The Smallest Spectral Radius of Graphs with a Given Clique Number

Hindawi Publishing Corporation e Scientific World Journal Volume 2014, Article ID 232153, 9 pages http://dx.doi.org/10.1155/2014/232153

Research Article The Smallest Spectral Radius of Graphs with a Given Clique Number Jing-Ming Zhang,1,2 Ting-Zhu Huang,1 and Ji-Ming Guo3 1

School of Mathematical Sciences, University of Electronic Science and Technology of China, Chengdu, Sichuan 611731, China College of Science, China University of Petroleum, Qingdao, Shandong 266580, China 3 College of Science, East China University of Science and Technology, Shanghai 200237, China 2

Correspondence should be addressed to Ting-Zhu Huang; [email protected] Received 4 May 2014; Revised 27 June 2014; Accepted 27 June 2014; Published 13 July 2014 Academic Editor: Luis Acedo Copyright © 2014 Jing-Ming Zhang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. The first four smallest values of the spectral radius among all connected graphs with maximum clique size 𝜔 ≥ 2 are obtained.

1. Introduction Let 𝐺 = (𝑉(𝐺), 𝐸(𝐺)) be a simple connected graph with vertex set 𝑉(𝐺) = {V1 , V2 , . . . , V𝑛 } and edge set 𝐸(𝐺). Its adjacency matrix 𝐴(𝐺) = (𝑎𝑖𝑗 ) is defined as 𝑛 × 𝑛 matrix (𝑎𝑖𝑗 ), where 𝑎𝑖𝑗 = 1 if V𝑖 is adjacent to V𝑗 and 𝑎𝑖𝑗 = 0, otherwise. Denote by 𝑑(V𝑖 ) or 𝑑𝐺(V𝑖 ) the degree of the vertex V𝑖 . It is well known that 𝐴(𝐺) is a real symmetric matrix. Hence, the eigenvalues of 𝐴(𝐺) can be ordered as 𝜆 1 (𝐺) ≥ 𝜆 2 (𝐺) ≥ ⋅ ⋅ ⋅ ≥ 𝜆 𝑛 (𝐺) ,

(1)

respectively. The largest eigenvalue of 𝐴(𝐺) is called the spectral radius of 𝐺, denoted by 𝜌(𝐺). It is easy to see that if 𝐺 is connected, then 𝐴(𝐺) is nonnegative irreducible matrix. By the Perron-Frobenius theory, 𝜌(𝐺) has multiplicity one and exists a unique positive unit eigenvector corresponding to 𝜌(𝐺). We refer to such an eigenvector corresponding to 𝜌(𝐺) as the Perron vector of 𝐺. Denote by 𝑃𝑛 and 𝐶𝑛 the path and the cycle on 𝑛 vertices, respectively. The characteristic polynomial of 𝐴(𝐺) is det(𝑥𝐼− 𝐴(𝐺)), which is denoted by Φ(𝐺) or Φ(𝐺, 𝑥). Let 𝑋 be an eigenvector of 𝐺 corresponding to 𝜌(𝐺). It will be convenient to associate with 𝑋 a labelling of 𝐺 in which vertex V𝑖 is labelled 𝑥𝑖 (or 𝑥V𝑖 ). Such labellings are sometimes called “valuation” [1]. The investigation on the spectral radius of graphs is an important topic in the theory of graph spectra. The recent

developments on this topic also involve the problem concerning graphs with maximal or minimal spectral radius, signless Laplacian spectral radius, and Laplacian spectral radius, of a given class of graphs, respectively. The spectral radius of a graph plays an important role in modeling virus propagation in networks [2]. It has been shown that the smaller the spectral radius, the larger the robustness of a network against the spread of viruses [3]. In [4], the first three smallest values of the Laplacian spectral radii among all connected graphs with maximum clique size 𝜔 are given. And, in [5], it is shown that among all connected graphs with maximum clique size 𝜔 the minimum value of the spectral radius is attained for a kite graph 𝑃𝐾𝑛−𝜔,𝜔 , where 𝑃𝐾𝑛−𝜔,𝜔 is a graph on 𝑛 vertices obtained from the path 𝑃𝑛−𝜔 and the complete graph 𝐾𝜔 by adding an edge between an end vertex of 𝑃𝑛−𝜔 and a vertex of 𝐾𝜔 (shown in Figure 1). Furthermore, in this paper, the first four smallest values of the spectral radius are obtained among all connected graphs with maximum clique size 𝜔. Let I𝑛,𝜔 be the set of all connected graphs of order 𝑛 with a maximum clique size 𝜔, where 2 ≤ 𝜔 ≤ 𝑛. It is easy to see that I𝜔,𝜔 = {𝐾𝜔 }. By direct calculation, we have 𝜌(𝐾𝜔 ) = 𝜔 − 1. If 𝐺 ∈ I𝜔+1,𝜔 , then, from the Perron-Frobenius theorem, the first 𝜔 − 1 smallest values of the spectral radius of I𝜔+1,𝜔 are 𝑃𝐾1,𝜔;𝑖 (0 ≤ 𝑖 ≤ 𝜔 − 2), respectively, where 𝑃𝐾1,𝜔;𝑖 is the graph obtained from 𝑃𝐾1,𝜔 by adding 𝑖 (0 ≤ 𝑖 ≤ 𝜔 − 2) edges. So in the following, we consider that 𝑛 ≥ 𝜔 + 2.

2

The Scientific World Journal 𝜔 .. . K𝜔

1

𝜔+1 𝜔+2 ···

G

n

k

2

··· k−1 2

G



ul

···

u3

u2

··· k k−1 2

1

1

u1

Gk+1,l−1

Figure 2: Grafting an edge.

2. Preliminaries In order to complete the proof of our main result, we need the following lemmas. Lemma 1 (see [6]). Let V be a vertex of the graph 𝐺. Then the inequalities

≥ ⋅ ⋅ ⋅ ≥ 𝜆 𝑛−1 (𝐺 − V) ≥ 𝜆 𝑛 (𝐺)

u1

Gk,l

Figure 1: Kite graph 𝑃𝐾𝑛−𝜔,𝜔 .

𝜆 1 (𝐺) ≥ 𝜆 1 (𝐺 − V) ≥ 𝜆 2 (𝐺) ≥ 𝜆 2 (𝐺 − V)



··· ul u2

Lemma 6 (see [12]). Let V be a vertex of 𝐺, let 𝜑(V) be the collection of circuits containing V, and let 𝑉(𝑍) denote the set of vertices in the circuit 𝑍. Then the characteristic polynomial Φ(𝐺) satisfies Φ (𝐺) = 𝑥Φ (𝐺 − V) − ∑Φ (𝐺 − V − 𝑤) 𝑤

(5)

(2)

− 2 ∑ Φ (𝐺 − 𝑉 (𝑍)) , 𝑍∈𝜑(V)

hold. If 𝐺 is connected, then 𝜆 1 (𝐺) > 𝜆 1 (𝐺 − V). For the spectral radius of a graph, by the well-known Perron-Frobenius theory, we have the following. Lemma 2. Let 𝐺 be a connected graph and 𝐻 a proper subgraph of 𝐺. Then 𝜌(𝐻) < 𝜌(𝐺). Lemma 3 (see [6, 7]). Let 𝐺 be a graph on 𝑛 vertices, then 𝜌 (𝐺) ≤ max {𝑑 (V) : V ∈ 𝑉 (𝐺)} .

(3)

The equality holds if and only if 𝐺 is a regular graph. Let V be a vertex of a graph 𝐺 and suppose that two new paths 𝑃 = V(V𝑘+1 )V𝑘 ⋅ ⋅ ⋅ V2 V1 and 𝑄 = V(𝑢𝑙+1 )𝑢𝑙 ⋅ ⋅ ⋅ 𝑢2 𝑢1 of lengths 𝑘 and 𝑙 (𝑘 ≥ 𝑙 ≥ 1) are attached to 𝐺 at V(= V𝑘+1 = 𝑢𝑙+1 ), respectively, to form a new graph 𝐺𝑘,𝑙 (shown in Figure 2), where V1 , V2 , . . . , V𝑘 and 𝑢1 , 𝑢2 , . . . , 𝑢𝑙 are distinct. Let 𝐺𝑘+1,𝑙−1 = 𝐺𝑘,𝑙 − 𝑢1 𝑢2 + V1 𝑢1 .

