- Email: [email protected]
On the multiplicity of eigenvalues of trees
Linear Algebra and its Applications 593 (2020) 180–187
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
On the multiplicity of eigenvalues of trees Dein Wong a,∗,1 , Qi Zhou a , Fenglei Tian b a
School of Mathematics, China University of Mining and Technology, Xuzhou 221116, China b School of Management, Qufu Normal University, Rizhao, 276826, Shandong, China
a r t i c l e
i n f o
Article history: Received 20 January 2020 Accepted 6 February 2020 Available online 7 February 2020 Submitted by R. Brualdi MSC: 05C50 Keywords: Multiplicity of an eigenvalue Nullity Tree
a b s t r a c t Let T be a tree of order n > 6 with μ = 1 as a positive eigenvalue of multiplicity k. Rowlinson [10] obtained a nice bound for k in terms of n, that is k ≤ n3 . It seems slightly imperfect that the bound n3 for k is not tight. Thus we are motivated to give a sharp upper bound for k in terms of n. Applying the theory of star set and star complement, we prove that k ≤ n−4 if μ2 is an integer at least 2, and the extremal 3 trees T satisfying k = n−4 are characterized completely. As 3 a corollary, we prove that if n ≥ 16, then k ≤ n−4 for an 3 arbitrary positive eigenvalue μ = 1. © 2020 Elsevier Inc. All rights reserved.
1. Introduction Throughout this paper we consider simple undirected graphs. For a graph G with vertex set V (G), the adjacency matrix of G is the matrix A(G) = (auv ), where auv = 1 if u, v are adjacent in G, and 0 otherwise. The eigenvalues of A(G) are called eigenvalues of G. If μ is an eigenvalue of G of multiplicity k, then a star set for μ in G is a subset * Corresponding author. 1
E-mail address: [email protected] (D. Wong). Supported by “the Fundamental Research Funds for the Central Universities (No. 2018ZDPY06)”.
https://doi.org/10.1016/j.laa.2020.02.004 0024-3795/© 2020 Elsevier Inc. All rights reserved.
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181
X of V (G) such that |X| = k and the induced subgraph G − X does not have μ as an eigenvalue. The induced subgraph G − X is called a star complement for μ in G. The theory on star complements originated independently in papers by Ellingham [6] and Rowlinson [9]. The theory of star complement has been used to study graphs with least eigenvalue −2 ([4], [5]), structure properties of strongly regular graphs ([13], [14]) and the multiplicities of graph eigenvalues, not only for trees but also for graphs that are regular, cubic, quartic, triangle-free, etc. (see [1], [2], [7]-[8], [10–12]). Denote by T a tree of order n > 6 with μ as a positive eigenvalue of multiplicity k. The motivation of this article comes mainly from [10] and [1]. Rowlinson (see [10], Theorem n−4 2.2) proved that k ≤ n3 if μ = 1; k ≤ n−3 2 if μ = 0 and n is odd; k ≤ 2 if μ = 0 and n is even. Bu et al. ([1], Theorem 3.5) pursued this topic and they proved that if the length of jπ each proper pendant path in T is at least s and μ ∈ / {2cos i+1 : j = 1, . . . , i, i = s, s + 1}, n then k ≤ s+2 . However, the bound n3 , obtained by Rowlinson [10], is not tight if μ = 1. Thus we are motivated to give a sharp upper bound for k in terms of n for the case μ = 1. We will prove that if μ2 is an integer at least 2 then k ≤ n−4 3 , and the equality holds if and only if T is a tree obtained from k + 1 copies of disjoint P3 by adding a vertex and k + 1 edges joining the vertex with exactly one arbitrarily chosen vertex of each P3 . As a corollary, we prove that if n > 16, then k ≤ n−4 3 for an arbitrary positive eigenvalue μ = 1. 2. Proof of the main result Firstly, we recall some properties on star sets and star complements from [10]. (i) Star sets and star complements exist for any eigenvalue of any graph. (ii) Suppose that G has μ as an eigenvalue of multiplicity k. If X is a star set for μ in G and if S is a proper subset of X then G − S has μ as an eigenvalue of multiplicity k − |S|. (iii) Let V (G) = {1, 2, . . . , n}, and let A be the adjacency matrix of G. Let P be the matrix which represents the orthogonal projection of Rn onto the eigenspace EA (μ) with respect to the standard orthonormal basis {e1 , e2 , . . . , en } of Rn . Then the subset X of V (G) is a star set for μ in G if and only if the vectors P ei , i ∈ X, form a basis for EA (μ). (iv) The matrix P of (iii) is a polynomial in A and μP ev = u∼v P eu . (v) If μ is an eigenvalue of the connected graph G then G has a connected star complement for μ. (vi) If u and v are adjacent vertices in a star set for G, then the edge uv is not a bridge of G. For bipartite graphs, if the square of an eigenvalue μ is not an integer then μ has an algebraic conjugate μ∗ such that μ, −μ, μ∗ , −μ∗ are distinct eigenvalues of multiplicity k, and so there is an upper bound for k, i.e., k ≤ n4√ . Thus we only focus on those eigenvalues μ of trees for which μ2 is an integer. Let S = { k : 1 = k ∈ Z+ }, the set of square roots
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Fig. 1. Graph Y6 .
