On the spectral radius of bicyclic graphs with n vertices and diameter d

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1 Linear Algebra and its Applications 4 (007) On the spectral radius of bicyclic graphs with n vertices and diameter d Shu-Guang Guo Department of Mathematics, Yancheng Teachers College, Yancheng 400, Jiangsu, PR China Received 6 February 006; accepted 8 September 006 Available online 3 October 006 Submitted by R.A. Brualdi Abstract Bicyclic graphs are connected graphs in which the number of edges equals the number of vertices plus one. In this paper we determine the graph with the largest spectral radius among all bicyclic graphs with n vertices and diameter d. As an application, we give first three graphs among all bicyclic graphs on n vertices, ordered according to their spectral radii in decreasing order. 006 Elsevier Inc. All rights reserved. AMS classification: 05C50 Keywords: Bicyclic graphs; Eigenvalues; Spectral radius; Diameter. Introduction Let G = (V (G), E(G)) be a (simple) graph with n vertices, and let A(G) be a (0, )-adjacency matrix of G. Since A(G) is symmetric, its eigenvalues are real. Without loss of generality, we can write them as λ (G) λ (G) λ n (G) and call them the eigenvalues of G. The characteristic polynomial of G is just det(λi A(G)), and is denoted by φ(g; λ). The largest Supported by the Basic Research Program of the Natural Science of the Universities of Jiangsu Province (06KJB0) and the Foundation for Professors and Doctors of Yancheng Teachers College. address: ychgsg@yahoo.com.cn /$ - see front matter ( 006 Elsevier Inc. All rights reserved. doi:0.06/j.laa

2 0 S.-G. Guo / Linear Algebra and its Applications 4 (007) 9 3 eigenvalue λ (G) is called the spectral radius of G, denoted by ρ(g). IfG is connected, then A(G) is irreducible, and by the Perron Frobenius theory of non-negative matrices, ρ(g) has multiplicity one and there exists a unique positive unit eigenvector corresponding to ρ(g). We shall refer to such an eigenvector as the Perron vector of G. The investigation on the spectral radius of graphs is an important topic in the theory of graph spectra. Recently, the problem of finding all graphs with maximal or minimal spectral radius among a given class of graphs has been studied extensively. For related results, one may refer to [ 9, 3,5 7] and the references therein. Bicyclic graphs are connected graphs in which the number of edges equals the number of vertices plus one. Chang et al. [3] determined graphs with the largest spectral radius among all the bicyclic graphs on n vertices with perfect matching. Yu and Tian [7] determined the graph with the largest spectral radius among all the bicyclic graphs on n vertices with a maximum matching of cardinality m. Guo et al. [7,3] determined the graph with the largest spectral radius among all the bicyclic graphs with n vertices and k pendant vertices. Simić[5] determined the bicyclic graphs on prescribed number of vertices with spectral radius minimal. The diameter of a connected graph is the maximum distance between pairs of its vertices. Very recently, Guo et al. [9] determined the graphs with the largest and the second largest spectral radius among all trees with n vertices and diameter d. Guo and Shao [6] characterized the first d + trees with n vertices and diameter d ordered according to their spectral radii. Guo [8] determined the graph with the largest spectral radius among all unicyclic graphs with n vertices and diameter d. In this paper, we study the spectral radius of bicyclic graphs with n vertices and diameter d, and determine the graph with the largest spectral radius among all bicyclic graphs with n vertices and diameter d. As an application, we give first three graphs among all bicyclic graphs on n vertices, ordered according to their spectral radii in decreasing order.. Preliminaries Denote by C n and P n the cycle and the path, respectively, each on n vertices. Let G uv denote the graph that arises from G by deleting the edge uv E(G). Similarly, G + uv is the graph that arises from G by adding the edge uv / E(G), where u, v V (G). For v V (G), N(v) denotes the set of all neighbors of vertex v in G, and d(v) = N(v) denotes the degree of vertex v in G. A pendant vertex of G is a vertex of degree. A pendant edge is an edge with which a pendant vertex is incident. For a real number x, we denote by x the greatest integer x, and by x the least integer x. We denote by B n,d the set of all bicyclic graphs with n vertices and diameter d. Let C p and C q be two vertex-disjoint cycles. Suppose that a is a vertex of C p and a l is a vertex of C q. Joining a and a l by a path a a a l of length l results in a graph B(p,l,q) (Fig. ) to be called an -graph, where l and l = means identifying a with a l. Let P l+, P p+ and P q+ be three vertexdisjoint paths, where l,p,q and at most one of them is. Identifying the three initial vertices and terminal vertices of them respectively results in a graph P(l,p,q) (Fig. ) to be called a θ-graph. Obviously B n,d consists of two types of graphs: one type, denoted by B n,d, are those graphs each of which is an -graph with trees attached; the other type, denoted by B θ n,d, are those graphs each of which is a θ-graph with trees attached. Then B n,d = B n,d Bθ n,d. In order to complete the proof of our main result, we need the following lemmas.