(4)

We call that 𝐺𝑘+1,𝑙−1 is obtained from 𝐺𝑘,𝑙 by grafting an edge (see Figure 2). Lemma 4 (see [8, 9]). Let 𝐺 be a connected graph on 𝑛 ≥ 2 vertices and V is a vertex of 𝐺. Let 𝐺𝑘,𝑙 and 𝐺𝑘+1,𝑙−1 (𝑘 ≥ 𝑙 ≥ 1) be the graphs as defined above. Then 𝜌(𝐺𝑘,𝑙 ) > 𝜌(𝐺𝑘+1,𝑙−1 ). Let V be a vertex of the graph 𝐺 and 𝑁(V) the set of vertices adjacent to V. Lemma 5 (see [10, 11]). Let 𝐺 be a connected graph, and let 𝑢, V be two vertices of 𝐺. Suppose that V1 , V2 , . . . , V𝑠 ∈ 𝑁(V) \ (𝑁(𝑢) ⋃{𝑢}) (1 ≤ 𝑠 ≤ 𝑑(V)) and 𝑥 = (𝑥1 , 𝑥2 , . . . , 𝑥𝑛 ) is the Perron vector of 𝐺, where 𝑥𝑖 corresponds to the vertex V𝑖 (1 ≤ 𝑖 ≤ 𝑛). Let 𝐺∗ be the graph obtained from 𝐺 by deleting the edges VV𝑖 and adding the edges 𝑢V𝑖 (1 ≤ 𝑖 ≤ 𝑠). If 𝑥𝑢 ≥ 𝑥V , then 𝜌(𝐺) < 𝜌(𝐺∗ ).

where the first summation extends over those vertices 𝑤 adjacent to V, and the second summation extends over all 𝑍 ∈ 𝜑(V). An internal path of a graph 𝐺 is a sequence of vertices V1 , V2 , . . . , V𝑘 with 𝑘 ≥ 2 such that (1) the vertices in the sequence are distinct (except possibly V1 = V𝑘 ); (2) V𝑖 is adjacent to V𝑖+1 , (𝑖 = 1, 2, . . . , 𝑘 − 1); (3) the vertex degrees 𝑑(V𝑖 ) satisfy 𝑑(V1 ) ≥ 3, 𝑑(V2 ) = ⋅ ⋅ ⋅ = 𝑑(V𝑘−1 ) = 2 (unless 𝑘 = 2) and 𝑑(V𝑘 ) ≥ 3. Let 𝑊𝑛 be the tree on 𝑛 vertices obtained from 𝑃𝑛−4 by attaching two new pendant edges to each end vertex of 𝑃𝑛−4 , respectively. Lemma 7 (see [13]). Suppose that 𝐺 ≠ 𝑊𝑛 is a connected graph and 𝑢V is an edge on an internal path of 𝐺. Let 𝐺𝑢V be the graph obtained from 𝐺 by subdivision of the edge 𝑢V. Then 𝜌(𝐺𝑢V ) < 𝜌(𝐺).

3. Main Results Let 𝐻1 be the graph obtained from 𝐾𝜔 and a path 𝑃4 : V1 V2 V3 V4 by joining a vertex of 𝐾𝜔 and a nonpendant vertex, say, V2 , of 𝑃4 by a path with length 2 and let 𝐻2 be the graph obtained from 𝐾𝜔 by attaching two pendant edges at two different vertices of 𝐾𝜔 (see Figure 3). Lemma 8. Let 𝐻1 and 𝐻2 be the graphs defined as above (see Figure 3). If 𝜔 ≥ 3, then 𝜌(𝐻2 ) > 𝜌(𝐻1 ).

The Scientific World Journal

3 𝜔 1

1 3

.. . K𝜔

4

.. . K𝜔

.. . K𝜔

2 H1

𝜔+1 𝜔+2 ···

2

n

n−1

···

i

𝑖 Figure 4: Graph 𝑃𝐾𝑛−𝜔,𝜔 , where 𝑖 = 𝜔 + 1, . . . , 𝑛 − 1.

H2

Figure 3: 𝐻1 and 𝐻2 .

𝜔

Proof. For 5 ≥ 𝜔 ≥ 3, by direct computations, we have 𝜌(𝐻2 ) > 𝜌(𝐻1 ). In the following, we suppose that 𝜔 ≥ 6. From Lemma 6, we have

1

.. . K𝜔

𝜔+1

𝜔+2

n ···

n−3

n−2

n−1

2 𝑛−2

Figure 5: Graph 𝑃𝐾𝑛−𝜔,𝜔 .

Φ (𝐻1 ) = (𝑥 + 1)𝜔−2 [𝑥7 − (𝜔 − 2) 𝑥6 − (𝜔 + 4) 𝑥5 + (5𝜔 − 10) 𝑥4 + (4𝜔 + 1) 𝑥3 − (5𝜔 − 10) 𝑥2 − (2𝜔 − 1) 𝑥 + 𝜔 − 2]

𝑖 Let 𝑃𝐾𝑛−𝜔,𝜔 be the graph obtained from the kite graph 𝑃𝐾𝑛−𝜔−1,𝜔 (see Figure 1) and an isolated vertex V𝑛 by adding an edge V𝑛 V𝑖 (𝜔 + 1 ≤ 𝑖 ≤ 𝑛 − 1) (see Figure 4). It is easy to see 𝜔+2 𝑛−1 that 𝑃𝐾5,𝜔 = 𝐻1 and 𝑃𝐾𝑛−𝜔,𝜔 = 𝑃𝐾𝑛−𝜔,𝜔 .

= (𝑥 − 𝜔 + 2)𝜔−2 𝑔1 (𝑥) . Φ (𝐻2 ) = (𝑥 + 1)𝜔−3 [𝑥5 − (𝜔 − 3) 𝑥4 − (2𝜔 − 1) 𝑥3

𝑛−2

+ (𝜔 − 5) 𝑥2 + (2𝜔 − 3) 𝑥 − 𝜔 + 3]

𝑛−2 Let 𝑃𝐾𝑛−𝜔,𝜔 = 𝑃𝐾𝑛−𝜔,𝜔 + V𝑛−1 V𝑛 (see Figure 5).

= (𝑥 + 1)𝜔−3 𝑔2 (𝑥) . (6)

(𝑛 ≥ 10) .

1 ) 𝜔2

= −𝜔3 + 2𝜔2 + 6𝜔 +

13 26 54 − + 𝜔 𝜔2 𝜔3

𝑛−2 ) < 𝜌 (𝑊𝑛 ) = 2 = 𝜌 (𝐶𝑛 ) . 𝜌 (𝑃𝑛 ) < 𝜌 (𝑃𝐾𝑛−2,2

66 166 416 + 2 − 3 𝜔 𝜔 𝜔

+

224 432 832 320 560 800 + 5 − 6 + 7 + 8 − 9 𝜔4 𝜔 𝜔 𝜔 𝜔 𝜔

+

160 384 320 128 + − + − 91 > 0; 𝜔10 𝜔11 𝜔12 𝜔14

𝑔2 (𝜔 − 1 + = −2𝜔 + −

𝑛−3 𝑛−3 Let 𝐺1 = 𝑃𝐾𝑛−3,3 − V𝑛−1 V𝑛−2 + V𝑛−3 V𝑛−1 , let 𝐺2 = 𝑃𝐾𝑛−3,3 + V𝑛−1 V𝑛 , and let 𝐶𝑛−1,1 be the graph obtained from 𝐶𝑛−1 and an isolated vertex by adding an edge between some vertex of 𝐶𝑛−1 and the isolated vertex (see Figure 6).

