Edge-Disjoint Spanning Trees and Eigenvalues of Regular Graphs

Transcription

1 Edge-Disjoint Spanning Trees and Eigenvalues of Regular Graphs Sebastian M. Cioabă and Wiseley Wong MSC: 05C50, 15A18, 05C4, 15A4 March 1, 01 Abstract Partially answering a question of Paul Seymour, we obtain a sufficient eigenvalue condition for the existence of k edge-disjoint spanning trees in a regular graph, when k {, 3}. More precisely, we show that if the second largest eigenvalue of a d-regular graph G is less than d k 1 d+1, then G contains at least k edge-disjoint spanning trees, when k {, 3}. We construct examples of graphs that show our bounds are essentially best possible. We conjecture that the above statement is true for any k < d/. 1 Introduction Our graph notation is standard (see West [1] for undefined terms). The adjacency matrix of a graph G with n vertices has its rows and columns indexed after the vertices of G and the (u, v)-entry of A is 1 if uv = {u, v} is an edge of G and 0 otherwise. If G is undirected, then A is symmetric. Therefore, its eigenvalues are real numbers, and we order them as λ 1 λ λ n. The Laplacian matrix L of G equals D A, where D is the diagonal degree matrix of G. The Laplacian matrix is positive semidefinite and we order its eigenvalues as 0 = µ 1 µ µ n. It is well known that if G is connected and d-regular, then µ i = d λ i for each 1 i n, λ 1 = d and λ i < d for any i 1 (see [3, 9]). Kirchhoff Matrix Tree Theorem [13] (see [3, Section 1.3.5] or [9, Section 13.] for short proofs) is one of the classical results of combinatorics. It states that the number of spanning trees of a graph G with n vertices is the principal minor of the Laplacian matrix L of the graph and consequently, equals spanning trees of G is n i= µ i n n i= (d λ i). n. In particular, if G is a d-regular graph, then the number of Department of Mathematical Sciences, University of Delaware, Newark, DE , cioaba@math.udel.edu. This work was partially supported by a grant from the Simons Foundation (#09309 to Sebastian M. Cioabă). Department of Mathematical Sciences, University of Delaware, Newark, DE , wwong@math.udel.edu. 1

2 Motivated by these facts and by a question of Seymour [19], in this paper, we find relations between the maximum number of edge-disjoint spanning trees (also called the spanning tree packing number or tree packing number; see Palmer [18] for a survey of this parameter) and the eigenvalues of a regular graph. Let σ(g) denote the maximum number of edge-disjoint spanning trees of G. Obviously, G is connected if and only if σ(g) 1. A classical result, due to Nash-Williams [16] and independently, Tutte [0] (see [1] for a recent short constructive proof), states that a graph G contains k edge-disjoint spanning trees if and only if for any partition of its vertex set V (G) = X 1 X t into t non-empty subsets, the following condition is satisfied: e(x i, X j ) k(t 1) (1) 1 i<j t A simple consequence of Nash-Williams/Tutte Theorem is that if G is a k-edge-connected graph, then σ(g) k (see Kundu [15]). Catlin [4] (see also [5]) improved this result and showed that a graph G is k-edge-connected if and only if the graph obtained from removing any k edges from G contains at least k edge-disjoint spanning trees. An obvious attempt to find relations between σ(g) and the eigenvalues of G is by using the relations between eigenvalues and edge-connectivity of a regular graph as well as the previous observations relating the edge-connectivity to σ(g). Cioabă [7] has proven that if G is a d-regular graph and r d is an integer such that λ < d (r 1), then G is d+1 r-edge-connected. While not mentioned in [7], it can be shown that the upper bound above is essentially best possible. An obvious consequence of these facts is that if G is a d-regular graph with λ < d (k 1) for some integer k, k d, then G is k-edge-connected and d+1 consequently, G contains k-edge-disjoint spanning trees. In this paper, we improve the bound above as follows. Theorem 1.1. If d 4 is an integer and G is a d-regular graph such that λ (G) < d 3 d+1, then G contains at least edge-disjoint spanning trees. We remark that the existence of edge-disjoint spanning trees in a graph implies some good properties (cf. [17]); for example, every graph G with σ(g) has a cycle double cover (see [17] for more details). The proof of Theorem 1.1 is contained in Section. In Section, we will also show that Theorem 1.1 is essentially best possible by constructing examples of d-regular graphs G d such that σ(g d ) < and λ (G d ) ( d 3, d 3 d+ d+3). In Section, we will answer a question of Palmer [18, Section 3.7, page 19] by proving that the minimum number of vertices of a d-regular graph with edge-connectivity and spanning tree number 1 is 3(d + 1). Theorem 1.. If d 6 is an integer and G is a d-regular graph such that λ (G) < d 5 d+1, then G contains at least 3 edge-disjoint spanning trees. The proof of this result is contained in Section 3. In Section 3, we will also show that Theorem 1. is essentially best possible by constructing examples of d-regular graphs H d such that σ(h d ) < 3 and λ (H d ) [ d 5, d 5 d+1 d+3). We conclude the paper with some final remarks and open problems. The main tools in our paper are Nash-Williams/Tutte Theorem stated above and eigenvalue interlacing described below (see also [3, 9, 10, 11]).

