On the least distance eigenvalue and its applications on the distance spread

On the least distance eigenvalue and its applications on the distance spread

Discrete Mathematics 338 (2015) 868–874 Contents lists available at ScienceDirect Discrete Mathematics journal homepage: www.elsevier.com/locate/dis...

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Discrete Mathematics 338 (2015) 868–874

Contents lists available at ScienceDirect

Discrete Mathematics journal homepage: www.elsevier.com/locate/disc

On the least distance eigenvalue and its applications on the distance spread Huiqiu Lin Department of Mathematics, East China University of Science and Technology, Shanghai, 200237, China

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Let G be a connected graph with order n and D(G) be its distance matrix. Suppose that λ1 (D) ≥ · · · ≥ λn (D) are the distance eigenvalues of G. In this paper, we give an upper bound √ on the least distance eigenvalue and characterize all the connected graphs with −1 − 2 √ ≤ λn (D) ≤ a where a is the smallest root of x3 − x2 − 11x − 7 = 0 and √ a ∈ (−1 − 2, −2). Furthermore, we show that connected graphs with λn (D) ≥ −1 − 2 are determined by their distance spectra. As applications, we give some lower bounds on the distance spread of graphs with given some parameters. In the end, we characterize connected graphs with the (k + 1)th smallest distance spread. © 2015 Elsevier B.V. All rights reserved.

Article history: Received 2 July 2014 Received in revised form 2 November 2014 Accepted 8 January 2015

Keywords: Distance spectra Distance spectral radius The least distance eigenvalue Distance spread

1. Introduction All graphs considered here are simple, undirected and connected. Let G be a graph with vertex set V (G) = {v1 , v2 , . . . , vn } and edge set E (G). Two vertices u and v are called adjacent if they are connected by an edge. Let dG (v) and NG (v) denote the degree and the neighbor set of a vertex v in G, respectively. Let ∆(G) and δ(G) denote the maximum degree and minimum degree of G, respectively. The distance between vertices u and v of a graph G is denoted by dG (u, v). The diameter of G, denoted by d or d(G), is the maximum distance between any pair of vertices of G. The induced subgraph G[X ] is the subgraph of G whose vertex set is X and whose edge set consists of all edges of G which have both ends in X . Let Gc denote the complement of the graph G. The complete product G1 ∨ G2 of graphs G1 and G2 is the graph obtained from G1 ∪ G2 by joining every vertex of G1 with every vertex of G2 . Let Ksr,t = Kr ∨ (Ks ∪ Kt ) with r ≥ 1. The distance matrix D(G) = (dij )n×n of a connected graph G is the matrix indexed by the vertices of G, where dij denotes the distance between the vertices vi and vj . The polynomial PD (λ) = det (λI − D(G)) is defined as the distance characteristic polynomial of the graph G. Let λ1 (D) ≥ λ2 (D) ≥ · · · ≥ λn (D) be the distance spectra of G, where λn (D) is called the least distance eigenvalue of G. Let Kn1 ,...,nk denote the complete k-partite graph and G(s, n1 , . . . , nk ) denote the graph (K1 ∪ K2 ) ∨ · · · ∨ (K1 ∪ K2 ) ∨ Kn1 ···,nk . Lin et al. [8] showed that λn (D(G)) = −2 if and only if G is a complete



 s



multipartite graph, and the result is independently proved by Yu [12] in a different way. In this paper, we first give the following result. 3 2 Theorem 1.1. Let G be √a connected graph and D be its distance matrix. Suppose that a is the smallest root √ of x − x − 11x − 7 = 0. Then λn (D) ∈ [−1 − 2, a] if and only if G = G(s, n1 , . . . , nk ) for s ≥ 1. Moreover, λn (D) = −1 − 2 with multiplicity s − 1.

E-mail address: [email protected]. http://dx.doi.org/10.1016/j.disc.2015.01.006 0012-365X/© 2015 Elsevier B.V. All rights reserved.