of all positive integers other than 1. For trees of order n ∈ [3, 6], it is easy to see which ones have an eigenvalue in S (see [3]). Observation 2.1. Let T be a tree of order n ∈ [3, 6]. If T has an eigenvalue μ in S, then μ and T are one of the following cases and μ is a simple eigenvalue of T . √ (1) μ = 2 and T is P3 . √ (2) μ = 3 and T is K1,3 or P5 . √ (3) μ = 5 and T is K1,5 . (4) μ = 2 and T is K1,4 or Y6 , where Y6 is the unique tree of order 6 with two vertices of degree 3 (see Fig. 1). For a star set X of a tree with μ ∈ S we have the following property. Lemma 2.2. Let T be a tree obtained from two vertex-disjoint trees H and K by identifying a vertex v of H with a vertex of K. Suppose that a pendant vertex x of H lies in a μ-star set X of T . If H has order n ∈ [3, 6], μ ∈ S and μ is an eigenvalue of H, then X = (X \ x) ∪ {v} is also a star set of T . Proof. In view of the observation, μ and H are one of the cases listed above. To complete the proof it suffices to prove that P ev is a nonzero multiple of P ex . Here, we only give a detailed proof for the case when H = Y6 and μ = 2. The proofs for the other cases are easier and are therefore omitted. Label the vertices of Y6 as in Fig. 1. The following cases are considered according to which vertex of Y6 is v. Case 1. v = y. For this case, P ev = 2P ex and we are done. Case 2. v = y1 . For this case, by P ey = 2Px = 2Px1 and 2P ey = P ex + P ex1 + P ey1 we have P ey1 = 2P ex , as desired. Case 3. v = x2 or v = x3 . For the case when v = x2 , from ⎫ ⎧ 2P ex = 2P ex1 = P ey ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ 2P ey = P ex + P ex + P ey ⎪ 1 1 ⎪ ⎪ ⎪ 2P ey1 = P ey + P ex2 + P ex3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 2P ex3 = P ey1 , it follows that P ev = P ex , as desired. The case when v = x3 is analogous and thus omitted.
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Case 4. v = x1 . We have P ev = P ex by the following equalities. ⎧ ⎪ ⎪ ⎪ ⎨
⎫ ⎪ 2P ex = P ey ⎪ ⎪ 2P ey = P ex + P ex1 + P ey1 ⎬ . ⎪ 2P ey1 = P ey + P ex2 + P ex3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 2P e = 2P e = P e ⎭ x3 x2 y1 In what follows, we denote by mG (μ) the multiplicity of μ as an eigenvalue of a graph G. Lemma 2.3. Let G be a graph obtained from a given graph H and P3 , disjoint with H, by √ √ adding an edge joining a vertex of P3 with a vertex z of H. Then mG ( 2) = mH−z ( 2). Proof. Arrange the vertices of P3 as P3 : v ∼ u ∼ w. √ If the vertex w is adjacent to z in G, 2I − A(G) can be arranged in the form as ⎛√
2 −1 0 ⎜ −1 √2 −1 ⎜ √ √ ⎜ 2I − A(G) = ⎜ 0 −1 2 ⎜ ⎝ 0 0 −1 0 0 0
0 0 −1 √ 2 −α
⎞ 0 ⎟ 0 ⎟ ⎟ 0 ⎟, ⎟ −α ⎠ √ 2I − C
where the first, the second, the third and the fourth row respectively corresponds to v, u, w and z in G, α means the transpose of α and C is the adjacency matrix of H − z. Let ⎛
⎞ 1 0 0 0 −α √ ⎜ 0 1 0 0 − 2α ⎟ ⎜ ⎟ ⎜ ⎟ Q = ⎜0 0 1 0 −α ⎟ . ⎜ ⎟ ⎝0 0 0 1 0 ⎠ 0 0 0 0 I Then ⎛√
2 ⎜ −1 ⎜ √ ⎜ Q ( 2I − A(G))Q = ⎜ 0 ⎜ ⎝ 0 0
−1 0 0 √ 2 −1 0 √ −1 2 −1 √ 0 −1 2 0 0 0
⎞ 0 ⎟ 0 ⎟ ⎟ 0 ⎟. ⎟ ⎠ 0 √ 2I − C
√ Since the rank of the left-upper principal submatrix of order 4 in Q ( 2I − A(G))Q is √ √ √ √ 4, we have rk( 2I − A(G)) = 4 + rk( 2I − C). Hence, mG ( 2) = mH−z ( 2).
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Fig. 2. A graph in T (P3 , 4).
