Spectral conditions for some graphical properties

Spectral conditions for some graphical properties∗ Lihua Feng, Pengli Zhang, Henry Liu, Weijun Liu, Minmin Liu, Yuqin Hu School of Mathematics and Statistics Central South University, New Campus Changsha, Hunan, 410083, China

December 12, 2016 Abstract By a unified approach, we present sufficient conditions based on spectral radius for a graph to be k-connected, k-edge-connected, k-Hamiltonian, k-edge-Hamiltonian, β-deficient and k-path-coverable.

Key words: Spectral radius; degree sequence; stability; graph properties.

1

Introduction

Let G be a graph with vertex set V (G), edge set E(G), order n = |V (G)|, and size e(G) = |E(G)|. For disjoint subsets A, B ⊂ V (G), we let e(A, B) denote the number of edges of G with one end-vertex in A and the other in B. Let dG (v) be the degree of a vertex v in G, and let δ(G) be the minimum degree of G. We use Gc to denote the complement of G, and Kn , En the complete graph and the empty graph of order n, respectively. Let Ks,t denote the complete bipartite graph whose partition classes have orders s and t. In particular, K1,t denotes the star graph with t edges. An (α, β)-biregular graph is a bipartite graph with bipartition A ∪ B, where all vertices of A have degree α, and all vertices of B have degree β. For two vertex-disjoint graphs G and H, we use G ∪ H and G ∨ H to denote the disjoint union and the join of G and H, respectively. The adjacency matrix of G is A(G) = (aij )n×n , whose entries satisfy aij = 1 if two vertices i and j are adjacent in G, and aij = 0 otherwise. The characteristic polynomial of G is PG (x) = det(xI − A(G)), and the eigenvalues of G are the roots of PG (x) (with multiplicities). Since A(G) is a symmetric matrix, the eigenvalues of G are real. The largest eigenvalue of G is called the spectral radius of G and is denoted by λ(G). The study of the relationship between graph properties and eigenvalues has attracted much attention. This is largely due to the following problem of Brualdi and Solheid [7]: Given a set G of graphs, find an upper bound for the spectral radii of the graphs of G, and characterize the graphs for which the maximal spectral radius is attained. This problem is ∗

E-mail addresses: [email protected] (L. Feng), [email protected] (P. Zhang), [email protected] (H. Liu), [email protected] (W. Liu, corresponding author), [email protected] (M. Liu), [email protected] (Y. Hu).

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studied for graphs with given number of cut vertices, chromatic number, matching number, and so on; see [3, 14, 15, 17, 21, 31, 40, 43]. One may refer to the recent comprehensive monograph of Stevanović [39] for more details. The study of eigenvalues and the matching number was initiated by Brouwer and Haemers [6], and then subsequently developed by Cioabˇa and many other researchers [11, 12]. An important paper relating connectivity and Laplacian eigenvalues of graphs is by Fiedler [18], and since then, this relationship has been well studied. Cioabˇ a and Gu [13] obtained a sufficient condition for a connected graph to be k-edge-connected in relation to the second largest eigenvalue. Other related results can be found in [10, 20, 32, 38]. Fiedler and Nikiforov [19] were the first to establish the spectral sufficient conditions for graphs to be Hamiltonian or traceable. Li [29] also reported several spectral sufficient conditions for Hamiltonian properties of graphs. Since then, many researchers have studied the analogous problems under various other conditions; see [22, 28, 33, 34, 35, 36, 37, 42, 44]. We now present some graph notations. The concept of stability was introduced by Bondy and Chv´ atal [5]. Let P be a property defined on all graphs of order n. Let k be a non-negative integer. The property P is said to be k-stable if whenever G + uv has property P and dG (u) + dG (v) ≥ k, where uv ∈ / E(G), then G itself has property P . Among all the graphs H of order n such that G ⊆ H and dH (u) + dH (v) < k for all uv ∈ / E(H), there is a unique smallest one with respect to size (for, if H1 and H2 have the above properties, then so does H1 ∩ H2 ). We shall call this graph the k-closure of G, and denote it by clk (G). Obviously, clk (G) can be obtained from G by recursively joining two non-adjacent vertices such that their degree sum is at least k. This concept plays a prominent role in many structural graph theory problems. A connected graph G is said to be k-connected (or k-vertex-connected ) if it has more than k vertices and remains connected whenever fewer than k vertices are deleted. Similarly, G is k-edge-connected if it has at least two vertices and remains connected whenever fewer than k edges are deleted. The deficiency of a graph G, denoted by def(G), is the number of vertices unmatched under a maximum matching in G. In particular, G has a 1-factor if and only if def(G) = 0. We say that G is β-deficient if def(G) ≤ β. A graph G is k-Hamiltonian if for all X ⊂ V (G) with |X| ≤ k, the subgraph induced by V (G)\X is Hamiltonian. Thus a graph G is 0-Hamiltonian is the same as G is Hamiltonian. A graph G is k-edge-Hamiltonian if any collection of vertex-disjoint paths with at most k edges altogether belong to a Hamiltonian cycle in G. A graph G is traceable if it contains a Hamiltonian path. More generally, G is k-pathcoverable if V (G) can be covered by k or fewer vertex-disjoint paths. In particular, a graph G is 1-path-coverable is the same as G is traceable. An integer sequence π = (d1 ≤ d2 ≤ · · · ≤ dn ) is called graphical if there exists a graph G having π as its vertex degree sequence, in which case, G is called a realization of π. If P is a graph property, such as Hamiltonian or k-connected, we call a graphical sequence π forcibly P if every realization of π has property P . A survey on degree sequences of graphs can be found in [1]. 2

