More eigenvalue problems of Nordhaus-Gaddum type Vladimir Nikiforov∗ and Xiying Yuan†‡ March 25, 2014 4 1 0 2 Abstract r a Let G be a graph of order n and let µ1(G) µn(G) be the eigenvalues of its adjacency M matrix. Thisnotestudieseigenvalueproblemsof≥No·r·d·h≥aus-Gaddumtype. LetGbethecomplement of a graph G. It is shown that if s 2 and n 15(s 1), then ≥ ≥ − 4 2 µs(G) + µs(G) n/p2(s 1) 1. | | | | ≤ − − s ] Also if s 1 and n 4 , then O ≥ ≥ C µn−s+1(G) + µn−s+1(G) n/√2s+1. | | | | ≤ . k h Ifs=2 +1forsomeintegerk,theseboundsareasymptoticallytight. Theseresultssettleinfinitely t many cases of a general open problem. a m AMS classification: 15A42, 05C50 [ Keywords: grapheigenvalues,complementarygraph,maximumeigenvalue,minimumeigenvalue, Nordhaus-Gaddum problems. 2 v 5 1 Introduction 6 3 Let G denote the complement of a graph G. A Nordhaus-Gaddum problem is of the type: 4 . 1 Given a graph parameter p(G), determine 0 4 max p(G)+p(G):v(G)=n or min p(G)+p(G):v(G)=n . 1 : (cid:8) (cid:9) (cid:8) (cid:9) v SincefirstintroducedbyNordhausandGaddumin[9],suchproblemshavebeenstudiedforahugevariety i ofgraphparameters;see [2] fora recentcomprehensivesurvey. The Nordhaus-Gaddumproblems attract X attention because they help to get deeper insights in extremal graphquestions. Also, these problems are r a the closest analog to Ramsey problems for non-discrete parameters p(G). In this note we shall be interested in the case when p(G) is a spectral graph parameter; thus, given a graphG of order n, let us index the eigenvalues of the adjacency matrix of G as µ (G) µ (G) 1 n ≥···≥ and set µ(G)=µ (G). 1 The first known spectral Nordhaus-Gaddum results belong to Nosal [10], and to Amin and Hakimi [1], who showed that for every graph G of order n, n 1 µ(G)+µ G <√2(n 1). (1) − ≤ − ∗Department of Mathematical Sciences, University o(cid:0)f M(cid:1)emphis, Memphis TN 38152, USA; email: vniki- forv@memphis.edu †Corresponding author. Department of Mathematics, Shanghai University, Shanghai 200444, China; email: xiy- ingyuan2007@hotmail.com ‡Research supported byNational Science Foundation of ChinagrantNo. 11101263, andbya grantof “TheFirst-class DisciplineofUniversitiesinShanghai”. 1 The lower bound in (1) is best possible and is attained if and only if G is a regular graph; however the upper bound can be improved significantly. A minor improvement has been shown in [8], but an essentially best possible bound has been found only recently, by Csikvari[4] and Terpai [11] who showed that µ(G)+µ G 4n/3 1. ≤ − A similar problem for other eigenvalues has been proposed in [8]: (cid:0) (cid:1) Given s and n, find or estimate the functions fs(n)= max µs(G) + µs G and fn−s(n)= max µn−s+1(G) + µn−s+1 G . v(G)=n| | v(G)=n| | (cid:12) (cid:0) (cid:1)(cid:12) (cid:12) (cid:0) (cid:1)(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) Several bounds have been proved in [8]; among these is a tight bound for f (n): 2 n n 32s, and G is a graph of order n, then ≥ s n 2 µ2 (G)+µ2 (G) +s . (4) n−i+1 n−i+1 ≤ 2 Xi=1(cid:0) (cid:1) (cid:16) (cid:17) From (4) we easily obtain another Nordhaus-Gaddum result, similar to Corollary 2. Corollary 6 If s 1, n>2s, and G is a graph of order n, then ≥ s n µn−i+1(G) + µn−i+1(G) +s √2s. | | | | ≤ 2 Xi=1(cid:0) (cid:1) (cid:16) (cid:17) We also can deduce the following corollary,whose short proof is in Section 2. Corollary 7 If s 1, n>4s, and G is a graph of order n, then ≥ 1 n 2 µ2 (G)+µ2 (G) +s . (5) n−s+1 n−s+1 ≤ s 2 (cid:16) (cid:17) Notethatthe rightsideof(5)includes loworderterms. Suchtermsmaybe reducedbutnotremoved completely, at least for some values of s : e.g., if s = 1, taking the complete balanced bipartite graph K , we see that n/2,n/2 n2 µ2 K +µ2 (K )= +1. n−s+1 n/2,n/2 n−s+1 n/2,n/2 4 We shall use Corollary 7 to obta(cid:0)in a new(cid:1)bound on fn−s(n) as well. Theorem 8 If s 1, n 4s, and G is a graph of order n, then ≥ ≥ n µn−s+1(G) + µn−s+1(G) +1. | | | | ≤ √2s All above bounds are essentially best possible whenever s=2k+1 and n is sufficiently large. Theorem 9 Let s=2k−1+1 for some integer k 1. There exists infinitely many graphs G such that if ≥ 2 i s, then ≤ ≤ v(G) v(G) µi(G) 1, µn−i+2(G) , ≥ 2 2(s 1) − ≤−2 2(s 1) − − v(G) v(G) µi G p 1, µn−i+2 G p . ≥ 2 2(s 1) − ≤−2 2(s 1) − − (cid:0) (cid:1) (cid:0) (cid:1) p p 2 Proofs For graphnotationandconcepts undefined here, we referthe readerto [3]. In particular,if G is a graph, we write v(G) for the number of vertices of G. For short we set µ :=µ (G), µ :=µ(G), µ :=µ G , i i i i and µ:=µ G . (cid:0) (cid:1) (cid:0) (cid:1) 2.1 Some useful observations Lemma 10 Let G be a graph of order n. If X 2,...,n , then ⊂{ } µ2(G) n2/4. i ≤ i∈X X 3 Proof Indeed, if A is the adjacency matrix of G, then n µ2+ µ2 µ2 =tr A2 =2e(G). i ≤ i i∈X i=1 X X (cid:0) (cid:1) Hence, in view of µ 2e(G)/n, we find that ≥ µ2 2e(G) µ2 2e(G) (2e(G)/n)2 n2/4, i ≤ − ≤ − ≤ i∈X X completing the proof. ✷ Lemma 11 Let n s 2, and let G be a graph of order n. If µ 0, then s ≥ ≥ ≤ n µ . s | |≤ 2√n s+1 − Proof Indeed, since µ (G) µ (G) 0, we see that n s ≤···≤ ≤ n n µ2 µ2 =(n s+1)µ2, i ≥ s − s i=s i=s X X and the assertion follows by Lemma 10. ✷ Theorem 12 Let k 0 and n 4k. If G is a graph of order n, then either ≥ ≥ µn−k+1(G) 1 and µn−k+1(G) 0 ≤− ≤ or µn−k+1(G) 1 and µn−k+1(G) 0. ≤− ≤ Proof A classical bound of Ramsey theory implies that every graph of order at least 4k contains either a complete graph on k+1 