2 ) 𝜔2

= 𝜔4 − 6𝜔3 + 7𝜔2 + 26𝜔 +

2 ) 𝜔2 12 18 8 48 48 − + − − 𝜔 𝜔2 𝜔3 𝜔4 𝜔5

8 64 32 32 + 7 − 8 + 10 < 0. 6 𝜔 𝜔 𝜔 𝜔

(9)

𝑛−3 7 ) ≥ 𝜌(𝑃𝐾8,2 )≈ For 𝑛 ≥ 10, from Lemma 2, we have 𝜌(𝑃𝐾𝑛−2,2 2.00659 > 𝜌(𝐶𝑛 ).

5 6 5 1 + 11 − 12 + 14 − 20 < 0; 10 𝜔 𝜔 𝜔 𝜔

𝑔1 (𝜔 − 1 +

(8)

2 Proof. Clearly, 𝑃𝑛 = 𝑃2𝑛−2,0 , 𝑃𝐾𝑛−2,2 = 𝑃2𝑛−3,1 . From Lemma 4, we have

26 34 54 20 20 25 + 4+ 5− 6+ 7+ 8− 9 𝜔 𝜔 𝜔 𝜔 𝜔 𝜔 +

𝑖 be the graphs defined as above (see FigLemma 9. Let 𝑃𝐾𝑛−𝜔,𝜔 ure 4). Then 𝑛−2 𝑛−3 ) < 𝜌 (𝐶𝑛 ) = 𝜌 (𝑊𝑛 ) < 𝜌 (𝑃𝐾𝑛−2,2 ), 𝜌 (𝑃𝑛 ) < 𝜌 (𝑃𝐾𝑛−2,2

By direct calculation, we have 𝑔1 (𝜔 − 1 +

From Lemmas 1 and 3, we have 𝜔 > 𝜌(𝐻1 ) ≥ 𝜌(𝐾𝜔 ) = 𝜔 − 1 ≥ 𝜆 2 (𝐻1 ) and 𝜔 > 𝜌(𝐻2 ) ≥ 𝜌(𝐾𝜔 ) = 𝜔 − 1. Then from (7) we have 𝜌(𝐻2 ) > 𝜔 − 1 + (2/𝜔2 ) > 𝜌(𝐻1 ).

(7)

Theorem 10. Among all connected graphs on 𝑛 vertices with maximum clique size 𝜔 = 2 and 𝑛 ≥ 10, the first four smallest 𝑛−2 spectral radii are exactly obtained for 𝑃𝑛 , 𝑃𝐾𝑛−2,2 , 𝐶𝑛 , 𝑊𝑛 , and 𝑛−3 𝑃𝐾𝑛−2,2 , respectively. Proof. Let 𝐺 be a connected graph with maximum clique size 𝜔 = 2 and 𝑛 ≥ 10 vertices. From Lemma 9, we have 𝑛−2 𝑛−3 ) < 𝜌(𝑊𝑛 ) = 𝜌(𝐶𝑛 ) < 𝜌(𝑃𝐾𝑛−2,2 ). Thus, we 𝜌(𝑃𝑛 ) < 𝜌(𝑃𝐾𝑛−2,2 𝑛−3 𝑛−2 only need to prove that 𝜌(𝐺) > 𝜌(𝑃𝐾𝑛−2,2 ) if 𝐺 ≠ 𝑃𝑛 , 𝑃𝐾𝑛−2,2 , 𝑛−3 𝑛−2 𝑊𝑛 , 𝐶𝑛 , 𝑃𝐾𝑛−2,2 . If 𝐺 is a tree, note that 𝐺 ≠ 𝑃𝑛 , 𝑃𝐾𝑛−2,2 , 𝑊𝑛 , 𝑛−3 𝑛−3 𝑃𝐾𝑛−2,2 , then, from Lemma 4, we have 𝜌(𝐺) > 𝜌(𝑃𝐾𝑛−2,2 ). If 𝐺 contains some cycle as a subgraph, then, from Lemmas 2 𝑛−3 and 7, we have 𝜌(𝐺) ≥ 𝜌(𝐶𝑛−1,1 ) > 𝜌(𝑃𝐾𝑛−2,2 ).

4

The Scientific World Journal n ···

···

G1

n−3

n−2

n−1

G2

g

𝜔

Cn−1

.. . K𝜔

Cn−1,1

H3

Figure 6: Graphs 𝐺1 , 𝐺2 , 𝐶𝑛−1,1 .

1 2

wk w2 ···

.. . K𝜔

w1 1

Fg

Cg 2 3

H4

Figure 7: Graphs 𝐻3 , 𝐻4 , and 𝐹𝑔 . 𝑛−2 𝑖 𝑃𝐾𝑛−𝜔,𝜔 , 𝑃𝐾𝑛−𝜔,𝜔 ,

Lemma 11. Let 𝐺1 and 𝐺2 be the graphs defined as above (see Figures 4, 5, and 6). Then 𝑛−2

𝑛−4 ) < min {𝜌 (𝑃𝐾𝑛−3,3 ) , 𝜌 (𝐺1 ) , 𝜌 (𝐺2 )} , 𝜌 (𝑃𝐾𝑛−3,3

(10)

(𝑛 ≥ 8) . Proof. For 8 ≤ 𝑛 ≤ 11, by direct calculation, we have 𝑛−4 𝜌(𝑃𝐾𝑛−3,3 ) < 𝜌(𝐺1 ). If 𝑛 ≥ 12, from Lemmas 2 and 7, we 𝑛−4 8 ) < 𝜌(𝑃𝐾9,3 ) < 2.23808. have 2.23601 < 𝜌(𝑃𝐾8,3 ) < 𝜌(𝑃𝐾𝑛−3,3 From Lemma 6, we have 𝑛−4 ) Φ (𝑃𝐾𝑛−3,3

= (𝑥5 − 4𝑥3 + 3𝑥) Φ (𝑃𝐾𝑛−8,3 ) − (𝑥4 − 2𝑥2 ) Φ (𝑃𝐾𝑛−8,3 − V𝑛−5 ) = 𝑓1 (𝑥) Φ (𝑃𝐾𝑛−8,3 ) − 𝑓2 (𝑥) Φ (𝑃𝐾𝑛−8,3 − V𝑛−5 ) , 5

3

Φ (𝐺1 ) = (𝑥 − 4𝑥 ) Φ (𝑃𝐾𝑛−8,3 )

𝑛−2

𝑛−4 𝜌 (𝑃𝐾𝑛−3,3 ) < 𝜌 (𝑃𝐾𝑛−3,3 ) ,

𝑛−4 𝜌 (𝑃𝐾𝑛−3,3 ) < 𝜌 (𝐺2 ) . (15)

Let 𝐻3 be the graph obtained from 𝐾𝜔 by attaching two pendant edges at some vertex of 𝐾𝜔 ; let 𝐻4 be the graph obtained from 𝐾𝜔 and 𝑃2 by adding two edges between two vertices of 𝐾𝜔 and two end vertices of 𝑃2 (see Figure 7). Theorem 12. Among all connected graphs on 𝑛 vertices with maximum clique size 𝜔 = 3 and 𝑛 ≥ 9, the first four smallest 𝑛−2 𝑛−3 spectral radii are exactly obtained for 𝑃𝐾𝑛−3,3 , 𝑃𝐾𝑛−3,3 , 𝑃𝐾𝑛−3,3 , 𝑛−4 𝑃𝐾𝑛−3,3 , respectively. Proof. Let 𝐺 be a connected graph with maximum clique size 𝜔 = 3 and 𝑛 ≥ 9 vertices. From Lemmas 2 and 7, we have 𝑛−4 𝑛−3 𝑛−2 𝜌 (𝑃𝐾𝑛−3,3 ) > 𝜌 (𝑃𝐾𝑛−3,3 ) > 𝜌 (𝑃𝐾𝑛−3,3 ) > 𝜌 (𝑃𝐾𝑛−3,3 ) . (16)

− (𝑥4 − 3𝑥2 ) Φ (𝑃𝐾𝑛−8,3 − V𝑛−5 ) = 𝑓3 (𝑥) Φ (𝑃𝐾𝑛−8,3 ) − 𝑓4 (𝑥) Φ (𝑃𝐾𝑛−8,3 − V𝑛−5 ) . (11) Then we have 𝑛−4 𝑓3 (𝑥) Φ (𝑃𝐾𝑛−3,3 ) − 𝑓1 (𝑥) Φ (𝐺1 )

= (𝑓1 (𝑥) 𝑓4 (𝑥) − 𝑓2 (𝑥) 𝑓3 (𝑥)) Φ (𝑃𝐾𝑛−8,3 − V𝑛−5 ) = (−𝑥7 + 7𝑥5 − 9𝑥3 ) Φ (𝑃𝐾𝑛−8,3 − V𝑛−5 )

(12)

= 𝑅1 (𝑥) Φ (𝑃𝐾𝑛−8,3 − V𝑛−5 ) .

