On the largest kth eigenvalues of trees with n ≡ 0 (modk)

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On the largest kth eigenvalues of trees with n ≡ 0 (modk)

174 Y. Zou et al. / Linear Algebra and its Applications 320 (2000) 173–182If T isatreeofordern, thenT is bipartite, and its eigenvalues satisfy the relationλ i (T ) =−λ n−i+1 (T ) (i = 1, 2,...,n). So, it suffices to study those eigenvaluesλ k (T ) for 1 k [n/2]. In this paper we always assume that 1 k [n/2].An interesting unsolved problem in the study of the spectra of trees is to find “thebest possible upper bound” for the kth eigenvalues of trees of order n.Inotherwords,letand letΓ n ={T | T is a tree of order n},¯λ k (n) = max{λ k (T ) | T ∈ Γ n }(1 k [n/2]).Then, the above problem asks to determine the function ¯λ k (n) and (if possible) findatreeT ∈ Γ n with λ k (T ) = ¯λ k (n).There have been considerable attempts in studying this problem, and the remainingunsolved case for ¯λ k (n) is the case n ≡ 0(modk), 7 k [n/2]. For this case,we write n = kt (t 2) and let¯Γ k,t ={T ∈ Γ kt | λ k (T ) = ¯λ k (kt)}.The trees in ¯Γ k,t are called the extremal trees.To be clear, we give the same definitions as those in [1] below.Definition 1. Let X k,t be the subset of trees in Γ kt which consists of k disjointstars S 1 ,...,S k of the order t(S 1∼ = S2 ∼ = ···∼ = Sk ∼ = K1,t−1 ) together with anotherk − 1 edges e 1 ,e 2 ,...,e k−1 such that the two end vertices of each edge e i (i =1, 2,...,k− 1) are noncentral vertices of different stars. We call S 1 ,...,S k the starsof this tree T ∈ X k,t , call the edges e 1 ,...,e k−1 the nonstar edges of T, and call theother edges the star edges of T.Definition 2. We define the condensed tree ̂T of T as V(̂T)= (S 1 ,S 2 ,...,S k ),andthere is an edge [S i ,S j ] (i /= j) in ̂T if and only if there exists some nonstar edge ofT with one end in S i and the other end in S j .Definition 3. Define X ′ k,tas the subset of X k,t which consists of those trees T inX k,t such that for any star S i of T, there is only one vertex in S i incident to somenonstar edges of T.A considerable necessary condition for extremal trees obtained in [1] is that if T ∈¯Γ k,t (k 2,t 5),thenT ∈ X k,t and Δ(̂T) 3, where Δ(̂T)is the maximal degreeof the condensed tree ̂T . In this paper, we establish a further necessary condition forextremal trees.


180 Y. Zou et al. / Linear Algebra and its Applications 320 (2000) 173–182and from (2.15)( )√t − 1 + λ 2 f 2cosπk+1 √ t − 1 + λ 2 (f 2 ). (3.9)Combining (3.6)–(3.8), we obtain (3.5). □Remark 1. We have also verified that (3.5) holds for k = 7, 8, 9. So, for k 7andt 4, if we denote byP k,t ={T ∈ X k,t | ̂T = P k },J k,t ={T ∈ X k,t | ̂T = J k },andL k,t ={T ∈ X k,t | ̂T = L k },then, from the previous results and Theorem 8, it suffices to find the extremal treesin P k,t ∪ J k,t ∪ L k,t .4. Some further discussionsIn this section, we establish some further results about the left problem of findingthe extremal trees in J k,t .Lemma 9. Let P k ,J k k 4 as in Fig. 2. We havei.e.λ i (J k ) = 2cos(2i − 1)π, i = 1, 2,...,k, (4.1)2k − 2λ i (J k ) = λ 2i−1 (P 2k−3 ). (4.2)Proof. From Fig. 2, we can write( ) A αA(J k ) =α T ,0where A is the adjacent matrix of P k−1 , α = (0, 1, 0,...,0) T .Let J k (λ) = det(λI − A(J k )). Then, we have the recursive relation as follows:J k (λ) = λJ k−1 (λ) − J k−2 (λ),since x 1,2 = λ ± √ λ 2 − 4/2 are the two roots of x 2 − λx + 1 = 0, we haveJ k (λ) = c 1 x1 k + c 2x2 k . (4.3)On the other hand,J 4 (λ) = λ 3 − 3λ 2 = c 1 x1 4 + c 2x2 4 , (4.4)


Y. Zou et al. / Linear Algebra and its Applications 320 (2000) 173–182 181J 5 (λ) = λ 5 − 4λ 3 + 2λ = c 1 x1 5 + c 2x2 5 . (4.5)Combining (4.4) and (4.5), we havec 1 = J 5(λ) − J 4 (λ)x 2x 5 1 − x4 1 x 2From (4.3) and (4.6), we haveJ k (λ) = J 5(λ) − J 4 (λ)x 2x k−4x 1 − x 2Let(2i − 1)πλ i = 2cos2k − 2 .Then the direct computations giveJ k (λ i ) = 0, i = 1, 2,...,k.Thus we obtain (4.1). Noticingλ i (P k ) = 2coswe have (4.2).□iπk + 1 ,, c 2 = J 5(λ) − J 4 (λ)x 1x2 5 − x4 2 x . (4.6)11+ J 5(λ) − J 4 (λ)x 1x 2 − x 1x k−42.Theorem 10. If there is no extremal tree in J k,t , then there is no extremal tree inJ s,t for k + 1 s 2k − 2.Proof. By Lemma 9, we haveλ 1 (J k ) = λ 1 (P 2k−3 ), λ 1 (J s )>λ 1 (J k ),andλ 1 (P s )>λ 1 (P k ), k + 1 s 2k − 2.So, from (2.15), we have√t − 1 + λ2 (f λ1 (J k ))


182 Y. Zou et al. / Linear Algebra and its Applications 320 (2000) 173–182References[1] Jia-yu Shao, On the largest kth eigenvalues of trees, Linear Algebra Appl. 221 (1995) 131–157.[2] D.M. Cretkvoic, M. Doob, H. Sachs, Spectra of Graphs, Academic, New York, 1980 (Appendix,Table 2).

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