Eigenvalues and edge-connectivity of regular graphs

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1 Eigenvalues and edge-connectivity of regular graphs Sebastian M. Cioabă University of Delaware Department of Mathematical Sciences Newark DE 19716, USA August 3, 009 Abstract In this paper, we show that if the second largest eigenvalue of a d-regular graph is less than d (k 1) d+1, then the graph is k-edge-connected. When k is or 3, we prove stronger results. Let ρ(d) denote the largest root of x 3 (d 3)x (3d )x = 0. We show that if the second largest eigenvalue of a d-regular graph G is less than ρ(d), then G is -edge-connected and we prove that if the second largest eigenvalue of G is less than d 3+ (d+3) 16, then G is 3-edge-connected. 1 Introduction Let κ(g) and κ (G) denote the vertex- and edge-connectivity of a connected graph G. If δ is the minimum degree of G, then 1 κ(g) κ (G) δ. Let L = D A be the Laplacian matrix of G, where D is the diagonal degree matrix and A is the adjacency matrix of G. We denote by 0 = µ 1 < µ µ n the eigenvalues of the Laplacian of G. The complement of G is denoted by G. A graph is called disconnected if it is not connected. The join G 1 G of two vertex-disjoint graphs G 1 and G is the graph formed from the union of G 1 and G by joining each vertex of G 1 to each vertex of G. A classical result in spectral graph theory due to Fiedler [7] states that κ(g) µ (G) (1) for any non-complete graph G. Fiedler called µ (G) the algebraic connectivity of G and his work stimulated a large amount of research in spectral graph theory over the last forty years (see [1, 11, 13, 15]). In [1], Kirkland, Molitierno, Neumann and Shader characterize the equality case in Fiedler s inequality (1). Research supported by a startup grant from the Department of Mathematical Sciences of the University of Delaware. 1

2 Theorem 1.1 (Kirkland-Molitierno-Neumann-Shader [1]). Let G be a non-complete connected graph on n vertices. Then κ(g) = µ (G) if and only if G = G 1 G where G 1 is a disconnected graph on n κ(g) vertices and G is a graph on κ(g) vertices with µ (G ) κ(g) n. Eigenvalue techniques have been also used recently by Brouwer and Koolen [5] to show that the vertex-connectivity of a distance-regular graph equals its degree (see also [, 4] for related results). In this paper, we study the relations between the edge-connectivity and the second largest eigenvalue of a d-regular graph. Let λ 1 (G) λ (G) λ n (G) denote the eigenvalues of the adjacency matrix of G. If G is d-regular graph, then λ i (G) = d µ i (G) for any 1 i n. Thus, λ 1 (G) = d and G is connected if and only if λ (G) < d. Chandran [6] proved that if G is a d-regular graph of order n and λ (G) < d 1 d, n d then κ (G) = d and the only disconnecting edge-cuts are trivial, i.e., d edges adjacent to the same vertex. Krivelevich and Sudakov [13] showed that λ (G) d implies κ (G) = d. If G is a d-regular graph on n vertices and n d + 1, then κ (G) = d regardless of the eigenvalues of G (see also the proof of Lemma 1.3). Both results are based on the following well-known lemma (see [15] for a short proof). Lemma 1.. If G = (V,E) is a connected graph of order n and S is a subset of vertices of G, then e(s,v \ S) µ S (n S ) n where e(s,v \ S) denotes the number of edges between S and V \ S. We extend and improve the previous results as follows. We prove the following sufficient condition for the k-edge-connectivity of a d-regular graph for any k d. Theorem 1.3. If d k are two integers and G is a d-regular graph such that λ (G) d (k 1)n, then (d+1)(n d 1) κ (G) k. When k = d, this result states that if λ (G) d (d 1)n, then (d+1)(n d 1) κ (G) = d. We get a small improvement for d even because κ (G) must be even in this case (see Lemma 3.1). (d )n When d is even, Theorem 1.3 shows that λ (G) d implies (d+1)(n d 1) κ (G) = d. A simple calculation reveals that these bounds improve the previous result of Chandran. When n d +, Theorem 1.3 implies that if G is a d-regular graph with λ (G) d + 4, then d+1 κ (G) = d. When d is even, the right hand-side of the previous inequality can be replaced by d + 6. These results improve the previous bound of Krivelevich and d+1 Sudakov. Note that there are many d-regular graphs G with d < λ (G) d + 4. d+1 An example is presented in Figure 1. For k {, 3} we further improve these results as shown below. Note that the edgeconnectivity of a connected, regular graph of even degree is even (see Lemma 3.1). Theorem 1.4. Let d 3 be an odd integer and ρ(d) denote the largest root of x 3 (d 3)x (3d )x = 0. If G is a d-regular graph such that λ (G) < ρ(d), then κ (G).