H. Lin / Discrete Mathematics 338 (2015) 868–874

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Two graphs are said to be D-cospectral if they have the same distance spectrum. A graph G is said to be determined by the D-spectra if there is no other nonisomorphic graph D-cospectral to G. It seems to be a difficult and interesting problem that which graphs are determined by their spectrum. This question was proposed by Dam and Haemers in [2]. In the paper, Dam and Haemers investigated the cospectrality of graphs up to order 11. Later, Dam et al. [3,4] provided two excellent surveys on this topic. Up to now, only a few families of graphs were shown to be determined by their spectra. In particular, there are much fewer results on which graphs are determined by their D-spectra. In [8], Lin et al. proved that the complete bipartite graph Kn1 ,n2 and the complete split graph Ka ∨ Kbc are determined by D-spectra, and conjectured that the complete k-partite graph Kn1 ,n2 ,...,nk is determined by its D-spectra. Recently, Jin and Zhang [6] confirmed the conjecture. Lin, Zhai and Gong [9] showed that the graph Ksr,t is determined by its distance spectra. In this paper, we show the following result. Theorem 1.2. Let G be a connected graph and D be its distance matrix. If λn (D) ≥ −1 − spectra.

√ 2, then G is determined by its distance

Let Sp(D) = λ1 (D) − λn (D) denote the distance spread of the connected graph G. In this paper, we will give some lower bounds on the distance spread with given some parameters. Furthermore, we will determine the extremal graphs with the (k + 1)th smallest distance spread. 2. On the least distance eigenvalue of graphs Lemma 2.1 (Cauchy Interlace Theorem). Let A be a Hermitian matrix with order n and let B be a principal submatrix of A with order m. If λ1 (A) ≥ λ2 (A) ≥ · · · ≥ λn (A) list the eigenvalues of A and µ1 (B) ≥ µ2 (B) ≥ · · · ≥ µm (B) list the eigenvalues of B, then λn−m+i (A) ≤ µi (B) ≤ λi (A) for i = 1, . . . , m. The following result is the complete theorem of Weyl and So. Lemma 2.2 ([11]). Let A and B be n × n Hermitian matrices and C = A + B. Then

λi (C ) ≤ λj (A) + λi−j+1 (B) (n ≥ i ≥ j ≥ 1), λi (C ) ≥ λj (A) + λi−j+n (B) (1 ≤ i ≤ j ≤ n). In either of these inequalities equality holds if and only if there exists a nonzero n-vector that is an eigenvector to each of the three involved eigenvalues. Suppose that K2,n−4 = 2K1 ∨ (n − 4)K1 and 2K1 = {u, v}. Let S2,n−2 denote the graph obtained from K2,n−4 by attaching two pendent vertices to each vertex of u and v , respectively. Let S2+,n−2 be the graph obtained by adding an edge between u and v of S2,n−2 . Theorem 2.3. Let G be a connected graph with diameter d and D be its distance matrix. Then we have the following statements. (i) If d = 2, then λn (D) ≤ −2 with√equality holding if and only if G is a complete multipartite graph. (ii) If d = 3, then λn (D) ≤ −2 − 2, moreover, the equality holds when G ∼ = S2+,n−2 .

√ (iii) If d = 4, then λn (D) ≤ −3 − 5, moreover, the equality holds when G ∼ = S2,n−2 . (iv) If d ≥ 5, then λn (D) ≤ −2d + 3. √ √ Proof. If d = 2,√ then the result follows from Lin√et al. [8]. Note that λ4 (D(P4 )) = −2 − 2 and λ5 (D(P5 )) = −3 − 5. Then λn (D) ≤ −2 − 2 if d = 3 and λn (D) ≤ −3 − 5 if d = 4. Suppose that d ≥ 5 and Pd+1 = v1 v2 · · · vd+1 is a diameter path of G. Let X = (x1 , x2 , . . . , xn )t be an n-vector with xi corresponding to vi . Assume that x1 = x2 = 1, xd = xd+1 = −1 and the other element of X is 0. Thus,

 λn (D) ≤

t

X DX XtX

=

vi ,vj ∈V (G)

dij xi xj

XtX

=

2(−4d + 4) + 4 4

= −2d + 3.