3 S.-G. Guo / Linear Algebra and its Applications 4 (007) 9 3 Fig.. B(p,l,q). Fig.. P(l,p,q). Lemma [6,]. Let G be a connected graph and let ρ(g) be the spectral radius of A(G). Let u, v be two vertices of G. Suppose v,v,...,v s N(v)\N(u)( s d(v)) and x = (x,x,...,x n ) is the Perron vector of A(G), where x i corresponds to the vertex v i ( i n). Let G be the graph obtained from G by deleting the edges vv i and adding the edges uv i ( i s). If x u x v, then ρ(g) < ρ(g ). By Lemma, we obtain easily Lemmas and 3 which may be regard as immediate consequences of Lemma. Since their proofs are similar, we only give the proof of Lemma. Lemma. Let G be a connected graph and let e = uv be a non-pendant edge of G with N(u) N(v) =. Let G be the graph obtained from G by deleting the edge uv, identifying u with v, and adding a pendant edge to u(= v). Then ρ(g) < ρ(g ). Proof. We use x u and x v to denote the components of the Perron vector of G corresponding to u and v respectively. Suppose that N(u) ={v, v,...,v s } and N(v) ={u, u,...,u t }. Since e = uv is a non-pendant edge of G, it follows that s, t. If x u x v, let G = G {vu,...,vu t }+{uu,...,uu t }. If x u <x v, let G = G {uv,...,uv s }+{vv,...,vv s }. Obviously, G = G = G. By Lemma,wehaveρ(G) < ρ(g ). This completes the proof. Lemma 3. Let G, G,G be three connected graphs disjoint in pairs. Suppose that u, v are two vertices of G, u is a vertex of G and u is a vertex of G. Let G be the graph obtained from G, G,G by identifying, respectively, uwith u and v with u. Let G be the graph obtained from G, G,G by identifying vertices u, u,u. Let G 3 be the graph obtained from G, G,G by identifying vertices v, u,u. Then either ρ(g )<ρ(g ) or ρ(g )<ρ(g 3 ). Let G be a connected graph, and uv E(G). The graph G u,v is obtained from G by subdividing the edge uv, i.e., adding a new vertex w and edges wu, wv in G uv. Hoffman and Smith define an internal path of G as a walk v 0 v v s (s ) such that the vertices v 0,v,...,v s are distinct, d(v 0 )>, d(v s )>, and d(v i ) =, whenever 0 <i<s(note, s is the length of the path). An internal path is closed if v 0 = v s. They proved the following result.

4 S.-G. Guo / Linear Algebra and its Applications 4 (007) 9 3 Fig. 3. W n. Lemma 4 [0]. Let uv be an edge of the connected graph G on n vertices. (i) If uv does not belong to an internal path of G, and G/= C n, then ρ(g u,v ) > ρ(g). (ii) If uv belongs to an internal path of G, and G/= W n, where W n is shown in Fig. 3, then ρ(g u,v ) < ρ(g). Lemma 5 [4]. The characteristic polynomial of P n satisfies the expression ( φ(p n ; λ) = x n+ λ x n+ ), 4 where x = ( ) λ + λ 4 and x = ( ) λ λ 4 are the roots of the equation x λx + = 0. Lemma 6 [,6]. Let u and v be two adjacent vertices of the connected graph G with d(u) > and d(v) >. For non-negative integers k and l, let G(k, l) denote