Historically, the vertex degrees of a graph have been used to provide sufficient conditions for the graph to have certain properties. Li et al. [30, 41] investigated the spectral sufficient conditions for some stable properties of graphs. In this paper, we will consider some graph properties from a different perspective, namely, in terms of the spectral radius. By borrowing ideas from [19, 30, 41], we shall utilize the degree sequence and the closure concepts to present several sufficient conditions for graphs to possess certain properties, including k-connected, k-edge-connected, k-Hamiltonian, k-edge-Hamiltonian, β-deficient, and k-path-coverable. Our results may be considered as new viewpoints for existing results.

2

Preliminaries

In this section, we present some lemmas that will be used later in order to prove our main results. Firstly, in the following lemma, we collect several results, with each item containing a sufficient graphical degree sequence condition implying the existence of a certain graph property. Lemma 2.1. Let π = (d1 ≤ d2 ≤ · · · ≤ dn ) be a graphical sequence. (1) [4] Suppose n ≥ k + 1. If di ≤ i + k − 2 ⇒ dn−k+1 ≥ n − i, for 1 ≤ i ≤ is forcibly k-connected.

n−k+1 , 2

then π

(2) [2] Suppose n ≥ k + 1, and d1 ≥ k ≥ 1. If di−k+1 ≤ i − 1 and di ≤ i + k − 2 ⇒ dn ≥ n − i + k − 1, for k + 1 ≤ i ≤ b n2 c, then π is forcibly k-edge-connected. In particular, if k ≥ b n2 c, then π is forcibly k-edge-connected. (3) [9] Suppose n ≥ 3, and 0 ≤ k ≤ n − 3. If di ≤ i + k ⇒ dn−i−k ≥ n − i, for 1 ≤ i < then π is forcibly k-Hamiltonian.

n−k 2 ,

(4) [25] Suppose n ≥ 3, and 0 ≤ k ≤ n − 3. If di−k ≤ i ⇒ dn−i ≥ n − i + k, for k + 1 ≤ i < n+k 2 , then π is forcibly k-edge-Hamiltonian. (5) [1, 26] Suppose 0 ≤ β ≤ n, and n ≡ β (mod 2). If di+1 ≤ i − β ⇒ dn+β−i ≥ n − i − 1, for 1 ≤ i ≤ n+β−2 , then π is forcibly β-deficient. 2 (6) [5, 27] Suppose k ≥ 1. If di+k ≤ i ⇒ dn−i ≥ n − i − k, for 1 ≤ i < forcibly k-path-coverable.

n−k 2 ,

then π is

Next, we consider stability results for the same graph properties. It is well known that the Hamiltonicity and traceability properties for graphs of order n are respectively n-stable and (n − 1)-stable. These two facts are included in the following lemma. Lemma 2.2. [5] The following stability results hold for graphs of order n. (1) The property that “G is k-connected” is (n + k − 2)-stable. (2) The property that “G is k-edge-connected ” is (n + k − 2)-stable. (3) The property that “G is k-Hamiltonian” is (n + k)-stable.