3 Figure 1: A 3-regular graph with 3 < λ = 1+ 5 < The value of ρ(d) is about d d+5. Theorem 1.5. Let d 3 be any integer. If G is a d-regular graph such that then κ (G) 3. λ (G) < d 3 + (d + 3) 16 The value of d 3+ (d+3) 16 is about d 4. d+3 We show that Theorem 1.4 and Theorem 1.5 are best possible in the sense that for each odd d 3, there exists a d-regular graph X d such that λ (X d ) = ρ(d) and κ (X d ) = 1. Also, for each d 3, there exists a d-regular graph Y d such that λ (Y d ) = d 3+ (d+3) 16 and κ (Y d ) =. The following result is an immediate consequence of Theorem 1.4 and Theorem 1.5. Corollary 1.6. If G is a 3-regular graph and λ (G) < 5.3, then κ (G) = 3. If G is 4-regular graph and λ < , then κ (G) = 4. Our results imply that if G is a 3-regular graph with λ (G) < 5, then κ(g) = 3. This is because for 3-regular graphs the vertex- and the edge-connectivity are the same. It is very likely that the eigenvalues results for κ(g) and κ (G) can be quite different. We comment on this in the last section of the paper. Proof of Theorem 1.3 In this section, we give a short proof of Theorem 1.3. Recall that a partition V 1 V l = V (G) of the vertex set of a graph G is called equitable if for all 1 i,j l, the number of neighbours in V j of a vertex v in V i is a constant b ij independent of v (see Section 9.3 of [8] or [9] for more details on equitable partitions). Proof of Theorem 1.3. We prove the contrapositive, namely we show that if G is a d-regular graph such that κ (k 1)n (G) k 1, then λ (G) > d. Let V (d+1)(n d 1) 1 V = V (G) be 3

4 a partition of G such that r = e(v 1,V ) k 1 d 1. Note that if V 1 d, then d 1 e(v 1,V ) V 1 (d V 1 + 1) d which is a contradiction. Thus, n i := V i d + 1 for each i = 1,. Since n 1 +n = n, this implies that n 1 n (d+1)(n d 1). The quotient matrix of the partition V 1 V = V (G) (see [8, 9, 10]) is [ d r A = r n n 1 r n 1 d r n The eigenvalues of A are d and d r n 1 r n. Eigenvalue interlacing (see [8, 9, 10]), r k 1 and n 1 n (d + 1)(n d 1) imply that ] () λ (G) d r n 1 r n d (k 1)n n 1 n d (k 1)n (d + 1)(n d 1) (3) We actually have strict inequality here. Otherwise, the partition V 1 V would be equitable. This would mean that each vertex of V 1 has the same number of neighbours in V which is impossible since there are vertices in V 1 without a neighbour in V. The proof is finished. 3 Proof of Theorem 1.4 In this section, we present the proof of Theorem 1.4. First, we show that any connected, regular graph of even degree has edge-connectivity larger than two. This follows from the next lemma. Lemma 3.1. Let G be a connected d-regular graph. If d is even, then κ (G) is even. Proof. Let V = A B be a partition of the vertex set of G such that e(a,b) = κ (G), where e(a,b) denotes the number of edges with one endpoint in A and one endpoint in B. Summing up the degrees of the vertices in A, we obtain that d A = e(a) + e(a,b) = e(a) + κ (G), where e(a) denotes the number of edges with both endpoints in A. Since d is even, this implies κ (G) is even which finishes the proof. For the rest of this section, we assume that d is an odd integer with d 3. We describe first the d-regular graph X d having κ (X d ) = 1 and λ (X d ) = ρ(d). If H 1,...,H k are pairwise vertex-disjoint graphs, then we define the product H 1 H H k recursively as follows. If k = 1, the product is H 1. If k =, then H 1 H is the usual join of H 1 and H, i.e. the graph formed from the union of H 1 and H by joining each vertex of H 1 to each vertex of H. If k 3, then H 1 H H k is the graph obtained by taking the union of H 1 H H k 1 and H k and joining each vertex of H k 1 to each vertex of H k. The extremal graph X d is defined as: X d = K M d 1 K 1 K 1 M d 1 K (4) where M d 1 is the unique (d 3)-regular graph on d 1 vertices (the complement of a perfect matching M d 1 on d 1 vertices). Figure shows X 3 and Figure 3 describes X 5. 4