If the center vertex u of P3 is adjacent to z in G, ⎛√ 2 0 ⎜ 0 √2 ⎜ √ ⎜ 2I − A(G) = ⎜ −1 −1 ⎜ ⎝ 0 0 0 0
√
2I − A(G) can be written as
−1 0 −1 0 √ 2 −1 √ 2 −1 0 −β
⎞ 0 ⎟ 0 ⎟ ⎟ 0 ⎟, ⎟ −β ⎠ √ 2I − C
where the first, the second, the third and the fourth row respectively corresponds to v, w, u and z in G, and C is the adjacency matrix of H − z. Let ⎛
1 ⎜ ⎜0 ⎜ Q1 = ⎜ 0 ⎜ ⎝0 0
0 1 0 0 0
0 0 1 0 0
√ ⎞ 0 − √22 β ⎟ 0 − 22 β ⎟ ⎟ 0 −β ⎟ . ⎟ 1 0 ⎠ 0 I
Then ⎛√ 2 0 −1 ⎜ 0 √2 −1 ⎜ √ √ ⎜ QT1 ( 2I − A(G))Q1 = ⎜ −1 −1 2 ⎜ ⎝ 0 0 −1 0 0 0
⎞ 0 0 ⎟ 0 0 ⎟ ⎟ −1 0 ⎟. √ ⎟ ⎠ 2 0 √ 0 2I − C
√ The left-upper principal submatrix of order 4 also has rank 4, thus mG ( 2) = √ mH−z ( 2). Let T (P3 , l) with l ≥ 2 the set of all trees obtained from l copies of disjoint P3 by adding a vertex v and l edges joining the vertex with exactly one arbitrarily chosen vertex of each P3 . The added vertex v is called a P3 -dominating vertex of T (P3 , l). See Fig. 2 for a graph in T (P3 , 4), where the black circle refers to the P3 -dominating vertex which is adjacent to two pendent vertices of two copies of P3 and adjacent to the central vertices of another two copies of P3 . Theorem 2.4. Let T be a tree of order n > 6 with μ ∈ S as an eigenvalue of multiplicity √ k. Then k ≤ n−4 2. 3 , with equality if and only if T ∈ T (P3 , k + 1) and μ =
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Proof. If T ∈ T (P3 , k + 1) with k ≥ 1, deleting a P3 together with the P3 -dominating vertex of T (P3 , k + 1) from T , the remaining subgraph consists of k disjoint copies of P3 , √ √ each of which has 2 as a simple eigenvalue. By Lemma 2.3, we have mT ( 2) = k. We proceed by induction on the order of T to prove the remaining assertion. If n = 7, √ then by direct computation one can see easily that μ = 2, k = 1 and T ∈ T (P3 , 2). Suppose that the assertion holds for any tree of order m ≥ 7 and consider the case when a tree T has order n > m. Let X be a star set for μ such that T − X is connected. Then the vertices in X are pendant vertices (see [10]). Let x be a vertex in X with neighbor x . We consider two cases for x . Case 1. x has degree 2. Let y be the other neighbor of x . Then the induced subgraph with vertices x, x , y is P3 and T can be viewed as a tree obtained by identifying a vertex of P3 , with a vertex of the subtree T − x − x of T . By μP ex = P ex and μP ex = P ex + P ey we have P ey = (μ2 − 1)P ex . Note that both x and y are not in X (since P ex and P ey are nonzero multiple of P ex ), thus x can be replaced by y to form a new star set, namely X = (X \ {x}) ∪ {y}, for μ By property (ii) of star sets, we know T − y has μ as an eigenvalue of multiplicity k − 1. Let T = T − P3 . Then mT (μ) = mT −y (μ) = k − 1. If a component H1 of T (when it exists) has order 1 or 2, then μ is not an eigenvalue of H1 . If a component H2 of T (when it exists) has order in [3, 6], then μ is also not an eigenvalue of H2 . Indeed, if μ is an eigenvalue of H2 , then some pendant vertex of H2 , say z, must lie in X (otherwise, T − X has a component H2 such that det(μI − A(H2 )) = 0, which contradicts to the fact μ is not an eigenvalue of T − X ). By Lemma 2.2, X = (X \ {z}) ∪ {v} is also a μ-star set of T , where v is the vertex of H2 adjacent to y. Thus a bridge yv is in the star set X , a contradiction to (vi) listed above. If a component H3 of T (when it exists) has order at least 7, then the induction hypothesis says mH3 (μ) ≤ |H33|−4 . Now, if two components of T have order at least 7, then we have mT (μ) = 1 + mT (μ) mH1 (μ) + mH2 (μ) + mH3 (μ) =1+ H1
=1+
H3
H2
mH3 (μ) ≤ 1 +
H3
|H3 | − 4 H3
3
,
where H1 (resp., H2 , H3 ) runs through all components of T of order in [1, 2] (resp., in [3, 6], in [7, ∞)) and |Hi | means the order of Hi . Since n ≥ 3 + H3 |H3 |, we have k ≤ n−8 3 . In this case, equality fails to hold.