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(4) The property that “G is k-edge-Hamiltonian” is (n + k)-stable. (5) The property that “G is β-deficient” is (n − β − 1)-stable. (6) The property that “G is k-path-coverable” is (n − k)-stable. From the definition of the k-closure of a graph G, we have the following. Lemma 2.3. [5] Let P be a property of a graph G. If P is k-stable and clk (G) has property P , then G itself has property P . Next, the following two results contain bounds on the spectral radius of a graph. Lemma 2.4. [24] Let G be a graph of order n with m edges, and with no isolated vertices. Then the spectral radius of G satisfies √ λ(G) ≤ 2m − n + 1. (1) Equality holds if and only if G consists of c components for some c ≥ 1, where c − 1 components are single edges, and the remaining component is either a complete graph or a star graph. Lemma 2.5. [23] Let G be a graph of order n, with degree sequence π = (d1 ≤ d2 ≤ · · · ≤ dn ). Then the spectral radius of G satisfies r 1X 2 λ(G) ≥ di . n Equality holds if and only if P each component of G is an r-regular graph, or an (r1 , r2 )biregular graph, where r2 = n1 d2i , and r1 , r2 satisfy r1 r2 = r2 . We have the following auxiliary lemma. Lemma 2.6. Let H be a graph of order n and G be any spanning subgraph of H. If for any pair of non-adjacent vertices u, v ∈ V (H) we have dH (u) + dH (v) ≤ ` , then n nλ2 (Gc ) e(H) ≥ − . 2 2n − 2 − ` Proof. From the assumption, we have that for any uv ∈ E(H c ), dH c (u) + dH c (v) = n − 1 − dH (u) + n − 1 − dH (v) ≥ 2n − 2 − `. Thus we obtain X

X

d2H c (x) =

x∈V (H c )

(dH c (u) + dH c (v)) ≥ (2n − 2 − `)e(H c ).

uv∈E(H c )

From Lemma 2.5, we have nλ2 (H c ) ≥

X

d2H c (x) ≥ (2n − 2 − `)e(H c ).

x∈V (H c )

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Since H c ⊆ Gc , we have e(H c ) ≤ Therefore it follows that

nλ2 (H c ) nλ2 (Gc ) ≤ . 2n − 2 − ` 2n − 2 − `

n nλ2 (Gc ) e(H) ≥ − . 2n − 2 − ` 2

This completes the proof. Finally, the following result concerns the spectral radius of the graph Ks ∨ Et . Lemma 2.7. For s, t ≥ 1, we have p (s − 1)2 + 4st λ(Ks ∨ Et ) = . 2 √ In particular, for the star graph K1,t , we have λ(K1,t ) = t. s−1+

(2)

Proof. Observe that Ks ∨ Et is the complete multipartite graph with s classes of order 1, and one class of order t. The spectral radius of a complete multipartite graph is well studied. For example, from Theorem 3.24 in [39], if G is a complete multipartite graph P ni whose classes have orders n1 , . . . , np , then λ(G) is the largest solution of pi=1 λ+n = 1. i Hence in particular, λ(Ks ∨ Et ) is the largest solution of s t + = 1, λ+1 λ+t from which (2) easily follows.

3

Main Results

We are now ready to present our main results. In each of the following results, we provide a condition on the spectral radius of a graph, or its complement, which would imply the existence of a certain property for the graph. We begin with the case when the property is k-connectedness. Theorem 3.1. Let G be a graph of order n ≥ k + 1 ≥ 2. √ (I) If G has no isolated vertices, and λ(G) ≥ n2 − 4n + 2k + 1, then G is k-connected. q (II) If λ(Gc ) ≤ (n−k)(n−k+1) , then G is k-connected, unless k = 1 and G = K1 ∪ Kn−1 . n Proof. We begin by proving the following claim. Claim. Let G be a graph of order n ≥ k + 1 ≥ 2. If e(G) ≥ n2 − n + k, then G is k-connected, unless e(G) = n2 − n + k and G = Kk−1 ∨ (K1 ∪ Kn−k ).

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Proof of Claim. Suppose that G is not k-connected. Then from Lemma 2.1 (1), there exists an integer 1 ≤ i ≤ n−k+1 such that di ≤ i + k − 2 and dn−k+1 ≤ n − i − 1. Thus we have 2 2e(G) ≤ i(i + k − 2) + (n − k − i + 1)(n − i − 1) + (k − 1)(n − 1) = n(n − 1) + 2i2 − 2(n − k + 1)i. Suppose f (x) = 2x2 − 2(n − k + 1)x with 1 ≤ x ≤ f ( n−k+1 ) 2

