Ordering trees by their largest eigenvalues

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1 Linear Algebra and its Applications 370 (003) Ordering trees by their largest eigenvalues An Chang a,, Qiongxiang Huang b a Department of Mathematics, Fuzhou University, Fuzhou, Fujian 35000, China b Department of Mathematics, Xinjiang University Urumuqi, Xinjiang , China Received November 00; accepted 7 January 003 Submitted by R.A. Brualdi Abstract The set of trees with n vertices is denoted by T n. Hofmeister has determined the first five values of the largest eigenvalue of trees in T n and the corresponding trees for these values [Linear Algebra Appl. 60 (997) 43]. In other words, an order of the first five trees in T n by their largest eigenvalues has been given. Focus on the same purpose, we shall give a partition for trees in T n first in the this paper and then extend this order to the eighth tree. 003 Elsevier Inc. All rights reserved. AMS classification: 05C05; 05C50 Keywords: Tree; Characteristic polynomial; Eigenvalue; Upper bound. Introduction Let G be a graph with vertex set {v,v,...,v n }. Its adjacency matrix A(G) = (a ij ) is defined to be the n n matrix (a ij ),wherea ij = ifv i is adjacent to v j,anda ij = 0 otherwise. The characteristic polynomial of G is just det(λi A(G)), which is denoted by P(G,λ).SinceA(G) is a real symmetric matrix, all of its eigenvalues are real; We assume, without loss of generality, that they are ordered in decreasing order, i.e. λ (G) λ (G) λ 3 (G) λ n (G) and call them the eigenvalues of G. IfG is a tree, then λ i (G) = λ n i+ (G), The research is supported by the National Natural Science Foundation of China (No. 0705) and NSFFJEC (JA045). Corresponding author. address: anchang@pub6.fz.fj.cn (A. Chang) /03/$ - see front matter 003 Elsevier Inc. All rights reserved. doi:0.06/s (03)

2 76 A. Chang, Q. Huang / Linear Algebra and its Applications 370 (003) i =,,..., n/, sinceg is a bipartite graph. Throughout the paper, we always denote T n the set of trees on n vertices. Cvetković has indicated twelve directions in further investigations of graph spectra in [5]. One of these directions but on which less progress has been made is Classifying and ordering graphs. Up to now, many results on the eigenvalues of trees in T n have been obtained (see [ 3]). In particular, for the largest eigenvalue of a tree T in T n, a classical upper bound is the following. Theorem.. Let T be a tree in T n.then λ (T ) n and the equality holds if and only if T = Sn, the star with n vertices. Hofmeister has refined this result in [] and obtained the following results. Theorem. []. Let T be a tree in T n \ { Sn} and n 4.Then λ (T ) n (n + 6n + 3) and the equality holds if and only if T = Sn which is shown in Fig.. Theorem.3 []. Let T be a tree in T n \ { Sn n},s and n 4. Then λ (T ) n (n + 0n + 33) and the equality holds if and only if T = Sn 3 which is shown in Fig.. Theorem.4 []. Let T be a tree in T n \ { Sn,S n n},s3 and n 5. Then λ (T ) n (n + 8n + 4) and the equality holds if and only if T = Sn 4 which is shown in Fig.. Theorem.5 []. Let T be a tree in T n \ { Sn,S n,s3 n n},s4 and n 6.Then λ (T ) n (n + 0n + 9) and the equality holds if and only if T = Sn 5, or T = S8 5, which are shown in Fig.. Fig.. The second to fifth trees in T n.