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If exactly one component, say H, of T has order at least 7, then we have mT (μ) = 1 + mT (μ) = 1 + mH (μ) |H| − 4 3 n−4 . ≤ 3 ≤1+
If mT (μ) = n−4 = H and mH (μ) = |H|−4 3 , then T √ 3 . The induction hypothesis applied to H implies that μ = 2 and H ∈ T (P3 , k). We claim that y is adjacent to the P3 -dominating vertex of T (P3 , k) (otherwise, by applying Lemma 2.3 we have mT (μ) = k − 1). In this case we have T ∈ T (P3 , k + 1), as required. If all components of T have orders at most 6, then mT (μ) = 0 and thus k = 1. Since n ≥ 7, we have k ≤ n−4 3 . If equality holds then n = 7, a contradiction to the assumption that n > 7.
Case 2. x has degree at least 3. Let T = T − x − x . Then T has at least two components. Since μP ex = P ex , we can replace x with x to form a new star set X = (X \ {x}) ∪ {x }. If two components of T have order at least 7, then by an argument analogous to that in case 1 we have mT (μ) = 1 + mT −x (μ) = 1 + mT (μ) ≤ 1 +
n−7 n − 10 = . 3 3
In this case, equality again fails to hold. If exactly one component of T , say H, has order at least 7, since mH (μ) = 0 for H of order at most 6 and mH (μ) ≤ |H|−4 (by induction hypothesis), we have 3 mT (μ) = 1 + mT (μ) = 1 + mH (μ) ≤ 1 +
|H| − 4 . 3
|H|−4 Recalling that n ≥ |H| + 3, we have k ≤ n−4 3 . If equality holds, then mH (μ) = 3 and T has just two components, one is H and the other component contains only one vertex, say z. Thus T is obtained from H and P3 (induced by x, x , z), by adding an edge joining x with a vertex of H. The induction hypothesis applied to H says that H ∈ T (P3 , k). We claim that x is adjacent to the P3 -dominating vertex of H and thus T ∈ T (P3 , k + 1). Indeed, if x is not adjacent to the P3 -dominating vertex of H, then by applying Lemma 2.3, we have mT (μ) = k − 1, a contradiction. If all components of T have order at most 6, then mT (μ) = 0 and thus k = 1. Also n−4 we have 1 = k ≤ n−4 3 . If 1 = k = 3 , then n = 7, a contradiction to the assumption that n > 7.
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Corollary 2.5. Let T be a tree of order n ≥ 16 and with μ = 1 as a positive eigenvalue of multiplicity k. Then k ≤ n−4 3 . Proof. If μ2 is an integer, the inequality has been proved in Theorem 2.4. If μ2 is not an n−4 integer then k ≤ n4 (see [10]). Since n4 ≤ n−4 3 , we have k ≤ 3 , as required. Declaration of competing interest There is no competing interest. Acknowledgement The authors would like to express their sincere gratitude to the referee for a very careful reading of the paper and for all his or her insightful comments and valuable suggestions, which make a number of improvements on this paper. References [1] C. Bu, X. Zhang, J. Zhou, A note on the multiplicities of graph eigenvalues, Linear Algebra Appl. 442 (2014) 69–74. [2] F.K. Bell, P. Rowlinson, On the multiplicities of graph eigenvalues, Bull. Lond. Math. Soc. 35 (2003) 401–408. [3] D. Cvetković, M. Doob, H. Sachs, Spectra of Graphs, 3rd edn., Johann Ambrosius Barth, Heidelberg, 1995. [4] D. Cvetković, M. Lepović, P. Rowlinson, S. Simić, The maximal exceptional graphs, J. Comb. Theory, Ser. B 86 (2002) 347–363. [5] D. Cvetković, P. Rowlinson, S. Simić, Graphs with least eigenvalue −2: the star complement technique, J. Algebraic Comb. 14 (2001) 5–16. [6] M.N. Ellingham, Basic subgraphs and graph spectra, Australas. J. Comb. 8 (1993) 247–265. [7] C.M. da Fonseca, A note on the multiplicities of the eigenvalues of a graph, Linear Multilinear Algebra 53 (2005) 303–307. [8] C.M. da Fonseca, A lower bound for the number of distinct eigenvalues of some real symmetric matrices, Electron. J. Linear Algebra 21 (2010) 3–11. [9] P. Rowlinson, Eutactic stars and graph spectra, in: R.A. Brualdi, S. Friedland, V. Klee (Eds.), Combinatorial and Graph-Theoretical Problems in Linear Algebra, Springer-Verlag, New York, 1993, pp. 153–164. [10] P. Rowlinson, On multiple eigenvalues of trees, Linear Algebra Appl. 432 (2010) 3007–3011. [11] P. Rowlinson, On eigenvalue multiplicity and the girth of a graph, Linear Algebra Appl. 435 (2011) 2375–2381. [12] P. Rowlinson, On graph with an eigenvalue of maximal multiplicity, Discrete Math. 313 (2013) 1162–1166. [13] P. Rowlinson, I. Sciriha, Some properties of the Hoffman–Singleton graph, Appl. Anal. Discrete Math. 1 (2007) 438–445. xević, A spectral proof of the uniqueness of a strongly regular graph with [14] D. Stevanović, M. Miloˇ parameters (81, 20, 1, 6), Eur. J. Comb. 30 (2009) 957–968.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
On the multiplicity of eigenvalues of trees Dein Wong a,∗,1 , Qi Zhou a , Fenglei Tian b a
School of Mathematics, China University of Mining and Technology, Xuzhou 221116, China b School of Management, Qufu Normal University, Rizhao, 276826, Shandong, China
a r t i c l e
i n f o
Article history: Received 20 January 2020 Accepted 6 February 2020 Available online 7 February 2020 Submitted by R. Brualdi MSC: 05C50 Keywords: Multiplicity of an eigenvalue Nullity Tree
a b s t r a c t Let T be a tree of order n > 6 with μ = 1 as a positive eigenvalue of multiplicity k. Rowlinson [10] obtained a nice bound for k in terms of n, that is k ≤ n3 . It seems slightly imperfect that the bound n3 for k is not tight. Thus we are motivated to give a sharp upper bound for k in terms of n. Applying the theory of star set and star complement, we prove that k ≤ n−4 if μ2 is an integer at least 2, and the extremal 3 trees T satisfying k = n−4 are characterized completely. As 3 a corollary, we prove that if n ≥ 16, then k ≤ n−4 for an 3 arbitrary positive eigenvalue μ = 1. © 2020 Elsevier Inc. All rights reserved.