vertices or an independent set on k+1 vertices. Suppose that G contains a complete graphonk+1vertices,andsoGcontainsanindependent setonk+1vertices. Foraninduced subgraph H of graph G, the Cauchy interlacing theorem implies that µm−i(H) µn−i(G) ≥ for all i=0,...,v(H) 1; therefore, − µn−k+1(G) µ2(Kk+1)= 1 and µn−k+1(G) µ2 Kk+1 =0 ≤ − ≤ as claimed. (cid:0) (cid:1) ✷ Using Weyl’s inequalities ([5], p. 181), we come up with the following pair of useful bounds: Lemma 13 (Weyl) If G is a graph of order n and 2 k n, then ≤ ≤ µk(G)+µn−k+2(G) 1, (6) ≤− µk(G)+µn−k+1(G) 1. (7) ≥− 4 2.2 Blow-ups of graphs For any graph G and integer t 1, write G(t) for the graph obtained by replacing each vertex u of G by ≥ a set V of t independent vertices and every edge u,v of G by a complete bipartite graph with parts u { } V andV . Usually G(t) is called a blow-upof G. Blow-upgraphshave a veryuseful algebraicrelationto u v G: thus, if A is the adjacency matrix of G, then the adjacency matrix A G(t) of G(t) is given by (cid:0) (cid:1) A G(t) =A J t ⊗ (cid:16) (cid:17) where is the Kronecker product and J is the all ones matrix of order t. This observation yields the t ⊗ following fact. Proposition 14 The eigenvalues of G(t) are tµ (G),...,tµ (G), together with n(t 1) additional 0’s. 1 n − We also wantto find the eigenvaluesof the complements of graphblow-ups. Given a graphG andan integer t > 0, set G[t] = G(t), i.e., G[t] is obtained from G(t) by joining all vertices within V for every u vertex u of G. We easily can check the following fact. Proposition 15 The eigenvalues of G[t] are tµ (G)+t 1,...,tµ (G)+t 1, together with n(t 1) 1 n − − − additional ( 1)’s. − 2.3 Proofs Proof of Theorem 1 Let 2 i s. First, we shall show that ≤ ≤ µ2 µ2 +µ2 . (8) i ≤ s+i−1 n−i+2 Indeed, if µ 0, then (6) implies that µ 0, and so i ≥ n−i+2 ≤ µ2 <(µ +1)2 µ2 . i i ≤ n−i+2 On the other hand, if µi <0, then µs+i−1 ≤µi <0 implies that µ2i ≤µ2s+i−1. So (8) is always true. Further, we obviously have n n µ2+ µ2+µ2+ µ2 =2e(G)+2e G =n(n 1). i i − i=2 i=2 X X (cid:0) (cid:1) Note that Weyl’s inequality (8) implies that s s 2s−1 n 2 µ2 µ2+ µ2+ µ2, (9) i ≤ i i i i=2 i=2 i=s+1 i=n−s+2 X X X X and by symmetry, s s 2s−1 n 2 µ2 µ2+ µ2+ µ2. (10) i ≤ i i i i=2 i=2 i=s+1 i=n−s+2 X X X X Further, the condition n 3s 2, implies that n s+2>2s 1 and so ≥ − − − s+1,...,2s 1 n s+2,...,n =∅. { − }∩{ − } Therefore, adding (9) and (10) together with µ2 and µ2, we see that s s n n µ2+2 µ2+µ2+2 µ2 µ2+ µ2 n(n 1). i i ≤ i i ≤ − i=2 i=2 i=1 i=1 X X X X 5 Finally, using (1) we find that µ2+µ2 (µ+µ)2/2 (n 1)2/2, and so ≥ ≥ − s s 1 1 n2 1 n2 µ2+ µ2 n(n 1) (n 1)2 = − < , i i ≤ 2 − − 4 − 4 4 i=2 i=2 X X completing the proof of Theorem 1. ✷ Proof of