3 A. Chang, Q. Huang / Linear Algebra and its Applications 370 (003) So from the above results, we know the ordering of the first five trees in T n by their largest eigenvalues. By introducing the definition of the edge-growing transformation of a tree in T n in this paper, we can give a partition for trees in T n,andthen determine the sixth to eighth trees in the above ordering.. Preliminaries We need some groundwork before giving our main results. First the following two results are often used to calculate the characteristic polynomials of trees. Lemma. [8]. Let G be the graph obtained by joining the vertex u of the graph G to the vertex v of the graph G by an edge. Then P(G,λ) = P(G, λ)p (G,λ) P(G \ u, λ)p (G \ v, λ). Lemma. [8]. Let v be a vertex of degree in the graph G and u be the vertex adjacent to v. Then P(G,λ) = λp (G \ v, λ) P(G\{u, v},λ). Lemma.3 [4]. Let G be a connected graph, and G be a proper subgraph of G. Then λ (G )<λ (G). Since the roots of the characteristic polynomial of a graph are real, we only consider polynomials with real roots in this paper. If f(x)is a polynomial in the variable x, the degree of f(x)is denoted by (f ), and the largest root of the equation f(x)= 0byλ (f ). Many of the discussions in the rest of the paper often involve comparing the largest root of a polynomial with that of another polynomial. The next result provides us a effective method to do this. Lemma.4 [9]. Let f(x), g(x) be two monic polynomials with real roots, and (f ) (g). Iff(x)= q(x)g(x) + r(x), where q(x) is also a monic polynomial, and (r) (g), λ (g) > λ (q). Then (i) when r(x) = 0, then λ (f ) = λ (g); (ii) when r(x) > 0 for any x satisfying x λ (g), then λ (f ) < λ (g); (iii) when r(λ (g)) < 0, then λ (f ) > λ (g). Definition.. Let T be a tree in T n,andn 3. Let e = uv be a nonpendent edge of T,andletT and T be the two components of T e, u T, v T. T 0 is the graph obtained from T in the following way. () Contract the edge e = uv (i.e. identify u of T with v of T ). () Add a pendent edge to the vertex u (= v).

4 78 A. Chang, Q. Huang / Linear Algebra and its Applications 370 (003) Fig.. The edge-growing transformation of a tree T. We call the procedures () and () the edge-growing transformation of T, or e.g.t of T for short (see Fig. ). If T is transformed into T 0 by one step of e.g.t, we denote this procedure by T T 0. Lemma.5 [6]. Let T be a tree with at least a nonpendent edge in T n, and n 3.If T T 0, then λ (T 0 )>λ (T ). One can see easily by Lemma.5 that for any tree T in T n, T can be transformed into the star Sn through carrying out e.g.t to T repeatedly. So Theorem. is a corollary of Lemma.5. It is evident that if there exists a nonpendent edge in a tree T in T n, one can always carry out one step of e.g.t to T. Furthermore we can give a partition for trees in T n in the following way. Let Tn i ={T T T n, and there exist exactly i nonpendent edges in T }. ThenT n = n 3 t=0 T n i. Obviously, the sets T 0 n and Tn n 3 contain only the star Sn and the tree P n, the path with n vertices, respectively. It is not difficult to see that for any tree T Tn i, i =,,...,n 3, one can transform T into Tn 0 by carrying out exactly i steps of e.g.t to T repeatedly. At the same time, we notice that the sets Tn and T n contain the following two kinds of trees as shown in Fig. 3, say Ti,j,T r,s,t, respectively, where i + j = n, i j n 3for Ti,j, and r + s + t = n 3, r s, t 0forT r,s,t. One can immediately find that trees Sn, S3 n are both in T n (S n = T,n 3, S3 n = T,n 4 ), and S4 n, S5 n in T n (S4 n = T,,n 5, S5 n = T,n 4,0 except for S5 8 = T 3,3 in the case n = 8). Also the set Tn 3 consists of the following two types of trees, say Tp,q,l,m and Tp,q,l,m, as shown in Fig. 4, where p + q + l + m = n 4, p 0, q,l,m andp + q + l + m = n 4, p,q 0, l,m. Two distinct edges in a graph G are independent if they are not adjacent in G. A matching in G is a set of pairwise independent edges in G, while a matching Fig. 3. Trees in T n and T n.