𝑛−4 Thus, we only need to prove that 𝜌(𝐺) > 𝜌(𝑃𝐾𝑛−3,3 ) if 𝑛−2 𝑛−3 𝑛−4 𝐺 ≠ 𝑃𝐾𝑛−3,3 , 𝑃𝐾𝑛−3,3 , 𝑃𝐾𝑛−3,3 , 𝑃𝐾𝑛−3,3 . We distinguish the following three cases.

Case 1. If there exist at least two vertices outside of 𝐾3 that are adjacent to some vertices of 𝐾3 , then we have that 𝐺 contains either 𝐻2 (𝜔 = 3) or 𝐻3 (𝜔 = 3) as a proper subgraph. If 𝐺 contains 𝐻2 (𝜔 = 3) as a proper subgraph, from Lemmas 2 and 7, we have 5 ) 𝜌 (𝐺) > 𝜌 (𝐻2 ) ≈ 2.30278 > 𝜌 (𝑃𝐾6,3

For 2.23601 < 𝑥 < 2.23808, we have 5

𝑛−4 Thus, 𝜌(𝑃𝐾𝑛−3,3 ) < 𝜌(𝐺1 ). By similar method, we have for 𝑛≥8

𝑛−4 ), ≈ 2.26542 > 𝜌 (𝑃𝐾𝑛−3,3

3

𝑓1 (𝑥) > 2.23601 − 4 × 2.23808 + 3

(13)

− 9 × 2.238083 ≈ 9 > 0. 𝑛−4 ) and Note that from Lemma 2, 𝜌(𝑃𝐾𝑛−8,3 − V𝑛−5 ) < 𝜌(𝑃𝐾𝑛−3,3 𝑛−4 2.23601 < 𝜌(𝑃𝐾𝑛−3,3 ) < 2.23808. Then, we have

𝑛−4 ) , 2.23808) . 𝑥 ∈ [𝜌 (𝑃𝐾𝑛−3,3

𝜌 (𝐺) > 𝜌 (𝐻3 ) ≈ 2.34292 5 𝑛−4 > 𝜌 (𝑃𝐾6,3 ) > 𝜌 (𝑃𝐾𝑛−3,3 ),

𝑅1 (𝑥) > −2.238087 + 7 × 2.236015

𝑛−4 ) > 𝑓1 (𝑥) Φ (𝐺1 ) , 𝑓3 (𝑥) Φ (𝑃𝐾𝑛−3,3

(17)

If 𝐺 contains 𝐻3 (𝜔 = 3) as a proper subgraph, from Lemmas 2 and 7, we have

× 2.23601 ≈ 17 > 0; 𝑓3 (𝑥) > 2.236015 − 4 × 2.238083 ≈ 11 > 0;

(𝜔 = 3) .

(14)

(𝜔 = 3) .

(18)

Case 2. Suppose that there exists a vertex, say, 𝑢, which does not belong to 𝐾3 , such that 𝑢 is adjacent to at least two vertices of 𝐾3 . Then 𝐺 contains 𝐶4∗ as a proper subgraph, where 𝐶4∗ is obtained from 𝐶4 by adding an edge between two disjoint vertices. From Lemmas 2 and 7, we have 5 𝑛−4 𝜌 (𝐺) > 𝜌 (𝐶4∗ ) ≈ 2.56155 > 𝜌 (𝑃𝐾6,3 ) > 𝜌 (𝑃𝐾𝑛−3,3 ) . (19)

The Scientific World Journal

5

Case 3. Suppose that there uniquely exists a vertex 𝑢 which does not belong to 𝐾3 such that 𝑢 is adjacent to a vertex of 𝐾3 . We distinguish the following two cases. Subcase 1. Suppose that 𝐺 − 𝑉(𝐾3 ) is a tree. If there exist two vertices 𝑢, 𝑟 ∈ 𝑉(𝐺 − 𝑉(𝐾3 )) such that 𝑑(𝑢) ≥ 3 and 𝑑(𝑟) ≥ 3, then, from Lemmas 2, 4, and 7, we have 𝜌(𝐺) > 𝑛−4 𝜌(𝑃𝐾𝑛−3,3 ). If there exists only one vertex 𝑢 ∈ 𝑉(𝐺 − 𝑉(𝐾3 )) such that 𝑑(𝑢) ≥ 4, then, from Lemmas 2, 7, and 11, we 𝑛−4 ). If there exists exactly one have 𝜌(𝐺) ≥ 𝜌(𝐺1 ) > 𝜌(𝑃𝐾𝑛−3,3 vertex 𝑢 ∈ 𝑉(𝐺 − 𝑉(𝐾3 )) such that 𝑑(𝑢) = 3, note that 𝑛−2 𝑛−3 𝑛−4 , 𝑃𝐾𝑛−3,3 , 𝑃𝐾𝑛−3,3 , then from Lemmas 2 and 7 we 𝐺 ≠ 𝑃𝐾𝑛−3,3 𝑛−4 have 𝜌(𝐺) > 𝜌(𝑃𝐾𝑛−3,3 ). Subcase 2. Suppose that 𝐺 − 𝑉(𝐾3 ) contains cycle 𝐶𝑔 as a subgraph. If 𝑔 = 3, 4, then, from Lemmas 2, 7 and 11, we 𝑛−2 𝑛−4 ) or 𝜌(𝐺) ≥ 𝜌(𝐺2 ) > have 𝜌(𝐺) ≥ 𝜌(𝑃𝐾𝑛−3,3 ) > 𝜌(𝑃𝐾𝑛−3,3 𝑛−4 𝜌(𝑃𝐾𝑛−3,3 ). If 𝑔 ≥ 5, then, from Lemma 2, we can construct a graph 𝐹𝑔 from 𝐺 by deleting vertices such that 𝜌(𝐺) ≥ 𝜌(𝐹𝑔 ), where 𝐹𝑔 is the graph obtained from 𝐾3 and a cycle 𝐶𝑔 by joining a vertex of 𝐾3 and a vertex of 𝐶𝑔 with a path and |𝑉(𝐹𝑔 )| ≤ 𝑛 (see Figure 7). Suppose that 𝐶𝑔 is labelled V1 , V2 , . . . , V𝑔 satisfying V𝑖 V𝑖+1 ∈ 𝐸(𝐶𝑔 ), (1 ≤ 𝑖 ≤ 𝑔 − 1), V1 V𝑔 ∈ 𝐸(𝐶𝑔 ), and 𝑑(V1 ) = 3. Then, from Lemmas 2 and 7, 𝑛−4 we have 𝜌(𝐹𝑔 − V2 V3 ) > 𝜌(𝑃𝐾𝑛−3,3 ). Thus, we have 𝜌(𝐺) > 𝑛−4 𝜌(𝑃𝐾𝑛−3,3 ). Lemma 13. Let

𝑖 𝑃𝐾𝑛−𝜔,𝜔

and

𝑛−2 𝑃𝐾𝑛−𝜔,𝜔

be the graphs defined as 𝑛−2

𝑛−3 above (see Figures 4 and 5). Then 𝜌(𝑃𝐾𝑛−𝜔,𝜔 ) > 𝜌(𝑃𝐾𝑛−𝜔,𝜔 ) (𝜔 ≥ 4). 𝑇

Proof. Let 𝑋 = (𝑥1 , 𝑥2 , . . . , 𝑥𝑛 ) be the Perron vector of 𝑛−2 𝑃𝐾𝑛−𝜔,𝜔 , where 𝑥𝑖 corresponds to V𝑖 . It is easy to prove that 𝑛−2