(n−k−1)2

that f (1) − = 2 fmax (x) = f (1) = −2(n − k), and

n−k+1 . 2

Then it is easy to see

≥ 0, with equality if and only if n = k + 1. Thus,

2e(G) ≤ n(n − 1) − 2(n − k). n

Hence e(G) = 2 − n + k, and all the inequalities in the proof above must be equalities. We have i = 1, and d1 = k − 1, d2 = d3 = · · · = dn−k+1 = n − 2, and dn−k+2 = dn−k+3 = · · · = dn = n − 1. It follows that G = Kk−1 ∨ (K1 ∪ Kn−k ), which is not k-connected. (I). Since G has no isolated vertices, from Lemma 2.4, we have p p n2 − 4n + 2k + 1 ≤ λ(G) ≤ 2e(G) − n + 1, which yields n e(G) ≥ − n + k. 2 Since Kk−1 ∨ (K1 ∪ Kn−k ) cannot be one of the graphs attaining equality in (1), by the claim above and Lemma 2.4, we have e(G) > n2 − n + k, and G is k-connected. (II). Suppose that G is not k-connected. From Lemma 2.2 (1), we can consider the closure H := cln+k−2 (G). By Lemma 2.3, H is not k-connected and obviously H 6= Kn . Thus for any pair of non-adjacent vertices u and v in H, we have dH (u) + dH (v) ≤ n + k − 3. Therefore from the assumption and Lemma 2.6, it follows that n nλ2 (Gc ) n e(H) ≥ − ≥ − n + k. 2 n−k+1 2 Since H is not k-connected, from the claim, H = Kk−1 ∨ (K1 ∪ Kn−k ). Since G ⊆ H, Gc contains the star K1,n−k . Using Lemma 2.7, we have r √ (n − k)(n − k + 1) c λ(G ) ≥ λ(K1,n−k ) = n − k ≥ . n By the assumption, we have equality above. Thus k = 1 and G = H = K1 ∪ Kn−1 , which is not connected. This completes the proof. Next, we consider the k-edge-connected property. Since every k-connected graph is also k-edge-connected, by Theorem 3.1, we have also obtained sufficient conditions for a graph to be k-edge-connected. However, by analogy with Theorem 3.1 (I), we are able to provide the sufficient condition with a smaller lower bound for λ(G), for large n. Theorem 3.2. Let G be a graph of order n ≥ k + 1. 6

p (I) If k ≥ 2, G has no isolated vertices, and λ(G) ≥ n2 − (k + 5)n + 2k 2 + 2k + 5, then G is k-edge-connected. q c (II) If k ≥ 1 and λ(G ) ≤ (n−k)(n−k+1) , then G is k-edge-connected, unless k = 1 and n G = K1 ∪ Kn−1 . For the case k = 1 in (I), since the 1-edge-connected and 1-connected properties are equivalent, √ we may use Theorem 3.1 (I) to obtain that, if G has no isolated vertices and λ(G) ≥ n2 − 4n + 3, then G is 1-edge-connected. For k ≥ 2, we remark that the lower bound for λ(G) in (I) is indeed smaller than the one in Theorem 3.1 (I), for n ≥ 2k. Finally, note that (II) is an instant corollary of Theorem 3.1 (II). Thus, we only have to prove (I). Proof. We first prove the following claim. Claim. Let G be a graph of order n ≥ k+1. If k ≥ 2 and e(G) ≥ 12 (n2 −(k+4)n+2k 2 +2k+4) then G is k-edge-connected. Proof of Claim. Suppose that G is not k-edge-connected. Then from Lemma 2.1 (2), there exists an integer k + 1 ≤ i ≤ n2 such that di−k+1 ≤ i − 1, di ≤ i + k − 2 and dn ≤ n − i + k − 2. In particular, note that n ≥ 2k + 2. We have 2e(G) ≤ (i − k + 1)(i − 1) + (k − 1)(i + k − 2) + (n − i)(n − i + k − 2) = n(n − 1) + (k − 1)(n + k − 1) + 2i2 − (2n + k − 1)i. Suppose f (x) = 2x2 − (2n + k − 1)x with k + 1 ≤ x ≤ n2 . Then it is easy to check that f (k + 1) − f ( n2 ) = (n−k−3)(n−2k−2) ≥ 0, with equality if and only if n = 2k + 2. It follows 2 that fmax (x) = f (k + 1) = −(k + 1)(2n − k − 3), and we have 2e(G) ≤ n(n − 1) + (k − 1)(n + k − 1) − (k + 1)(2n − k − 3) = n2 − (k + 4)n + 2k 2 + 2k + 4. Thus e(G) = 12 (n2 − (k + 4)n + 2k 2 + 2k + 4), and we have i = k + 1, and d1 = d2 = k, d3 = · · · = dk+1 = 2k − 1, dk+2 = · · · = dn = n − 3. We show that if G has this degree sequence, then it must be k-edge-connected, which will be a contradiction. It suffices to show that for any partition V (G) = A ∪ B, where |A| + |B| = n and 1 ≤ |A| ≤ |B|, we have e(A, B) ≥ k. Since G has minimum degree δ(G) = k, if |A| = 1, then e(A, B) ≥ k, and if |A| = 2, then e(A, B) ≥ 2(k − 1) ≥ k. Now, let |A| ≥ 3. Note that n ≥ 2k + 2 and |A| ≤ |B| imply |B| ≥ k + 1. Firstly, if |B| = k + 1, then we also have |A| = k + 1. Thus by the symmetry of A and B, we may assume that A has k vertices with degree at least 2k − 1 in G. Then, each such vertex in A has at least (2k − 1) − k ≥ 1 neighbour in B, and hence e(A, B) ≥ k. Secondly, suppose that |B| ≥ k + 2. If A contains a vertex v with dG (v) = n − 3, then v has at least |B| − 2 ≥ k neighbours in B, and e(A, B) ≥ k. Otherwise, all vertices with degree n − 3 in G lie in B. Since |A| ≥ 3, each such vertex in B has at least |A| − 2 ≥ 1 neighbour in A. Since there are n − (k + 1) ≥ (2k + 2) − (k + 1) > k such vertices, it follows again that e(A, B) ≥ k. This completes the proof of the claim. 7