1. Introduction Throughout this paper we consider simple undirected graphs. For a graph G with vertex set V (G), the adjacency matrix of G is the matrix A(G) = (auv ), where auv = 1 if u, v are adjacent in G, and 0 otherwise. The eigenvalues of A(G) are called eigenvalues of G. If μ is an eigenvalue of G of multiplicity k, then a star set for μ in G is a subset * Corresponding author. 1
E-mail address: [email protected] (D. Wong). Supported by “the Fundamental Research Funds for the Central Universities (No. 2018ZDPY06)”.
https://doi.org/10.1016/j.laa.2020.02.004 0024-3795/© 2020 Elsevier Inc. All rights reserved.
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181
X of V (G) such that |X| = k and the induced subgraph G − X does not have μ as an eigenvalue. The induced subgraph G − X is called a star complement for μ in G. The theory on star complements originated independently in papers by Ellingham [6] and Rowlinson [9]. The theory of star complement has been used to study graphs with least eigenvalue −2 ([4], [5]), structure properties of strongly regular graphs ([13], [14]) and the multiplicities of graph eigenvalues, not only for trees but also for graphs that are regular, cubic, quartic, triangle-free, etc. (see [1], [2], [7]-[8], [10–12]). Denote by T a tree of order n > 6 with μ as a positive eigenvalue of multiplicity k. The motivation of this article comes mainly from [10] and [1]. Rowlinson (see [10], Theorem n−4 2.2) proved that k ≤ n3 if μ = 1; k ≤ n−3 2 if μ = 0 and n is odd; k ≤ 2 if μ = 0 and n is even. Bu et al. ([1], Theorem 3.5) pursued this topic and they proved that if the length of jπ each proper pendant path in T is at least s and μ ∈ / {2cos i+1 : j = 1, . . . , i, i = s, s + 1}, n then k ≤ s+2 . However, the bound n3 , obtained by Rowlinson [10], is not tight if μ = 1. Thus we are motivated to give a sharp upper bound for k in terms of n for the case μ = 1. We will prove that if μ2 is an integer at least 2 then k ≤ n−4 3 , and the equality holds if and only if T is a tree obtained from k + 1 copies of disjoint P3 by adding a vertex and k + 1 edges joining the vertex with exactly one arbitrarily chosen vertex of each P3 . As a corollary, we prove that if n > 16, then k ≤ n−4 3 for an arbitrary positive eigenvalue μ = 1. 2. Proof of the main result Firstly, we recall some properties on star sets and star complements from [10]. (i) Star sets and star complements exist for any eigenvalue of any graph. (ii) Suppose that G has μ as an eigenvalue of multiplicity k. If X is a star set for μ in G and if S is a proper subset of X then G − S has μ as an eigenvalue of multiplicity k − |S|. (iii) Let V (G) = {1, 2, . . . , n}, and let A be the adjacency matrix of G. Let P be the matrix which represents the orthogonal projection of Rn onto the eigenspace EA (μ) with respect to the standard orthonormal basis {e1 , e2 , . . . , en } of Rn . Then the subset X of V (G) is a star set for μ in G if and only if the vectors P ei , i ∈ X, form a basis for EA (μ). (iv) The matrix P of (iii) is a polynomial in A and μP ev = u∼v P eu . (v) If μ is an eigenvalue of the connected graph G then G has a connected star complement for μ. (vi) If u and v are adjacent vertices in a star set for G, then the edge uv is not a bridge of G. For bipartite graphs, if the square of an eigenvalue μ is not an integer then μ has an algebraic conjugate μ∗ such that μ, −μ, μ∗ , −μ∗ are distinct eigenvalues of multiplicity k, and so there is an upper bound for k, i.e., k ≤ n4√ . Thus we only focus on those eigenvalues μ of trees for which μ2 is an integer. Let S = { k : 1 = k ∈ Z+ }, the set of square roots
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Fig. 1. Graph Y6 .