3 Theorem 1.3. Let λ j (M) be the j-th largest eigenvalue of a matrix M. If A is a real symmetric n n matrix and B is a principal submatrix of A with order m m, then for 1 i m, λ i (A) λ i (B) λ n m+i (A). () This theorem implies that if H is an induced subgraph of a graph G, then the eigenvalues of H interlace the eigenvalues of G. If S and T are disjoint subsets of the vertex set of G, then we denote by E(S, T ) the set of edges with one endpoint in S and another endpoint in T. Also, let e(s, T ) = E(S, T ). If S is a subset of vertices of G, let G[S] denote the subgraph of G induced by S. The previous interlacing result implies that if A and B are two disjoint subsets of a graph G such that e(a, B) = 0, then the eigenvalues of G[A B] interlace the eigenvalues of G. As the spectrum of G[A B] is the union of the spectrum of G[A] and the spectrum of G[B] (this follows from e(a, B) = 0), it follows that λ (G) λ (G[A B]) min(λ 1 (G[A]), λ 1 (G[B])) min(d(a), d(b)), (3) where d(s) denotes the average degree of G[S]. Consider a partition V (G) = V 1... V s of the vertex set of G into s non-empty subsets. For 1 i, j s, let b i,j denote the average number of neighbors in V j of the vertices in V i. The quotient matrix of this partition is the s s matrix whose (i, j)-th entry equals b i,j. A theorem of Haemers (see [10] and also, [3, 9]) states that the eigenvalues of the quotient matrix interlace the eigenvalues of G. The previous partition is called equitable if for each 1 i, j s, any vertex v V i has exactly b i,j neighbors in V j. In this case, the eigenvalues of the quotient matrix are eigenvalues of G and the spectral radius of the quotient matrix equals the spectral radius of G (see [3, 9, 10] for more details). Eigenvalue condition for edge-disjoint spanning trees In this section, we give a proof of Theorem 1.1 showing that if G is a d-regular graph such that λ (G) < d 3, then G contains at least edge-disjoint spanning trees. We show that d+1 the bound d 3 is essentially best possible by constructing examples of d-regular graphs G d+1 d having σ(g d ) < and d 3 < λ d+ (G d ) < d 3. d+3 Proof of Theorem 1.1. We prove the contrapositive. Assume that G does not contain -edgedisjoint spanning trees. We will show that λ (G) d 3. d+1 By Nash-Williams/Tutte Theorem, there exists a partition of the vertex set of G into t subsets X 1,..., X t such that e(x i, X j ) (t 1) 1 = t 3. (4) 1 i<j t It follows that t r i 4t 6 (5) i=1 3

4 where r i = e(x i, V \ X i ). Let n i = X i for 1 i t. It is easy to see that r i d 1 implies n i d + 1 for each 1 i 3. If t =, then e(x 1, V \X 1 ) = 1. By results of [7], it follows that λ (G) > d > d 3 d+4 d+1 and this finishes the proof of this case. Actually, we may assume r i for every 1 i t since r i = 1 and results of [7] would imply λ (G) > d > d 3. d+4 d+1 If t = 3, then r 1 + r + r 3 6 which implies r 1 = r = r 3 =. The only way this can happen is if e(x i, X j ) = 1 for every 1 i < j 3. Consider the partition of G into X 1, X and X 3. The quotient matrix of this partition is d 1 1 n 1 n 1 n 1 A 3 = 1 n d 1 n n. 1 1 n 3 n 3 d n 3 The largest eigenvalue of A 3 is d and the second eigenvalue of A 3 equals d , n 1 n n 3 n 1 n n 3 n 1 n n n 3 n 3 n 1 which is greater than d 1 n 1 1 n 1 n 3. Thus, eigenvalue interlacing and n i d+1 for 1 i 3 imply λ (G) λ (A 3 ) d 3. This finishes the proof of the case t = 3. d+1 Assume t 4 from now on. Let a denote the number of r i s that equal and b denote the number of r j s that equal 3. Using equation (5), we get 4t 6 t r i a + 3b + 4(t a b) = 4t a b, i=1 which implies a + b 6. Recall that d(a) denotes the average degree of the subgraph of G induced by the subset A V (G). If a = 0, then b 6. This implies that there exist two indices 1 i < j t such that r i = r j = 3 and e(x i, X j ) = 0. Eigenvalue interlacing (3) implies λ (G) λ (G[X i X j ]) min(λ 1 (G[X i ]), λ 1 (G[X j ])) min(d(x i ), d(x j ) min(d 3 n i, d 3 n j ) d 3 d+1. If a = 1, then b 4. This implies there exist two indices 1 i < j t such that r i =, r j = 3 and e(x i, X j ) = 0. Eigenvalue interlacing (3) implies λ (G) λ (G[X i X j ]) min(λ 1 (G[X i ]), λ 1 (G[X j ])) min(d(x i ), d(x j )) min(d n i, d 3 n j ) d 3 d+1. If a =, then b. If there exist two indices 1 i < j t such that r i = r j = and e(x i, X j ) = 0, then eigenvalue interlacing (3) implies λ (G) λ (G[(X i X j ]) min(λ 1 (G[X i ]), λ 1 (G[X j ])) min(d(x i ), d(x j )) min(d n i, d n j ) d > d 3. d+1 d+1 Otherwise, there exist two indices 1 p < q t such that r p =, r q = 3 and e(x p, X q ) = 0. By a similar eigenvalue interlacing argument, we get λ (G) d 3 in this case as well. d+1 If a = 3, then if there exist two indices 1 i < j t such that r i = r j = and e(x i, X j ) = 0, then as before, eigenvalue interlacing (3) implies λ (G) d > d 3. d+1 d+1 This finishes the proof of Theorem