We only have to prove (I). Since G has no isolated vertices, by Lemma 2.4, we obtain p p n2 − (k + 5)n + 2k 2 + 2k + 5 ≤ λ(G) ≤ 2e(G) − n + 1, implying that 1 e(G) ≥ (n2 − (k + 4)n + 2k 2 + 2k + 4). 2 Thus, (I) follows from the claim. Now, we consider the k-Hamiltonian and k-edge-Hamiltonian properties. When k = 0, the 0-Hamiltonian and 0-edge-Hamiltonian properties are both equivalent to the Hamiltonian property. This case was studied by Fiedler and Nikiforov [19]. Thus, in the next two results, we assume that k ≥ 1. Observe that if G has order n and is k-Hamiltonian, then we clearly require δ(G) ≥ k + 2. For if there exists a vertex v ∈ V (G) with dG (v) ≤ k + 1, then we can delete at most k neighbours of v to obtain a subgraph of G which is not Hamiltonian. Theorem 3.3. Let k ≥ 1, and let G be a graph of order n ≥ k + 6. p (I) If G has no isolated vertices, and λ(G) ≥ (n − 2)2 + 2k + 1, then G is kHamiltonian. q (II) If λ(Gc ) ≤ (n−k−1)(n−k−2) , then G is k-Hamiltonian. n Proof. We first prove the following claim. Claim. Let k ≥ 1, and let G be a graph of order n ≥ k + 6. If e(G) ≥ n2 − n + k + 2, then G is k-Hamiltonian, unless e(G) = n2 − n + k + 2 and G = Kk+1 ∨ (K1 ∪ Kn−k−2 ). Proof of Claim. Suppose that G is not k-Hamiltonian. From Lemma 2.1 (3), there exists such that di ≤ i + k and dn−i−k ≤ n − i − 1. We have an integer 1 ≤ i ≤ n−k−1 2 2e(G) ≤ i(i + k) + (n − 2i − k)(n − i − 1) + (i + k)(n − 1) = n(n − 1) + 3i2 − (2n − 2k − 1)i. Suppose f (x) = 3x2 − (2n − 2k − 1)x with 1 ≤ x ≤ n−k−1 , and 1 ≤ k ≤ n − 3. 2 (n−k−3)(n−k−5) n−k−1 We can easily find that f (1) − f ( 2 ) = > 0. Thus, fmax (x) = f (1) = 4 −2(n − k − 2), and 2e(G) ≤ n(n − 1) − 2(n − k − 2). Hence e(G) = n2 − n + k + 2, and from the proof above, we have i = 1, and d1 = k + 1, d2 = · · · = dn−k−1 = n − 2, dn−k = · · · = dn = n − 1. Thus G = Kk+1 ∨ (K1 ∪ Kn−k−2 ), which is not k-Hamiltonian, since the minimum degree of this graph is less than k + 2. (I). Since G has no isolated vertices, by Lemma 2.4, we have p p (n − 2)2 + 2k + 1 ≤ λ(G) ≤ 2e(G) − n + 1, implying that n e(G) ≥ − n + k + 2. 2 8

Since Kk+1 ∨ (K1 ∪ Kn−k−2 ) cannot be one of the graphs attaining equality in (1), by the claim above and Lemma 2.4, we have e(G) > n2 − n + k + 2, and G is k-Hamiltonian. (II). Suppose that G is not k-Hamiltonian. From Lemma 2.2 (3), we can consider the closure H := cln+k (G). By Lemma 2.3, H is also not k-Hamiltonian, and H 6= Kn . Thus for any pair of non-adjacent vertices q u and v in H, we have dH (u) + dH (v) ≤ n + k − 1. From the initial assumption λ(Gc ) ≤

