The distance spectral radius of graphs with given number of odd vertices

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1 Electronic Journal of Linear Algebra Volume 31 Volume 31: (016 Article The distance spectral radius of graphs with given number of odd vertices Hongying Lin South China Normal University, lhongying0908@16.com Bo Zhou South China Normal University, zhoubo@scnu.edu.cn Follow this and additional works at: Part of the Algebra Commons, and the Discrete Mathematics and Combinatorics Commons Recommended Citation Lin, Hongying and Zhou, Bo. (016, "The distance spectral radius of graphs with given number of odd vertices", Electronic Journal of Linear Algebra, Volume 31, pp DOI: This Article is brought to you for free and open access by Wyoming Scholars Repository. It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository. For more information, please contact scholcom@uwyo.edu.

2 THE DISTANCE SPECTRAL RADIUS OF GRAPHS WITH GIVEN NUMBER OF ODD VERTICES HONGYING LIN AND BO ZHOU Abstract. The graphs with smallest, respectively largest, distance spectral radius among the connected graphs, respectively trees with a given number of odd vertices, are determined. Also, the graphs with the largest distance spectral radius among the trees with a given number of vertices of degree 3, respectively given number of vertices of degree at least 3, are determined. Finally, the graphs with the second and third largest distance spectral radius among the trees with all odd vertices are determined. Key words. Distance spectral radius, Distance matrix, Distance Perron vector, Odd vertex, Maximum degree, Graph, Tree. AMS subject classifications. 05C50, 15A Introduction. Throughout this paper, we consider simple graphs. Let G be aconnectedgraphwith vertexsetv(g andedgesete(g. Thedistancebetweenvertices u,v V(G, denoted by d G (u,v, is the length of a shortest path between them. The distance matrix of G, denoted by D(G, is the matrix D(G = (d G (u,v u,v V (G. Since D(G is real and symmetric, its eigenvalues are real. The distance spectral radius of G, denoted by ρ(g, is the largest eigenvalue of D(G. Since D(G is irreducible, we have by the Perron-Frobenius theorem that ρ(g is simple, and there is a unique positive unit eigenvector x(g of D(G corresponding to ρ(g, which is called the distance Perron vector of G. The study of eigenvalues of the distance matrix of a connected graph dates back to the classicalworkofgraham and Pollack[5], Grahamand Lovász[4], and Edelberg et al. []. For more details on spectra of distance matrices and especially on distance spectral radius, one may refer to the recent survey of Aouchiche and Hansen [1]. A vertex is an odd vertex (respectively, even vertex if its degree is odd (respectively, even. It is well known that the number of odd vertices in a graph is always even. A vertex in a tree with degree at least 3 is known as a branch vertex. The Received by the editorson January 19, 015. Accepted forpublication on April7, 016. Handling Editor: Bryan L. Shader. School ofmathematical sciences, South China NormalUniversity, Guangzhou , P.R.China (lhongying0908@16.com, zhoubo@scnu.edu.cn. Supported by the Specialized Research Fund for the Doctoral Program of Higher Education of China (no and the Scientific Research Fund of the Science and Technology Program of Guangzhou, China (no

3 The Distance Spectral Radius of Graphs 87 number of branch vertices may be used to analyze graph structures, see, e.g. [3, 7]. In this paper, we determine the graphs with smallest, respectively largest, distance spectral radius among the connected graphs, respectively trees with a given number of odd vertices. Also, we determine the graphs with the largest distance spectral radius among the trees with a given number of vertices of degree 3, respectively given number of vertices of degree at least 3. Finally, we determine the graphs with the second and third largest distance spectral radius among the trees with all odd vertices.. Preliminaries. Let G be a connected graph with V(G = {v 1,...,v n }. A column vector x = (x v1,...,x vn R n (whether it is the distance Perron vector of G or not can be considered as a function defined on V(G which maps vertex v i to x vi, i.e., x(v i = x vi for i = 1,...,n. Then x D(Gx = {u,v} V (G d G (u,vx u x v, and λ is an eigenvalue of D(G with corresponding eigenvector x if and only if x 0 and for each u V(G, (.1 λx u = v V (G d G (u,vx v. We call (.1 the (λ,x-eigenequation for G at u. For a unit column vector x R n with at least one nonnegative entry, by Rayleigh s principle, we have ρ(g x D(Gx with equality if and only if x is the distance Perron vector of G. For a connected graph G with v V(G, let δ G (v be the degree of v in G, and let N G (v be the set of neighbors of v in G. Let P n, C n, S n and K n be respectively the path, the cycle, the star and the complete graph on n vertices. A caterpillar is a tree such that the deletion of all pendant vertices yields a path. Obviously, S n and P n are caterpillars. Let G be a connected graph. For V 1 V(G, G V 1 denotes the graph obtained from G by deleting all vertices of V 1 (and the incident edges. If V 1 = {u}, then we write G u for G {u}. For E 1 E(G, G E 1 denotes the graph obtained from G by deleting all edges of E 1. If E 1 = {uv}, then we write G uv for G {uv}. If E is a subset of edges of the complement of G, then GE denotes the graph obtained from G by inserting all edges of E. If E = {uv}, then we write Guv for G{uv}.