the graph obtained from G by adding pendant paths of length k at u and length l at v. If k l, then ρ(g(k, l)) > ρ(g(k +,l )). Lemma 7 [,6]. Let G be a connected graph, and G be a proper spanning subgraph of G. Then ρ(g) > ρ(g ) and φ(g ; λ)>φ(g; λ) for all λ ρ(g). The following two results are often used to calculate the characteristic polynomials of graphs. Lemma 8 [4,4]. Let e = uv be an edge of G, and C(e) be the set of all cycles containing e. The characteristic polynomial of G satisfies φ(g; λ) = φ(g e; λ) φ(g u v; λ) φ(g\v(z); λ). Z C(e) Lemma 9 [4,4]. Let u be a vertex of G, and let C(u) be the set of all cycles containing u. The characteristic polynomial of G satisfies φ(g; λ) = λφ(g u; λ) φ(g u v; λ) φ(g\v(z); λ). v N(u) Z C(u) As usual, we assume that the characteristic polynomial of an empty graph is equal to. Let d and i d. We denote by Pd+ (i) the graph obtained from a path P d+ : v v v d+ and isolated vertices v d+,...,v n by adding edges v i v d+,...,v i v n. Denote by Pd+ (i) the graph obtained from Pd+ (i) by adding edges v d+v d+3 and v d+4 v d+5,bypd+ (i) the graph

5 S.-G. Guo / Linear Algebra and its Applications 4 (007) obtained from Pd+ (i) by adding edges v i v d+ and v d+3 v d+4,bypd+ (i) the graph obtained from Pd+ (i) by adding edges v i v d+ and v i+ v d+3, and by P + d+ (i) the graph (Fig. 7) obtained from Pd+ (i) by adding edges v i v n and v i+ v n. Lemma 0. Let d 3. Then ρ(pd+ (i)) ρ(p d+ (i)) ρ(p d+ (i)) with the first equality if and only if i = d and the second equality if and only if i =. Proof. Clearly, Pd+ (d) = P d+ (d), P d+ () = P d+ (). Denote by P d+ (i) the graph obtained from Pd+ (i) by adding edge v i v d+. For i<d, applying Lemma 8 to edge v i+ v d+3 of Pd+ (i) and edge v d+3v d+4 of Pd+ (i) respectively, we have φ(pd+ (i); λ) = φ(p d+ (i); λ) φ(p d+ (i) v i+ v d+3 ; λ) λ n d 3 φ(p i ; λ)φ(p d i ; λ), φ(pd+ (i); λ) = φ(p d+ (i); λ) φ(p d+ (i) v d+3 v d+4 ; λ) λ n d 4 φ(p i ; λ)φ(p d i+ ; λ). Note that Pd+ (i) v i+ v d+3 is a proper spanning subgraph of Pd+ (i) v d+3 v d+4, and K P d i is a proper spanning subgraph of P d i+. By Lemma 7 we have φ(p d+ (i); λ) φ(p (i); λ) d+ = φ(p d+ (i) v i+ v d+3 ; λ) φ(p d+ (i) v d+3 v d+4 ; λ) + λ n d 4 φ(p i ; λ)[λφ(p d i ; λ) φ(p d i+ ; λ)] > 0 for all λ ρ(pd+ (i)). Thus ρ(p d+ (i)) > ρ(p d+ (i)). For 3 i d, applying Lemma 8 to edge v i v d+ of Pd+ (i) and edge v d+v d+3 of P respectively, by similar reasoning as above, we have φ(p d+ (i); λ) φ(p (i); λ) d+ = φ(p d+ (i) v i v d+ ; λ) φ(p d+ (i) v d+ v d+3 ; λ) + λ n d 5 (λ )φ(p d i+ ; λ)[λφ(p i ; λ) φ(p i ; λ)] > 0 for all λ ρ(pd+ (i)). Thus ρ(p d+ (i)) > ρ(p d+ (i)). This completes the proof. d+ (i) Lemma. Let G (i), G (i) and Pd+ θ (i), shown in Figs. 4 6, belong to B n,d. Then ρ(g (i)) < ρ(g (i)) ρ(pd+ θ (i)), and the equality holds if and only if i =. Fig. 4. G (i).