of all positive integers other than 1. For trees of order n ∈ [3, 6], it is easy to see which ones have an eigenvalue in S (see [3]). Observation 2.1. Let T be a tree of order n ∈ [3, 6]. If T has an eigenvalue μ in S, then μ and T are one of the following cases and μ is a simple eigenvalue of T . √ (1) μ = 2 and T is P3 . √ (2) μ = 3 and T is K1,3 or P5 . √ (3) μ = 5 and T is K1,5 . (4) μ = 2 and T is K1,4 or Y6 , where Y6 is the unique tree of order 6 with two vertices of degree 3 (see Fig. 1). For a star set X of a tree with μ ∈ S we have the following property. Lemma 2.2. Let T be a tree obtained from two vertex-disjoint trees H and K by identifying a vertex v of H with a vertex of K. Suppose that a pendant vertex x of H lies in a μ-star set X of T . If H has order n ∈ [3, 6], μ ∈ S and μ is an eigenvalue of H, then X = (X \ x) ∪ {v} is also a star set of T . Proof. In view of the observation, μ and H are one of the cases listed above. To complete the proof it suffices to prove that P ev is a nonzero multiple of P ex . Here, we only give a detailed proof for the case when H = Y6 and μ = 2. The proofs for the other cases are easier and are therefore omitted. Label the vertices of Y6 as in Fig. 1. The following cases are considered according to which vertex of Y6 is v. Case 1. v = y. For this case, P ev = 2P ex and we are done. Case 2. v = y1 . For this case, by P ey = 2Px = 2Px1 and 2P ey = P ex + P ex1 + P ey1 we have P ey1 = 2P ex , as desired. Case 3. v = x2 or v = x3 . For the case when v = x2 , from ⎫ ⎧ 2P ex = 2P ex1 = P ey ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ 2P ey = P ex + P ex + P ey ⎪ 1 1 ⎪ ⎪ ⎪ 2P ey1 = P ey + P ex2 + P ex3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 2P ex3 = P ey1 , it follows that P ev = P ex , as desired. The case when v = x3 is analogous and thus omitted.
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Case 4. v = x1 . We have P ev = P ex by the following equalities. ⎧ ⎪ ⎪ ⎪ ⎨
⎫ ⎪ 2P ex = P ey ⎪ ⎪ 2P ey = P ex + P ex1 + P ey1 ⎬ . ⎪ 2P ey1 = P ey + P ex2 + P ex3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 2P e = 2P e = P e ⎭ x3 x2 y1 In what follows, we denote by mG (μ) the multiplicity of μ as an eigenvalue of a graph G. Lemma 2.3. Let G be a graph obtained from a given graph H and P3 , disjoint with H, by √ √ adding an edge joining a vertex of P3 with a vertex z of H. Then mG ( 2) = mH−z ( 2). Proof. Arrange the vertices of P3 as P3 : v ∼ u ∼ w. √ If the vertex w is adjacent to z in G, 2I − A(G) can be arranged in the form as ⎛√
2 −1 0 ⎜ −1 √2 −1 ⎜ √ √ ⎜ 2I − A(G) = ⎜ 0 −1 2 ⎜ ⎝ 0 0 −1 0 0 0
0 0 −1 √ 2 −α
⎞ 0 ⎟ 0 ⎟ ⎟ 0 ⎟, ⎟ −α ⎠ √ 2I − C
where the first, the second, the third and the fourth row respectively corresponds to v, u, w and z in G, α means the transpose of α and C is the adjacency matrix of H − z. Let ⎛
⎞ 1 0 0 0 −α √ ⎜ 0 1 0 0 − 2α ⎟ ⎜ ⎟ ⎜ ⎟ Q = ⎜0 0 1 0 −α ⎟ . ⎜ ⎟ ⎝0 0 0 1 0 ⎠ 0 0 0 0 I Then ⎛√
2 ⎜ −1 ⎜ √ ⎜ Q ( 2I − A(G))Q = ⎜ 0 ⎜ ⎝ 0 0
−1 0 0 √ 2 −1 0 √ −1 2 −1 √ 0 −1 2 0 0 0
⎞ 0 ⎟ 0 ⎟ ⎟ 0 ⎟. ⎟ ⎠ 0 √ 2I − C
√ Since the rank of the left-upper principal submatrix of order 4 in Q ( 2I − A(G))Q is √ √ √ √ 4, we have rk( 2I − A(G)) = 4 + rk( 2I − C). Hence, mG ( 2) = mH−z ( 2).
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Fig. 2. A graph in T (P3 , 4).