By a simple calculation, we have PD(S + ) (x) = (x + 2)n−5 (x2 + 4x + 2)[x3 − (2n − 6)x2 − (2n + 6)x − 20 + 2n] and 2,n−4 PD(S2,n−4 ) (x) = (x + 2)n−5 (x2 + 6x + 4)[x3 −(2n − 4)x2 +(2n − 28)x − 48 + 8n]. Let f1 (x) = x3 −(2n − 6)x2 −(2n + 6)x − 20 + 2n



and f2 (x) = x3 − (2n − 4)x2 + (2n − 28)x − 48 + 8n. If d = 3 and n = 4, then λ4 (D(P4 )) = −2 − 2 and the equality holds immediately. So we may assume that √ n ≥ 5. Since f1 (1) = −19 − 2n < 0, f1 (−1) = 2n − 9 > 0 and f1 (−3) = 25 − 10n < 0, we have λn (D(S2+,n−4 )) = −2 − 2. If d = 4 and note that f2 (−3) = 45 − 16n < 0, f2 (−1) = 4n − 17 > 0 and f2 (3) = −69 − 4n < 0, we have λn (D(S2,n−4 )) = −3 −



5. Therefore we complete the proof.



By the above theorem, we have the following result, which is a conjecture due to Aouchiche and Hansen [1]. Theorem 2.4 ([1], Conjecture 7.13). Let G be a connected graph and D be its distance matrix. Then λn (D) ≤ −d where d is the diameter of G and the equality holds if and only if G is a complete multipartite graph.

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H. Lin / Discrete Mathematics 338 (2015) 868–874

Fig. 1. The forbidden subgraphs Hi for i = 1, 2, 3.

Lemma 2.5. Let G = G(s, n√ 1 , . . . , nk ) for s ≥ 1 and D be its distance matrix. Then we have the following statements. (i) If s = 1, then −1 − 2 ≤ λn (√ D) ≤ a, where a is the smallest root of x3 − x2 − 11x − 7 = 0. (ii) If s ≥ 2, then λn (D) = −1 − 2 with multiplicity s − 1. Proof. Since diam(G) = 2, we have D(G) = J − I + A(Gc ) and Gc = sP3 ∪

k

i=1

Kni . Then by Lemma 2.2, we obtain

λn (J − I ) + λi (A(G )) ≤ λi (D) ≤ λ2 (J − I ) + λi−1 (A(G )) for 2 ≤ i ≤ n. (1) √ √ If s = 1, then −1 − 2 ≤ λn (D) < −2. If λn (D) = −1 − 2, then by Lemma 2.2, there exists a unit vector X = (x1 , . . . , xn )t k n that is an eigenvector to each of the eigenvalues λn (D), λn (P3 ∪ i=1 Kni ) and λn (Jn − In ). The latter implies that i=1 xi = 0 as X is orthogonal to the eigenspace of λ1 (Jn − In ), which is Span(jn ). Let P3 =√v1 v2 v3 and xi correspond to vi for i = 1, . . . , n. √ n k Since λn (P3 ∪ i=1 Kni ) = − 2, so xi = 0 for vi ̸∈ V (P3 ) and x1 = x3 = − 22 x2 for vi ∈ V (P3 ). It follows that i=1 xi ̸= 0, √ which is a contradiction. Hence λn (D((K1 ∪ K2 ) ∨ Kn1 ,···,nk )) > −1 − 2. Note that G contains (K1 ∪ K2 ) ∨ K1 as an induced subgraph. Then λn (D) ≤ λ4 (D((K1 ∪ K2 ) ∨ K1 )) = a where a is the smallest root of x3 − x2 −√ 11x − 7 = √0. Suppose that s ≥ 2 and n1 ≥ n2 ≥ · · · ≥ nt ≥ 3 > nt +1 ≥ · · · ≥ nk . Note that spec (Ac ) = {n1 − 1, . . . , nt − 1, 2, . . . , 2, nt +1 − 1, . . . , nk −    √ √ √s 1, 0, . . . , 0, −1, . . . , −1, − 2, . . . , − 2}. Thus by (1), we have −2 ≥ λn−s+1 (D) ≥ −1 − 2 = λn−s+2 = · · · = λn . If          s s n−3s−k √ X = (x1 , . . . , xn )t is the eigenvector of D corresponding to −1 − 2, then X is also the eigenvector of Jn − In corresponding √ n X is the eigenvector of A(Gc ) to −1 and the eigenvector of Ac corresponding to − 2. Thus we have i =1 x i = √ √ 0. Since √ corresponding to − 2, xi = 0 if vi not belongs to some vertex in P3 and xi2 = − 2xi1 = − 2xi3 if P3 = vi1 vi2 vi3 . So we √ √ s may assume that X = (−a1 , 2a1 , −a1 , . . . , −as , 2as , −as , 0, . . . , 0)t and i=1 ai = 0, that is a1 = −a2 − · · · − as . Let √ √ √ √ X1 = (1, − 2, 1, −1, 2, −1, 0, . . . , 0, 0, . . . , 0)t , . . . , Xs−1 = (1, − 2, 1, 0, . . . , 0, −1, 2, −1, 0, . . . , 0)t . Obviously,       3s √ 3s X1 , . . . , Xs−1 are linear independent eigenvectors of D corresponding to −1 − 2. Therefore, X = a2 X1 + · · · + as Xs−1 , which √ implies that the multiplicity of −1 − 2 is s − 1.  c