4 88 H. Lin and B. Zhou For a subgraph H of a connected graph G, let σ G (H be the sum of the entries of the distance Perron vector of G corresponding to the vertices in V(H. Lemma.1. [9] Let G be a connected graph with u,v V(G. If uv E(G, then ρ(g > ρ(guv. Lemma.. [10] Let G be a connected graph on n vertices with u,v V(G, and let u and v be pendant neighbors of u and v, respectively. Let x = x(g. Then x u x v = ρ(g ρ(g (x u x v. Lemma.3. [11, 1] Let G be a connected graph and u a cut vertex of G. Suppose that G u consists of vertex disjoint subgraphs G 1, G and G 3. Let G 3 be the subgraph of G induced by V(G 3 {u}. For v V(G, let G = G { uw : w N G 3 (u } { vw : w N G 3 (u }. If σ G (G 1 σ G (G, then ρ(g > ρ(g. Lemma.4. [10] Let G be a connected graph and uv a non-pendant cut edge of G. Let G be the graph obtained from G by contracting uv to a vertex u and attaching a pendant vertex v to u. Then ρ(g < ρ(g. 3. Distance spectral radius of graphs with given number of odd vertices. For integers n and k with 0 k n, let G(n,k be the set of connected graphs with n vertices and k odd vertices, and let K n (k be the graph obtained from K n by deleting k pairwise disjoint edges. In particular, K n (0 = K n. Theorem 3.1. Let G G(n,k, where n 3 and 0 k n. (i If n is odd, then ρ(g ρ(k n (k with equality if and only if G = K n (k. (ii If n is even, then ρ(g ρ ( K n ( n k with equality if and only if G = K n ( n k. Proof. Let G be the graph in G(n,k with minimum distance spectral radius. If n is odd and k = 0, or n is even and k = n, then by Lemma.1, G = K n (0. If n is even and k = 0, then since G is a spanning subgraph of K n ( n, we have by Lemma.1 that G = K n ( n. Suppose 1 k < n. For z V(G, let N z = V(G\(N G (z {z}. Obviously, N z = n 1 δ G (z. Let V 1 (respectively, V be the set of odd (respectively, even vertices of G. Suppose that there are vertices u V 1 and v V such that uv / E(G. Let G = Guv. Note that δ G (u = δ G (u 1 is even and δ G (v = δ G (v 1 is odd. We have G G(n,k. By Lemma.1, ρ(g < ρ(g, a contradiction. Thus, each

5 vertex of V 1 is adjacent to each vertex of V. Case 1. n is odd. The Distance Spectral Radius of Graphs 89 Supposethat thereis avertexu V with δ G (u < n 1. Then δ G (u n 3. Let t = n 1 δg(u. Then t is a positive integer, and N u = n 1 δ G (u = t. Let N u = {u 1,...,u t }. Suppose that u 1 is not adjacent to u t. Let G = G{uu 1,uu t,u 1 u t }. Obviously, G G(n,k. By Lemma.1, ρ(g < ρ(g, a contradiction. Thus, u 1 u t E(G. Let G = G u 1 u t {uu 1,uu t }. Obviously, G G(n,k. Let H 1 and H be the subgraphs of G induced by V 1 and V \(N u {u}, respectively. Let x = x(g. From (.1 for G at u and u 1, we have Thus, ρ(g x u = σ G (H 1 σ G (H x u 1 x u x u t 1 x u t, ρ(g x u 1 σ G (H 1 σ G (H x u x u t 1 x u t x u. (ρ(g 1 ( x u x u 1 σg (H 1 x u 1 x u x u t 1 x u > 0, which implies that x u x u 1 > 0. Similarly, x u x u t > 0. As we pass from G to G, the distance between u 1 and u t is increased by 1, the distance between u and u 1 is decreased by 1, the distance between u and u t is decreased by 1, and the distance between any other vertex pair remains unchanged. Therefore, 1 (ρ(g ρ(g 1 x (D(G D(G x = x u( x u1 x u t x u1 x u t = 1 (( x u x u t x u1 ( x u x u 1 x ut > 0. This leads to the contradiction that ρ(g > ρ(g. Thus, the degree of each vertex in V is n 1. Suppose that there is a vertex u V 1 with δ G (u < n. Then δ G (u n 4. Let t = n δg(u. Then t is a positive integer, and N u = t 1. Let N u = {u 1,...,u t1 }. Arguing as above we see u 1 u t1 E(G. Let G = G u 1 u t1 {uu 1,uu t1 }. Obviously, G G(n,k. As above, we have ρ(g > ρ(g, a contradiction. Thus, the degree of each vertex in V 1 is n. Since each even degree is n 1 and each odd degree is n, we have G = K n (k.