If the center vertex u of P3 is adjacent to z in G, ⎛√ 2 0 ⎜ 0 √2 ⎜ √ ⎜ 2I − A(G) = ⎜ −1 −1 ⎜ ⎝ 0 0 0 0
√
2I − A(G) can be written as
−1 0 −1 0 √ 2 −1 √ 2 −1 0 −β
⎞ 0 ⎟ 0 ⎟ ⎟ 0 ⎟, ⎟ −β ⎠ √ 2I − C
where the first, the second, the third and the fourth row respectively corresponds to v, w, u and z in G, and C is the adjacency matrix of H − z. Let ⎛
1 ⎜ ⎜0 ⎜ Q1 = ⎜ 0 ⎜ ⎝0 0
0 1 0 0 0
0 0 1 0 0
√ ⎞ 0 − √22 β ⎟ 0 − 22 β ⎟ ⎟ 0 −β ⎟ . ⎟ 1 0 ⎠ 0 I
Then ⎛√ 2 0 −1 ⎜ 0 √2 −1 ⎜ √ √ ⎜ QT1 ( 2I − A(G))Q1 = ⎜ −1 −1 2 ⎜ ⎝ 0 0 −1 0 0 0
⎞ 0 0 ⎟ 0 0 ⎟ ⎟ −1 0 ⎟. √ ⎟ ⎠ 2 0 √ 0 2I − C
√ The left-upper principal submatrix of order 4 also has rank 4, thus mG ( 2) = √ mH−z ( 2). Let T (P3 , l) with l ≥ 2 the set of all trees obtained from l copies of disjoint P3 by adding a vertex v and l edges joining the vertex with exactly one arbitrarily chosen vertex of each P3 . The added vertex v is called a P3 -dominating vertex of T (P3 , l). See Fig. 2 for a graph in T (P3 , 4), where the black circle refers to the P3 -dominating vertex which is adjacent to two pendent vertices of two copies of P3 and adjacent to the central vertices of another two copies of P3 . Theorem 2.4. Let T be a tree of order n > 6 with μ ∈ S as an eigenvalue of multiplicity √ k. Then k ≤ n−4 2. 3 , with equality if and only if T ∈ T (P3 , k + 1) and μ =
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Proof. If T ∈ T (P3 , k + 1) with k ≥ 1, deleting a P3 together with the P3 -dominating vertex of T (P3 , k + 1) from T , the remaining subgraph consists of k disjoint copies of P3 , √ √ each of which has 2 as a simple eigenvalue. By Lemma 2.3, we have mT ( 2) = k. We proceed by induction on the order of T to prove the remaining assertion. If n = 7, √ then by direct computation one can see easily that μ = 2, k = 1 and T ∈ T (P3 , 2). Suppose that the assertion holds for any tree of order m ≥ 7 and consider the case when a tree T has order n > m. Let X be a star set for μ such that T − X is connected. Then the vertices in X are pendant vertices (see [10]). Let x be a vertex in X with neighbor x . We consider two cases for x . Case 1. x has degree 2. Let y be the other neighbor of x . Then the induced subgraph with vertices x, x , y is P3 and T can be viewed as a tree obtained by identifying a vertex of P3 , with a vertex of the subtree T − x − x of T . By μP ex = P ex and μP ex = P ex + P ey we have P ey = (μ2 − 1)P ex . Note that both x and y are not in X (since P ex and P ey are nonzero multiple of P ex ), thus x can be replaced by y to form a new star set, namely X = (X \ {x}) ∪ {y}, for μ By property (ii) of star sets, we know T − y has μ as an eigenvalue of multiplicity k − 1. Let T = T − P3 . Then mT (μ) = mT −y (μ) = k − 1. If a component H1 of T (when it exists) has order 1 or 2, then μ is not an eigenvalue of H1 . If a component H2 of T (when it exists) has order in [3, 6], then μ is also not an eigenvalue of H2 . Indeed, if μ is an eigenvalue of H2 , then some pendant vertex of H2 , say z, must lie in X (otherwise, T − X has a component H2 such that det(μI − A(H2 )) = 0, which contradicts to the fact μ is not an eigenvalue of T − X ). By Lemma 2.2, X = (X \ {z}) ∪ {v} is also a μ-star set of T , where v is the vertex of H2 adjacent to y. Thus a bridge yv is in the star set X , a contradiction to (vi) listed above. If a component H3 of T (when it exists) has order at least 7, then the induction hypothesis says mH3 (μ) ≤ |H33|−4 . Now, if two components of T have order at least 7, then we have mT (μ) = 1 + mT (μ) mH1 (μ) + mH2 (μ) + mH3 (μ) =1+ H1
=1+
H3
H2
mH3 (μ) ≤ 1 +
H3
|H3 | − 4 H3
3
,
where H1 (resp., H2 , H3 ) runs through all components of T of order in [1, 2] (resp., in [3, 6], in [7, ∞)) and |Hi | means the order of Hi . Since n ≥ 3 + H3 |H3 |, we have k ≤ n−8 3 . In this case, equality fails to hold.