c

So, we are ready to present the proof of Theorem 1.1 (see Fig. 1). Proof of Theorem 1.1. If G = G(s, n1 , . . . , nk ) for s ≥ 1, then by Lemma 2.5, the result holds. For the converse, we suppose √ that −1 − 2 ≤ λn (D(G)) ≤ a. If Gc contains P4 as an induced subgraph, then G contains P4 as an induced subgraph as well. Note that diam(G) = √ 2 by Theorem 2.3, thus G contains one of {H1 , H2 , H3 } as √ an induced subgraph. By a simple calculation, √ √ λ5 (D(H1 )) < −1 − 2, λ5 (D(H2 )) < −1 − 2 and λ5 (D(H3 )) < −1 − 2, so by Lemma 2.1, λn (D) < −1 − 2, a c contradiction. Thus G does not contain P4 as an induced subgraph. We will distinguish the fact by the following two cases. Case 1. Gc is connected. Since Gc does not contain P4 as an induced subgraph, we have ∆(Gc ) ≥ 3. So we can find four vertices, say {v1 , v2 , v3 , v4 } ∈ V (Gc ), such that Gc [v1 , v2 , v3 , v4 ]  K4 and {v1 v2 , v1 v3 , v1 v4 } ⊆ E (Gc ) since G is connected. By Theorem 2.3, we have diam(G) = 2. Then D(G) contains one of {A1 , A2 , A3 } as a principle submatrix.



0 2 A1 =  2 2

2 0 2 2

2 2 0 1



2 2 , 1 0



0 2 A2 =  2 2

2 0 1 2

2 1 0 1



2 2 , 1 0





0 2 A3 =  2 2

By a simple calculation, we have √ λ4 (A1 ) < −1 − 2, λ4 (A2 ) < −1 − we have λn (D) ≤ λ4 (Ai ) < −1 − 2, which is a contradiction.

2 0 1 1



2 1 0 1



2 1 . 1 0

2 and λ4 (A3 ) < −1 −

√ 2. Then by Lemma 2.1,

Case 2. Gc is not connected. Since G is not a complete multipartite graph, Gc contains some non-complete components. Then we have the following claim. Claim. Each non-complete component of Gc is P3 .

H. Lin / Discrete Mathematics 338 (2015) 868–874

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If not, then we may assume that G1 is a non-complete component of Gc with order at least 4. Since Gc does not contain P4 as an induced subgraph, we√ have ∆(G1 ) ≥ 3. Similar to Case 1, D(G) contains one of {A1 , A2 , A3 } as a principle submatrix. Thus, we have λn (D) < −1 − 2, which is a contradiction. This completes the proof. 