6 4 S.-G. Guo / Linear Algebra and its Applications 4 (007) 9 3 Fig. 5. G (i). Fig. 6. P θ d+ (i). Proof. Applying Lemma 8 to edge v n v n of G (i) and edge v i v n of G (i) respectively, we have φ(g (i); λ) = φ(g (i) v n v n ; λ) φ(g (i) v n v n ; λ) φ(g (i) v n v n v i ; λ) φ(g (i) v n v n v n ; λ), φ(g (i); λ) = φ(g (i) v i v n ; λ) φ(g (i) v i v n ; λ) φ(g (i) v i v n v n ; λ) φ(g (i) v i v n v n ; λ). Note that G (i) v n v n = G (i) v i v n and G (i) v n v n v i = G (i) v i v n v n. Also note that G (i) v i v n, G (i) v i v n v n are proper spanning subgraphs of G (i) v n v n, G (i) v n v n v n respectively. By Lemma 7 we have φ(g (i) v i v n ; λ)>φ(g (i) v n v n ; λ) for all λ ρ(g (i) v n v n ) and φ(g (i) v i v n v n ; λ)>φ(g (i) v n v n v n ; λ) for all λ ρ(g (i) v n v n v n ). These imply that φ(g (i); λ)>φ(g (i); λ) for all λ ρ(g (i)). Thus ρ(g (i)) < ρ(g (i)). Clearly, G () = Pd+ θ (). For 3 i d, applying Lemma 8 to edge v n v n of G (i) and edge v i v n of Pd+ θ (i) respectively, we have φ(g (i); λ) = φ(g (i) v n v n ; λ) φ(g (i) v n v n ; λ) λ n d 3 φ(p i ; λ)φ(p d i+ ; λ) λ n d 4 φ(p i ; λ)φ(p d i+ ; λ), φ(pd+ θ (i); λ) = φ(pθ d+ (i) v i v n ; λ) φ(pd+ θ (i) v i v n ; λ) λ n d φ(p i ; λ)φ(p d i+ ; λ) λ n d 3 φ(p i ; λ)φ(p d i+ ; λ). Applying Lemma 8 to edge v n v n of G (i) v n v n and edge v i v n of Pd+ θ (i) v i v n respectively, by similar arguments to the proof of Lemma 0 we can show φ(g (i) v n v n ; λ)>φ(pd+ θ (i) v i v n ; λ)

7 S.-G. Guo / Linear Algebra and its Applications 4 (007) Fig. 7. P + d+ (i). Fig. 8. G 3 (i). for all λ ρ(g (i)). Note that Pd+ θ (i) v i v n is proper spanning subgraph of G (i) v n v n. By Lemma 7 we have φ(pd+ θ (i) v i v n ; λ)>φ(g (i) v n v n ; λ) for all λ ρ(g (i) v n v n ). Moreover, since λφ(p i ; λ) φ(p i ; λ) = φ(p i 3 ; λ) > 0 for all λ ρ(g (i)) >, it follows that λ n d φ(p i ; λ)φ(p d i+ ; λ) > λ n d 3 φ(p i ; λ)φ(p d i+ ; λ) and λ n d 3 φ(p i ; λ)φ(p d i+ ; λ) > λ n d 4 φ(p i ; λ)φ(p d i+ ; λ) for all λ ρ(g (i)). These imply that φ(g (i); λ)>φ(pd+ θ (i); λ) for all λ ρ(g (i)), Thus ρ(g (i)) < ρ(pd+ θ (i)). This completes the proof. ( Lemma. If n d + 4 and d 4 is even, then ρ Pd+ θ Proof. Let a = d. Denote P d+ θ Applying Lemma 9 several times we have ( P θ d+ ( )) ( d+4 <ρ Pd+ θ ( ) d+4 by G(a +,a), and Pd+ θ ( )) d+. ( ) d+ by G(a, a + ). ( ) ) ( ( ) ) d + 4 d + φ ; λ φ Pd+ θ ; λ = φ(g(a +,a); λ) φ(g(a, a + ); λ) = λφ(g(a, a); λ) φ(g(a,a); λ) λφ(g(a, a); λ) + φ(g(a, a ); λ) = φ(g(a, a ); λ) φ(g(a,a); λ) = =φ(g(, 0); λ) φ(g(0, ); λ) = λφ(g(0, 0); λ) φ(k,n d ; λ) λφ(g(0, 0); λ) + λ n d 3 φ(k, ; λ) > 0 ( ( )) ( ( )) ( ( )) for all λ ρ Pd+ θ d+4. Thus ρ Pd+ θ d+4 <ρ Pd+ θ d+. This completes the proof. Lemma 3. If n d + 3 and i d i +, then ρ ( P θ d+ (i)) <ρ ( P + d+ (i)).

