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A NEW UPPER BOUND ON THE LARGEST NORMALIZED LAPLACIAN EIGENVALUE

O SCAR ROJO

AND

R ICARDO L. S OTO

(Communicated by Richard A. Brualdi) Abstract. Let G be a simple undirected connected graph on n vertices. Suppose that the vertices of G are labelled 1,2,... ,n. Let di be the degree of the vertex i. The Randić matrix of G , if the vertices i and j are denoted by R, is the n × n matrix whose (i, j) − entry is √ 1 di d j

IN

T

adjacent and 0 otherwise. The normalized Laplacian matrix of G is L = I − R, where I is the n × n identity matrix. In this paper, by using an upper bound on the maximum modulus of the subdominant Randić eigenvalues of G , we obtain an upper bound on the largest eigenvalue of L . We also obtain an upper bound on the largest modulus of the negative Randić eigenvalues and, from this bound, we improve the previous upper bound on the largest eigenvalue of L .

1. Introduction

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Let G = (V, E) be a simple undirected graph on n vertices. Some matrices on G are the adjacency matrix A, the Laplacian matrix L = D − A and the signless Laplacian matrix Q = D + L, where D is the diagonal matrix of vertex degrees. It is well known that L and Q are positive semidefinite matrices and that (0, 1) is an eigenpair of L where 1 is the all ones vector. Fiedler [16] proved that G is a connected graph if and only if the second smallest eigenvalue of L is positive. This eigenvalue is called the algebraic connectivity of G . The signless Laplacian matrix has recently attracted the attention of several researchers. Recent papers on this matrix are [5, 6, 7, 8, 9] and some of its basic properties [6] are:

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1. For a connected graph, the smallest eigenvalue of Q is equal to 0 if and only if the graph is bipartite. In this case, 0 is a simple eigenvalue. Then, for a connected graph, the smallest eigenvalue of Q is positive if and only if the graph is not bipartite. 2. If G is a bipartite graph then Q and L have the same characteristic polynomial.

Mathematics subject classification (2010): 05C50, 15A48. Keywords and phrases: normalized Laplacian matrix, Randić matrix, upper bound, largest eigenvalue, subdominant eigenvalue. This research was supported by Project Fondecyt 1100072, Chile. It was finished when the first author was a visitor at the Department of Mathematics, Universidade de Aveiro, Aveiro, Portugal. c D l , Zagreb

Paper OaM-0594

1

2

O. ROJO AND R. L. S OTO

Other matrices on the graph G are the normalized Laplacian matrix and the Randić matrix of G . Suppose that the vertices of G are labelled 1, 2, . . . , n. Let di be the degree 1 of the vertex i. Let D− 2 be the diagonal matrix whose diagonal entries are 1 1 1 √ , √ ,..., √ d1 d2 dn 1

whenever di 6= 0. If di = 0 for some i then the corresponding diagonal entry of D− 2 is defined to be 0. The normalized Laplacian matrix of G , denoted by L , was introduced by F. Chung [15] as 1

1

1

1

L = D− 2 LD− 2 = I − D− 2 AD− 2 .

(1)

The eigenvalues of L are called the normalized Laplacian eigenvalues of G . From (1) , we have 1

1

1

1

and thus

D 2 L D 2 1 = L1 = 01. 1

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Hence 0 is an eigenvalue of L with eigenvector D 2 1. We recall the following results on L [15] :

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D2 L D2 = D−A = L

1. The eigenvalues of L lie in the interval [0, 2].

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2. 0 is a simple eigenvalue of L if and only if G is connected.

3. 2 is an eigenvalue of L if and only if a connected component of G is bipartite and nontrivial. Among papers on L , we mention [10, 11, 13, 14] and [17]. From now on, we assume that G is connected graph. Then di > 0 for all i. The 1 1 notation i ∼ j means that the vertices i and j are adjacent. The matrix R = D− 2 AD− 2 1 in (1) is the Randić matrix of G in which the (i, j)-entry is √ if i ∼ j and 0 di d j

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otherwise. Moreover

I − L = R.