𝑥𝑛 = 𝑥𝑛−1 . From 𝐴𝑋 = 𝜌(𝑃𝐾𝑛−𝜔,𝜔 )𝑋, we have 𝑛−2

(20)

𝑛−2

𝑥𝑛−3 = [𝜌 (𝑃𝐾𝑛−𝜔,𝜔 ) (𝜌 (𝑃𝐾𝑛−𝜔,𝜔 ) − 1) − 2] 𝑥𝑛 . 𝑛−2

From Lemma 2, for 𝜔 ≥ 4 we have 𝜌(𝑃𝐾𝑛−𝜔,𝜔 ) ≥ 𝜌(𝐾𝜔 ) = 𝜔 − 1 ≥ 3. Then 𝑛−2

𝑛−3 ) − 𝜌 (𝑃𝐾𝑛−𝜔,𝜔 ) 𝜌 (𝑃𝐾𝑛−𝜔,𝜔 𝑛−2

𝑛−3 ) 𝑋 − 𝑋𝑇 𝐴 (𝑃𝐾𝑛−𝜔,𝜔 ) 𝑋 ≥ 𝑋𝑇 𝐴 (𝑃𝐾𝑛−𝜔,𝜔

𝑛−2

𝑛−2

= 2 [𝜌 (𝑃𝐾𝑛−𝜔,𝜔 ) (𝜌 (𝑃𝐾𝑛−𝜔,𝜔 ) − 2) − 2] 𝑥𝑛

1

M𝜔2

Figure 8: Graph 𝑀𝜔2 .

Let 𝑀𝜔2 (𝜔 ≥ 4) be the graph as shown in Figure 8. Theorem 14. Among all connected graphs on 𝑛 vertices with maximum clique size 𝜔 ≥ 4 and 𝑛 ≥ 𝜔 + 5, the first four smallest spectral radii are exactly obtained for 𝑃𝐾𝑛−𝜔,𝜔 , 𝑛−2

𝑛−2 𝑛−3 𝑃𝐾𝑛−𝜔,𝜔 , 𝑃𝐾𝑛−𝜔,𝜔 , 𝑃𝐾𝑛−𝜔,𝜔 , respectively.

Proof. Let 𝐺 be a connected graph with maximum clique size 𝜔 ≥ 4 and 𝑛 ≥ 𝜔 + 5 vertices. Suppose that 𝐾𝜔 is a maximum clique of 𝐺. From Lemmas 2, 4, and 13, we have 𝑛−2

𝑛−3 𝑛−2 𝜌 (𝑃𝐾𝑛−𝜔,𝜔 ) > 𝜌 (𝑃𝐾𝑛−𝜔,𝜔 ) > 𝜌 (𝑃𝐾𝑛−𝜔,𝜔 ) > 𝜌 (𝑃𝐾𝑛−𝜔,𝜔 ) . (22) 𝑛−3 Thus, we only need to prove that 𝜌(𝐺) > 𝜌(𝑃𝐾𝑛−𝜔,𝜔 ) if 𝑛−2

𝑛−2 𝑛−3 𝐺 ≠ 𝑃𝐾𝑛−𝜔,𝜔 , 𝑃𝐾𝑛−𝜔,𝜔 , 𝑃𝐾𝑛−𝜔,𝜔 , 𝑃𝐾𝑛−𝜔,𝜔 . We distinguish the following three cases.

Case 1. If there exist at least two vertices outside of 𝐾𝜔 that are adjacent to some vertices of 𝐾𝜔 , then 𝐺 contains either 𝐻2 or 𝐻3 as a proper subgraph. If 𝐺 contains 𝐻2 as a proper subgraph, from Lemmas 2, 7, and 8, we have 𝑛−3 ). 𝜌 (𝐺) > 𝜌 (𝐻2 ) > 𝜌 (𝐻1 ) ≥ 𝜌 (𝑃𝐾𝑛−𝜔,𝜔

(23)

If 𝐺 contains 𝐻3 as a proper subgraph, from Lemmas 2, 5, 7, and 8, we have (24)

Case 2. Suppose that there exists a vertex, say, 𝑢, which does not belong to 𝐾𝜔 , such that 𝑢 is adjacent to at least two vertices of 𝐾𝜔 . From Lemmas 2, 7, and 8, we have 𝑛−3 𝜌 (𝐺) > 𝜌 (𝑀𝜔2 ) > 𝜌 (𝐻4 ) > 𝜌 (𝐻2 ) > 𝜌 (𝐻1 ) ≥ 𝜌 (𝑃𝐾𝑛−𝜔,𝜔 ). (25)

Case 3. Suppose that there uniquely exists a vertex 𝑢 which does not belong to 𝐾𝜔 such that 𝑢 is adjacent to a vertex of 𝐾𝜔 . 𝑛−2 𝑛−3 If 𝐺−𝑉(𝐾𝜔 ) is a tree, note that 𝐺 ≠ 𝑃𝐾𝑛−𝜔,𝜔 , 𝑃𝐾𝑛−𝜔,𝜔 , 𝑃𝐾𝑛−𝜔,𝜔 , 𝑛−3 then, from Lemmas 2, 4, and 7, we have 𝜌(𝐺) > 𝜌(𝑃𝐾𝑛−𝜔,𝜔 ). Suppose that 𝐺−𝑉(𝐾𝜔 ) contains cycle 𝐶𝑔 as a subgraph. If 𝑔 = 𝑛−2

(21)

= 2𝑥𝑛 (𝑥𝑛−3 − 𝑥𝑛−2 − 𝑥𝑛 )

2

𝑛−3 𝜌 (𝐺) > 𝜌 (𝐻3 ) > 𝜌 (𝐻2 ) > 𝜌 (𝐻1 ) ≥ 𝜌 (𝑃𝐾𝑛−𝜔,𝜔 ).

𝑥𝑛−2 = (𝜌 (𝑃𝐾𝑛−𝜔,𝜔 ) − 1) 𝑥𝑛 , 𝑛−2

.. . K𝜔

3, note that 𝐺 ≠ 𝑃𝐾𝑛−𝜔,𝜔 , then, from Lemmas 2 and 7, we have 𝑛−3 𝑛−3 𝜌(𝐺) > 𝜌(𝐺∗ ) > 𝜌(𝑃𝐾𝑛−𝜔,𝜔 ), where 𝐺∗ = 𝑃𝐾𝑛−𝜔,𝜔 + V𝑛−1 V𝑛 . If 𝑔 ≥ 4, then by the similar reasoning as that of Subcase 2 of 𝑛−3 Case 3 of Theorem 12, we have 𝜌(𝐺) > 𝜌(𝑃𝐾𝑛−𝜔,𝜔 ). Lemma 15. Let 𝐻3 and 𝐻4 be the graphs defined as above (see Figure 7). Then

≥ 2𝑥𝑛 > 0. 𝑛−2

𝑛−3 So, 𝜌(𝑃𝐾𝑛−𝜔,𝜔 ) > 𝜌(𝑃𝐾𝑛−𝜔,𝜔 ).

𝜌 (𝐻4 ) > 𝜌 (𝐻3 )

(𝜔 ≥ 3) .

(26)

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Proof. Let 𝑋 = (𝑥1 , 𝑥2 , . . . , 𝑥𝑛 )𝑇 be the Perron vector of 𝐻3 , where 𝑥𝑖 corresponds to V𝑖 . From 𝐴𝑋 = 𝜌(𝐻3 )𝑋, we have 𝜌 (𝐻3 ) 𝑥1 = 𝑥2 , 𝜌 (𝐻3 ) 𝑥2 = 2𝑥1 + (𝜔 − 1) 𝑥𝜔 ,

(27)

𝜌 (𝐻3 ) 𝑥𝜔 = (𝜔 − 2) 𝑥𝜔 + 𝑥2 .