8 6 S.-G. Guo / Linear Algebra and its Applications 4 (007) 9 3 Fig. 9. G +. Proof. Applying Lemma 8 to edge v i v n of Pd+ θ (i) and edge v nv i+ of P + d+ (i), and then applying Lemma 9 to the highest degree vertices of Pd+ θ (i) v i v n and P + d+ (i) v n v i+ respectively, we have φ ( Pd+ θ (i); λ) φ ( P + d+ (i); λ) = λ n d 3 φ(p i ; λ)φ(p d i ; λ) + λ n d 3 [λ + λ (n d )] [φ(p i ; λ)φ(p d i ; λ) φ(p i ; λ)φ(p d i+ ; λ)]. Case. i <d i +. Applying Lemma 9 several times, we have further φ(p i ; λ)φ(p d i ; λ) φ(p i ; λ)φ(p d i+ ; λ) =[λφ(p i ; λ) φ(p i 3 ; λ)]φ(p d i ; λ) φ(p i ; λ)[λφ(p d i ; λ) φ(p d i ; λ)] = φ(p i ; λ)φ(p d i ; λ) φ(p i 3 ; λ)φ(p d i ; λ) = =φ(p ; λ)φ(p d i+3 ; λ) λφ(p d i+4 ; λ) = λφ(p d i+ ; λ) φ(p d i+3 ; λ) = φ(p d i+ ; λ) 0 for all λ ρ(p d i+3 ). It follows that φ ( Pd+ θ (i); λ) φ ( P + d+ (i); λ) > 0 for all λ ρ(p θ d+ (i)) > n d +. Thus ρ(p θ d+ (i)) < ρ ( P + d+ (i)). Case. i = d i +. Then d i + 3 = 0. Similarly to Case, we have φ(p i ; λ)φ(p d i ; λ) φ(p i ; λ)φ(p d i+ ; λ) = φ(p ; λ) λ =. From Lemma 5 we can easily get that φ(p n ; λ) is equal to sinh((n+)θ) sinh(θ) after putting λ = cosh(θ). Making use of this fact, we easily get that φ(p i ; λ)φ(p d i ; λ) λ, for λ and d i + 3 (follows from the fact that sinh(x) = sinh(x) cosh(x) and that cosh and sinh are increasing functions in the interval [0, )). Combining the above arguments, in this case we have φ ( Pd+ θ (i); λ) φ ( P + d+ (i); λ) = λ n d 3 [φ(p i ; λ)φ(p d i ; λ) λ λ + n d ] λ n d 3 [λ λ λ + n d ] =λ n d 3 (λ λ + n d ) >0 for all λ ρ(pd+ θ (i)) > n d +. Thus ρ(pd+ θ (i)) < ρ ( P + d+ (i)). This completes the proof. Lemma 4. Let d 3 and n d + 3. Then ρ(g 3 (i)) < ρ ( P + d+ (i)). Proof. Let a = i, b = d i and G 0 (i) = G 3 (i) {v d+,...,v n }. Applying Lemma 9 to vertices v d+,..., v n of G 3 (i) and to vertices v d+,..., v n of P + d+ (i) respectively, we have

9 S.-G. Guo / Linear Algebra and its Applications 4 (007) φ(g 3 (i); λ) = λ n d φ(g 0 (i); λ) (n d )λ n d 3 φ(p b ; λ)φ( P a ; λ), φ ( P + d+ (i); λ) = λ n d φ(g 0 (i); λ) (n d )λ n d 3 φ(p d+ ; λ), where P a is the graph obtained from a cycle C 3 : u u u 3 and a path P a by joining u 3 and an end vertex of P a. Applying Lemma 9 to v a+3 of P d+ and u of P a respectively, we have φ(g 3 (i); λ) φ ( P + d+ (i); λ) = (n d )λ n d 3 [φ(p d+ ; λ) φ(p b ; λ)φ( P a ; λ)] = (n d )λ n d 3 [(λ + )φ(p a ; λ)φ(p b ; λ) φ(p a+ ; λ)φ(p b ; λ)]. From this we can see that it is sufficient to show for all λ ρ(g 3 (i)) (λ + )φ(p a ; λ)φ(p b ; λ) φ(p a+ ; λ)φ(p b ; λ) > 0. By Lemma 5, wehavex + x = λ, x x = and (λ + )φ(p a ; λ)φ(p b ; λ) φ(p a+ ; λ)φ(p b ; λ) = [( x λ + x + )( x a+ x a+ )( x b+ x b+ ) ( x a+3 x a+3 )( x b 4 x b )] = [ x d λ 4 + x d + xd + x d + x a+3 x b + xb xa+3 x a+ x b+ x b+ x a+ x a+ x b+ x b+ x a+ x a+ x b+ x b+ x a+ ]. Since ρ(g 3 (i)) > n d +, it follows that x > and x > 0 when λ ρ(g 3 (i)). Let r>. It is easy to see that function f(x)= r x + r x is increasing strictly on interval [0, + ). We consider the following two cases. Case. a b. Since a + b = d, it follows that d a b +. For all λ ρ(g 3 (i)), byx x = we have further (λ + )φ(p a ; λ)φ(p b ; λ) φ(p a+ ; λ)φ(p b ; λ) = [ x d λ 4 + x d + xd + x d + x a b+3 + x a b+3 x a b+ x a b+ x a b x a b x a b x a b ] = [ ( x d λ 4 + x d ) ( x a b + x (a b) ) ( + x d + x (d ) ) ( x a b+ + x (a b+) ) + ( x a b+3 + x (a b+3) Case. a b. By x x = we have further ) ( x a b + x (a b ) (λ + )φ(p a ; λ)φ(p b ; λ) φ(p a+ ; λ)φ(p b ; λ) = [ x d λ 4 + x d + xd + x d + x b a 3 x b a + x b a 3 x b a x b a+ ] = λ 4 x b a+ x b a x b a [ [( x d + x d ) ( x b a + x (b a) ( x b a 3 + x (b a 3) )] + [( x d + x (d ) )] [( x b a+ + x (b a+) ) ) ( x b a + x (b a ) )]]. )] > 0.