(n−k−1)(n−k−2) n

and Lemma 2.6, it follows that

n n nλ2 (Gc ) ≥ − n + k + 2. e(H) ≥ − n−k−1 2 2 Since H is not k-Hamiltonian, from the claim above, H = Kk+1 ∨ (K1 ∪ Kn−k−2 ). Since G ⊆ H, Gc contains the star K1,n−k−2 . Using Lemma 2.7, we have r √ (n − k − 1)(n − k − 2) c λ(G ) ≥ λ(K1,n−k−2 ) = n − k − 2 > . n This contradicts the assumption in (II), and the proof is complete. Next, we consider the k-edge-Hamiltonian property, with k ≥ 1. Theorem 3.4. Let k ≥ 1, and let G be a graph of order n ≥ k + 6. p (I) If G has no isolated vertices, and λ(G) ≥ (n − 2)2 + 2k + 1, then G is k-edgeHamiltonian. q (II) If λ(Gc ) ≤ (n−k−1)(n−k−2) , then G is k-edge-Hamiltonian. n Proof. We first prove the following claim. Claim. Let k ≥ 1, and let G be a graph of order n ≥ k + 6. If e(G) ≥ n2 − n + k + 2, then G is k-edge-Hamiltonian, unless e(G) = n2 − n + k + 2 and G = Kk+1 ∨ (K1 ∪ Kn−k−2 ). Proof of Claim. Suppose that G is not k-edge-Hamiltonian. From Lemma 2.1 (4), there exists an integer k + 1 ≤ i ≤ n+k−1 such that di−k ≤ i and dn−i ≤ n − i + k − 1. Hence 2 2e(G) ≤ (i − k)i + (n + k − 2i)(n − i + k − 1) + i(n − 1) = (n + k)(n + k − 1) + 3i2 − (2n + 4k − 1)i. Suppose f (x) = 3x2 − (2n + 4k − 1)x with k + 1 ≤ x ≤ n+k−1 . Since n ≥ k + 6, it 2 (n−k−3)(n−k−5) n+k−1 is easy to obtain f (k + 1) − f ( 2 ) = > 0. So fmax (x) = f (k + 1) = 4 −(k + 1)(2n + k − 4), and we have 2e(G) ≤ (n + k)(n + k − 1) − (k + 1)(2n + k − 4) = n(n − 1) − 2(n − k − 2). Hence e(G) = n2 −n+k+2, and from the proof above, we have i = k+1, and d1 = k+1, d2 = d3 = · · · = dn−k−1 = n − 2, dn−k = · · · = dn = n − 1. Thus G = Kk+1 ∨ (K1 ∪ Kn−k−2 ), which is not k-edge-Hamiltonian, since in this graph, no Hamilton cycle contains a path of length k within the Kk+1 . 9