5 A. Chang, Q. Huang / Linear Algebra and its Applications 370 (003) Fig. 4. Trees in T 3 n. with maximum cardinality is called a maximum matching in G. IfM(G) is a maximum matching in G, the number of edges in M(G) is denoted by M(G) called the matching number of G. Lemma.6 [7]. Let T be a tree in T n, and suppose M(T) =i. Then λ (T ) (n i + + (n i + ) 4(n i + )) and the equality holds if and only if T = Ti which is shown in Fig. 5. We find that M(T) = for a tree T in Tn, M(T) =3 for a tree T in T n and M(T) =4for a tree T in Tn 3. So by Lemma.6, we have Corollary.. Let T be a tree in Tn 3.Then λ (T ) n (n 3 + 0n + 37) and the equality holds if and only if T = T4 = Tn 7,,,. Corollary.. Let T be a tree in T n, and T Tn,T T n.then λ (T ) n (n 3 + 0n + 37) and the equality holds if and only if T = Tn 7,,,. Proof. If T Tn 3, the result follows by Corollary.; If T T i n with i 4, then T can be transformed into a tree in Tn by carrying out at least one step of e.g.t to T. So by Lemma.5, we have λ (T ) < (n 3 + n 0n + 37)). The proof is completed. Fig. 5. Tree T i.

6 80 A. Chang, Q. Huang / Linear Algebra and its Applications 370 (003) Results Theorem 3.. The order of trees in Tn by their largest eigenvalues is as follows: λ T,n 3 >λ T,n 4 >λ T 3,n 5 >λ T 4,n 6 > >λ T n n,, where x, x denote the largest integer not greater than x and the smallest integer greater than x, respectively. Proof. By Lemma., it is easy to see that the characteristic polynomial of the tree Ti,j in T n is P ( Ti,j,λ) = λ n 4 [λ 4 (n )λ + ij ]. Thus λ ( T i,j ) = (n + (n ) 4ij ). Obviously, λ T i,j will decrease strictly when ij increases. So the result follows because the function f(x)= x(n x) is a strictly increasing function in [, n ] since ij = i(n i). The proof is completed. Theorem 3.. Let T be a tree in Tn \{S4 n,s5 n }, and n 8. Then λ (T ) γ, where γ is the largest root of the equation λ 6 (n )λ 4 + (3n 3)λ (n 6) = 0, and the equality holds if and only if T = T,,n 6. Proof. By Lemmas. and., it is not difficult to get the characteristic polynomial of the tree Tr,s,t in T n is P(T r,s,t,λ)= λn 6 [λ 6 (n )λ 4 + (rs + rt + st + r + s)λ rst]. We distinguish the following two cases. Case. r =, s 3andt /= 0. Thus n t + 7 8sincen 4 = s + t t + 3. The characteristic polynomials of trees T,,n 6 and T,s,t are P(T,,n 6,λ)= λ n 6 [λ 6 (n )λ 4 + (3n 3)λ (n 6)] and P(T,s,t,λ)= λn 6 [λ 6 (n )λ 4 + (s + t + st + )λ st], respectively. Let f (λ) = λ 6 (n )λ 4 + (s + t + st + )λ st, and g(λ) = λ 6 (n )λ 4 + (3n 3)λ (n 6). Obviously, when n 6, λ (T,,n 6 ), λ (T,s,t ) are the largest roots of the equations g(λ) = 0andf (λ) = 0, respectively. Moreover, f (λ) = g(λ) + (st + s n + 0)λ (n 6) st. Let r (λ) = (st + s n + 0)λ + (n 6) st. Then r (λ) = (st + s n + 0)λ.Since st + s n + 0 = n 4 t + (n 4 t)t n + 0, = (t )n t 5t + 6, = (t )(n t 6).