10 8 S.-G. Guo / Linear Algebra and its Applications 4 (007) 9 3 By a + b = d, we have d (b a) = a +. For r>, since f (x) = (ln r) (r x + r x )>0 for all x in [0, + ), it follows that f(x)is concave up on (0, + ). Therefore f(d) f(b a)>f(b a + ) f(b a ), f(d) f(b a)>f(b a ) f(b a 3) and so [f(d) f(b a)] >f(b a + ) f(b a 3). This implies that ) ( x b a + x (b a) )] ( > x b a+ + x (b a+) ) ( x b a 3 + x (b a 3) ) [( x d + x d for λ ρ(g 3 (i)). Hence, for all λ ρ(g 3 (i)),wehave (λ + )φ(p a ; λ)φ(p b ; λ) φ(p a+ ; λ)φ(p b ; λ) > 0. Combining Cases and, the proof follows. Lemma 5. Let i d i +. Then ρ ( P + d+ (i)) <ρ ( P + d+ (i )). Proof. Let a = i and b = d i. Denote P + d+ (i) by G(a, b). Similarly, denote P + d+ (i ) by G(a,b+ ). Applying Lemma 9 several times we have φ ( P + d+ (i); λ) φ ( P + d+ (i ); λ) = φ(g(a, b); λ) φ(g(a,b+ ); λ) = λφ(g(a,b); λ) φ(g(a,b); λ) λφ(g(a,b); λ) + φ(g(a,b ); λ) = φ(g(a,b ); λ) φ(g(a,b); λ) = =φ(g(a b, 0); λ) φ(g(a b, ); λ) = λ n d (λ + ) φ(p a b ; λ) > 0 for all λ ρ(p + d+ (i)). Thus ρ ( P + d+ (i)) <ρ(p + d+ (i )). This completes the proof. 3. Main results Theorem. Let n d + 4 and G B n,d. If d 4, then ( ( )) d + ρ(g) ρ P + d+ ( ) with equality if and only if G = P + d+ d+ ; if d = 3, then ρ(g) ρ(p4 θ (3)) with equality if and only if G = P4 θ (3). Proof. Choose G B n,d such that the spectral radius of G is as large as possible. Denote the vertex set of G by {v,v,...,v n } and the Perron vector of G by x = (x,x,...,x n ), where x i corresponds to the vertex v i ( i n). We first prove the following fact.

11 S.-G. Guo / Linear Algebra and its Applications 4 (007) Fact. There is not an internal path of G with length greater than unless the path lies on a cycle of length 3. Proof. Assume, on the contrary, that P k+ : v i v i+ v i+k is an internal path of G with length k and P k+ does not lie on a cycle of length 3. Let G = G {v i v i+,v i+ v i+ }+{v i v i+ }. If the diameter of G v i+ is d, it is easy to see that there is v V(G v i+ ) such that G = G +{vv i+ } B n,d. If the diameter of G v i+ is d, then any shortest path of G between two vertices with length d contains v i v i+ v i+k as a part. Let v be an initial vertex of such a path and let G = G +{vv i+ }, then G B n,d. By Lemmas 4 and 7, in both cases we have ρ(g )>ρ(g), a contradiction. This completes the proof of Fact. Let P d+ be a shortest path between two vertices of G with length d. Since B n,d = B n,d B θ n,d, it follows that G B n,d or G Bθ n,d. Now we distinguish two cases to determine G. Case. Suppose that G B n,d. Let B(p,l,q) be the -graph in G. We first prove that V(P d+ ) V(C p ) or V(P d+ ) V(C q ). Assume, on the contrary, that V(P d+ ) V(C p ) =0and V(P d+ ) V(C q ) =0. Let P k : u u u k be a shortest path such that u V(P d+ ) and u k V(C p ) V(C q ). Then k. Applying Lemma to the edge u u,we get a graph G B n,d with ρ(g )>ρ(g), a contradiction. Hence V(P d+ ) V(C p ) or V(P d+ ) V(C q ). Let V = V(P d+ ) V(B(p,l,q))and G = G[V ] be the induced subgraph of G. Then G is G with some trees attached. Applying Lemma to the non-pendant edges, we can similarly prove that all these attached trees are stars with centers in V. That is to say that G is G with some pendant edges attached. Applying Lemma 3, we can further prove that all these pendant edges are attached at the same vertex of G. From Fact, we can see that p = q = 3. Let P d+ : v v i v i+s v d+, where v i+k V(B(p,l,q)), k = 0,,...,s. We claim that the path a a l in B(p,l,q) lies on P d+. Otherwise, if l, we may assume a a / E(P d+ ). Applying