5 The graph X d is d-regular, has d + 4 vertices and its edge-connectivity is 1. It has an equitable partition of its vertices into six parts which is described by the following quotient matrix X d. The sizes of the parts in this equitable partition are,d 1, 1, 1,d 1,. 1 d d X d = 0 d d 1 0 (5) d d 1 1 Next, we show that the second largest eigenvalue of X d equals the second largest eigenvalue of Xd. This will enable us to get sharp estimates for λ (X d ) which will be used later in this section. Figure : X 3 is 3-regular with λ (X 3 ) = ρ(3).7784 Lemma 3.. For each odd integer d 3, we have that λ (X d ) = λ ( X d ) = ρ(d). Proof. Since the partition with quotient matrix X d is equitable, it follows that the spectrum of X d contains the spectrum of Xd. Using Maple, we determine the characteristic polynomial of Xd : P Xd (x) = (x d)(x 1)(x + )(x 3 (d 3)x (3d )x ) (6) The roots of P Xd are d, 1, and the three roots of Q(x) = x 3 (d 3)x (3d )x. Denote by λ λ 3 λ 4 the roots of Q(x). Since these roots sum up to d 3 0 and their product is, it follows that λ is positive and both λ 3 and λ 4 are negative. Because Q(1) = 4d + 4 < 0, we deduce that λ > 1. Thus, λ ( X d ) = λ = ρ(d). Let W R d+4 be the subspace of vectors which are constant on each part of the six part equitable partition. The lifted eigenvectors corresponding to the six roots of P Xd form a basis for W. The remaining eigenvectors in a basis of eigenvectors for X d can be chosen to be perpendicular to the vectors in W. Thus, they may be chosen to be perpendicular to the characteristic vectors of the parts in the six-part equitable partition since these characteristic vectors form a basis for W. This implies that these eigenvectors will correspond to the non-trivial eigenvalues of the graph obtained as a disjoint union of K and M d 1. The corresponding eigenvalues will be ( ) (d 3), ( 1) (), 0 (d 1), where the exponent denotes the multiplicity of each eigenvalue. These d eigenvalues together with the six roots of P Xd form the spectrum of the graph X d. Thus, λ (X d ) = λ ( X d ) = ρ(d). 5