5 a 1 a b 1 b a 3 b 3 Figure 1: The 4-regular graph G 4 with σ(g 4 ) = 1 and 3.5 = 4 3 < λ 4+ (G 4 ) < We show that our bound is essentially best possible by presenting a family of d-regular graphs G d with d 3 < λ d+ (G d ) < d 3 and σ(g d+3 d) = 1, for every d 4. For d 4, consider three vertex disjoint copies G 1, G, G 3 of K d+1 minus one edge. Let a i and b i be the two non adjacent vertices in G i for 1 i 3. Let G d be the d-regular graph obtained by joining a 1 with a, b and b 3 and a 3 and b 1. The graph G d has 3(d+1) vertices and is d-regular. The partition of the vertex set of G d into V (G 1 ), V (G ), V (G 3 ) has the property that the number of edges between the parts equals 3. By Nash-Williams/Tutte Theorem, this implies σ(g d ) <. For d 4, denote by θ d the largest root of the cubic polynomial P 3 (x) = x 3 + ( d)x + (1 d)x + d 3. (6) Lemma.1. For every integer d 4, the second largest eigenvalue of G d is θ d. Proof. Consider the following partition of the vertex set of G d into nine parts: V (G 1 ) \ {a 1, b 1 }, V (G ) \ {a, b }, V (G 3 ) \ {a 3, b 3 }, {a 1 }, {b 1 }, {a }, {b }, {a 3 }, {b 3 }. This is an equitable partition whose quotient matrix is the following d d d d A 9 = d (7) 0 d d d d The characteristic polynomial of A 9 is P 9 (x) = (x d)(x + 1) [x 3 + ( d)x + (1 d)x + d 3]. (8) 5

6 Let λ λ 3 λ 4 denote the solutions of the equation x 3 +( d)x +(1 d)x+d 3 = 0. Because the above partition is equitable, it follows that d, λ, λ 3, λ 4 and 1 are eigenvalues of G d, and the multiplicity of each of them as an eigenvalue of G d is at least. We claim the spectrum of G d is d (1), λ (), λ () 3, λ () 4, ( 1) (3d 4). (9) It suffices to obtain 3d 4 linearly independent eigenvectors corresponding to 1. Consider two distinct vertices u 1 and u in V (G 1 ) \ {a 1, b 1 }. Define an eigenvector where the entry corresponding to u 1 is 1, the entry corresponding to u is 1, and all the other entries are 0. We create d eigenvectors by letting u to be each of the d vertices in V (G 1 )\{a 1, b 1, u 1 }. This can also be done to two vertices u 1, u V (G ) \ {a, b } or two vertices u 1, u V (G 3 ) \ {a 3, b 3 }. This way, we obtain a total of 3d 6 linearly independent eigenvectors corresponding to 1. Furthermore, define an vector with entries at three fixed vertices u 1 V (G 1 ) \ {a 1, b 1 }, u 1 V (G ) \ {a, b }, u 1 V (G 3 ) \ {a 3, b 3 } equal to 1, with entries at a 1, b, a 3 equal to 1 and with entries 0 everywhere else. It is easy to check this is an eigenvector corresponding to 0. To obtain the final eigenvector, define a new vector by setting the entries at three fixed vertices u 1 V (G 1 ) \ {a 1, b 1 }, u 1 V (G ) \ {a, b }, u 1 V (G 3 ) \ {a 3, b 3 } to be 1, the entries at b 1, a, and b 3 to be 1 and the remaining entries to be 0. It is easy to check all these 3d 4 vectors are linearly independent eigenvectors corresponding to eigenvalue 1. Having obtained the entire spectrum of G d, the second largest eigenvalue of G d must be θ d. Lemma.. For every integer d 4, d 3 d + < θ d < d 3 d + 3. Proof. We find that for d 4, ( P 3 d 3 ) = 3 (9 + d ( + d + d )) d + ) = P 3 ( d 3 d + 3 ( + d) 3 < 0, d (3 + d) 3 > 0, and P 3(x) > 0 beyond x = 1 3 ( 1 + d) < d. Hence, 3 d+3 for every d 4. d 3 d + < θ d < d 3 d + 3 (10) Palmer [18] asked whether or not the graph G 4 has the smallest number of vertices among all 4-regular graphs with edge-connectivity and spanning tree number 1. We answer this question affirmatively below. Proposition.3. Let d 4 be an integer. If G is a d-regular graph such that κ (G) = and σ(g) = 1, then G has at least 3(d + 1) vertices. The only graph with these properties and 3(d + 1) vertices is G d. 6

7 Proof. As σ(g) = 1 <, by Nash-Williams/Tutte theorem, there exists a partition V (G) = X 1 X t such that e(x 1,..., X t ) t 3. This implies r 1 + +r t 4t 6. As κ (G) =, it means that r i for each 1 i t which implies 4t 6 t and thus, t 3. If t = 3, then r i = for each 1 i 3 and thus, e(x i, X j ) = 1 for each 1 i j 3. As d 4 and r i =, we deduce that X i d + 1. Equality happens if and only if X i induces a K d+1 without one edge. Thus, we obtain that V (G) = X 1 + X + X 3 3(d + 1) with equality if and only if G = G d. If t 4, then let α denote the number of X i s such that X i d + 1. If α 3, then V (G) > 3(d + 1) and we are done. Otherwise, α. Note that if X i d, then r i d. Thus, 4t 6 r r t α + d(t α) = dt (d )α which implies (d )α (d 4)t + 6. As α and t 4, we obtain (d ) (d 4)4 + 6 which is equivalent to d 6, contradiction. This finishes our proof. 3 Eigenvalue condition for 3 edge-disjoint spanning trees In this section, we give a proof of Theorem 1. showing that if G is a d-regular graph such that λ (G) < d 5, then G contains at least 3 edge-disjoint spanning trees. We show that d+1 the bound d 5 is essentially best possible by constructing examples of d-regular graphs d+1 H d having σ(h d ) < 3 and d 5 λ d+1 (H d ) < d 5. d+3 Proof of Theorem 1.. We prove the contrapositive. We assume that G does not contain 3-edge-disjoint spanning trees and we prove that λ (G) d 5. d+1 By Nash-Williams/Tutte Theorem, there exists a partition of the vertex set of G into t subsets X 1,..., X t such that e(x i, X j ) 3(t 1) 1 = 3t 4. 1 i<j t It follows that t i=1 r i 6t 8, where r i = e(x i, V \ X i ). If r i for some i between 1 and t, then by results of [7], it follows that λ (G) d 4 > d+3 d 5. d+1 Assume r i 3 for each 1 i t from now on. Let a = {i : 1 i t, r i = 3}, b = {i : 1 i t, r i = 4} and c = {i : 1 i t, r i = 5}. We get that which implies 6t 8 r r t 3a + 4b + 5c + 6(t a b c) 3a + b + c 8. (11) If for some 1 i < j t, we have e(x i, X j ) = 0 and max(r i, r j ) 5, then eigenvalue interlacing (3) implies λ (G) λ (G[X i X j ]) min(λ (G[X i ]), λ (G[X j ])) min(d(x i ), d(x j )) d 5 and we would be done. Thus, we may assume that d+1 e(x i, X j ) 1 (1) 7