Lemma to a a we get a graph G B n,d with ρ(g )>ρ(g), a contradiction. If l = and a / V(P d+ ), applying Lemma 3 to a and v i we get a graph G B n,d with ρ(g )>ρ(g), a contradiction. Hence the path a a l lies on P d+. We distinguish the following four cases. Subcase. V(P d+ ) V(C p ) = V(P d+ ) V(C q ) =. Applying Lemma 3, wehave G = Pd+ (i). This contradicts Lemma 0. Subcase. V(P d+ ) V(C p ) = and V(P d+ ) V(C q ) =. We may assume that v i v i E(C p ) and v j V(C q ). By Lemma 3, all the pendant edges, not lying on P d+,of G must be attached at v j. So we may further assume that i j. ByFactwehavej i +. If j = i +, applying Lemma to v i and v i+, by similar reasoning as the proof of Lemma we can obtain a graph G B n,d such that ρ(g )>ρ(g), a contradiction. Thus j = i, and so G = Pd+ (i). This contradicts Lemma 0 when i<d.fori = d, it is easy to see that Pd+ (d) = P d+ (d). Applying Lemma to vertices v d and v d+ of Pd+ (d), we have either ρ(pd+ (d)) < ρ(p d+ θ ()) or ρ(p d+ (d)) < ρ(p d+ θ (d)), a contradiction. Subcase 3. V(P d+ ) V(C p ) = and V(P d+ ) V(C q ) =. By similar reasoning as Subcase, we can obtain a contradiction. Subcase 4. V(P d+ ) V(C p ) = V(P d+ ) V(C q ) =. We may assume that v i v i C p, v j v j+ C q, and j i. By Fact, we have either j i + orj = i +, d(v i+ )>. If j = i +, or j = i +, d(v i+ )>, applying Lemma to vertices v i and v i+ we can obtain a graph G B n,d such that ρ(g )>ρ(g), a contradiction. Thus j = i. Applying Lemma to

12 30 S.-G. Guo / Linear Algebra and its Applications 4 (007) 9 3 vertices v i and v i+ we can obtain a graph G B θ n,d such that ρ(g )>ρ(g), a contradiction. Case. Suppose that G B θ n,d. Let P(l,p,q)be the θ-graph in G. By similar reasoning as Case, we can prove that V(P d+ ) V(P(l,p,q)). Let V = V(P d+ ) V(P(l,p,q)) and G = G[V ] be the induced subgraph of G. By similar reasoning as Case, we can further prove that G is G with some pendant edges attached at one vertex. This implies that at most 5 vertices of G have degree greater than. By the definition of P(l,p,q), we have that l,p,q and at most one of them is. Without loss of generality, we may assume that l p q. We claim that l = and p = q =. Indeed, if l, by Fact and the fact that at most 5 vertices of G have degree greater than, we have l = p = q = and the two vertices of degree of P(l,p,q)lie on P d+, the third vertex, denoted by w, of degree of P(l,p,q)is attached by some pendant edges. Applying Lemma to aw in G (a is as given in Fig. ), we obtain a graph G B θ n,d such that ρ(g )>ρ(g), a contradiction. So l =. Similarly, by Fact we can show that p q 3 and that if q = 3 then p =. If q = 3, denote P q+ : auvb, where a and b are shown in Fig.. By Fact, we have d(u) > and d(v) >. In the case when neither au nor vb lies on P d+, applying Lemma to bv, we obtain a graph G B θ n,d such that ρ(g )>ρ(g), a contradiction. So we many assume that au lies on P d+. If neither uv nor ab lies on P d+, applying Lemma to bv, we obtain similarly a contradiction. Otherwise, G must be the graph G + shown in Fig. 9. Apply Lemma to u and v, we obtain a graph G B θ n,d such that ρ(g )>ρ(g), a contradiction. So, in further, l =, p = q =. Subcase. V(P d+ ) V(P(l,p,q)) =. Applying Lemma 3, we can prove that G = G (i) or G (i). By Lemma we have G = G () = Pd+ θ (). Since d 3, by Lemma 6 we have ρ(g ())<ρ(pd+ θ (3)), a contradiction. Subcase. V(P d+ ) V(P(l,p,q)) =. If one edge of P p+ or P q+ lies on P d+,wemay assume that P p+ : aub and au lies on P d+. Applying Lemma (and Lemma 6, if necessary) to u and b we can