6 90 H. Lin and B. Zhou Case. n is even. Similarly to the proof in Case 1, we have that each odd degree is n 1 and each even degree is n. Thus, G = K n ( n k. Lemma 3.. Let T be a tree with u V(T, and let N T (u = {u 1,...,u k }, where k 3. Let T i be the component of T u containing u i for 1 i k. Let T = T {uu i : i t}{wu i : i t}, where t k 1 and w V(T k. If σ T (T 1 σ T (T k, then ρ(t > ρ(t. Proof. Let x = x(t. As we pass from T to T, the distance between a vertex of V(T V(T t and a vertex of V(T 1 {u} is increased by d T (u,w, the distance between a vertex of V(T V(T t and a vertex of V(T k is decreased by at most d T (u,w, and the distance between any other vertex pair is increased or remains unchanged. Thus, 1 (ρ(t ρ(t 1 x (D(T D(Tx t d T (u,w σ T (T i (σ T (T 1 σ T (T k x u Therefore, ρ(t > ρ(t. > 0. i= Let G 1 (s,t be the graph shown in Fig. 1, where G 1 is a nontrivial connected graph, and s,t 1. G 1 v v s 1 v s v s1 v s v s3 v st u 1... u u s 1 u s u s1... u s u s3 u st u st1 Fig. 1. Graph G 1 (s,t. Lemma 3.3. Let G 1 be a nontrivial connected graph. For s t, we have ρ(g 1 (s1,t 1 > ρ(g 1 (s,t. Proof. LetG = G 1 (s,t. LetG and G 3 bethe componentsofg u s1 containing u 1 and u st1, respectively. Let G = G {u s v s,u s1 v s1 }{u s v s1,u s1 v s }.

7 Obviously, G 1 (s1,t 1 = G. Let x = x(g. Claim 1. σ G (G 1 x vs > 0. The Distance Spectral Radius of Graphs 91 Choose z V(G 1 such that d G (z,v s1 = max v V (G1d G (v,v s1. Let d = d G (z,v s1. Since V(G 1, d 1, and z v s1. From (.1 for G at v s1, z and v s, we have ρ(gx vs1 = dx z 3x vs d G (v s1,wx w w V(G 1\{z,v s1} w V(G\(V(G 1 {v s} ρ(gx z = dx vs1 (d3x vs w V(G\(V(G 1 {v s} ρ(gx vs = 3x vs1 (d3x z w V(G\(V(G 1 {v s} d G (v s1,wx w, w V(G 1\{z,v s1} d G (z,wx w, w V(G 1\{z,v s1} d G (v s,wx w. d G (z,wx w (d G (v s1,w3x w Note that for w V(G\(V(G 1 {v s }, d G (v s1,wd G (z,w d G (v s,w 0. Thus, ρ(g ( x vs1 x z x vs (d 3xvs1 3x z (d6x vs (d G (z,w 3x w, w V(G 1\{z,v s1} and (ρ(g3 ( ( ( σ G (G 1 x vs ρ(g xvs1 x z x vs 3 σg (G 1 x vs (d 3x vs1 3x z (d6x vs (d G (z,w 3x w w V(G 1\{z,v s1} 3 x vs1 x z x w x vs w V(G 1\{z,v s1} = dx vs1 (d3x vs

8 9 H. Lin and B. Zhou d G (z,wx w w V(G 1\{z,v s1} > 0. Therefore, Claim 1 follows. Claim. σ G (G σ G (G 3. Let y k = x uk x vk for k st, y 1 = x u1, and y st1 = x ust1. Suppose s t (3.1 y i < y s1i. i=1 i=1 From (.1 for G at u k with 1 k st1, we have ρ(g ( s t (3. x us x us = y j j=1 j=1 y s1j and (3.3 ρ(g ( ( x us1i x us1 i ρ(g xusi x us i s t = y s1 j j=i s = y j j=1 j=1 j=i y s1j t i 1 i 1 y s1j y s1 j y s1j j=1 j=1 for i t 1. We now prove that x us1i x us1 i < 0 for 1 i t by induction on i. If i = 1, then from (3.1 and (3., we have x us x us < 0, and by Lemma., y s = x us x vs < x us x vs = y s. Suppose i t 1andx us1j x us1 j < 0 for1 j i 1. Inparticular,x usi x us i < 0. ByLemma., y s1j = x us1j x vs1j < x us1 j x vs1 j = y s1 j. Thus, i 1 j=1 y s1j i 1 j=1 y s1 j < 0. Now from (3.1 and (3.3, we have x us1i x us1 i < x usi x us i < 0 for i t. It follows that x us1 i x us1i > 0 for 1 i t. Thus, s y i i=1 t y s1i i=1 t y i i=1 t y s1i > 0, which leads to the contradiction that s i=1 y i t i=1 y s1i. Hence, s i=1 y i t i=1 y s1i 0, from which Claim follows. i=1 As we pass from G to G, the distance between a vertex of V(G 1 and a vertex of V(G {u s1 } is increased by 1, the distance between a vertex of V(G 1 and a