8 for every 1 i < j t with max(r i, r j ) 5. Similar arguments imply for example that a + b + c 6, a + b 5, a 4. (13) For the rest of the proof, we have to consider the following cases: Case 1. a. The inequality 1 i<j t e(x i, X j ) 3t 4 implies t 3. As a = {i : r i = 3}, assume without loss of generality that r 1 = r = 3. Because G is connected, this implies e(x 1, X ) < 3. Otherwise, e(x 1 X, V (G) \ (X 1 X )) = 0, contradiction. If e(x 1, X ) =, then e(x 1 X, V (G) \ (X 1 X )) =. Using the results in [7], this implies λ (G) d 4 > d 5 and finishes the proof. d+ d+1 Thus, we may assume e(x 1, X ) = 1. Let Y 3 = V (G) \ (X 1 X ). As r 1 = r = 3, we deduce that e(x 1, Y 3 ) = e(x, Y 3 ) =. This means e(y 3, V (G) \ Y 3 ) = 4 and since d 6, this implies := Y 3 d + 1. Consider the partition of the vertex set of G into three parts: X 1, X and Y 3. The quotient matrix of this partition is d 3 1 n 1 n 1 n 1 B 3 = 1 n d 3 n n. d 4 The largest eigenvalue of B 3 is d. Eigenvalue interlacing and n 1, n, d + 1 imply λ (G) λ (B 3 ) tr(b 3) d d d 3 n 1 3 n 3 (d + 1) 3 (d + 1) d + 1 = d 5 d + 1. This finishes the proof of this case. Case. a = 1. Inequalities (11) and (13) imply b + c 5 b + c. Actually, because we assumed that e(x i, X j ) 1 for every 1 i j t with max(r i, r j ) 5, we deduce that b + c 3. Otherwise, if b + c 4, then there exists i j such that r i = 3, r j {4, 5} and e(x i, X j ) = 0. The only solution of the previous inequalities is b = and c = 1. Without loss of generality, we may assume r 1 = 3, r = r 3 = 4 and r 4 = 5. Using the facts of the previous paragraph, we deduce that e(x 1, X j ) = 1 for each j 4 and e(x i, X j ) 1 for each i j 4. If e(x, X 3 ) 3, then e(x, X 4 ) = 0 which is a contradiction with the first paragraph of this subcase. If e(x, X 3 ) =, then t 5 and e(x 1 X X 3 X 4, V (G) \ (X 1 X X 3 X 4 )) =. Using results from [7], it follows that λ (G) d 4 > d 5 which finishes the proof of d+ d+1 this subcase. If e(x, X 3 ) = 1, then there are some subcases to consider: 1. If e(x, X 4 ) = e(x 3, X 4 ) = 1, then t 5. If Y 5 := V (G) \ (X 1 X X 3 X 4 ), then e(x 4, Y 5 ) =, e(x 3, Y 5 ) = e(x, Y 5 ) = 1. These facts imply e(y 5, V (G) \ Y 5 ) = 4 and 8

9 e(x 1, Y 5 ) = 0. As d 6, it follows that n 5 := Y 5 d + 1. Eigenvalue interlacing (3) implies λ (G) λ (G[X 1 Y 5 ]) min(λ 1 (G[X 1 ]), λ 1 (G[Y 5 ])) min(d(x 1 ), d(y 5 )) ( min d 3, d 4 ) d 4 n 1 n 5 d + 1 > d 5 d + 1 which finishes the proof of this subcase.. If e(x, X 4 ) = and e(x 3, X 4 ) = 1, then t 5. If Y 5 := V (G) \ (X 1 X X 3 X 4 ), then e(x 4, Y 5 ) = e(x 3, Y 5 ) = 1. These facts imply e(y 5, V (G) \ Y 5 ) =. Using results in [7], we obtain λ (G) > d 4 > d 5 which finishes the proof of this subcase. d+ d+1 3. If e(x, X 4 ) = 1 and e(x 3, X 4 ) =, then the proof is similar to the previous case and we omit the details. 4. If e(x, X 4 ) = e(x 3, X 4 ) =, then t = 4. Consider the partition of the vertex set of G into three parts: X 1, X, X 3 X 4. The quotient matrix of this partition is d 3 1 n 1 n 1 n 1 C 3 = 1 n d 4 3 n n 3 d 5 where = X 3 X 4 = X 3 + X 4 (d + 1). The largest eigenvalue of C 3 is d. Eigenvalue interlacing and n 1, n d+1, (d+1) imply λ (G) λ (C 3 ) tr(c 3) d d d 3 n 1 n 5 3 (d + 1) d (d + 1) d 4.75 d + 1 > d 5 d + 1. Case 3. a = 0. Inequalities (11) and (13) imply b + c 8, b + c 6, b 5. If b = 0, then c 8 and c 6 which is a contradiction that finishes the proof of this subcase. If b = 1, then c 6 and c 5 which is a contradiction that finishes the proof of our subcase. If b =, then c 4 which implies that there exists i j such that e(x i, X j ) = 0 and r i = 4 and r j {4, 5}. This contradicts (1) and finishes the proof. If b = 3, then c. Assume that c = first. Without loss of generality, assume r 1 = r = r 3 = 4 and r 4 = r 5 = 5. (1) implies that e(x i, X j ) = 1 for each 1 i < j 5 except when i = 4 and j = 5 where e(x 4, X 5 ) =. 9