obtain a graph G B θ n,d such that ρ(g )>ρ(g), a contradiction. If P l+ lies on P d+, by Fact, Lemmas and we can similarly prove that all the pendant edges, not lying on P d+,ofgmust be at one of a and b. That is to say that G = Pd+ θ (i).ford 5, by Lemmas 6, and 3, we have further ( )) ( ( )) d + 3 d + 3 ρ(pd+ (P θ (i)) ρ d+ θ <ρ P + d+, a contradiction. For d = 4, applying Lemmas 6 and 9, by direct calculation we have ρ(p5 θ ()) < ρ(p5 θ (3))<ρ(P+ 5 (3)), and by Lemma 3 we have ρ(pθ 5 (4))<ρ(Pθ 5 (3))<ρ(P+ 5 (3)). That is to say that ρ(p5 θ (i)) < ρ(p + 5 (3)), a contradiction. For d = 3, by Lemma 9 we have ρ(pθ 4 ()) < ρ(p4 θ (3)). Hence G = P 4 θ (3). Subcase 3. V(P d+ ) V(P(l,p,q)) =3. Then G must be G 0 (i) (see Lemma 4) with n d pendant edges attached at v j, where i, j d. By Fact we may assume that i j i +. If j = i +, applying Lemma to v j and v i, we can obtain a graph G B θ n,d such that ρ(g )>ρ(g), a contradiction. So G must be P + d+ (i) or G 3(i) for some i. By ( ) Lemma 4, wehaveg = P + d+ (i). By Lemma 5, we have further G = P + d+ d+.for d = 3, applying Lemma 9, by direct calculation we have ρ(p + 4 ())<ρ(pθ 4 (3)), a contradiction.

13 S.-G. Guo / Linear Algebra and its Applications 4 (007) Combining Cases and, we have G = P + d+ 3. This completes the proof. ( ) d+ for d 4 and G = P4 θ (3) for d = ( ( )) ( ( )) By Lemma 6,wehaveρ P + d+ d+ <ρ P d + d+ for d 5. Moreover, it is easy to see that P3 () and P + 3 () are all bicyclic graphs with n vertices and diameter. Applying Lemma 9, by direct calculation we can show when n 9 ρ(p + 3 ())>ρ(p 3 ())>ρ(p4 θ (3))>ρ(P+ 5 (3)). Combining these inequalities and Theorem, we have the following two corollaries. Corollary. Let d 4, and G be a bicyclic graph on n vertices with diameter not less than d. Then ( ( )) d + ρ(g) ρ P + d+, ( ) and the equality holds if and only if G = P + d+ d+. Corollary. Let n 9. The first three graphs among all bicyclic graphs on n vertices, ordered according to their spectral radii in decreasing order, are P + 3 (), P 3 () and P4 θ (3). Acknowledgments I am grateful to the anonymous referees for valuable suggestions and corrections which result in an improvement of the original manuscript. References [] A. Berman, X.D. Zhang, On the spectral radius of graphs with cut vertices, J. Combin. Theory Ser. B 83 (00) [] R.A. Brualdi, E.S. Solheid, On the spectral radius of connected graphs, Publ. Inst. Math. (Beograd) 39 (53) (986) [3] A. Chang, F. Tian, Aimei Yu, On the index of bicyclic graphs with perfect matchings, Discrete Math. 83 (004) [4] D. Cvetković, M. Doob, H. Sachs, Spectra of Graphs, Academic Press, New York, 980, nd revised ed.: Barth, Heidelberg, 995. [5] D. Cvetković, P. Rowlinson, The largest eigenvalues of a graph: a survey, Linear Multilinear Algebra 8 (990) [6] J.M. Guo, J.Y. Shao, On the spectral radius of trees with fixed diameter, Linear Algebra Appl. 43 () (006) [7] S.G. Guo, The spectral radius of unicyclic and bicyclic graphs with n vertices and k pendant vertices, Linear Algebra Appl. 408 (005) [8] S.G. Guo, On the spectral radius of unicyclic graphs with n vertices and diameter d, Australas. J. Combin., submitted for publication. [9] S.G. Guo, G.H. Xu, Y.G. Chen, On the spectral radius of trees with n vertices and diameter d, Adv. Math. (China) 34 (6) (005) (in Chinese). [0] A.J. Hoffman, J.H. Smith, On the spectral radii of topologically equivalent graphs, in: Fiedler (Ed.), Recent Advances in Graph Theory, Academia Praha, New York, 975, pp

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