.. . K𝜔

3

5

2

4

3

2

.. . K𝜔

1

1 6

𝜔+1 PK3,𝜔

4

.. . K𝜔

5

H5

Figure 9: Graphs

H6

𝜔+1 𝑃𝐾3,𝜔 ,

𝐻5 , and 𝐻6 .

From above equations, we have 𝜌3 (𝐻3 ) − (𝜔 − 2) 𝜌2 (𝐻3 ) − (𝜔 + 1) 𝜌 (𝐻3 ) + 2𝜔 − 4 = 0. (28) Let 𝑟1 (𝑥) = 𝑥3 − (𝜔 − 2) 𝑥2 − (𝜔 + 1) 𝑥 + 2𝜔 − 4.

(29)

Then 𝑟1 (𝜔 − 1) = −2 < 0.

(30)

For 𝑥 > 𝜔 − 1 and 𝜔 ≥ 3, we have 𝑟1󸀠 (𝑥) = 3𝑥2 − 2 (𝜔 − 2) 𝑥 − (𝜔 + 1) > 0.

(31)

Note that 𝜌(𝐻3 ) > 𝜌(𝐾𝜔 ) = 𝜔 − 1. From (30) and (31), we have 𝜌(𝐻3 ) which is the largest root of equation 𝑟1 (𝑥) = 0. Similarly, we have 𝜌(𝐻4 ) which is the largest root of equation 𝑟2 (𝑥) = 𝑥3 − (𝜔 − 1) 𝑥2 − 2𝑥 + 2𝜔 − 4 = 0.

𝜔+1

Lemma 17. Let 𝑃𝐾3,𝜔 and 𝐻5 be the graphs defined as above (see Figure 9). Then 𝜔+1

𝜌 (𝐻5 ) > 𝜌 (𝑃𝐾3,𝜔 ) ,

(35)

𝜔+1

𝑥5 . From 𝐴𝑋 = 𝜌(𝑃𝐾3,𝜔 )𝑋, we have 𝜔+1

𝜌 (𝑃𝐾3,𝜔 ) 𝑥1 = 𝑥1 + 𝑥2 , 𝜔+1

𝜌 (𝑃𝐾3,𝜔 ) 𝑥2 = 2𝑥1 + 𝑥3 , (36)

𝜔+1

𝜌 (𝑃𝐾3,𝜔 ) 𝑥3 = 𝑥2 + (𝜔 − 1) 𝑥4 , (33)

Thus, we have 𝜌(𝐻3 ) < 𝜌(𝐻4 ). Theorem 16. Let 𝐺 be a graph on 𝑛 vertices with maximum clique size 𝜔 ≥ 3 and 𝑛 = 𝜔 + 2. Let 𝑃𝐾2,𝜔 , 𝐻2 , 𝐻3 , and 𝐻4 be the graphs defined as above (see Figures 1, 3 and 7). The first four smallest spectral radii are obtained for 𝑃𝐾2,𝜔 , 𝐻2 , 𝐻3 , 𝐻4 , respectively.

𝜔+1

𝜌 (𝑃𝐾3,𝜔 ) 𝑥4 = 𝑥3 + (𝜔 − 2) 𝑥4 . From above equations, we have 𝜔+1

𝑥2 = (𝜌 (𝑃𝐾3,𝜔 ) − 1) 𝑥1 , 𝜔+1

𝑥4 =

(34)

Thus, we only need to prove that, for 𝐺 ≠ 𝑃𝐾2,𝜔 , 𝐻2 , 𝐻3 , and 𝐻4 , 𝜌(𝐺) > 𝜌(𝐻4 ). We distinguish the following two cases.

𝜔+1

𝜌 (𝑃𝐾3,𝜔 ) − 𝜔 + 2

(37) 𝑥1 .

Then for 𝜔 ≥ 4, we have 𝜔+1

𝜔+1

𝜌 (𝐻5 ) − 𝜌 (𝑃𝐾3,𝜔 ) ≥ 𝑋𝑇 𝐴 (𝐻5 ) 𝑋 − 𝑋𝑇 𝐴 (𝑃𝐾3,𝜔 ) 𝑋 = 2𝑥1 (𝑥4 − 𝑥2 − 𝑥1 )

Case 1. Suppose that there exists exactly one vertex outside of 𝐾𝜔 that is adjacent to at least two vertices of 𝐾𝜔 . Then 𝐺 contains 𝑀𝜔2 (see Figure 8) as a subgraph. From Lemmas 2 and 7, we have 𝜌(𝑀𝜔2 ) > 𝜌(𝐻4 ). Case 2. Suppose that the two vertices outside of 𝐾𝜔 that are all adjacent to some vertices of 𝐾𝜔 . Note that 𝐺 ≠ 𝐻2 , 𝐻3 , 𝐻4 . Then 𝐺 contains one of graphs 𝐻3 and 𝑀𝜔2 as a subgraph, where 𝐻3 is obtained from 𝐻3 by adding an edge between two pendant vertices. From Lemma 5, we have 𝜌(𝐺) ≥ 𝜌(𝐻3 ) > 𝜌(𝐻4 ). From Lemmas 2 and 7, 𝜌(𝐺) > 𝜌(𝑀𝜔2 ) > 𝜌(𝐻4 ).

𝜔+1

𝜌2 (𝑃𝐾3,𝜔 ) − 𝜌 (𝑃𝐾3,𝜔 ) − 2

Proof. From Lemmas 2, 5, 8, and 15, we have 𝜌 (𝐻4 ) > 𝜌 (𝐻3 ) > 𝜌 (𝐻2 ) > 𝜌 (𝐻1 ) > 𝜌 (𝑃𝐾2,𝜔 ) .

(𝜔 ≥ 4) .

Proof. Let 𝑋 = (𝑥1 , 𝑥2 , . . . , 𝑥𝑛 )𝑇 be the Perron vector of 𝜔+1 𝑃𝐾3,𝜔 , where 𝑥𝑖 corresponds to V𝑖 . It is easy to see that 𝑥1 =

(32)

Then we have, for 𝑥 > 𝜔 − 1, 𝑟1 (𝑥) − 𝑟2 (𝑥) = 𝑥2 − (𝜔 − 1) 𝑥 > 0.

Let 𝐻5 be the graph obtained from 𝐻2 and an isolated vertex by adding an edge between a pendant vertex of 𝐻2 and 𝜔+1 the isolated vertex; let 𝑃𝐾3,𝜔 and 𝐻6 be the graphs as shown in Figure 9.

𝜔+1

=2

(𝜔 − 3) 𝜌 (𝑃𝐾3,𝜔 ) − 2 𝜔+1

𝜌 (𝑃𝐾3,𝜔 ) − 𝜔 + 2

𝑥1 > 0. (38)

The result follows. Lemma 18. Let 𝐻5 and 𝐻6 be the graphs defined as above (see Figure 9). Then 𝜌 (𝐻6 ) > 𝜌 (𝐻5 ) ,

(𝜔 ≥ 4) .

(39)

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Proof. For 𝜔 = 4, by direct calculation, we have 𝜌(𝐻6 ) > 𝜌(𝐻5 ). In the following, we suppose that 𝜔 ≥ 5. Then, from Lemmas 2 and 3, we have 𝜔 > 𝜌(𝐻5 ) > 𝜌(𝐾𝜔 ) = 𝜔 − 1 ≥ 4. Let 𝑋 = (𝑥1 , 𝑥2 , . . . , 𝑥𝑛 )𝑇 be the Perron vector of 𝐻5 , where 𝑥𝑖 corresponds to V𝑖 . From 𝐴𝑋 = 𝜌(𝐻5 )𝑋, we have

.. . K𝜔

.. . K𝜔

𝜔

H7

2

.. . K𝜔

.. . K𝜔

H8

𝜔+1

H10

H9

Figure 10: Graphs 𝐻7 , 𝐻8 , 𝐻9 , 𝐻10 .