The eigenvalues of R are called the Randić eigenvalues of G . Clearly L and R are both real symmetric matrices. The Randić matrix was earlier studied in connection with the Randić index [1, 2, 18] and [19]. Two recent papers on the Randić matrix are [3] and [4]. Throughout this paper 0 = λn 6 λn−1 6 . . . 6 λ1 and

ρn 6 ρn−1 6 . . . 6 ρ1

3

B OUND ON THE LARGEST NORMALIZED L APLACIAN EIGENVALUE

are the normalized Laplacian eigenvalues and the Randić eigenvalues of G , respectively. It follows that λi = 1 − ρn−i+1 (1 6 i 6 n) . If M is a nonnegative matrix then, by the Perron-Frobenius Theorem, M has an eigenvalue equal to its spectral radius, called the Perron root of M. In addition, if M is irreducible then the Perron root of M is a simple eigenvalue with a corresponding positive eigenvector, called the Perron vector of M. Since G is a connected 1 graph, Randić matrix of G is a irreducible nonnegative matrix. Let v = D 2 1. Then √ √ √ T v = d1 , d2 , . . . , dn . An easy computation shows that Rv = v.

T HEOREM 1. [17] Let G be a connected graph. Then qn qn 6 λ1 6 2 − . δ ∆

IN

2−

T

Hence, 1 and v are the Perron root and the Perron vector of R, respectively. Let ∆ and δ be the largest and smallest vertex degrees of G , respectively, and let qn be the smallest eigenvalue of Q. A recent result involving the largest eigenvalue of L and the smallest eigenvalue of Q is

(2)

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We may consider 2 − q∆n as an upper bound on λ1 . Observe that 2 − q∆n = 2 if and only if G is a bipartite graph. In this paper, we search for a new upper bound on λ1 not exceeding the trivial upper bound 2. 2. Searching for an upper bound on λ1

Since ∑ni=1 ρi = tr (R) = 0, it follows that ρn < 0. We have

λ1 = 1 − ρn = 1 + |ρn | .

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In order to find an upper bound on λ1 not exceeding 2, we look for an upper bound on |ρn | not exceeding 1. An eigenvalue of a nonnegative matrix M which is different from the Perron root is called a subdominant eigenvalue of M. Let ξ (M) be the maximum modulus of the subdominant eigenvalues of M. Special attention has been devoted to find upper bounds on ξ (M) . In [20], we can find a unified presentation of results concerning upper bounds on ξ (M) . These upper bounds are important because ξ (M) plays a major role in convergence properties of powers of M. Since

λ1 6 1 + ξ (R) ,

(3)

we focus our attention on upper bounds on ξ (R) . We recall the result [12, p. 295] :

4


T HEOREM 2. If M = (mi, j ) > 0 of order n × n has a positive eigenvector w = [w1 , w2 , . . . , wn ]T corresponding to the spectral radius ρ (M) of M then n mi,k m j,k 1 . ξ (M) 6 max ∑ wk − 2 i< j k=1 wi wj

where the maximum is taken over all pairs (i, j) , 1 6 i < j 6 n. In order to apply Theorem 2, it is convenient to observe that the Randić matrix of G is diagonally similar to the row stochastic matrix 1

1

S = D− 2 RD 2 .

(4)

The following lemma gives some immediate properties of S.

Ni .

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L EMMA 1. 1. The (i, j) − entry of S is d1i if j ∼ i and 0 otherwise. 2. S1 = 1 where 1 is the all ones vector. 3. u is an eigenvector for R corresponding to the eigenvalue α if and only if 1 D− 2 u is an eigenvector for S corresponding to the eigenvalue α . 4. If G is an r− regular graph then S = R. Let Ni be the set of neighbours of the vertex vi and let |Ni | be the cardinality of

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T HEOREM 3. Let G be a simple undirected connected graph. If λ1 is the largest eigenvalue of L then ( ) Ni ∩ N j (5) |λ1 | 6 2 − min i< j max di , d j where the minimum is taken over all pairs (i, j) , 1 6 i < j 6 n.

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Proof. We know that the Randić matrix of G is similar to the row stochastic matrix S defined in (4) . Then ξ (R) = ξ (S) . The eigenvector corresponding to the spectral of S is w = 1 . Applying Theorem 2 to S = (si, j ) , we have

ξ (S) 6

n 1 max ∑ si,k − s j,k 2 i< j k=1

! 1 1 1 1 ∑ di + ∑ d j + ∑ di − d j k∈N j −Ni k∈Ni ∩N j k∈Ni −N j ! Ni − N j N j − Ni 1 1 1 = max + + ∑ − 2 i< j di dj dj k∈Ni ∩N j di ! Ni ∩ N j N j ∩ Ni 1 1 1 = max 2 − − + ∑ − . 2 i< j di dj dj k∈Ni ∩N j di 1 = max 2 i< j