√ 3. Graphs with the least distance eigenvalue at least −1 −

2 are determined by their distance spectra

By the result of Lin et al. [8], we know that λn (D(G)) = −2 if and only if G is a complete multipartite graph. In the same paper, they conjectured that the complete multipartite graphs are determined by their distance spectra. Recently, Jin and Zhang [6] confirmed the conjecture. Let a be the smallest root of x3 − x2 − 11x − 7 = 0. By Theorem 1.1, √ λn (D(G)) ∈ [−1 − 2, a] if and only if G ∼ = G(s, n1 , . . . , nk ) for s ≥ 1 and there are no graphs with λn (D(G)) ∈ (a, −2). In order to prove Theorem 1.2, it suffices to show that G(s, n1 , . . . , nk ) is determined by its distance spectra. Before proceeding, the following lemmas are needed. Let mλ (D) denote the multiplicity of the eigenvalue λ of the distance matrix D of G. Lemma 3.1. Let u, v be two non-adjacent vertices of a connected graph G. If u, v have the same neighborhood, then m−2 (D(G)) = 1 + m−2 (D((G − u))). Proof. Note that m−2 (D(G)) = n − rank(−2I − D) = n − rank(2I + D). Let (0 2 α) and (2 0 α) be the row vector of the distance matrix of G corresponding to the vertices u and v , respectively. Then the uth row (column) of D + 2I is equal to the v th row (column) of D + 2I. Then by letting the v th row (column) of D + 2I minus the uth row (column) of D + 2I, we obtain that rank(D + 2In ) = rank(D((G − {u})) + 2In−1 ). Thus, we have m−2 (D(G − {u})) = n − 1 − rank(D((G − {u})) + 2In−1 ) = n − 1 − rank(D((G − {u})) + 2In−1 ) = m−2 (D(G)) − 1. This completes the proof.  Lemma 3.2. Let G = G(s, n1 , . . . , nk ) for s ≥ 1. Then PD(G) (λ) = (λ + 2)n−3s−k (λ2 + 2λ − 1)s−1 (λ + 1)s [(λ2 − (3s − 2)λ −

7s − 1)

k

i=1

(λ − ni + 2) − (λ2 + 2λ − 1)

k

i =1

ni

k

j=1,j̸=i

(λ − ni + 2)].

Proof. The distance matrix of G has the form

 D=

D1 Jn−3s,3s

J3s,n−3s D2



,

where

0 2 2 

.  1 

2 0 1

2 1 0

1 1 1

··· ··· ··· ··· ··· ··· ···

D1 = ..

.. .

.. .

.. .

1 1

1 1 1

1 1 1

1 1 1

1 1 1

1 1 1

.. .

.. .

0 2 2

2 0 1

1 1 1 

..  .  2 

2Jn1 − 2In1 Jn2 ×n1

Jn1 ×n2 2Jn2 − 2In2

.

.. .

··· ··· .. .

.. .

Jnk ×n1

Jnk ×n2

···

2Jnk − 2Ink



and D2 =  ..

1 0

Jn1 ×nk Jn2 ×nk

Then det (λI − D) = (λ + 2)n−3s−k (λ2 + 2λ − 1)s−1 (λ + 1)s |S |.

 2 λ − λ − 8 − (s − 1)(3λ + 7)  −s(3λ + 7)  |S | = −s(3λ + 7) .. . −s(3λ + 7)   λ2 − λ − 8 − (s − 1)(3λ + 7)   =   −s(3λ + 7)  = (λ2 − (3s − 2)λ − 7s − 1)

k 

         −n 1 −n 2 · · · λ − 2(nk − 1)  k k     ni (λ − nj + 2)   i=1 j=1,j̸=i  k k k      (λ − ni + 2) − ni (λ − nj + 2)  i=1 i=1 j=1,j̸=i

−n 1 λ − 2(n1 − 1) −n 1 .. .

−n 2 −n 2 λ − 2(n2 − 1) .. .

(λ − ni + 2) − (λ2 + 2λ − 1)

i =1

Therefore, the proof is completed.



Lemma 3.3. Let G = G(s, n1 , . . . , nk ). Then m−2 (D(G)) = n − 3s − k.

k  i=1

ni

··· ··· ··· .. .

k  j=1,j̸=i

−nk −nk −nk .. .

(λ − nj + 2).

  . 

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H. Lin / Discrete Mathematics 338 (2015) 868–874

Proof. If s = 0, then G ∼ = Kn1 ,...,nk and the result follows from Lin et al. [8]. If s ≥ 1, then by Lemma 3.1, we have m−2 (D(G)) = m−2 (D(G(s, 1, . . . , 1))) + n − 3s − k. It is sufficient to show that m−2 (D(G(s, 1, . . . , 1))) = 0. By Lemma 3.2, PD(G(s,1,...,1)) (λ) = (λ2 + 2λ − 1)s−1 (λ + 1)s+k−1 [(λ2 − (3s − 2)λ − (7s + 1))(λ + 1) − (λ2 + 2λ − 1)k]. Let f (λ) = [λ2 − (3s − 2)λ − 7s − 1](λ + 1) − (λ2 + 2λ − 1)k. It is clear that f (−2) = 4s − 1 + k > 0, which implies that m−2 (D(G(s, 1, . . . , 1))) = 0. This completes the proof.  Now, we present the main result in this section. Theorem 3.4. The graph G = G(s, n1 , . . . , nk ) is determined by its distance spectra. Proof. If s = 0, then G ∼ = Kn1 ,...,nk and the result follows from Jin and Zhang [6]. If s ≥ 1, then by Lemmas 2.5 and 3.3, the graph cospectral to G must be G′ = G(s, n′1 , . . . , n′k ). By Lemma 3.2, PD(G) (λ) = (λ + 2)n−3s−k (λ2 + 2λ − 1)s−1 (λ +

