√m2 − (d(G)− 1)

)√2(Zg(G)− 2m).

Finally, we conclude that the above inequalities give the range whereLE+(G) may vary in function of the parameter

√2(Zg(G)− 2m).

4 A bound for the Laplacian energy for trees

In this section, denoting the number of edges of a graph G by m(G), weprove a proposition which is an application of the following results given in[10].

Theorem 9 [10] Let H be an induced subgraph of G and δ = m(G)−m(H)−m(G−H). Then, we have

(a) L (H) and L (G−H) are induced subgraphs of L (G);

(b) E (L (H)) + E (L (G−H)) + 2δ ≤ E (L (G)) .

Theorem 10 [10] If G is a graph with n ≥ 1 vertices and m ≥ 1 edges, thenthe following statements hold:

(a) If m < n then LE+(G)− 4mn

(n−m) ≤ E(L(G)) < LE+(G). The lefthand equality is attained if and only if G is the direct sum of m copiesof complete K2;


(b) If m > n then LE+(G) < E(L(G)) < LE+(G) + 4(m− n);

(c) Finally, LE+(G) = E(L(G)) if and only if m = n.

Based on the above results, we may prove the following theorem.

Theorem 11 Let T be a tree on n vertices and H a k-vertex connectedsubgraph of T . Then

LE(H) + LE(T −H) < LE(T ) + 2(2 + δ),

where δ is the number of connected components of T −H.

Proof. Since the number of edges of T is m = n − 1 and the subgraph His connected (therefore a subtree) of order k, it follows that the number ofconnected components of T −H is

δ = n− 1− (k − 1)−m(T −H) = n− k −m(T −H).

Notice that δ also coincides with the number of edges between the verticesof T which are in and out of the subtree H. Hence, m(T −H) = n− k − δ.Taking into account that T is bipartite, then LE+(T ) = LE(T ). By usingTheorem 10-(a), we obtain

LE(T )− 4 (n− 1)

n≤ E(L(T )) < LE(T ), (24)

where the left hand inequality is attained as equality if and only if T is K2.By using Theorem 9-(b) applied to the subtree H, we have

E (L (H)) + E (L (T −H)) ≤ E (L (T ))− 2δ.

Now, applying Theorem 10-(a) to the subtree H and to the subgraph T −H,it follows

LE(H)− 4 (k − 1)

k≤ E(L(H)) < LE(H)

and

LE(T −H)− 4 (n− k − δ)n− k

δ ≤ E(L(T −H)) < LE(T −H).

By adding the above inequalities, we arrive at


LE(H) + LE(T −H)− 4(k − 1)(n− k) + (n− k − δ)kδ

(n− k)k≤ E(L(H)) + E(L(T −H))

≤ E (L (T ))− 2δ

< LE(T )− 2δ.

This implies,

LE(H) + LE(T −H) < LE(T ) + 4(k − 1)(n− k) + (n− k − δ)kδ

(n− k)k− 2δ

= LE(T ) + 2(2 + δ)− 4n− k + kδ2

(n− k)k

< LE(T ) + 2(2 + δ).

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