7 A. Chang, Q. Huang / Linear Algebra and its Applications 370 (003) When t, we have st + s n + 0 > 0 since n t + 7 >t+ 6. Hence, r (λ) > 0whenλ>0. So the function r (λ) is an increasing function in (0, + ). Moreover, r () = (st + s n + 0) + (n 6) st = s > 0. Thus we have r (λ) > 0whenλ. When t =, since s = n 4 t = n 5, we have r (λ) = (s n + 0)λ + (n 6) s, = (n 6) (n 5) = n 7 > 0 since n t + 7 = 8. Hence, when λ λ T,,n 6 (>) and n 8, we always have r (λ) > 0. Therefore, by Lemma.4, we have λ (T,s,t )<λ (T,,n 6 ) when n 8. Case. r, s. Thus n t + 7sincer + s + t = n 3 Let f (λ) = λ 6 (n )λ 4 + (rs + rt + st + r + s)λ rst.then f (λ) = g(λ) + (rs + rt + st + r + s 3n + 3)λ + (n 6) rst. Let r (λ) = (rs + rt + st + r + s 3n + 3)λ + (n 6) rst. Thenr (λ) = (rs + rt + st + r + s 3n + 3)λ.Since rs + rt + st + r + s 3n + 3 = rs + (n 3 t)t + (n 3 t) 3n + 3, = rs + (t )n t 4t + 0, = rs + (t )(n t 6). When t, we have rs + rt + st + r + s 3n + 3 rs > 0sincer, s, and n t + 7. Hence, r (λ) > 0whenλ>0. So the function r (λ) is an increasing function in (0, + ). SinceTr,s,t contains S t+, the star with t + vertices, as its proper subgraph, we know λ T r,s,t >λ S t+ = t by Lemma.3. But r ( t) =[rs + (t )(n t 6)]t + (n 6) rst ( =[t(t ) ) + ](n t 6) >0sincen t + 7. Thus, r (λ) > 0foranyλ λ T,,n 6 (> t).soby Lemma.4, we have λ T r,s,t <λ T,,n 6. When t = 0, we have rs + rt + st + r + s 3n + 3 = s(n 3 s) (n 6), = (s )n s 3s + 0, = (s )(n s 5) 0 since s, n s + 5. Hence, r (λ) 0whenλ>0. So the function r (λ) is a nondecreasing function in (0, + ). But r () = ( rs + r ) + s 3n (n 6) = rs > ( 0. Thus, ) r (λ) > 0foranyλ λ T,,n 6 (>). So by Lemma.4, we have λ T 0,s,t < λ T,,n 6.

8 8 A. Chang, Q. Huang / Linear Algebra and its Applications 370 (003) When t =, we have rs + rt + st + r + s 3n + 3 = s(n 4 s) (n 7), = (s )n s 4s + 5, = (s )(n s 5) >0 since s, n s + 6. Hence, r (λ) > 0whenλ>0. So the function r (λ) is an increasing function in (0, + ). But r () = (rs + r + s 3n + 3) + (n 6) rs = n 7 > 0 since n s Thus, r (λ) > 0foranyλ λ (T,,n 6 )(> ). So by Lemma.4, we have λ T,s,t <λ T,,n 6. Similar to the proof of Theorem 3., we can get the following result. Theorem 3.3. Let T be a tree in Tn \ { Sn 4,S5 n,,n 6},T, and n 8. Then λ (T ) γ, where γ is the largest root of the equation λ 6 (n )λ 4 + (3n 3)λ (n 5) = 0, and the equality holds if and only if T = T,n 5,. T n In fact, Theorems 3. and 3.3 give the third and fourth tree in the order of trees in by their largest eigenvalues. Lemma 3.. For trees T3,n 5,T,,n 6 and T n 7,,,, we have λ T 3,n 5 >λ T,,n 6, n 6, λ T,,n 6 >λ T n 7,,,, n 8. Proof. We have seen that when n 6, λ ( T 3,n 5 ), λ ( T,,n 6 ) are the largest roots of the equations λ 4 (n )λ + 3(n 5) = 0 and λ 6 (n )λ 4 + (3n 3)λ (n 6) = 0, respectively. Moreover, λ 6 (n )λ 4 + (3n 3)λ (n 6) = λ [λ 4 (n )λ + 3(n 5)]+λ (n 6). By Lemma.3, λ T 3,n 5 > n 5sinceT 3,n 5 contains the star with n 4 vertices as its proper subgraph. Hence, we have λ (n 6) >0foranyλ λ T 3,n 5 (> n 5). So the first inequality holds by Lemma.4. We can get the characteristic polynomial of Tn 7,,, is P ( Tn 7,,,,λ) = λ n 8 (λ [λ 4 )(n 3)λ + (n 7)] by Lemmas. and.. Obviously, when n 8, λ T n 7,,, is the largest root of the equation λ 4 (n 3)λ + (n 7) = 0. Since λ 6 (n )λ 4 + (3n 3)λ (n 6) = (λ )[λ 4 (n 3)λ + (n 7)] andλ T n 7,,, > whenn 8, the second inequality holds by Lemma.4.