9 The Distance Spectral Radius of Graphs 93 vertex of V(G 3 \{v s } is decreased by 1, the distance between v s and a vertex of V(G 3 \{v s } is increased by 1, the distance between v s and a vertex of V(G {u s1 } is decreased by 1, and the distance between any other vertex pair remains unchanged. Thus, 1 (ρ(g ρ(g 1 x (D(G D(Gx = σ G (G 1 ( σ G (G x us1 σg (G 1 ( σ G (G 3 x vs x vs ( σg (G 3 x vs xvs ( σg (G x us1 = ( σ G (G σ G (G 3 x vs x us1 ( σg (G 1 x vs, which, together with Claims 1,, implies that ρ(g > ρ(g. For b a 0 with (ab n, let C(n,a,b be the tree obtained from the path P n a b with consecutive vertices u 0,u 1,...,u by attaching a pendant vertex v i to vertex u i for i {1,...,a} {n a b 1,...,n a b }. In particular, C(n,0,0 is just the path P n. Lemma 3.4. [10] Let T be a tree with n 4 vertices. If T = Pn and C(n,0,1, then ρ(t < ρ(c(n,0,1 < ρ(p n. Lemma 3.5. If a 0, b a, and (ab < n, then ρ(c(n,a1,b 1 > ρ(c(n,a,b. Proof. Let T = C(n,a,b, p = n a b, and q = n a b. Let x = x(t. For 0 i, let s i = x ui if u i has no pendant vertex, and s i = x ui x vi otherwise. Claim 1. x ui1 x un a b i > x ui x u i > 0 for 0 i a. First we prove that x up > x uq1. Suppose p s i i=q1 s i. We prove that x up i x uq1i for 0 i p by induction on i. For i = 0, we have ρ(t ( p (3.4 x up x uq1 = (q 1 p s i, i=q1 and thus, x up x uq1. Suppose i 1 and x up j x uq1j for 0 j i 1. If δ T (u p j =, then s p j = x up j x uq1j s q1j. If δ T (u p j = 3, then by Lemma., x vp j x vq1j, and thus, s p j = x up j x vp j x uq1j x vq1j = s q1j. In either case, s p j s q1j. Hence, ρ(t ( x up i x uq1i ρ(t ( xup1 i x uqi s i

10 94 H. Lin and B. Zhou = = 0, j=q1i j=q1 p i s j s j j=0 p j=0 s j s j i 1 (s q1j s p j and thus, x up i x uq1i x up1 i x uqi. It follows that x up i x uq1i for 0 i p, and thus, s p i s q1i for 0 i p. Since b a, there exists an i with 0 i p such that δ T (u p i = and δ T (u q1i = 3, and thus, s p i = x up i x uq1i < s q1i. This leads to the contradiction that p s i < i=q1 s i. Hence, p s i < i=q1 s i, and by (3.4, we have x up > x uq1. Suppose x u0 x u. We prove that x ui x u i for 0 i p by induction on i. Suppose i 1 and x uj x u j for 0 j i 1. As above, we have by Lemma. that s j s j. It follows that ρ(t ( x ui x u i ρ(t ( xui 1 x un a b i i 1 = (s j s j 0. j=0 Hence, x ui x u i x ui 1 x un a b i 0. Thus, x ui x u i for 0 i p. In particular, x up x uq1, which is a contradiction. It follows that x u0 > x u, and as above, we have by induction that x ui1 x un a b i > x ui x u i for 0 i a. Claim. x vn a b 1 < x un a b x vn a b. Let V = V(T\{v n a b 1, u n a b,v n a b }. Since for u V, j=0 d T (u n a b,ud T (v n a b,u d T (v n a b 1,u 0, we have from (.1 for T at u n a b, v n a b and v n a b 1 that ρ(t ( x un a b x vn a b x vn a b 1 = xun a b x vn a b 5x vn a b 1 u V (d T (u n a b,ud T (v n a b,u d T (v n a b 1,ux u x un a b x vn a b 5x vn a b 1.

11 The Distance Spectral Radius of Graphs 95 Thus, (ρ(t ( x un a b x vn a b x vn a b 1 xun a b 3x vn a b 1 > 0, from which Claim follows. Claim 3. x ua1i x un a b 1 i > x uai x un a b i > 0 for 0 i n 1 b a 1. It is sufficient to prove that x ui x un b i > x ui1 x un b 1 i > 0 for a 1 i n 1 b. Let t = n 1 and t 1 = n 1. By the proof of Claim 1, t b j=0 s j < j=q1 s j. Since t b q 1, we have t b s j < j=0 j=q1 s j j=t b s j j=n b t We prove that x ui > x un b i for a1 i t b by induction on i. For i = t b, we have ρ(t ( t b x ut b x un b t = (t1 1 t s j > 0. j=n b t Hence, x ut b > x un b t. Suppose a 1 i t b 1 and x uj > x un b j for i1 j t b. Then ρ(t ( x ui x un b i ρ(t ( xui1 x un b 1 i = = > 0, j=n b i j=n b t s j i j=0 t b s j j=0 s j s j t b j=i1 s j. j=0 s j (s n b j s j and thus, x ui x un b i > x ui1 x un b 1 i > 0. This proves Claim 3. Claim 4. x un a b 1 < x un a b. By Claim 1, x ui > x u i for 0 i a. As above, we have by Lemma. that s i > s i. Thus, a s i > a s i = i=n a b 1 s i. Let m = b a and m 1 = b a. Suppose n a b m s i i=n a b 1 m s i. We prove that x un a b i x un a b 1 i for 1 i m by induction on i. For i = m, we have ρ(t ( x un a b m x un a b 1 m