6 Figure 3: X 5 is 5-regular with λ (X 5 ) = ρ(5) Using Maple, we find that ρ(3).7784 and ρ(5) A simple algebraic manipulation shows that ρ(d) satisfies the equation d ρ(d) = d ρ (d) + 3ρ(d) + > d d + 3d +, (7) where the last inequality follows since ρ(d) < d. Using (7), we deduce that d d + 4 < ρ(d) < d d + 5 for d 5. We are now ready to prove Theorem 1.4. Proof of Theorem 1.4. We will prove the contrapositive, namely we will show that if G is a d-regular graph with edge connectivity 1 (containing a bridge), then λ (G) λ (X d ) = ρ(d). We also show that equality happens if and only if G = X d. Consider a connected d-regular graph G that contains a bridge x 1 x. Deleting the edge x 1 x partitions V (G) into two connected components G 1 and G such that x i V (G i ) for i = 1,. Let n i = V (G i ) for i = 1,. Without loss of generality, we assume from now on that n 1 n. The graph G 1 contains n 1 1 vertices of degree d and one vertex of degree d 1. Because d is odd, this implies that n 1 is odd. By symmetry, n is also odd. Thus, n n 1 d +. Note that G = X d if and only if n 1 = n = d +. Let A be the quotient matrix of the partition of V (G) into V (G 1 ) and V (G ). Then [ d 1 ] A = n 1 1 n 1 1 n d 1 n The eigenvalues of A are: d and λ (A ) = d 1 n 1 1 n. Let A 3 be the quotient matrix of the partition of V (G) into V (G 1 ), {x },V (G ) \ {x }. Then d a a 0 A 3 = 1 0 d 1 (10) 0 b d b where a = 1 n 1 and b = d 1 n 1 3 are d,λ (A 3 ) = d a b+ (d a+b) +4(a b),λ 3 (A 3 ) = d a b (d a+b) +4(a b). Taking partial derivatives with respect to a and to b respectively, we 6 (8) (9)

7 find that when d > 1, the eigenvalue λ (A 3 ) is strictly monotone decreasing with respect to both a and b and so is strictly monotone increasing with respect to both n 1 and n. We consider now a partition of V (G) into the following six parts: V (G 1 )\(x 1 N(x 1 )),N(x 1 )\ {x }, {x 1 }, {x },N(x ) \ {x 1 } and V (G ) \ (x N(x )). Here N(u) denotes the neighbourhood of vertex u in G. Let e 1 denote the number of edges between N(x 1 ) \ {x } and V (G 1 ) \ (x 1 N(x 1 )). Let e denote the number of edges between N(x ) \ {x 1 } and V (G ) \ (x N(x )). Note that e i (d 1)(n i d) for i = 1,. Let A 6 be the quotient matrix of the previous partition of V (G) into six parts. Then d e 1 e 1 n 1 d n d e 1 d 1 e d 1 d 1 A 6 = 0 d d 1 0 (11) d 1 e Eigenvalue interlacing (see [8, 9, 10]) implies that d 1 e d 1 e n d d e n d λ (G) max(λ (A ),λ (A 3 ),λ (A 6 )). (1) We will use this as well as the inequalities (7) and (8) to show that λ (G) λ (X d ) = ρ(d) and that equality happens if and only if G = X d. Recall that n n 1 d + and that d,n 1 and n are all odd. If n 1 d+6 and d 5, then we use the partition of V (G) into two parts whose quotient matrix A is given in (9). Using inequality (8), we have that λ (G) λ (A ) = d 1 n 1 1 n d d + 6 > ρ(d). If n 1 d+6 and d = 3, then we use the partition of V (G) into three parts whose quotient matrix A 3 is given in (10). We have that (3 λ (G) λ (A 3 ) ( 9 8) ) = >.84 > ρ(3). 7 If n n 1 = d + 4, then we use the partition of V (G) into three parts whose quotient matrix A 3 is shown in (10). We have that (d λ (G) λ (A 3 ) d 1 d 1 + d+4 d ) d 1 ( d+4 d ) d 1 d+4 d+3 = d3 + 6d + 8d d d d d 3 + 8d 96d (d + 7d + 1) = d3 + 6d + 8d (d 3 + 8d + 1d 9) + 8d + 10d (d + 7d + 1) 7