  (λ − ni + 2) − (λ2 + 2λ − 1) ki=1 ni kj=1,j̸=i (λ − nj + 2)] and PD(G′ ) (λ) =    (λ+ 2)n−3s−k (λ2 + 2λ− 1)s−1 (λ+ 1)s [(λ2 −(3s − 2)λ− 7s − 1) ki=1 (λ− n′i + 2)−(λ2 + 2λ− 1) ki=1 n′i kj=1,j̸=i (λ− n′j + 2)]. k k k k k+2−i Suppose that f (λ) = (λ2 −(3s − 2)λ− 7s − 1) i=1 (λ− ni + 2)−(λ2 + 2λ− 1) i=1 ni j=1,j̸=i (λ− nj + 2) = i=0 αi λ     k k k k ′ k+2−i ′ ′ ′ 2 2 . Let and g (λ) = (λ − (3s − 2)λ − 7s − 1) i=1 (λ − ni + 2) − (λ + 2λ − 1) i=1 ni j=1,j̸=i (λ − nj + 2) = i=0 αi λ k   s1 = n , s = n n , . . . , s = n · · · n , . . . , s = n · · · n . Suppose that f ( s , . . . , s i ji k 1 k i 1 i) = j =1 j 2 1≤j1
k

i=1

4. Lower bounds on the distance spread of graphs Obviously, the star K1,n−1 is the unique tree with the smallest distance spread among all trees. In the following, we first give the unique tree with the second smallest distance spread among all trees. Let Ga,b (a + b = n − 2) denote the double star obtained by adding pendent vertices to one endvertex of P2 and b pendent vertices to the other. The following result can be found in Lin and Zhou [10]. Lemma 4.1. If a ≥ b ≥ 2, then λn (D(Ga,b )) < λn (D(Ga+b−1,1 )) Let Tn,d be the tree obtained from Pd+1 = v0 v1 · · · vd by attaching n − d − 1 pendent vertices on the vertex v⌊d/2⌋ . Theorem 4.2. The tree T1,n−3 attains the second smallest distance spread among all trees. Proof. Let T  K1,n−1 . Note that G1,n−3 attains the second smallest distance spectral radius among all trees (see [5] for details). If diam(T ) = 3, then by Lemma 4.1, we have Sp(D(T )) ≥ Sp(D(G1,n−3 )) with equality if and only if T ∼ = G1,n−3 . Note that PD(G1,n−3 ) (x) = (x + 2)n−4 [x4 − (2n − 8)x3 − (14n − 36)x2 − (16n − 32)x − 4n + 4]. So we may assume that f (x) = x4 − (2n − 8)x3 − (14n − 36)x2 − (16n − 32)x − 4n + 4. Since f (2n − 1) = 84n2 − 40n + 1 > 0, f (2n − 2) = −4(n − 1)(2n2 − 18n + 9) < 0, f (0) = −4n + 4 < 0, f (−1) = 1, f (−2) = 36 − 12n < 0 and f (−6)= 676 + 20n > 0,

we have Sp(D(G1,n−3 )) < 2n − 1 + 6 = 2n + 5. If diam(T ) = 4, then λ1 (T ) ≥ λ1 (Tn,4 ) ≥ mini,j