12 96 H. Lin and B. Zhou = (m 1 1 m 0. ( i=n a b 1 m n a b m s i s i Hence, x un a b m x un a b 1 m. Suppose 1 i m 1 and x un a b j x un a b 1 j for i 1 j m. By Lemma., s n a b j s n a b 1 j for i1 j m. Hence, ρ(t ( ( x un a b i x un a b 1 i ρ(t xun a b 1i x un a b i = n a b i s j = 0, j=n a b 1 i j=n a b 1 m j=0 s j n a b m s j s j j=0 m j=i1 (s n a b 1 j s n a b j and thus, x un a b i x un a b 1 i x un a b 1i x un a b i 0. It follows that for 1 i m, x un a b i x un a b 1 i. As above, s n a b i s n a b 1 i. Thus, m i=1 s n a b i m i=1 s n a b 1 i = n a b i=n a b 1 m s i, and n a b m s i > = a m s i i=1 i=n a b 1 s n a b i s i i=n a b 1 m s i, n a b i=n a b 1 m a contradiction. Thus, n a b m s i > i=n a b 1 m s i. This proves Claim 4. Let T = C(n,a1,b 1. It is easily seen that s i (3.5 1 (ρ(t ρ(t 1 x (D(T D(Tx = x vn a b 1 W, where

13 W = r and r = n a b. The Distance Spectral Radius of Graphs 97 a (s i s i r r 1 b a 1 i=1 s n a b 1i (r i ( x un a b 1 i x ua1i, From Claim 1, s 0 s = x u0 x ( u < x ua1 x un a b. By Lemma. and Claim 1, s i s i 1 ρ(t (xua1 x un a b < ( x ua1 x un a b for 1 i a. Let { 0 if b = a, F = By Claims, 3 and 4, r b a 1 i= s n a b1i if b > a. ρ(t ( x ua1 x un a b 1 = W rxvn a b 1 < W r ( x un a b x vn a b = W r r 1 (r i ( x ua1i x un a b 1 i a (s i s i F < W r 1 (r i ( x ua1 x un a b 1 r(a1 ( x ua1 x un a b Hence, (3.6 < W r 1 (r i(a1r ( x ua1 x un a b 1. r 1 W > ρ(t (r i(a1r ( x ua1 x un a b 1. Claim 5. The minimum rowsum ofd(t is largerthan r 1 (r i(a1r.

14 98 H. Lin and B. Zhou For 0 j a, u V(T d T (u j,u > > n a b i=a1 n a b i=n a b 1 n a b i=1 d T (u a,u i d T (u a,u (d T (u a,u i d T (u a,v i ir n a b i=n a b 1 r If a 1, then for 1 j a, > r 1 (r i(a1r. u V (T d T (v j,u > For a1 j n a b, u V (T d T (u j,u > > = u V(T n a b i=a1 d T (u j,u > r 1 (r i(a1r. d T (u j,u i d T (u j,u 0 a (d T (u j,u i d T (u j,v i d T (u j,u i i=1 d T (u j,v i d T (u j,u n a b i=a1 r 1 For n a b 1 j, u V (T d T (u j,u > d T (u a r1,u i (r i(a1r. n a b i=a1 (a1r d T (u n a b 1,u i d T (u n a b 1,u 0 a (d T (u n a b 1,u i d T (u n a b 1,v i i=1

15 The Distance Spectral Radius of Graphs 99 > > n a b i=1 r 1 For n a b 1 j n a b, u V (T Thus, Claim 5 follows. d T (v j,u > u V(T i(a1r (r i(a1r. d T (u j,u > r 1 (r i(a1r. Since ρ(t is bounded below by the minimum row sum of D(T [6, p. 4], we have by Claim 5 that ρ(t > r 1 (r i(a1r. Now by (3.6 and Claim 3, W > 0, and thus by (3.5, ρ(t ρ(t > 0. Let (G be the maximum degree of a graph G. Lemma 3.6. Let T be a caterpillar with n vertices and k pendant vertices, where k 3. If (T = 3, then ρ(t ρ ( C ( n, k, k with equality if and only if T = C ( n, k, k. Proof. If n is even and k = n 1, then the result is trivial. If k = 3, then the result follows from Lemma 3.4. Suppose 4 k n. Let T be a caterpillar with maximum distance spectral radius satisfying the hypothesis in the lemma. Let U be the set of vertices of degree in T. Then k U 3(n k U = (n 1, and thus, U = n k > 0, i.e., U. Obviously, the diameter of T is n (k 1= n k1. Let u 0 u 1...u n k1 be a diametrical path of T. Assume without loss of generality that δ T (u 1 δ T (u n k. Then δ T (u 1 δ T (u n k 3. Suppose δ T (u n k =. Then δ T (u 1 = and there is u j with j n k 1 such that δ T (u j = 3. Let v j be the pendant neighbor of u j. Let T 1 and T be the nontrivial components of T u j containing u 0 and u n k1, respectively. Assume without loss of generality that σ T (T 1 σ T (T. Let T = T u j v j u n k v j. Obviously, T is a caterpillar with n vertices and k pendant vertices, and (T = 3. By Lemma.3, ρ(t < ρ(t, a contradiction. Thus, δ T (u n k = 3. Suppose δ T (u 1 =. If T U has exactly one nontrivial component, then T = C(n,0,k, and by Lemma 3.5, ρ(t = ρ(c(n,0,k < ρ ( C ( n, k, k, a contradiction. If T U has at least two nontrivial components, then there are vertices