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If exactly one component, say H, of T has order at least 7, then we have mT (μ) = 1 + mT (μ) = 1 + mH (μ) |H| − 4 3 n−4 . ≤ 3 ≤1+
If mT (μ) = n−4 = H and mH (μ) = |H|−4 3 , then T √ 3 . The induction hypothesis applied to H implies that μ = 2 and H ∈ T (P3 , k). We claim that y is adjacent to the P3 -dominating vertex of T (P3 , k) (otherwise, by applying Lemma 2.3 we have mT (μ) = k − 1). In this case we have T ∈ T (P3 , k + 1), as required. If all components of T have orders at most 6, then mT (μ) = 0 and thus k = 1. Since n ≥ 7, we have k ≤ n−4 3 . If equality holds then n = 7, a contradiction to the assumption that n > 7.
Case 2. x has degree at least 3. Let T = T − x − x . Then T has at least two components. Since μP ex = P ex , we can replace x with x to form a new star set X = (X \ {x}) ∪ {x }. If two components of T have order at least 7, then by an argument analogous to that in case 1 we have mT (μ) = 1 + mT −x (μ) = 1 + mT (μ) ≤ 1 +
n−7 n − 10 = . 3 3
In this case, equality again fails to hold. If exactly one component of T , say H, has order at least 7, since mH (μ) = 0 for H of order at most 6 and mH (μ) ≤ |H|−4 (by induction hypothesis), we have 3 mT (μ) = 1 + mT (μ) = 1 + mH (μ) ≤ 1 +
|H| − 4 . 3
|H|−4 Recalling that n ≥ |H| + 3, we have k ≤ n−4 3 . If equality holds, then mH (μ) = 3 and T has just two components, one is H and the other component contains only one vertex, say z. Thus T is obtained from H and P3 (induced by x, x , z), by adding an edge joining x with a vertex of H. The induction hypothesis applied to H says that H ∈ T (P3 , k). We claim that x is adjacent to the P3 -dominating vertex of H and thus T ∈ T (P3 , k + 1). Indeed, if x is not adjacent to the P3 -dominating vertex of H, then by applying Lemma 2.3, we have mT (μ) = k − 1, a contradiction. If all components of T have order at most 6, then mT (μ) = 0 and thus k = 1. Also n−4 we have 1 = k ≤ n−4 3 . If 1 = k = 3 , then n = 7, a contradiction to the assumption that n > 7.
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Corollary 2.5. Let T be a tree of order n ≥ 16 and with μ = 1 as a positive eigenvalue of multiplicity k. Then k ≤ n−4 3 . Proof. If μ2 is an integer, the inequality has been proved in Theorem 2.4. If μ2 is not an n−4 integer then k ≤ n4 (see [10]). Since n4 ≤ n−4 3 , we have k ≤ 3 , as required. Declaration of competing interest There is no competing interest. Acknowledgement The authors would like to express their sincere gratitude to the referee for a very careful reading of the paper and for all his or her insightful comments and valuable suggestions, which make a number of improvements on this paper. References [1] C. Bu, X. Zhang, J. Zhou, A note on the multiplicities of graph eigenvalues, Linear Algebra Appl. 442 (2014) 69–74. [2] F.K. Bell, P. Rowlinson, On the multiplicities of graph eigenvalues, Bull. Lond. Math. Soc. 35 (2003) 401–408. [3] D. Cvetković, M. Doob, H. Sachs, Spectra of Graphs, 3rd edn., Johann Ambrosius Barth, Heidelberg, 1995. [4] D. Cvetković, M. Lepović, P. Rowlinson, S. Simić, The maximal exceptional graphs, J. Comb. Theory, Ser. B 86 (2002) 347–363. [5] D. Cvetković, P. Rowlinson, S. Simić, Graphs with least eigenvalue −2: the star complement technique, J. Algebraic Comb. 14 (2001) 5–16. [6] M.N. Ellingham, Basic subgraphs and graph spectra, Australas. J. Comb. 8 (1993) 247–265. [7] C.M. da Fonseca, A note on the multiplicities of the eigenvalues of a graph, Linear Multilinear Algebra 53 (2005) 303–307. [8] C.M. da Fonseca, A lower bound for the number of distinct eigenvalues of some real symmetric matrices, Electron. J. Linear Algebra 21 (2010) 3–11. [9] P. Rowlinson, Eutactic stars and graph spectra, in: R.A. Brualdi, S. Friedland, V. Klee (Eds.), Combinatorial and Graph-Theoretical Problems in Linear Algebra, Springer-Verlag, New York, 1993, pp. 153–164. [10] P. Rowlinson, On multiple eigenvalues of trees, Linear Algebra Appl. 432 (2010) 3007–3011. [11] P. Rowlinson, On eigenvalue multiplicity and the girth of a graph, Linear Algebra Appl. 435 (2011) 2375–2381. [12] P. Rowlinson, On graph with an eigenvalue of maximal multiplicity, Discrete Math. 313 (2013) 1162–1166. [13] P. Rowlinson, I. Sciriha, Some properties of the Hoffman–Singleton graph, Appl. Anal. Discrete Math. 1 (2007) 438–445. xević, A spectral proof of the uniqueness of a strongly regular graph with [14] D. Stevanović, M. Miloˇ parameters (81, 20, 1, 6), Eur. J. Comb. 30 (2009) 957–968.