Since the inequalities for λ(G) and λ(Gc ) in Theorem 3.4 are the same as those in Theorem 3.3, we may prove Theorem 3.4 with almost identical arguments as in Theorem 3.3, but using the above claim instead. Our next task will be to consider the β-deficient property. Theorem 3.5. Let G be a graph of order n ≥ 11. Let 0 ≤ β ≤ n with n ≡ β (mod 2). p (I) If G has no isolated vertices, and λ(G) ≥ n2 − 6n − 4β + 11, then G is β-deficient. q (II) If λ(Gc ) ≤ (n+β)(2n+2β−5) , then G is β-deficient. n Proof. We first prove the following claim. Claim. Let G be a graph of order n ≥ 11. Let 0 ≤ β ≤ n with n ≡ β (mod 2). If e(G) ≥ 12 (n2 − 5n − 4β + 10), then G is β-deficient, unless e(G) = 12 (n2 − 5n − 4β + 10), and β = 0 and G = K1 ∨ (E2 ∪ Kn−3 ), or β = 1 and G = E2 ∪ Kn−2 . Proof of Claim. Suppose that G is not β-deficient. Then from Lemma 2.1 (5), there exists such that di+1 ≤ i − β and dn+β−i ≤ n − i − 2. We have an integer 1 ≤ i ≤ n+β−2 2 2e(G) ≤ (i + 1)(i − β) + (n + β − 2i − 1)(n − i − 2) + (i − β)(n − 1) = (n − 1)(n − 2) − 2β + 3i2 − (2n + 2β − 5)i. Suppose f (x) = 3x2 − (2n + 2β − 5)x with 1 ≤ x ≤ n+β−2 . Since n ≥ 11, we easily have 2 (n+β−4)(n+β−10) n+β−2 f (1) − f ( 2 ) = > 0. Thus fmax (x) = f (1) = −2(n + β − 4), and 4 2e(G) ≤ (n − 1)(n − 2) − 2β − 2(n + β − 4) = n2 − 5n − 4β + 10. Hence e(G) = 12 (n2 − 5n − 4β + 10), and from the proof above, we have i = 1. Since i − β ≥ di+1 ≥ 0, we have β = 0 or β = 1. If β = 0, then d1 = d2 = 1, d3 = · · · = dn−1 = n − 3, dn = n − 1. Thus G = K1 ∨ (E2 ∪ Kn−3 ), which is not 0-deficient. If β = 1, then d1 = d2 = 0, d3 = · · · = dn = n − 3. Thus G = E2 ∪ Kn−2 , which is not 1-deficient. (I). Since G has no isolated vertices, from Lemma 2.4, we have p p n2 − 6n − 4β + 11 ≤ λ(G) ≤ 2e(G) − n + 1, which yields 1 e(G) ≥ (n2 − 5n − 4β + 10). 2 Since K1 ∨ (E2 ∪ Kn−3 ) and E2 ∪ Kn−2 cannot be the graphs attaining equality in (1), by the claim above and Lemma 2.4, we have e(G) > 12 (n2 − 5n − 4β + 10), and G is β-deficient. (II). Suppose that G is not β-deficient. From Lemma 2.2 (5), we can consider the closure H := cln−β−1 (G). By Lemma 2.3, H is not β-deficient, and H 6= Kn . Thus for any pair of non-adjacent verticesq u and v in H, we have dH (u) + dH (v) ≤ n − β − 2. From the initial

assumption λ(Gc ) ≤

(n+β)(2n+2β−5) n

and Lemma 2.6, it follows that

n nλ2 (Gc ) 1 e(H) ≥ − ≥ (n2 − 5n − 4β + 10). 2 n+β 2 10

Since H is not β-deficient, from the claim, either β = 0 and H = K1 ∨ (E2 ∪ Kn−3 ), or β = 1 and H = E2 ∪ Kn−2 . Since G ⊆ H, if the former holds, then Gc contains K2 ∨ En−3 as a subgraph, and by Lemma 2.7, p 1 + 1 + 8(n − 3) √ c > 2n − 5, λ(G ) ≥ λ(K2 ∨ En−3 ) = 2 which contradicts the assumption of the theorem. If the latter holds, then Gc contains K2 ∨ En−2 as a subgraph, and thus r p 1 + 1 + 8(n − 2) (n + 1)(2n − 3) c > . λ(G ) ≥ λ(K2 ∨ En−2 ) = 2 n It is easy to check that the last inequality holds for n ≥ 11. Again we have a contradiction, and this completes the proof. Finally, we consider the k-path-coverable property. When k = 1, the 1-path-coverable property is equivalent to the traceable property, and this case was studied in [19]. Thus in the following theorem, we assume that k ≥ 2. Theorem 3.6. Let k ≥ 2, and let G be a graph of order n ≥ 5k + 6. p (I) If G has no isolated vertices, and λ(G) ≥ (n − k − 2)2 + k + 1, then G is k-pathcoverable. q c , then G is k-path-coverable. (II) If λ(G ) ≤ (k+1)(n+k−1)(2n−k−4) 2n Proof. We first prove the following claim. Claim. Let k ≥ 2, and let G be a graph of order n ≥ 5k + 6. If 1 e(G) ≥ (n2 − (2k + 3)n + k 2 + 5k + 4), 2

(3)

then G is k-path-coverable, unless equality holds in (3) and G = K1 ∨ (Ek+1 ∪ Kn−k−2 ). Proof of Claim. Suppose that G is not k-path-coverable. From Lemma 2.1 (6), there exists such that di+k ≤ i and dn−i ≤ n − i − k − 1. Therefore we have an integer 1 ≤ i ≤ n−k−1 2 2e(G) ≤ (i + k)i + (n − k − 2i)(n − i − k − 1) + i(n − 1) = (n − k)(n − k − 1) + 3i2 − (2n − 4k − 1)i. Suppose f (x) = 3x2 − (2n − 4k − 1)x with 1 ≤ x ≤ n−k−1 . It is easy to see that 2 (n−k−3)(n−5k−5) n−k−1 f (1) − f ( 2 ) = > 0, since n ≥ 5k + 6. Hence fmax (x) = f (1) = 4 −2(n − 2k − 2), and 2e(G) ≤ (n − k)(n − k − 1) − 2(n − 2k − 2) = n2 − (2k + 3)n + k 2 + 5k + 4. Thus e(G) = 21 (n2 − (2k + 3)n + k 2 + 5k + 4), and from the proof above, we have i = 1, and d1 = · · · = dk+1 = 1, dk+2 = dk+3 = · · · = dn−1 = n − k − 2, dn = n − 1. Thus G = K1 ∨ (Ek+1 ∪ Kn−k−2 ), which is not k-path-coverable. 11