Di Dj ≥ 2n −

3 , 2n

the first inequality is due to√Du et al. [5] and the second inequality is due to Lin et al. [7]. Therefore, by Theorem 2.3, 3 Sp(D(T )) ≥ 2n − 2n −(−3 − 5) > 2n + 5 > Sp(D(G1,n−3 )) for n ≥ 7. If diam(T ) ≥ 5, then by Theorem 2.3, λn (D(T )) ≤ −7, it follows that Sp(D(T )) > Sp(D(G1,n−3 )). This completes the proof.  Let Sn denote the graph obtained from K1,n−1 by adding an edge. Yu et al. [13] showed that Sn attains the minimal distance √ spectral radius among all unicyclic graphs if n ≥ 6. Later, Lin and Zhou [10] determined −2 − 2 < λn (D(Sn )) < −2. Hence we have the following result. Theorem 4.3. Let G be the graph that attains the minimal distance spread among all unicyclic graphs. Then we have (i) If n ≥ 6, then G ∼ = Sn . (ii) If n = 4, 5, then G ∼ = Cn . Proof. If n = 4, 5, then by a simple calculation, we have Sp(D(G)) ≥ Sp(D(Cn )) with equality if and only if G ∼ = Cn . So in ∼ the following, we may assume that n ≥ 6 . If diam ( G ) = 2, then obviously G S . If diam ( G ) ≥ 3, then by Theorem 2.3, we = n √ √ have λn (D(G)) ≤ −2 − 2 and thus Sp(D(G)) > λ1 (D(Sn )) − (−2 − 2) > Sp(D(Sn )). This completes the proof. 

H. Lin / Discrete Mathematics 338 (2015) 868–874

873

Lemma 4.4 ([14]). Let G be a connected graph with clique number ω(G). Then

λ1 (D) ≥

1 2

[(n + 2k − 3) +

 (n + 2k + 1)2 − 4k(k + 1)(ω + 1)],

where k = ⌊ ωn ⌋ and the equality holds if and only if G is a Turán graph. Then we get the following corollary immediately. Corollary 4.5. Let G be a connected graph with clique number ω(G). Then Sp(D) ≥

1 2

[(n + 2k + 1) +

 (n + 2k + 1)2 − 4k(k + 1)(ω + 1)],

where k = ⌊ ωn ⌋ and the equality holds if and only if G is a Turán graph. Note that α K1 ∨ Kn−α attains the minimal distance spectral radius among all the graphs with independent number α and α K1 ∨ Kn−α is also the complete multipartite graph. Therefore we have the following result. Corollary 4.6. Let G be a connected graph with independent number α . Then Sp(D) ≥

1 2

[n + α + 1 +



(n + α − 3)2 − 8(α − 1)(n − 2α − 1)],

and the equality holds if and only if G ∼ = α K1 ∨ Kn−α . Let M (n, d) be the graph obtained from a complete graph with n − d + 2 vertices by removing an edge, adding a pendant path of ⌈ 2d ⌉ − 1 edges to one endvertex of the removed edge, and adding a pendant path of ⌊ 2d ⌋ − 1 edges to its other endvertex. Problem 4.7. Does the graph M (n, d) attain the minimal distance spread among all graphs with diameter d? Let T (n, k) be the graph obtained by adding k pendent vertices to a vertex of complete graph with order n − k. Problem 4.8. Does the graph T (n, k) attain the minimal distance spread among all graphs with k pendent vertices? 5. On the k + 1th smallest distance spread of graphs Theorem 5.1. Let G be a graph with order n and size m = graph which attains the minimum distance spectral radius.

n(n−1) 2

− k where k ≤ ⌊ 2n ⌋. Then the graph Kn − kK2 is the unique

Proof. Let G = Kn − kK2 = Kn − {u1 v1 , . . . , uk vk }, where {u1 v1 , . . . , uk vk } is an independent edge set of Kn . Let X = (x1 , . . . , xn )t be the Perron vector of G and xv1 , . . . , xvk , xu1 , . . . , xuk be the coordinates of X corresponding to v1 , . . . , vk , u1 , . . . , uk . Then we have the following claim. Claim. xvi = xvj and xvi = xui for i ̸= j. Note that

λ1 (D)xvi =

n 

xi − xvi + xui ,

i=1

and

λ1 (D)xui =

n 

xi − xui + xvi .