16 300 H. Lin and B. Zhou u i and u j with i < j n k 1 such that δ T (u i = 3 and δ T (u j =. Let v i be the pendant neighbor of u i. Let T 1 and T be the components of T u i containing u 0 and u n k1, respectively. Assume without loss of generality that σ T (T 1 σ T (T. Let T = T u i v i u j v i. Obviously, T is a caterpillar with n vertices and k pendant vertices, and (T = 3. By Lemma.3, ρ(t < ρ(t, a contradiction. Thus, δ T (u 1 = 3, and T U has at least two nontrivial components. Suppose that T U has at least three nontrivial components. There are three vertices u i,u j,u l with i < j < l n k 1 in T such that δ T (u j = 3 and {u i,u l } U. Let T 1 and T be the nontrivial components of T u j containing u i and u l, respectively. Let v j be the pendant neighbor of u j. Assume without loss of generality that σ T (T 1 σ T (T. Let T = T u j v j u l v j. By Lemma.3, ρ(t < ρ(t, a contradiction. Thus, T U contains exactly two nontrivial components, implying that T = C(n,a,b, where a b = k, and a,b 1. By Lemma 3.5, T = C ( n, k, k. For integers n and k with 1 k n, let T(n,k be the set of trees with n vertices and k odd vertices. Theorem 3.7. Let T T(n,k, where 1 k n. Then ( ( k 1 k 1 ρ(t ρ C n,, with equality if and only if T = C ( n, k 1, k 1. Proof. If k = 1,, then the result follows from Lemma 3.4. Suppose k 3. Let T be a tree in T(n,k with maximum distancespectral radius. Suppose that the maximum odd degree is larger than 3. Then δ T (u = t1 for some u V(T and t. Let N T (u = {u 1,...,u t1 }. Let T i be the component of T u containing u i, where 1 i t 1. Assume without loss of generality that σ T (T 1 σ T (T t1. Let w be a pendant vertex of T in V(T t1. Let T = T {uu i : 3 i t}{wu i : 3 i t}. Note that the degrees of u and w remain odd in T. Then T T(n,k. By Lemma 3., ρ(t > ρ(t, a contradiction. Thus, the maximum odd degree is 3. If n is even, and k = n, then by Lemma 3.3, T = C ( n, k 1, k 1. Suppose k < n. Let U be the set of even vertices of T. Then U 1. Suppose that the maximum even degree is larger than. Then δ T (u = t for some u V(T and t. Let N T (u = {u 1,...,u t }. Let T i be the component of T u containing u i, where 1 i t. Assume without loss of generality that σ T (T 1 σ T (T t. Let w be a pendant vertex of T in V(T t. Let T = T uu wu. Note that the degree of u is odd and the degree of w is even in T. Then T T(n,k. By Lemma.3,

17 The Distance Spectral Radius of Graphs 301 ρ(t > ρ(t, a contradiction. Thus, the maximum even degree is, and each vertex in U is of degree in T. Suppose that T is not a caterpillar. Since (T = 3, there is a vertex u of degree 3 in the graph obtained from T by deleting all pendant vertices. Let N T (u = {u 1,u,u 3 }. Obviously, δ T (u i for i = 1,,3. Let T i be the component of T u containing u i, where 1 i 3. Claim. U V(T i for some i with i = 1,,3. Otherwise, there are two vertices of degree in T, one in V(T i and the other in V(T j, where 1 i < j 3. Assume without loss of generality that δ T (v 1 = δ T (v 3 = with v 1 V(T 1 and v 3 V(T 3, and that σ T (T 1 σ T (T 3. Let T = T uu v 3 u. Obviously, δ T (u is even and δ T (v 3 is odd. Thus, T T(n,k. By Lemma.3, ρ(t > ρ(t, a contradiction. This proves the Claim. Since (T = 3, we have by the Claim that T = G 1 (s,t for some s and t with s t. Obviously, G 1 (s1,t 1 T(n,k. By Lemma 3.3, ρ(g 1 (s1,t 1 > ρ(t, a contradiction. Thus, T is a caterpillar with (T = 3. By Lemma 3.6, we have T = C ( n, k 1, k Distance spectral radius of trees with given number of vertices of degree 3 or of degree at least 3. Let T be a tree with n vertices, in which k vertices of degree at least 3. Let r be the number of pendant vertices in T. Then r(n r k3k (n 1, i.e., r k. This implies that k kr n, and thus, k n 1. As an application of Theorem 3.7, we have Theorem 4.1. Let T be a tree on n vertices with k vertices of degree 3, where n, and 0 k n 1. Then ρ(t ρ( C ( n, k, k with equality if and only if T = C ( n, k, k. Proof. If k = 0,1, then the result follows from Lemma 3.4. Suppose k. Let T be a tree with maximum distance spectral radius on n vertices with k vertices of degree 3. Let = (T. Case Let u V(T and N T (u = {u 1,...,u }. Let T i be the componentoft ucontainingu i, where1 i. Assumewithout lossofgenerality that σ T (T 1 σ T (T. Let w be a pendant vertex oft in V(T. Let T = T {uu i : i 1}{wu i : i 1}. Note that the number of vertices of degree 3 in T remains k. By Lemma 3., ρ(t > ρ(t, a contradiction. Case. = 4. Let u V(T and N T (u = {u 1,u,u 3,u 4 }. Let T i be the component of T u containing u i, where 1 i 4. Assume without loss of generality that σ T (T 1 σ T (T 4. Let v be a pendant vertex of T in V(T 4, and T = T uu