10 Consider the partition of the vertex set of G into three parts: X 1, X X 3, and X 4 X 5. The quotient matrix of this partition is d 4 n 1 n 1 n 1 D 3 = n d 6 n 4 4 n d 6 where n = X X 3 = X + X 3 (d + 1) and = X 4 X 5 = X 4 + X 5 (d + 1). The largest eigenvalue of D 3 is d. Eigenvalue interlacing and n 1 d + 1, n, (d + 1) imply λ (G) λ (D 3 ) tr(d 3) d d 3 3 n 1 n d d (d + 1) 3 (d + 1) = d 5 d + 1. This finishes the proof of this subcase. If c 3, then since b = 3, it follows that there exists i j such that e(x i, X j ) = 0 and r i = 4 and r j {4, 5}. This contradicts (1) and finish the proof of this subcase. If b = 4, we have inequality (13) implies c. If c =, then there exist i j such that e(x i, X j ) = 0, r i = 4 and r j {4, 5}. This contradicts (1) and finishes the proof of this subcase. Suppose c = 0. Without loss of generality, assume that r i = 4 for 1 i 4. If t = 4, then (1) implies that the graph G is necessarily of the form shown in Figure. Figure : The structure of G when a = 0, b = 4, c = 0, and t = 4. Consider the partition of the vertex set of G into three parts: X 1, X, X 3 X 4. quotient matrix of this partition is d 4 n 1 n 1 n 1 E 3 = n d 4 n n d 4 The where = X 3 X 4 = X 3 + X 4 (d + 1). 10

11 The largest eigenvalue of E 3 is d. Eigenvalue interlacing and n 1, n d + 1, (d + 1) imply λ (G) λ (E 3 ) tr(e 3) d d n 1 n d d + 1 d + 1 (d + 1) = d 5 d + 1. If t 5, then there are two possibilities: either e(x i, X j ) = 1 for each 1 i < j 4 or without loss of generality, e(x i, X j ) = 1 for each 1 i < j 4 except for i = 1 and j = where e(x 1, X ) =. In the first situation, if Y 5 := V (G) \ (X 1 X X 3 X 4 ), then e(x i, Y 5 ) = 1 for each 1 i 4 and thus, e(y 5, V (G) \ Y 5 ) = 4. This implies Y 5 d + 1. Consider the partition of V (G) into three parts X 1, X X 3, X 4 Y 5. The quotient matrix of this partition is F 3 = d 4 n 1 n n 1 n 1 4 n d 6 n 4 d 6 where n = X X 3 = X + X 3 (d + 1) and = X 4 Y 5 = X 4 + Y 5 (d + 1). The largest eigenvalue of F 3 is d. Eigenvalue interlacing and n 1 d + 1, n, (d + 1) imply λ (G) λ (F 3 ) tr(f 3) d d 3 3 n 1 n d d (d + 1) 3 (d + 1) = d 5 d + 1, which finishes the proof of this subcase. Figure 3: The structure of G when a = 0, b = 4, c = 1, and t 5. In the second situation, if Y 5 := V (G)\(X 1 X X 3 X 4 ) then e(x 1, Y 5 ) = e(x, Y 5 ) = 0 and e(x 3, Y 5 ) = e(x 4, Y 5 ) = 1. This implies e(y 5, V (G) \ Y 5 ) =. By results of [7], we deduce that λ (G) d 4 > d 5 which finishes the proof of this subcase. d+ d+1 Assume that c = 1. Without loss of generality, assume that r i = 4 for 1 i 4, and r 5 = 5. Our assumption (1) implies that the graph is necessarily of the form shown in Figure 3, where Y is a component that necessarily joins to X 5. By results of [7], it follows that λ (G) > d > d 5 and this finishes the proof of this case. d+4 d+1 11

12 If b = 5, then c = 0 by (1). Also, by (1), it follows that t = 5 and e(x i, X j ) = 1 for each 1 i < j 5. Consider the partition of the vertex set of G into three parts: X 1, X X 3, X 4 X 5. The quotient matrix of this partition is d 4 n 1 n 1 n 1 G 3 =, n d 6 n 4 4 n d 6 which is identical to the quotient matrix F 3 in a previous case, which yields λ (G) d 5 d+1. If b > 5, then (1) will yield a contradiction. This finishes the proof of Theorem 1.. We show that our bound is essentially best possible by presenting a family of d-regular graphs H d with d 5 λ d+1 (H d ) < d 5 and σ(h d+3 d) =, for every d 6. For d 6, consider the graph obtained from K d+1 by removing two disjoint edges. Consider now 5 vertex disjoint copies H 1, H, H 3, H 4, H 5 of this graph. For each copy H i, 1 i 5, denote the two pairs of non-adjacent vertices in H i by a i, c i and b i, d i. Let H d be the d-regular graph whose vertex set is 5 i=1v (H i ) and whose edge set is the union 5 i=1e(h i ) with the following set of 10 edges: {b 1 a, b a 3, b 3 a 4, b 4 a 5, b 5 a 1, c 1 d 3, c 3 d 5, c 5 d, c d 4, c 4 d 1 }. b1 a a1 c1 d d1 b c b5 c5 d3 a3 d5 c3 a5 b3 c4 d4 b4 a4 Figure 4: The 10-regular graph H 10 with σ (H 10 ) = and < λ 10+1 (H 10 ) < The graph H d is d-regular and has 5(d + 1) vertices. The partition of the vertex set of H d into the five parts: V (H 1 ), V (H ), V (H 3 ), V (H 4 ), V (H 5 ) has the property that the number of edges between the parts equals 10 < 1 = 3(5 1). By Nash-Williams/Tutte Theorem, this implies σ(h d ) < 3. 1