Theorem 3 If µ (G) 0 and µ (G) 0, then s s ≥ ≥ s µ2+µ2 (s 1) µ2+µ2 , i i ≥ − s s i=2 X(cid:0) (cid:1) (cid:0) (cid:1) and inequality (3) follows by Theorem 1. Next, if µ <0 and µ <0, then Lemma 11 implies that s s n2 n2 µ2+µ2 < , s s ≤ 2(n s+1) 4(s 1) − − so (3) follows in this case as well. Finally, assume that µs <0 and µs ≥0. Then µ2s−1 ≤···≤µs+1 ≤µs <0 and so, (s 1)µ2 µ2 + +µ2 . (11) − s ≤ s+1 ··· 2s−1 Since µ µ 0, inequality (6) implies that µ 0, and so 2 ≥···≥ s ≥ n−i+2 ≤ µ2 µ2 <(µ +1)2 µ2 s ≤ i i ≤ n−i+2 for every i=2,...,s. Therefore (s 1)µ2 µ2 + +µ2 . (12) − s ≤ n ··· n−s+2 Since the condition n 3s 2, implies that n s+2>2s 1, we see that ≥ − − − s+1,...,2s 1 n s+2,...,n =∅. { − }∩{ − } Hence, settingX := s+1,...,2s 1 n s+2,...,n , inequalities(11), (12)andLemma 10imply { − }∪{ − } that 2s−1 n n2 (s 1) µ2+µ2 µ2+ µ2 = µ2 , − s s ≤ i i i ≤ 4 i=s+1 i=n−s+2 i∈X (cid:0) (cid:1) X X X completing the proof of Theorem 3. ✷ Proof of Theorem 4 Assume that s 2, n 15(s 1), and G is a graph of order n. As mentioned ≥ ≥ − above, using the AM - QM inequality and Theorem 3, we always have n µ (G) + µ (G) . (13) s s | | | | ≤ 2(s 1) − To the end of the proofwe shallshow that we canadd a p1 to the rightside of this inequality. Thus, let − G be a graph of order n with µ (G) + µ (G) = f (n), s s s | | | | and assume for a contradiction that n f (n)> 1, (14) s 2(s 1) − − p 6 Ourfirstaimisto showthatµ >0 andµ >0.Indeed, if bothµ andµ arenon-positive,thenLemma s s s s 11 implies that n n µ (G) + µ (G) < 1. s s | | | | √n s+1 ≤ 2(s 1) − − − Now, let µs >0 and µs ≤0. Then, Lemmas 10 and 11 imply tphat n n µ and µ , | s|≤ 2√s 1 | s| ≤ 2√n s+1 − − and so, in view of n 15(s 1), we see that ≥ − n n n + < 1, 2√s 1 2√n s+1 2(s 1) − − − − p contradicting (14). Therefore µ >0 and µ >0. s s Now, let t be a positive integer and set H := G(t). Since µ >0 and µ >0, Propositions 14 and 15 s s imply that [t] µ (H)=tµ and µ H =µ (G )=tµ +t 1, s s s s s − and therefore (cid:0) (cid:1) µ (H) + µ H =tf (n)+t 1. s s s | | | | − On the other hand, (13) implies that (cid:0) (cid:1) tn µ (H) + µ H ; s s | | | | ≤ 2(s 1) − (cid:0) (cid:1) hence p n t 1 f (n) − . s ≤ 2(s 1) − t − Now, letting t tend to , we obtain a contradipction to (14), and thus complete the proof of Theorem 4. ✷ ∞ Proof of Theorem 5 We start with the obvious fact n (µ +µ )= µ µ. (15) i i − − i=2 X For i=2,...,n, set w :=µ +µ . Rearranging (15) and using (6), we find that i i n−i+2 s+1 n−s (wi+wn−i+2)= µ µ wi µ µ+n 2s 1. − − − ≥− − − − i=2 i=s+2 X X On the other hand, µ2 = µ w 2 =µ2 2w µ +w2 >µ2 2w µ . i n−i+2− i n−i+2− i n−i+2 i n−i+2− i n−i+2 (cid:0) (cid:1) Since w <0 and µ n/2 (by Lemma 11), we see that i n−i+2 ≥− µ2 µ2 +nw . i ≥ n−i+2 i 7 Therefore, s+1 s+1 s+1 µ2i +µ2i ≥ µ2n−i+2+µ2n−i+2 +n (wi+wn−i+2) i=2 i=2 i=2 X(cid:0) (cid:1) X(cid:0) (cid:1) X s s+1 = µ2n−i+1+µ2n−i+1 +n (wi+wn−i+2) i=1 i=2 X(cid:0) (cid:1) X s µ2 +µ2 +n(n 2s 1 µ µ). ≥ n−i+1 n−i+1 − − − − i=1 X(cid:0) (cid:1) Using that n>2s, we see that s+1 s µ2+µ2+ (µ2+µ2)+ (µ2 +µ2 ) n(n 1), i i n−i+1 n−i+1 ≤ − i=2 i=1 X X and so, s µ2+µ2+2 (µ2 +µ2 )+n(n 2s 1 µ µ) n(n 1). n−i+1 n−i+1 − − − − ≤ − i=1 X Rearranging this inequality, we find that s 2 (µ2 +µ2 ) 2sn+n(µ+µ) µ2 µ2 n−i+1 n−i+1 ≤ − − i=1 X 2sn+n(µ+µ) (µ+µ)2/2 ≤ − 2sn+n2/2, ≤ completing the proof of Theorem 5. ✷ Proof of Corollary 7 Since Theorem 12 implies that µn−s+1 ≤0 and µn−s+1 ≤0, we get s µ2 +µ2 s µ2 +µ2 , n−i+1 n−i+1 ≥ n−s+1 n−s+1 i=1 X(cid:0) (cid:1) (cid:0) (cid:1) and the assertion follows by Theorem 5. ✷ Proof of Theorem 8 Assume that s 1, n 4s, and G is a graph of order n with ≥ ≥ µn−s+1(G) + µn−s+1(G) = fn−s(n). | | | | To begin with, using the AM - QM inequality and Corollary 7, we see that n fn−s(n) +√2s. ≤ √2s Note that this inequality is almost what we need, as the main term is the correct one, but the constant term is larger than desired. Thus, to the end of the proof we shall show that we can make the additive term equal to 1. We shall use the same techniques as in the proof of Theorem 4. Assume for a contradiction that n fn−s(n)> +1, (16) √2s 8 First, Theorem12 implies that µ 0, and so Lemma 11 implies that µ n/(2√s). Hence, n−s+1 ≤ n−s+1 ≤ n n n (cid:12) (cid:12) |µn−s+1|> √2s +1− µn−s+1 > √2s +1− 2√s(cid:12)>1, (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) and, by symmetry, µ >1. That is to say, n−s+1 (cid:12)(cid:12) (cid:12)(cid:12) µn−s+1 <−1 and µn−s+1 <−1. Let t be a positive integer and set H :=G(t). Since µn−s+1 <−1 and µn−s+1 <−1, Propositions 14 and 15 imply that [t] µtn−s+1(H)=tµn−s+1 and µtn−s+1 H =µtn−s+1(G )=tµn−s+1+t−1, and therefore (cid:0) (cid:1) µtn−s+1(H) + µtn−s+1 H =tfn−s(n) t+1. | | | | − On the other hand, Corollary 7 implies that (cid:0) (cid:1) tn µtn−s+1(H) + µtn−s+1(H) +√2s; | | | | ≤ √2s hence n √2s t 1 fn−s(n) + + − . ≤ √2s t t Now, letting t tend to , we obtain a contradiction to (16), and thus complete the proof of Theorem 8. ✷ ∞ 3 Lower bounds on f (n) and f (n) s n s − Define an infinite sequence of square (0,1) matrices A ,A ,... as follows. Let 1 2 1 0 1 1 A1 = 0 1 , B = 1 −1 , (cid:20) (cid:21) (cid:20) − − (cid:21) and for every k =1,2,... set 1 Ak+1 = 2((2Ak−J2k)⊗B+J2k+1). First note that A is a (0,1) symmetric matrix of order 2k+1. k+1 To give some properties of the eigenvalues of the matrices A we first point out a fact without a k+1 proof. Lemma 16 LetM beasymmetricrealmatrixofordernwithallrow-sumsequaltor,andr,µ (M),...,µ (M) 2 n be the eigenvalues of M. If a and b are real numbers, then the eigenvalues of the matrix aM +bJ are n ar+bn,aµ (M),...,aµ (M). 