(I). Since G has no isolated vertices, by Lemma 2.4, we have p p (n − k − 2)2 + k + 1 ≤ λ(G) ≤ 2e(G) − n + 1, which implies 1 e(G) ≥ (n2 − (2k + 3)n + k 2 + 5k + 4). 2 Since K1 ∨ (Ek+1 ∪ Kn−k−2 ) cannot be one of the graphs attaining equality in (1), by the claim above and Lemma 2.4, we have e(G) > 21 (n2 − (2k + 3)n + k 2 + 5k + 4), and G is k-path-coverable. (II). Suppose that G is not k-path-coverable. From Lemma 2.2 (6), we can consider the closure H := cln−k (G). By Lemma 2.3, H is not k-path-coverable, and H 6= Kn . Thus for any pair of non-adjacent vertices u and v in H, we have q dH (u) + dH (v) ≤ n − k − 1. By

Lemma 2.6, together with the initial assumption λ(Gc ) ≤ (k+1)(n+k−1)(2n−k−4) , it follows 2n that n 1 n nλ2 (Gc ) ≥ − (k + 1)(2n − k − 4) e(H) ≥ − n+k−1 2 2 2 1 = (n2 − (2k + 3)n + k 2 + 5k + 4). 2

Since H is not k-path-coverable, from the claim above, H = K1 ∨ (Ek+1 ∪ Kn−k−2 ). Since G ⊆ H, Gc contains Kk+1 ∨ En−k−2 as a subgraph, and thus by Lemma 2.7, p k + k 2 + 4(k + 1)(n − k − 2) c λ(G ) ≥ λ(Kk+1 ∨ En−k−2 ) = . 2 To complete the proof, we show that r p k + k 2 + 4(k + 1)(n − k − 2) (k + 1)(n + k − 1)(2n − k − 4) > , (4) 2 2n q which implies λ(Gc ) > (k+1)(n+k−1)(2n−k−4) , a contradiction to the assumption in (II). It 2n is easy to see that (4) is equivalent to p k k 2 + 4(k + 1)(n − k − 2) − k 2 − 6k − 4 n > (k + 1) (k − 6)n − (k − 1)(k + 4) . (5) Since n ≥ 5k + 6 and k ≥ 2, we have k 2 + 4(k + 1)(n − k − 2) ≥ k 2 + 4(k + 1)(4k + 4) > 16k 2 , and since 3k 2 − 6k − 4 > (k + 1)(k − 6), we have p k k 2 + 4(k + 1)(n − k − 2) − k 2 − 6k − 4 n > (3k 2 − 6k − 4)n > (k + 1)(k − 6)n. This implies (5), and thus (4), as required. The proof is complete.

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4

Further Remarks

The matrix L(G) = D(G) − A(G) is called the Laplacian matrix of G, where D(G) = diag(d1 , . . . , dn ) is the diagonal matrix of the vertex degrees of G. We call the matrix Q(G) = D(G) + A(G) the signless Laplacian matrix of G, and its largest eigenvalue is denoted by q(G). The results in this paper can be formulated in terms of the signless Laplacian spectral radius, by using the following lemmas and the ideas from [44]. A similar problem for the Hamiltonicity of graphs was studied in [42]. Lemma 4.1. [16] Let G be a connected graph of order n with m edges. Then q(G) ≤

2m + n − 2. n−1

Equality holds if and only if G = K1,n−1 or G = Kn . Lemma 4.2. [44] Let G be a graph of order n with m edges, and degree sequence (di ). Then q(G) ≥

1 X 2 di . m

Also, when the minimum degree is involved, the results in Section 3 would become more interesting. We will present these in the near future.

Acknowledgements L. Feng was supported by NSFC (No. 11301302), and Mathematics and Interdisciplinary Sciences Project of CSU. H. Liu was supported by China Postdoctoral Science Foundation (Nos. 2015M580695, 2016T90756), and International Interchange Plan of CSU. W. Liu was supported by NSFC (Nos. 11671402, 11371207), and Hunan Provincial Natural Science Foundation (2016JJ2138).

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