9 A. Chang, Q. Huang / Linear Algebra and its Applications 370 (003) By comparing directly λ T 4,n 6 = (n + n 8n + 97)) with λ T n 7,,, = (n 3 + n 0n + 37)), and by Lemma 3. and Theorem 3., we can get the following result. Lemma 3.. Let T be a tree in T n \ { S n,s3 n,t 3,n 5}, and n. Thenλ (T ) < λ ( T n 7,,, ). Lemma 3.3. Let T be a tree in Tn \ { Sn 4,S5 n,,n 6},T, and n 0.Thenλ (T ) < λ T n 7,,,. Proof. By Lemma 3. and Theorem 3.3, the result follows by comparing λ T,n 5, with λ T n 7,,,. Since λ T,n 5, is the largest root of the equation λ 6 (n )λ 4 + (3n 3)λ (n 5) = 0whenn 6, and λ 6 (n )λ 4 + (3n 3)λ (n 5) = (λ )[λ 4 (n 3)λ + (n 7)]+(n 9), wehave the result by Lemma.4. The proof is completed. From the proof of Lemma 3.3, we find λ T,n 5, = λ T n 7,,, when n = 9. Denote T3,n 5, T,,n 6 and T n 7,,, by S6 n, S7 n and S8 n, respectively. We are now in a position to state our main results which follow immediately from Corollary. and Lemmas Theorem 3.4. Let T be a tree in T n \ { S n,s n,s3 n,s4 n,s5 n} and n. Then λ (T ) (n + n 4n + 6) and the equality holds if and only if T = S 6 n. Theorem 3.5. Let T be a tree in T n \ { S n,s n,s3 n,s4 n,s5 n,s6 n} and n. Then λ (T ) γ, where γ is the largest root of the equation λ 6 (n )λ 4 + (3n 3)λ (n 6) = 0, and the equality holds if and only if T = Sn 7. Theorem 3.6. Let T be a tree in T n \ { S n,s n,s3 n,s4 n,s5 n,s6 n,s7 n} and n.then λ (T ) (n 3 + n 0n + 37) and the equality holds if and only if T = S 8 n.

10 84 A. Chang, Q. Huang / Linear Algebra and its Applications 370 (003) Conclusion From the table of the spectra of all trees with n vertices ( n 0) in [8], we can easily determine the order of trees in T n by their largest eigenvalues. When n, combining Theorem with the results in [], we show the distribution of the first eight trees in the order of trees in T n by their largest eigenvalues in terms of the partition T n = n 3 t=0 T n i in the following table. Tn 0 Sn / / / / Tn Sn Sn 3 Sn 6 T4,n 6 Tn Sn 4 Sn 5 Sn 7 T,n 5, Tn 3 Sn 8 Tn 4. Acknowledgement The authors would like to thank the referee for giving a lot of comments on improving this paper. References [] M. Hofmeister, On the two largest eigenvalues of trees, Linear Algebra Appl. 60 (997) [] A. Neumaier, The second largest eigenvalue of a tree, Linear Algebra Appl. 46 (98) 9 5. [3] J.Y. Shao, Bounds on the kth eigenvalues of trees and forests, Linear Algebra Appl. 49 (99) [4] Q. Li, K.Q. Feng, On the largest eigenvalue of graphs, Acta Math. Appl. Sinica (979) (in Chinese). [5] D. Cvetković, Some possible directions in further investigations of graph spectra, in: Algebra Methods in Graph Theory, vol., North-Holland, Amsterdam, 98, pp [6] G.H. Xu, On the spectral radius of trees with perfect matchings, in: Combinatorics and Graph Theory, World Scientific, Singapore, 997. [7] J.M. Guo, S.W. Tan, On the spectral radius of trees, Linear Algebra Appl. 39 (00) 8. [8] D. Cvetković, M. Doob, H. Sachs, Spectra of Graph Theory and Applications, Academic Press, New York, 980. [9] An Chang, On the largest eigenvalue of a tree with perfect matchings, Discrete Mathematics, in press.

On the second Laplacian eigenvalues of trees of odd order

Linear Algebra and its Applications 419 2006) 475 485 www.elsevier.com/locate/laa On the second Laplacian eigenvalues of trees of odd order Jia-yu Shao, Li Zhang, Xi-ying Yuan Department of Applied Mathematics,