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Suppose di = max di , d j . In this case Ni ∩ N j N j ∩ Ni 1 1 2− − + ∑ − di dj dj k∈Ni ∩N j di Ni ∩ N j N j ∩ Ni 1 1 Ni ∩ N j − + − = 2− di dj d j di 2 Ni ∩ N j = 2− . di Similarly, if d j = max di , d j then Ni ∩ N j N j ∩ Ni 1 1 2− − + ∑ − di dj dj k∈Ni ∩N j di 2 N j ∩ Ni . = 2− dj Hence

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IN

n 1 max ∑ si,k − s j,k 2 i< j k=1 ( ) 2 N j ∩ Ni 1 = max 2 − 2 i< j max di , d j ( ) Ni ∩ N j = 1 − min i< j max di , d j

ξ (S) 6

Since λ1 6 1 + ξ (R) = 1 + ξ (S), the upper bound in (5) follows.

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R EMARK 1. If G is a bipartite graph then Ni ∩ N j = 0, for some i < j, and consequently the upper bound in (5) is equal to 2. This is sufficient condition but it is not a necessary condition. In fact, there are other instances in which Ni ∩ N j = 0 for some i < j. One of them is given by a nonbipartite graph having a bridge. However, if mini< j Ni ∩ N j > 1 and qn < 1 then

In fact and

( ) Ni ∩ N j q < 2− n. 2 − min i< j ∆ max di , d j qn < 1 6 Ni ∩ N j for i < j

1 qn for i < j. 6 ∆ max di , d j

(6)

6


Then

It follows

Ni ∩ N j qn for i < j. < ∆ max di , d j

Ni ∩ N j qn . 2 − > 2 − min i< j max di , d j ∆

Hence, if mini< j Ni ∩ N j > 1 and qn < 1 then (5) gives a better upper bound for λ1 than the second inequality in (2) does. 3. Improving the upper bound on λ1 We have

λ1 = u1 + |qn | 6 1 + ξ (R) = 1 + ξ (S) .

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The upper bound on λ1 in (5) was obtained by using an upper bound on ξ (R) . In this section, in order to get an improved upper bound on λ1 , we search for an upper bound on |qn | , that is, on the largest modulus of the negative Randić eigenvalues.

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T HEOREM 4. Let G be a simple undirected connected graph. If ρn is eigenvalue with the largest modulus among the negative Randić eigenvalues of G then ( ) Ni ∩ N j |ρn | 6 1 − min i∼ j max di , d j

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where the minimum is taken over all pairs (i, j) , 1 6 i < j 6 n, such that the vertices i and j are adjacent. Proof. Let ρn be the largest modulus of the negative Randić eigenvalues of G . Let

x = [x1 , x2 , . . . , xn ]T

be such that

Sx = ρn x.

(7)

− 21

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From Lemma 1, we have x =D u where Ru = ρn u. Since u is orthogonal to the Per√ √ √ T ron vector v = d1 , d2 , . . . , dn , the vector u has at least one positive component 1

and at least one negative component. Since x =D− 2 u, this is also true for the vector x. Let max {x1 , x2 , . . . , xn } = xi and let

x j = min {xk : k ∼ i} .

Since x has at least one positive component, xi > 0. Let S = (si, j ) . From (7) n

ρn x j =

1

∑ s j,k xk = d j ∑ xk

k=1

k∈N j

(8)


7

and n

ρ n xi =

1

∑ si,k xk = di ∑ xk .

k=1

(9)

k∈Ni

Subtracting (9) from (8) , we get

ρn (x j − xi ) =

1 dj

1

∑ xk − d i ∑ xk .

k∈N j

k∼i

Then qn (x j − xi ) 1 1 1 1 = xk + xk − xk − ∑ ∑ ∑ xk . d j k∈N∑ d d d j k∈N j ∩Ni i k∈Ni −N j i k∈N j ∩Ni j −Ni

(10)

xk 6 N j − Ni xi

∑

xk 6 − Ni − N j x j .

and

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k∈Ni −N j

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∑

k∈N j −Ni

−

T

By definition, x j 6 xk for all k ∼ i and xk 6 xi for all k . Hence

Replacing the inequalities (11) and (12) in (10) , we obtain qn (x j − xi ) 1 1 1 1 6 N j − Ni xi − |Ni − Ni | x j + ∑ xk . − dj di di k∈N j ∩Ni d j

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Thus

qn (x j − xi ) 6

1 1 1 1 N j − Ni (xi − x j ) + Ni − N j (xi − x j ) 2 dj 2 di 1 1 1 + N j − Ni − Ni − N j (xi + x j ) 2 dj di 1 1 + ∑ xk . − di k∈Ni ∩N j d j

Clearly 1 1 N j − Ni − Ni − N j = dj di

1 1 − di d j

Ni ∩ N j .