18 30 H. Lin and B. Zhou vu. Let T 4 be the component of T u containing u 4. Note that δ T (u = 3 and u is a cut vertex. Assume without loss of generality that σ T (T 1 σ T (T 4. Let w be a pendant vertex of T in V(T 4 and T = T uu 3 wu 3. Note that the number of vertices of degree 3 in T remains k. By Lemma.3, ρ(t > ρ(t > ρ(t, a contradiction. Now we have proven that = 3. Let r be the number of pendant vertices in T. Since r(n k r3k = (n 1, we have r = k, and thus, T T(n,k1. By Theorem 3.7, we have T = C ( n, k, k. Theorem 4.. Let T be a tree with n vertices and k vertices of degree at least 3, where 0 k n 1. Then ρ(t ρ( C ( n, k, k with equality if and only if T = C ( n, k, k. Proof. If k = 0, then the result follows from Lemma 3.4. Suppose k 1. Let T be a tree with maximum distance spectral radius among trees with n vertices and k vertices of degree at least 3. Let = (T. Suppose 4. Let u V(T and N T (u = {u 1,...,u }. Let T i be the componentoft ucontainingu i, where1 i. Assumewithout lossofgenerality that σ T (T 1 σ T (T. Let w be a pendant vertex of T in V(T. Let T = T uu wu. Obviously, T is a tree with n vertices and k vertices of degree at least 3. By Lemma 3., ρ(t > ρ(t, a contradiction. Hence, 3, implying that T is a tree with k vertices of degree 3. By Theorem 4.1, T = C ( n, k, k. 5. Distance spectral radius of trees with all odd vertices. For n, let T(n be the set of all trees with n vertices, which are all odd. Let E n = C(n,0,n 1. Let T be a tree with n vertices. Then ρ(t ρ(s n with equality if and only if T = S n, see [9]. By this result and Theorem 3.7, we have Theorem 5.1. Let T T(n. Then ρ(s n ρ(t ρ(e n with left equality if and only if T = S n and right equality if and only if T = E n. For integers n and a with 1 a n, let D n,a be the double star obtained by adding an edge between the center u of S a1 and the center v of S n a 1. Lemma 5.. [8] For a, ρ(d n,a > ρ(d n,a 1. Theorem 5.3. Let T T(n and T S n, where n 3. Then ρ(t ρ(d n, with equality if and only if T = D n,. Proof. Let T be the tree with minimum distance spectral radius in T(n\{S n }. Let t be the diameter of T. Obviously, t 3. Let u 1...u t1 be a diametrical path of

19 The Distance Spectral Radius of Graphs 303 T. Suppose t 4. Let T = T {vu 4 : v N T (u 4 \{u 3 }}{vu 3 : v N T (u 4 \{u 3 }}. Obviously, T T(n\{S n }. By Lemma.4, ρ(t > ρ(t, a contradiction. Thus, t = 3, i.e., T is a double star. By Lemma 5., T = D n,. For n 3 and i n1, let B(n,i be the caterpillar obtained from the path P n with consecutive vertices u 1,...,u n by attaching a pendant vertex v j to u j for j n 1 with j i and attaching three pendant vertices w 1,w,w 3 to u i. For n 5 and 3 i n1, let F(n,i be the caterpillar obtained from the path P n with consecutive vertices u 1,...,u n by attaching a pendant vertex v j to u j for j n 1 with j i and adding an edge between u i and the center of S 3. Lemma 5.4. For n 5, and i n1, ρ(f(n,i1 > ρ(b(n,i. Proof. Let G = B(n,i and G = B(n,i {u i w,u i w 3 } {v i1 w,v i1 w 3 }. Obviously, G = F(n,i1. Let x = x(g. As we pass from G to G, the distance between a vertex of {w,w 3 } and a vertex of {u 1,...,u i,v,...,v i 1,w 1 } is increased by, the distance between a vertex of {w,w 3 } and v i1 is decreased by, and the distance between any other vertex pair remains unchanged. Hence, 1 (ρ(g ρ(g 1 x (D(G D(Gx From (.1 for G at u i, w 1 and v i1, we have = (x w x w3 ( x ui x w1 x vi1. i 1 ρ(gx ui = x w1 x w x w3 x vi1 d G (u i,u k x uk d G (u i,v k x vk i 1 k= n 1 k=i n k=i1 (d G (u i1,v k 1x vk, k=1 (d G (u i1,u k 1x uk i 1 ρ(gx w1 = x ui x w x w3 3x vi1 (d G (u i,u k 1x uk (d G (u i,v k 1x vk i 1 k= n 1 k=i (d G (u i1,v k x vk, k=1 n (d G (u i1,u k x uk k=i1 i 1 ρ(gx vi1 = x ui 3x w1 3x w 3x w3 (d G (u i,u k x uk k=1