i =1

Then we have xvi = xui . And note that λ1 (D)xvi = i=1 xi + xvi and λ1 (D)xvj = i=1 xi + xvj , it follows that xvi = xvj . This completes the proof. It is well-known that each graph with edges k ≤ ⌊ 2n ⌋ can be gotten from kK2 by identifying some vertices. Let G⋆ be an ar-

n

n(n−1)

n

bitrary graph with order n and size m = 2 − k where k ≤ ⌊ 2n ⌋ and V (G) = V (G⋆ ). Then G⋆ can be gotten from G by deleting some edges in {v1 , . . . , vk , u1 , . . . , uk } and adding the same number of edges in {u1 v1 , . . . , uk vk }. Therefore, by the claim, we have λ1 (D(G⋆ )) ≥ X t D(G⋆ )X = X t D(G)X = λ1 (D(G)). If λ1 (D(G⋆ )) = λ1 (D(G)), then λ1 (D(G⋆ ))X = λ1 (D(G))X , that is D(G⋆ )X = D(G)X . If G⋆  G, then there exists a vertex in {v1 , . . . , vk , u1 , . . . , uk }, say v1 in V (G⋆ ) such that dG⋆ (v1 ) = n − 1. Then

λ1 (D(G⋆ ))xv1 =

 v∈V (G⋆ )

xv − xv1 <



xv + xu1 − xv1 = λ1 (D(G))xv1 ,

v∈V (G)

which is a contradiction. This completes the proof.



874

H. Lin / Discrete Mathematics 338 (2015) 868–874

Corollary 5.2. Let G be a connected graph with order n and size m =



n−1+

(n−1)2 +8k 2

n(n−1) 2

− k where k ≤ ⌊ 2n ⌋. Then λ1 (D(G)) ≥

and equality if and only if G ∼ = Kn − kK2 .

Proof. Let G = Kn − kK2 where k ≤ ⌊ 2n ⌋. Since xvi = xvj and xvi = xui , let xvi = xui = x1 . By symmetry, we know that xu = xv = x2 , for u, v ∈ V (G) \ {v1 , . . . , vk , u1 , . . . , uk }. Therefore,

λ1 (D)x1 = 2x1 + (2k − 2)x1 + (n − 2k)x2 , and

λ1 (D)x2 = 2kx1 + (n − 2k − 1)x2 . √ n−1+ (n−1)2 +8k Then we have λ1 (D(G)) = . By Theorem 5.1, the result holds. 2



Lemma 5.3. Let G = (K1 ∪ K2 ) ∨ Kn−3 with n ≥ 4 and D be its distance matrix. Then 1  2

[ (n − 1)2 + 12 + n − 1] < λ1 (D) < n + 1

and



2 < λn (D) ≤ a.

−1 −

Proof. Since PD (x) = (x + 1)n−3 [x3 − (n − 3)x2 − (2n + 3)x + n − 11], λ1 (D) and λn (D) are the largest and smallest roots of the function f (x) = x3 − (n − 3)x2 − (2n + 3)x + n − 11 = 0, respectively. Note that 2 f (n + 1) = 2n + 4n − 10 > 0,  f ( 21 [ (n − 1)2 + 12 + n − 1]) = − (n − 1)2 + 12 − 4 < 0, f (1) = −2n − 10 < 0, f (−1) = 2n − 6 > 0, f (−2) =√n − 1, √  √ f (−1 − 2) = 4 2 − 6. It follows that 21 [ (n − 1)2 + 12 + n − 1] < λ1 (D) < n + 1 and −1 − 2 < λn (D) ≤ a and

a ∈ (−1 −



2, −2).

 (n−1)2 +12

Theorem 5.4. For 0 ≤ k ≤ connected graphs with order n.

6

+ 3/2, the graph Kn − kK2 attains the (k + 1)th smallest distance spread among all √

Proof. Note that Sp(D(Kn − kK2 )) =

n−1+

(n−1)2 +8k

+ 2 by Corollary 5.2 and Theorem 2.3. If G is not a complete multipartite √ (n−1)2 +12+n−1 graph, then λ1 (D(G)) ≥ λ1 (D((K1 ∪ K2 ) ∨ Kn−3 )) > and λn (D(G)) ≤ a by Lemma 5.3. Note that 2 √ √ √ √ (n−1)2 +12+n−1 (n−1)2 +12 (n−1)2+8k+n−1 a ∈ (−1 − 2, −2), then Sp(D(G)) > − a > + 2 for k ≤ + 3/2. It is clear that 2 2 6 λ1 (D(Kn − kK1 )) > λ1 (D(Kn − (k − 1)K1 )). Thus we complete the proof.  2

Acknowledgment The author is very grateful to the anonymous referees for valuable comments, corrections and suggestions which result in a great improvement of the original paper. The author was supported by the National Natural Science Foundation of China (No. 11401211) and the China Postdoctoral Science Foundation (No. 2014M560303). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14]

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