13 For d 6, denote by γ d the largest root of the polynomial x 10 + (8 d)x 9 + (d 16d + 30)x 8 + (8d 50d + 58)x 7 + (0d 66d + 36)x 6 + (8d + 18d 70)x 5 + ( 9d + 140d 146)x 4 + ( 0d + 57d 1)x 3 + (14d 83d + 109)x + (4d 13d + 5)x d + 5d 5. Lemma 3.1. For every integer d 6, the second largest eigenvalue of H d is γ d. Proof. Consider the following partition of the vertex set of H d into 5 parts: 5 parts of the form V (H i ) \ {a i, b i, c i, d i }, i = 1,, 3, 4, 5. The remaining 0 parts consist of the 0 individual vertices {a i }, {b i }, {c i }, {d i }, i = 1,, 3, 4, 5. This partition is equitable and the characteristic polynomial of its quotient matrix (which is described in Section 5) is P 5 (x) = (x d)(x 1)(x + 1) (x + 3)[x 10 + (8 d)x 9 + (d 16d + 30)x 8 + (8d 50d + 58)x 7 + (0d 66d + 36)x 6 + (8d + 18d 70)x 5 + ( 9d + 140d 146)x 4 + ( 0d + 57d 1)x 3 + (14d 83d + 109)x + (4d 13d + 5)x d + 5d 5]. Let λ λ 3... λ 11 denote the solutions of the degree 10 polynomial P 10 (x). Because the partition is equitable, it follows that these 10 solutions, d, 1, 1, and 3 are eigenvalues of H d, including multiplicity. We claim the spectrum of H d is d (1), 1 (1), 3 (1), 1 (5d 18), λ () i for i =, 3,..., 11. (14) It suffices to obtain 5d 18 linearly independent eigenvectors corresponding to 1. Consider two distinct vertices u 1 1 and u 1 in V (H 1 ) \ {a 1, b 1, c 1, d 1 }. Define a vector where the entry corresponding to u 1 1 is 1, the entry corresponding to u 1 is 1, and all other entries are 0. This is an eigenvector corresponding to the eigenvalue 1. We can create d 4 eigenvectors by letting u 1 to be each of the d 4 vertices in V (H 1 ) \ {a 1, b 1, c 1, d 1, u 1 1}. This can also be applied to vertices u i 1, u i in V (H i ) \ {a i, b i, c i, d i }, for i =, 3, 4, 5. This way, we obtain a total of 5d 0 linearly independent eigenvectors corresponding to the eigenvalue 1. Furthermore, define a vector whose entry at some fixed vertex u i 1 V (H i ) \ {a i, b i, c i, d i } is, whose entries at a i and d i are 1, for each 1 i 5 and whose remaining entries are 0. Define another vector whose entries at a fixed vertex u i 1 V (H i ) \ {a i, b i, c i, d i } is, whose entries at b i and c i are 1, for each 1 i 5 and whose remaining entries are 0. These last two vectors are also eigenvectors corresponding to the eigenvalue 1. It is easy to check that all these 5d 18 vectors we have constructed are linearly independent eigenvectors corresponding to the eigenvalue 1. By obtaining the entire spectrum of H d, we conclude that the second largest eigenvalue of H d is γ d. Lemma 3.. For every integer d 6, d 5 d + 1 γ d < d 5 d

14 Proof. The lower bound follows directly from Theorem 1. as σ(h d ) < 3. Moreover, by some technical calculations (done in Mathematica and included in Section 5) ( P (n) 10 d 5 ) > 0, for n = 0, 1,..., 10. d + 3 Descartes Rule of Signs implies γ d < d 5 d+3. Hence, for every d 6. 4 Final Remarks d 5 d + 1 γ d < d 5 d + 3 In this paper, we studied the relations between the eigenvalues of a regular graph and its spanning tree packing number. Based on the results contained in this paper, we make the following conjecture. Conjecture 4.1. Let d 8 and 4 k d be two integers. If G is a d-regular graph such that λ (G) < d k 1, then G contains at least k edge-disjoint spanning trees. d+1 Let ω(h) denote the number of components of the graph H. The vertex-toughness of G is defined as min S, where the minimum is taken over all subsets of vertices S whose ω(g\s) removal disconnects G. Alon [1] and independently, Brouwer [] have found close relations between the eigenvalues of a regular graph and its vertex-toughness. These connections were used by Alon in [1] to disprove a conjecture of Chvátal that a graph with sufficiently large vertex-toughness is pancyclic. For c 1, the higher order edge-toughness τ c (G) is defined as τ c (G) := min X ω(g \ X) c where the minimum is taken over all subsets X of edges of G with the property ω(g \ X) > c (see Chen, Koh and Peng [6] or Catlin, Lai and Shao [5] for more details). The Nash- Williams/Tutte Theorem states that σ(g) = τ 1 (G). Cunningham [8] generalized this result and showed that if τ 1 (G) p for some natural numbers p and q, then G contains p spanning q trees (repetitions allowed) such that each edge of G lies in at most q of the p trees. Chen, Koh and Peng [6] proved that τ c (G) k if and only if G contains at least c edge-disjoint forests with exactly c components. It would be interesting to find connections between the eigenvalues of the adjacency matrix (or of the Laplacian) of a graph G and τ c (G). Another question of interest is to determine sufficient eigenvalue condition for the existence of nice spanning trees in pseudorandom graphs. A lot of work has been done on this problem in the case of random graphs (see Krivelevich [14] for example). Acknowledgments We thank the referee for some useful remarks. 14 (15)