20 304 H. Lin and B. Zhou and thus, (d G (u i,v k x vk i 1 k= n 1 k=i (d G (u i1,v k 1x vk, n (d G (u i1,u k 1x uk k=i1 (ρ(g ( i 1 x ui x w1 x vi1 = xui 3x vi1 (d G (u i,u k 1x uk i 1 > 0, k=1 (d G (u i,v k 1x vk k= n k=i1 n 1 k=i (d G (u i1,u k x uk (d G (u i1,v k x vk implying that x ui x w1 x vi1 > 0. Therefore, ρ(g > ρ(g, i.e., ρ(f(n,i1 > ρ(b(n,i. Theorem 5.5. Let T T(n and T E n, where n 3. (i For n = 3,4, ρ(t ρ(b(n, with equality if and only if T = B(n,; (ii For n 5, ρ(t ρ(f(n,3 with equality if and only if T = F(n,3. Proof. For n = 3,4, we have T(n = {E n,b(n,}, and thus, the result follows from Theorem 5.1. Suppose n 5. Let T be a tree with maximum distance spectral radius in T(n\{E n }. Let = (T. Suppose 7. Let u V(T and N T (u = {u 1,...,u }. Let T i be the componentoft ucontainingu i, where1 i. Assumewithout lossofgenerality that σ T (T 1 σ T (T. Let w be a pendant vertex of T in V(T. Let T = T {uu,uu 3 }{wu,wu 3 }. Then T T(n and T E n. By Lemma 3., ρ(t > ρ(t, a contradiction. Thus, = 3 or 5. Suppose = 5. By similar argument as above, there is exactly one vertex of degree 5. Let u 1 u t1 be a longest path passing the vertex of degree 5, say u i, where i t. Let v 1,v,v 3 be the other three neighbors of u i outside the above path. We claim that v 1,v,v 3 are pendant vertices. Otherwise, suppose that v 1 is not

21 The Distance Spectral Radius of Graphs 305 apendantvertex. LetT 1 andt bethecomponentsoft u i containingu i 1 andu i1, respectively. Assume without loss of generality that σ T (T 1 σ T (T. Let T = T {u i v,u i v 3 }{u t1 v,u t1 v 3 }. Obviously, T T(n and T E n. By Lemma 3., ρ(t > ρ(t, a contradiction. Since each vertex different from u i is of degree 1 or 3 in T, we have by Lemma 3.3 that T = B(n,i with i n1. Suppose 3 i n1. Let T = B(n,i {u i v,u i v 3 }{u v,u v 3 } if σ T (T 1 σ T (T, and T = B(n,i {u i v,u i v 3 }{u n 1 v,u n 1 v 3 } if σ T (T 1 > σ T (T. It is easily seen that T = B(n,. By Lemma 3., ρ(t = ρ(b(n,i < ρ(t = ρ(b(n,, a contradiction. Then i =, and thus, T = B(n,. Suppose = 3. ByLemma3.3, T = F(n,iwith3 i n1. ByLemma3.3, ρ(f(n,3 > ρ(f(n,4 > > ρ ( F ( n, n1. Thus, T = F(n,3. By Lemma 5.4, ρ(b(n, < ρ(f(n,3. Thus, T = F(n,3. Acknowledgment. The authors thank the referee for careful review and helpful comments. REFERENCES [1] M. Aouchiche and P. Hansen. Distance spectra of graphs: A survey. Linear Algebra Appl., 458: , 014. [] M. Edelberg, M.R. Garey, and R.L. Graham. On the distance matrix of a tree. Discrete Math., 14:3 39, [3] L. Gargano, M. Hammar, P. Hell, L. Stacho, and U. Vaccaro. Spanning spiders and light-splitting switches. Discrete Math., 85:83 95, 004. [4] R.L. Graham and L. Lovász. Distance matrix polynomials of trees. Adv. Math., 9:60 88, [5] R.L. Graham and H.O. Pollack. On the addressing problem for loop switching. Bell System Tech. J., 50: , [6] H. Minc. Nonnegative Matrices. John Wiley & Sons, New York, [7] H. Matsuda, K. Ozeki, and T. Yamashita. Spanning trees with a bounded number of branch vertices in a claw-free graph. Graphs Combin., 30:49 437, 014. [8] M. Nath and S. Paul. On the distance spectral radius of trees. Linear Multilinear Algebra, 61: , 013. [9] S.N. Ruzieh and D.L. Powers. The distance spectrum of the path P n and the first distance eigenvector of connected graphs. Linear Multilinear Algebra, 8:75 81, [10] Y. Wang and B. Zhou. On distance spectral radius of graphs. Linear Algebra Appl., 438: , 013. [11] R. Xing, B. Zhou, and F. Dong. The effect of a graft transformation on distance spectral radius. Linear Algebra Appl., 457:61 75, 014. [1] G. Yu, H. Jia, H. Zhang, and J. Shu. Some graft transformations and its applications on the distance spectral radius of a graph. Appl. Math. Lett., 5: , 01.