(11)

(12)

8


Hence

ρn (x j − xi ) 6

=

1 1 1 1 N j − Ni (xi − x j ) + Ni − N j (xi − x j ) 2 dj 2 di 1 1 1 1 1 1 + (xk + xk ) − − Ni ∩ N j (xi + x j ) + 2 di d j 2 k∈N∑ d j di i ∩N j 1 1 1 1 N j − Ni (xi − x j ) + Ni − N j (xi − x j ) 2 dj 2 di 1 1 1 + (xi − xk + x j − xk ) . − 2 k∈N∑ di d j i ∩N j

Moreover

Therefore

1 1 1 1 N j − Ni (xi − x j ) + Ni − N j (xi − x j ) 2 dj 2 di 1 1 1 + di − d j (xi − x j ) . 2 k∈N∑ i ∩N j

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ρn (x j − xi ) 6

IN

T

1 1 (xi − xk + x j − xk ) − ∑ dj k∈Ni ∩N j di 1 1 1 1 6 ∑ − (xi − xk ) + ∑ − (xk − x j ) dj dj k∈Ni ∩N j di k∈Ni ∩N j di 1 1 = ∑ − (xi − x j ) . dj k∈Ni ∩N j di

(13)

If x j = xi then xk = xi for all k ∼ i. Consequently, from Sx = ρn x, we have q n xi =

xi 1 1 xi = d i = xi . xk = ∑ d d d i k∈Ni i k∈Ni i

∑

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Thus ρn = 1, which is a contradiction. Hence xi − x j > 0. Dividing both sides of (13) by (xi − x j ) , we obtain 11 1 1 Ni − N j + 1 ∑ 1 − 1 . N j − Ni + (14) −ρn 6 2 dj 2 di 2 k∈Ni ∩N j di d j As in the proof of Theorem 3, we get

11 1 1 Ni − N j + 1 ∑ 1 − 1 N j − Ni + 2 dj 2 di 2 k∈Ni ∩N j di d j Ni ∩ N j . = 1− max di , d j


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Consequently Ni ∩ N j . |ρn | 6 1 − max di , d j

Observe that the vertices vi and v j are adjacent. Hence ( ( ) ) Ni ∩ N j Ni ∩ N j = 1 − min . |ρn | 6 max 1 − i∼ j i∼ j max di , d j max di , d j

The proof is complete.

Finally, we have T HEOREM 5. Let G be a simple undirected connected graph. If λ1 is the largest normalized Laplacian eigenvalue of G then ( ) Ni ∩ N j λ1 6 2 − min i∼ j max di , d j

T

where the minimum is taken over all pairs (i, j) , 1 6 i < j 6 n, such that the vertices i and j are adjacent.

E XAMPLE 1. G :

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Proof. Since λ1 = 1 − ρn = 1 + |ρn | , the proof is immediate using the upper bound on |ρn | given by Theorem 4. 5

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6

3

1

Let

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For this graph

4

2

Ni ∩ N j b (i, j) = max di , d j b (1, 2) =

1 2

2 1 b (1, 3) = b (2, 3) = b(3, 4) = b (3, 5) = , b (3, 6) = 5 5 1 b (4, 6) = b (5, 6) = . 3 Then mini∼ j b (i, j) = 51 . Hence the largest modulus of the negative Randić eigenvalues is bounded above by 45 and the largest normalized Laplacian eigenvalue is bounded above by 59 = 1.8. To four decimal places the smallest signless Laplacian eigenvalue of = 1.8518. G is 0.7411. Since ∆ = 5, the upper bound in (2) becomes 2 − 0.7411 5

10

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(Received November 19, 2011)

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Oscar Rojo Department of Mathematics Universidad Católica del Norte Antofagasta, Chile e-mail: [email protected] Ricardo L. Soto Department of Mathematics Universidad Católica del Norte Antofagasta, Chile e-mail: [email protected]