15 References [1] N. Alon, Tough Ramsey graphs without short cycles, J. Algebraic Combin. 4 (1995), no. 3, [] A.E. Brouwer, Toughness and spectrum of a graph, Linear Algebra Appl. 6/8 (1995), [3] A.E. Brouwer and W.H. Haemers, Spectra of Graphs, Springer Universitext 01, 50pp monograph, available online at aeb/math/ipm.pdf. [4] P. Catlin, Edge-connectivity and edge-disjoint spanning trees, available online at hjlai/. [5] P.A. Catlin, H.-J. Lai and Y. Shao, Edge-connectivity and edge-disjoint spanning trees, Discrete Math. 309 (009), [6] C.C. Chen, M.K. Koh, Y.-H. Peng, On the higher-order edge toughness of a graph, Graph theory and combinatorics (Marseille-Luminy, 1990), Discrete Math. 111 (1993), no. 1-3, [7] S.M. Cioabă, Eigenvalues and edge-connectivity of regular graphs, Linear Algebra Appl. 43 (010), [8] W. H. Cunningham, Optimal attack and reinforcement of a network. J. Assoc. Comp. Mach. 3 (1985), [9] C. Godsil and G. Royle, Algebraic Graph Theory, Graduate Texts in Mathematics 07, Springer-Verlag, New York, 001. [10] W.H. Haemers, Interlacing eigenvalues and graphs, Linear Algebra Appl. 6/8 (1995), [11] R.A. Horn and C.A. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, xiii+561 pp. [1] T. Kaiser, A short proof of the tree-packing theorem, Discrete Math., to appear, available at [13] G. Kirchhoff, Über die Auflösung der Gleichungen, auf welche man bei der untersuchung der linearen verteilung galvanischer Ströme geführt wird, Ann. Phys. Chem. 7, 1847, [14] M. Krivelevich, Embedding spanning trees in random graphs, SIAM J. Discrete Math. 4 (010), [15] S. Kundu, Bounds on the number of disjoint spanning trees, Journal of Combinatorial Theory, Series B 17 (1974),

16 [16] C. Nash-Williams, Edge-disjoint spanning trees of finite graphs, J. London Math. Soc. 36 (1961) [17] K. Ozeki and T. Yamashita, Spanning trees: a survey, Graphs Combin. 7 (011), no. 1, 1 6. [18] E.M. Palmer, On the spanning tree packing number of a graph: a survey, Discrete Math. 30 (001), [19] P. Seymour, Private communication to the 1st author, April 010. [0] W. T. Tutte, On the problem of decomposing a graph into n connected factors, J. London Math. Soc. 36 (1961) [1] D.B. West, Introduction to Graph Theory, Prentice Hall, Inc., Upper Saddle River, NJ, xvi+51 pp. 5 Calculations for Lemma Justify characteristic polynomial in 5 parts The following is the characteristic polynomial of the equitable partition in 5 parts: Factor[CharacteristicPolynomial[ d d d d d d d d d d d d d d d d d d d d d d d d d , x]] (d x)( 1 + x)(1 + x) (3 + x)( 5 + 5d d + 5x 13dx + 4d x + 109x 83dx + 14d x 1 x dx 3 0d x 3 146x dx 4 9d x 4 70x dx 5 + 8d x x 6 66 x 6 + 0d x x 7 50dx 7 + 8d x x 8 16dx 8 + d x 8 + 8x 9 dx 9 + x 10 ) 5. Justify P (n) n=0 ( d 5 d+3) > 0, for n = 0, 1,..., d d + 5x 13dx + 4d x + 109x 83dx + 14d x 1x 3 + Factor 57dx 3 0d x 3 146x dx 4 9d x 4 70x dx 5 + 8d x x 6 66dx 6 + 0d x x 7 50dx 7 + 8d x 7 /.x d 5/(d + 3) + 30x 8 16dx 8 + d x 8 + 8x 9 dx 9 + x 10 5 ( d+45996d d d d d d d d d d d 1) () 10 16

17 Looking at the numerator, d d d d d d d d d d d d d+45996d d d 4 ( ) d 3 ( ) d d d d 9 ( ) d 7 ( ) d 8 ( ) d n=1 Apart FullSimplify D = d d d d d d d 9 > d d + 5x 13dx + 4d x + 109x 83dx + 14d x 1x dx 3 0d x 3 146x dx 4 9d x 4 70x dx 5 + 8d x x 6 66dx 6 + 0d x x 7 50dx 7 + 8d x x 8 16dx 8 + d x 8, x + /.x d 5/(d + 3) 8x 9 dx 9 + x d+935d 1605d 3 80d 4 +40d () () () () () () () () Looking at [ the fraction terms, ] Together () () () () () () () () ( d d d d d d d d 8) () 9 The expression is positive. The only concern now are the terms d + 935d 1605d 3 80d d 5. Direct calculations for d = 6 and 7 yield the values 145 and 1840, respectively. For d 8, ( d + 935d ) 1605d 3 80d d 5 = 935d + (d 1)(935)d 665d d 4 + (d )(40)d 4 80d d 3 935d + (7)(935)(8) 665d d 4 + (6)(40)(8)d 3 80d d n= = ( )d 3 + ( )d + ( ) > d d + 5x 13dx + 4d x + 109x 83dx + Apart FullSimplify D 14d x 1x dx 3 0d x 3 146x dx 4 9d x 4 70x dx 5 + 8d x 5 +, {x, } 36x 6 66dx 6 + 0d x x 7 50dx 7 + 8d x 7 /.x d 5/(d + 3) + 30x 8 16dx 8 + d x 8 + 8x 9 dx 9 + x d 3758d 480d d 4 344d 5 60d d 7 + d () () () () () () () Looking at the fraction terms and d 8, [ ] Together () () () () () () () d d d d d d 5 1 () d d d d d d d d 13 +5d d 15 + d 16 By comparing terms, the expression is positive. The only concern now are the terms d 3758d 480d d 4 344d 5 60d d 7. Direct calculations for d = 6 and 7 yield the values and , respectively. Clearly we have for the first two terms that d > 0. Now assume d 8. Looking at the next three terms, 3758d 480d d 4 = 4944d 3 + (d )(47)d 3 480d d For the final three terms, 4944d 3 + (6)(47)(8)d 480d d > 0 344d 5 60d d 7 = 64d (d 4)d 6 60d 6 344d 5 64d (4)(8)d 5 60d 6 344d 5 > 0. 17