2 n In the following lemma and its proof we shall use a[b] to indicate an eigenvalue a of multiplicity b. Lemma 17 If k 1, then the row-sums of A are equal to 2k and its spectrum is k+1 ≥ 2k, 2k/2 [2k−1],0[2k−1], 2k/2 [2k−1]. − (cid:16) (cid:17) (cid:16) (cid:17) 9 Proof We shall prove the lemma by induction on k. If k =1, we see that 1 0 0 1 1 0 0 1 1 A2 = 2((2A1−J2)⊗B+J4)= 0 1 1 0 .  1 1 0 0      Therow-sumsofA areequalto2,andits eigenvaluesare2,√2,0and √2.Assume thatthe statement 2 holds for A ; in particular, the row-sums of A are equal to 2k−1. Hen−ce, the row-sums of both 2A k k k − J2k and (2Ak −J2k)⊗B are zero, and so the row-sums of Ak+1 are 2k, proving the first part of the statement. Further, the spectrum of A is k 2k−1, 2(k−1)/2 [2k−2], 0[2k−1−1], 2(k−1)/2 [2k−2]. − (cid:16) (cid:17) (cid:16) (cid:17) Thus, by Lemma 16 the spectrum of 2Ak−J2k is 2(k+1)/2 [2k−2], 0[2k−1], 2(k+1)/2 [2k−2]. − (cid:16) (cid:17) (cid:16) (cid:17) Since the eigenvalues of B are √2 and −√2, the spectrum of (2Ak−J2k)⊗B is 2(k+2)/2 [2k−1], 0[2k], 2(k+2)/2 [2k−1]. − (cid:16) (cid:17) (cid:16) (cid:17) Finally, the row-sums of (2Ak−J2k)⊗B are zero, so by Lemma 16 the spectrum of Ak+1 is 2k, 2k/2 [2k−1], 0[2k−1], 2k/2 [2k−1], − (cid:16) (cid:17) (cid:16) (cid:17) completing the induction step and the proof of Lemma 17. ✷ IfP is a Hermitian matrixofsize n, weindex the eigenvaluesof P asµ (P) µ (P). Observe 1 n ≥···≥ the following special case of Weyl’s inequalities that we shall need in the proof of Theorem 9. If P and Q are Hermitian matrices of size n, then for each 1 s n ≤ ≤ µ (P Q) µ (P) µ (Q) µ (P Q). n s s 1 − ≤ − ≤ − Proof of Theorem 9 Let A (t) be matrix obtained from A J by zeroing all diagonal entries. k+1 k+1 t Clearly A (t) is the adjacency matrix of a graph G of order n=2⊗k+1t. For s=2k−1+1, and for each k+1 i s we have ≤ v(G) µ (G) µ (G) µ (A J ) µ (A J A (t))=2k/2t 1= 1. i s s k+1 t 1 k+1 t k+1 ≥ ≥ ⊗ − ⊗ − − 2 2(s 1) − − Next,itisnothardtoseethattheadjacencymatrixofGisobtainedfrom(J2k+1 −pAk+1)⊗Jt byzeroing all diagonal entries. Since Ak+1 and J2k+1 −Ak+1 have the same spectrum, we also find that v(G) µi G ≥µs G ≥µs((J2k+1 −Ak+1)⊗Jt)−1=2k/2t−1= 2 2(s 1) −1. − (cid:0) (cid:1) (cid:0) (cid:1) Finally, we have p v(G) µn−i+2(G) µn−s+2(G) µn−s+2(Ak+1 Jt) µn(Ak+1 Jt Ak+1(t))= 2k/2t= , ≤ ≤ ⊗ − ⊗ − − −2 2(s 1) − and the same bound holds also for µn−i+2 G . The proof of Theorem 9 is completed. p ✷ (cid:0) (cid:1) 10

More eigenvalue problems of Nordhaus-Gaddum type

Similar More eigenvalue problems of Nordhaus-Gaddum type