𝜌 (𝐻5 ) 𝑥1 = 𝑥2 , Case 1. There exists exactly one vertex outside of 𝐾𝜔 that is adjacent to only one vertex of 𝐾𝜔 . Then 𝐺 must be one of 𝜔+1 𝜔+1 , and 𝑃𝐾3,𝜔 . graphs 𝑃𝐾3,𝜔 , 𝑃𝐾3,𝜔

𝜌 (𝐻5 ) 𝑥2 = 𝑥1 + 𝑥3 , 𝜌 (𝐻5 ) 𝑥3 = 𝑥2 + 𝑥4 + (𝜔 − 2) 𝑥6 , (40) 𝜌 (𝐻5 ) 𝑥4 = 𝑥3 + 𝑥5 + (𝜔 − 2) 𝑥6 , 𝜌 (𝐻5 ) 𝑥5 = 𝑥4 , 𝜌 (𝐻5 ) 𝑥6 = 𝑥3 + 𝑥4 + (𝜔 − 3) 𝑥6 . From above equations, we have for 𝜔 > 𝜌(𝐻5 ) > 𝜔 − 1 ≥ 4, 𝑥6 =

𝜌2 (𝐻5 ) − 1 𝑥 𝜌 (𝐻5 ) − 𝜔 + 3 1 +

>

(𝜌2 (𝐻5 ) + 𝜌 (𝐻5 )) (𝜌2 (𝐻5 ) − 1) − 𝜌2 (𝐻5 ) (𝜌 (𝐻5 ) − 𝜔 + 3) (𝜌2 (𝐻5 ) + 𝜌 (𝐻5 ) − 1)

𝑥1 (41)

𝜌2 (𝐻5 ) − 1 𝑥1 > 𝜌 (𝐻5 ) 𝑥1 = 𝑥2 . 3

Case 3. If there exactly exist two vertices outside of 𝐾𝜔 that are adjacent to some vertices of 𝐾𝜔 , then 𝐺 contains 𝐻5 or 𝐻9 (see Figures 9 and 10) as a subgraph. If 𝐺 contains 𝐻9 as a subgraph, then, from Lemmas 2 and 5, we have 𝜌(𝐺) ≥ 𝜌(𝐻9 ) > 𝜌(𝐻5 ). If 𝐺 contains 𝐻5 as a subgraph, note that 𝐺 ≠ 𝐻5 , then, from Lemma 2, we have 𝜌(𝐺) > 𝜌(𝐻5 ). Case 4. If there exist three vertices outside of 𝐾𝜔 that are adjacent to some vertices of 𝐾𝜔 , then 𝐺 contains one of graphs 𝐻6 , 𝐻7 , and 𝐻8 (see Figures 9 and 10) as a subgraph. From Lemmas 5 and 18, we have 𝜌(𝐻8 ) > 𝜌(𝐻7 ) > 𝜌(𝐻6 ) > 𝜌(𝐻5 ). Then, from Lemma 2, we have 𝜌(𝐺) > 𝜌(𝐻5 ). 𝜔+2

Then, from Lemma 5, we have 𝜌(𝐻6 ) = 𝜌(𝐻5 − V1 V2 + V1 V6 ) > 𝜌(𝐻5 ). Let 𝐻7 be the graph obtained from 𝐻3 and an isolated vertex by adding an edge between V𝜔 and the isolated vertex; let 𝐻8 be the graph obtained from 𝐻3 and an isolated vertex by adding an edge between V2 and the isolated vertex; let 𝐻9 be the graph obtained from 𝐻3 and an isolated vertex by adding an edge between one pendant vertex and the isolated 𝜔+1 vertex; and let 𝐻10 be the graph obtained from 𝑃𝐾3,𝜔 and an isolated vertex by adding an edge between V𝜔+1 and the isolated vertex (see Figure 10). 𝜔+1 , 𝑃𝐾3,𝜔 , 𝑃𝐾3,𝜔

Case 2. There exists one vertex outside of 𝐾𝜔 that is adjacent to at least two vertices of 𝐾𝜔 . Then 𝐺 contains 𝑀𝜔2 (see Figure 8) as a proper subgraph. From Lemmas 2 and 7, we have 𝜌(𝐺) > 𝜌(𝑀𝜔2 ) > 𝜌(𝐻5 ).

𝜔+1 𝑃𝐾3,𝜔 ,

and 𝐻5 be the graphs Theorem 19. Let defined as above (see Figures 1, 4, 5, and 9). Among all connected graphs on 𝑛 vertices with maximum clique size 𝜔 and 𝑛 = 𝜔 + 3 (𝜔 ≥ 4), the first four smallest spectral radii 𝜔+1 𝜔+1 , 𝑃𝐾3,𝜔 , and 𝐻5 , respectively. are obtained for 𝑃𝐾3,𝜔 , 𝑃𝐾3,𝜔

𝑖 Lemma 20. Let 𝑃𝐾𝑛−𝜔,𝜔 and 𝑃𝐾4,𝜔 be the graphs defined as above (see Figures 4 and 5). Then 𝜔+2

𝜔+1 ) > 𝜌 (𝑃𝐾4,𝜔 ) , 𝜌 (𝑃𝐾4,𝜔

(𝜔 ≥ 4) .

(43)

Proof. From Lemma 6, we have 𝜔+2

Φ (𝑃𝐾4,𝜔 ) = (𝑥4 − 4𝑥2 − 2𝑥 + 1) Φ (𝐾𝜔 ) − (𝑥3 − 3𝑥 − 2) Φ (𝐾𝜔−1 ) = 𝑓5 (𝑥) Φ (𝐾𝜔 ) − 𝑓6 (𝑥) Φ (𝐾𝜔−1 ) ; 𝜔+1 Φ (𝑃𝐾4,𝜔 ) = (𝑥4 − 3𝑥2 + 1) Φ (𝐾𝜔 )

(44)

− (𝑥3 − 𝑥) Φ (𝐾𝜔−1 ) = 𝑓7 (𝑥) Φ (𝐾𝜔 ) − 𝑓8 (𝑥) Φ (𝐾𝜔−1 ) . Then, we have

Proof. From Lemmas 2, 4, and 17, we have 𝜌 (𝐻5 ) >

𝜔+1 𝜌 (𝑃𝐾3,𝜔 )

𝜔+1 > 𝜌 (𝑃𝐾3,𝜔 ) > 𝜌 (𝑃𝐾3,𝜔 ) .

𝜔+2

𝜔+1 𝑓7 (𝑥) Φ (𝑃𝐾4,𝜔 ) − 𝑓5 (𝑥) Φ (𝑃𝐾4,𝜔 )

(42)

Thus, we only need to prove that 𝜌(𝐺) > 𝜌(𝐻5 ) if 𝐺 ≠ 𝑃𝐾3,𝜔 , 𝜔+1

𝜔+1 𝑃𝐾3,𝜔 , 𝑃𝐾3,𝜔 , and 𝐻5 . We distinguish the following four cases.

= (𝑓5 (𝑥) 𝑓8 (𝑥) − 𝑓6 (𝑥) 𝑓7 (𝑥)) Φ (𝐾𝜔−1 ) = (𝑥5 − 5𝑥3 − 4𝑥2 + 2𝑥 + 2) Φ (𝐾𝜔−1 ) = 𝑅2 (𝑥) Φ (𝐾𝜔−1 ) .

(45)

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For 𝑥 > 𝜔 − 1 (𝜔 ≥ 4), we have 𝑓5 (𝑥) > 0, 𝑅2 (𝑥) > 0,

𝑓7 (𝑥) > 0, Φ (𝐾𝜔−1 ) > 0.