18 5..4 n= d d + 5x 13dx + 4d x + 109x 83dx + Apart FullSimplify D 14d x 1x dx 3 0d x 3 146x dx 4 9d x 4 70x dx 5 + 8d x 5 +, {x, 3} 36x 6 66dx 6 + 0d x x 7 50dx 7 + 8d x 7 /.x d 5/(d + 3) + 30x 8 16dx 8 + d x 8 + 8x 9 dx 9 + x d 7180d d d d d d () () () () () () Looking at the fraction terms and 48d 7, [ Together () () () () () () () d d d d d d d d d d d d d d d 7 ] Looking at the numerator, d d d d d d d d d d d d 1 +84d 13 +4d d d d d d d d (6 8) ( ) ( d ) d ( ) d 3 ( ) d 4 ( ) d 5 ( ) d 6 = d d d d d d d 7 > 0. The only concern now are the terms d 7180d d d d d 6. Direct calculations for d = 6 and 7 yield the values and , respectively. We ignore the first positive constant, and assume d 8. Looking at the next three terms, 933d 7180d d 3 = 81888d + (d )(40944)d 7180d 933d 81888d + (6)(40944)(8)d 7180d 933d > 0 For the final three terms, 5340d d d 6 = 1680d 5 + (d 5)336d d d n=4 Apart FullSimplify D 1680d 5 + (3)336(8)d d d 4 > d d + 5x 13dx + 4d x + 109x 83dx + 14d x 1x dx 3 0d x 3 146x dx 4 9d x 4 70x dx 5 + 8d x x 6 66dx 6 + 0d x x 7 50dx 7 + 8d x x 8 16dx 8 + d x 8, {x, 4} + /.x d 5/(d + 3) 8x 9 dx 9 + x d d 4180d d d d () () () () () Looking at the fraction terms and 67d 6, [ ] Together 67d () () () () () ( d d d d d d d d d d 10 +5d d 1) () 6 This expression is positive. The only concern now are the terms d d 4180d d d 5. Direct calculations for d = 6 and 7 yield the values and , respectively. Now assume d 8. Looking at the first 3 terms, d d = d + (d 3)39604d d d + (5)39604(8) d > 0. For the final three terms, 4180d d d 5 = 0160d 4 + (d 5)403d d d d 4 + (3)403(8)d d d 3 > 0. 18

19 5..6 n= d d + 5x 13dx + 4d x + 109x 83dx + 14d x 1x dx 3 0d x 3 146x dx 4 9d x 4 70x dx 5 + 8d x x 6 66dx 6 + 0d x x 7 50dx 7 + 8d x x 8 16dx 8 + d x 8, {x, 5} + /.x d 5/(d + 3) 8x 9 dx 9 + x d 15400d 16000d d d () () () () Apart FullSimplify D Looking at the fraction terms, [ ] Together () () () () ( d d d d 4) () 5 This expression is positive. The only concern now are the terms d 15400d 16000d d d 5. Clearly for the first two terms we have d > 0 for d 6. Looking at the four remaining terms, 15400d 16000d d d d 16000d (36)d +670(36)d 3 = 15400d 16000d d +4190d 3 > n= d d + 5x 13dx + 4d x + 109x 83dx + 14d x 1x dx 3 0d x 3 146x dx 4 9d x 4 70x dx 5 + 8d x x 6 66dx 6 + 0d x x 7 50dx 7 + 8d x x 8 16dx 8 + d x 8 + 8x 9 dx 9 + x d d d d () () () Apart FullSimplify D, {x, 6} /.x d 5/(d + 3) Looking at the fraction terms and 50400d d, [ Together () () () ] d d 1440 ( d+947d d d d d 6 +40d 7 +35d 8) () 4 This expression is clearly positive for d 6. The only terms left are d d 3, and we get 5..8 n=7 Apart FullSimplify D d d d (6)d d d > d d + 5x 13dx + 4d x + 109x 83dx + 14d x 1x dx 3 0d x 3 146x dx 4 9d x 4 70x dx 5 + 8d x x 6 66dx 6 + 0d x x 7 50dx 7 + 8d x x 8 16dx 8 + d x 8 + 8x 9 dx 9 + x d d + 840d () () , {x, 7} /.x d 5/(d + 3) Looking at the fraction terms and 8840d 3, [ ] Together 8840d () () ( d+0790d d d d d 6) () 3 This expression is clearly positive for d 6. The only remaining terms are d d. We have 5..9 n= d d d (6)d = d d > d d + 5x 13dx + 4d x + 109x 83dx + 14d x 1x 3 + Apart FullSimplify D 57dx 3 0d x 3 146x dx 4 9d x 4 70x dx 5 + 8d x x 6 66dx 6 + 0d x x 7 50dx 7 + 8d x x 8 16dx 8 + d x 8, {x, 8} + /.x d 5/(d + 3) 8x 9 dx 9 + x d d () At d = 6, the value is Clearly the expression is increasing for d 6, and hence always positive for d n= d d + 5x 13dx + 4d x + 109x 83dx + 14d x 1x 3 + Apart FullSimplify D 57dx 3 0d x 3 146x dx 4 9d x 4 70x dx 5 + 8d x x 6 66dx 6 + 0d x x 7 50dx 7 + 8d x x 8 16dx 8 + d x 8 + 8x 9 dx 9 + x d At d = 6, the value is Clearly the expression is increasing for d 6, and hence always positive for d n=10 The value will be 10! > 0. 19, {x, 9} /.x d 5/(d + 3)