Proof. Let 𝐺 be a connected graph with maximum clique size 𝜔 ≥ 4 and 𝑛 = 𝜔 + 4 vertices. Suppose that 𝐾𝜔 is a maximum clique of 𝐺. From Lemmas 2, 4, and 20, we have (46)

𝜔+2 𝜌(𝑃𝐾4,𝜔 )

> 𝜌(𝐾𝜔 ) = 𝜔 − 1 and From Lemma 2, we have 𝜔+1 ) > 𝜌(𝐾𝜔 ) = 𝜔 − 1. Thus, for 𝑥 > 𝜔 − 1 (𝜔 ≥ 4), we 𝜌(𝑃𝐾4,𝜔 𝜔+2

𝜔+1 𝜔+1 have 𝑓7 (𝑥)Φ(𝑃𝐾4,𝜔 )−𝑓5 (𝑥)Φ(𝑃𝐾4,𝜔 ) > 0. Then 𝜌(𝑃𝐾4,𝜔 )> 𝜔+2

𝜌(𝑃𝐾4,𝜔 ), (𝜔 ≥ 4). 𝑖 Lemma 21. Let 𝑃𝐾𝑛−𝜔,𝜔 and 𝐻2 be the graphs defined as above (see Figures 3 and 4). Then 𝜔+1 ), 𝜌 (𝐻2 ) > 𝜌 (𝑃𝐾4,𝜔

(𝜔 ≥ 3) .

(47)

Proof. For 𝜔 = 3, 4, 5, by direct calculation, we have 𝜌(𝐻2 ) > 𝜔+1 𝜌(𝑃𝐾4,𝜔 ). In the following, we suppose that 𝜔 ≥ 6. From Lemma 6, we have 𝜔+1 Φ (𝑃𝐾4,𝜔 ) = (𝑥 + 1)𝜔−2 [𝑥6 − (𝜔 − 2) 𝑥5 − (𝜔 + 3) 𝑥4

+ (4𝜔 − 8) 𝑥3 + (3𝜔 − 1) 𝑥2 − (2𝜔 − 4) 𝑥 − 𝜔 + 1] = (𝑥 + 1)

𝜔−2

(48)

𝑔3 (𝑥) .

1 ) 𝜔2

13 17 24 6 14 4 = −𝜔 − + − + + + 𝜔 𝜔2 𝜔3 𝜔4 𝜔5 𝜔6 16 2 5 4 1 + + − + + 7 < 0; 𝜔7 𝜔8 𝜔9 𝜔10 𝜔12

𝑔3 (𝜔 − 1 +

2 ) 𝜔2

= 𝜔3 − 5𝜔2 + 2𝜔 − +

𝜔+1 ) if 𝐺 ≠ Thus, we only need to prove that 𝜌(𝐺) > 𝜌(𝑃𝐾4,𝜔 𝜔+2

𝜔+2 𝜔+1 𝑃𝐾4,𝜔 , 𝑃𝐾4,𝜔 , 𝑃𝐾4,𝜔 , 𝑃𝐾4,𝜔 . We distinguish the following three cases.

Case 1. There exists exactly one vertex outside of 𝐾𝜔 that is adjacent to one vertex of 𝐾𝜔 . Subcase 1. Suppose that 𝐺 − 𝑉(𝐾𝜔 ) is a tree. If 𝐺 contains exactly one pendant vertex, then 𝐺 = 𝑃𝐾4,𝜔 . If 𝐺 contains 𝜔+1 𝜔+2 exactly two pendant vertices, then 𝐺 = 𝑃𝐾4,𝜔 or 𝐺 = 𝑃𝐾4,𝜔 . If 𝐺 contains three pendant vertices, then 𝐺 = 𝐻10 (see Fig𝜔+1 ure 10). From Lemma 4, we have 𝜌(𝐻10 ) > 𝜌(𝑃𝐾4,𝜔 ). Subcase 2. Suppose that 𝐺 − 𝑉(𝐾𝜔 ) contains a cycle. If 𝐺 − 𝑉(𝐾𝜔 ) contains 𝐶4 , then 𝐺 contains 𝐻11 as a subgraph, where 𝜔+1 𝐻11 is obtained from 𝑃𝐾4,𝜔 by adding an edge between two pendant vertices. From Lemma 2, we have 𝜌(𝐻11 ) > 𝜔+2 𝜔+1 ). If 𝐺 − 𝑉(𝐾𝜔 ) does not contain 𝐶4 , then 𝐺 = 𝑃𝐾4,𝜔 𝜌(𝑃𝐾4,𝜔 or 𝐺 contains 𝑃𝐾3,𝜔 as a proper subgraph. From Lemmas 2 𝜔+1

𝜔+1 and 7, we have 𝜌(𝑃𝐾3,𝜔 ) > 𝜌(𝐻11 ) > 𝜌(𝑃𝐾4,𝜔 ). Note that 𝜔+2

𝜔+1 𝐺 ≠ 𝑃𝐾4,𝜔 . Thus, we have 𝜌(𝐺) > 𝜌(𝑃𝐾4,𝜔 ).

Case 2. There exists at least one vertex outside of 𝐾𝜔 that is adjacent to at least two vertices of 𝐾𝜔 . Then 𝐺 contains 𝑀𝜔2 (see Figure 8) as a subgraph. From Lemmas 2, 7, and 21, we 𝜔+1 have 𝜌(𝐺) > 𝜌(𝑀𝜔2 ) > 𝜌(𝐻2 ) > 𝜌(𝑃𝐾4,𝜔 ).

2

−

(50)

𝜔+1

For 𝜔 ≥ 6, we have 𝑔3 (𝜔 − 1 +

𝜔+2

𝜔+1 𝜔+2 𝜌 (𝑃𝐾4,𝜔 ) > 𝜌 (𝑃𝐾4,𝜔 ) > 𝜌 (𝑃𝐾4,𝜔 ) > 𝜌 (𝑃𝐾4,𝜔 ) .

Case 3. There exist at least two vertices outside of 𝐾𝜔 that are adjacent to some vertices of 𝐾𝜔 . Then 𝐺 contains 𝐻2 or 𝐻3 as a subgraph (see Figures 3 and 7). From Lemmas 2, 5, and 21, 𝜔+1 we have 𝜌(𝐻3 ) > 𝜌(𝐻2 ) > 𝜌(𝑃𝐾4,𝜔 ). Thus, from Lemma 2, 𝜔+1 we have 𝜌(𝐺) > 𝜌(𝑃𝐾4,𝜔 ).

4. Conclusion 58 20 108 192 48 + 3 − 4 + 5 + 𝜔 𝜔2 𝜔 𝜔 𝜔

192 256 32 160 128 64 − 7 + 8 + 9 − 10 + 12 + 24 > 0. 𝜔6 𝜔 𝜔 𝜔 𝜔 𝜔 (49)

In this paper, the first four graphs, which have the smallest values of the spectral radius among all connected graphs of order 𝑛 with maximum clique size 𝜔 ≥ 2, are determined.

Conflict of Interests The authors declare that they have no conflict of interests.

𝜔+1 𝜌(𝑃𝐾4,𝜔 )

From Lemmas 1 and 3, we have 𝜔 > ≥ 𝜌(𝐾𝜔 ) = 𝜔+1 𝜔 − 1 ≥ 𝜆 2 (𝑃𝐾4,𝜔 ). Then from (49) we have 𝜔 − 1 + 2/𝜔2 > 𝜔+1 𝜌(𝑃𝐾4,𝜔 ) > 𝜔 − 1 + 1/𝜔2 . From the proof of Lemma 8, we have 𝜌(𝐻2 ) > 𝜔 − 1 + 2/𝜔2 (𝜔 ≥ 6). The result follows. Theorem 22. Among all connected graphs on 𝑛 vertices with maximum clique size 𝜔 and 𝑛 = 𝜔 + 4 (𝜔 ≥ 4), the first four 𝜔+2 𝜔+2 , 𝑃𝐾4,𝜔 , smallest spectral radii are obtained for 𝑃𝐾4,𝜔 , 𝑃𝐾4,𝜔 𝜔+1 and 𝑃𝐾4,𝜔 (see Figures 1, 4, and 5), respectively.

Authors’ Contribution All authors completed the paper together. All authors read and approved the final paper.

Acknowledgments This research is supported by NSFC (nos. 10871204, 61370147, and 61170309) and Chinese Universities Specialized Research Fund for the Doctoral Program (20110185110020).

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