8 If d = 3, the right hand side of the previous inequality equals >.796 > ρ(3). When d 5, from the previous inequality and (8) we obtain that λ (G) d3 + 6d + 8d (d 3 + 8d + 1d 9) + 8d + 10d (d + 7d + 1) > d3 + 6d + 8d (d 3 + 8d + 1d 9) = d3 + 7d + 10d 4 (d + 7d + 1) d + 7d + 1 d + 4 = d d + 7d + 1 > d d + 5 > ρ(d) If n 1 = d + and n d + 6, then we use the partition of V (G) into three parts whose quotient matrix is is given in (10). We obtain that (d λ (G) λ (A 3 ) d 1 d 1 + d+ d ) d 1 ( d+ d ) d 1 d+ d+5 = d3 + 6d + 8d 3 + d d d d 3 4d + 56d + 39 (d + 7d + 10) When d = 3, the right hand side equals >.7835 > ρ(3). If 5 d 17, then from the previous identity, we deduce that λ (G) d3 + 6d + 8d 3 + (d 3 + 8d + 8d 5) 8d + 136d (d + 7d + 10) > d3 + 6d + 8d 3 + (d 3 + 8d + 8d 5) (d + 7d + 10) = d3 + 7d + 8d 4 d + 7d + 10 = d d + 5 > ρ(d). When d 19, from the previous inequality we obtain that λ (G) d3 + 6d + 8d 3 + (d 3 + 8d + 8d 6) + d 3 + 8d + 15d + 93 (d + 7d + 10) > d3 + 6d + 8d 3 + (d 3 + 8d + 8d 6) = d3 + 7d + 8d 4.5 (d + 7d + 10) d + 7d + 10 d = d d + 7d + 10 > d d d + 3d + > ρ(d). The only case which remains to consider is n 1 = d + and n = d + 4. In this case, e 1 = d 4,n d = 4 and e is an even integer with 4d 1 e 4d 4. Thus, 1 d d A 6 = 0 d d d 1 e e d 1 d 1 e d e 4 8

9 Let P A6 (x) denote the characteristic polynomial of A 6. Since each row sum of A 6 is d, it follows that d is a root of P A6 (x). Thus, P A6 (x) = (x d)p 5 (x). The second largest root of P A6 (x) is the largest root of P 5 (x). Using the results of [3] page 130, we observe that P 5 (x) equals the characteristic polynomial of the following matrix: d 1 d d 1 0 e d d e e d 1 n d Using Maple to divide P 5 (x) by the polynomial x 3 + (3 d)x + ( 3d)x, we obtain the quotient Q(x) = x 4d 4d de 3e x e and the remainder R(x) = (d e 4(d 1) d 1 )x + de +e +4d 4d. Thus, P 5 (x) = (x 3 + (3 d)x + ( 3d)x )Q(x) + (d e )x + de + 4d + e 4d. Because ρ(d) satisfies the equation x 3 + (3 d)x + ( 3d)x = 0, it follows that P 5 (ρ(d)) = (d e )ρ(d) + de + 4d + e 4d ( = e ρ(d) + d + 1 ) + (d )(ρ(d) d). Because d > ρ(d) > d 1 d+1, we deduce that P 5 (ρ(d)) < 0. (13) Assume that ρ(d) λ (A 6 ). Then ρ(d) is larger than any root of P 5 (x). This implies that P 5 (ρ(d)) = θ root of P 5 (ρ(d) θ) 0 which is a contradiction with (13). Thus, λ (G) λ (A 6 ) > ρ(d). Hence, λ (G) > ρ(d) whenever n d + 4. The last case to be considered is n 1 = n = d +. This means that G = X d and λ (G) = ρ(d) which finishes the proof. 4 Proof of Theorem 1.5 In this section we present the proof of Theorem 1.5. We describe first the d-regular graph Y d having κ (Y d ) = and λ (Y d ) = d 3+ (d+3) 16. Consider the graph H d = K d 1 K. It has d 1 vertices of degree d and two vertices of degree d 1. We construct Y d by taking two disjoint copies of K d 1 K and adding two disjoint edges between the vertices of degree d 1 in different copies of K d 1 K. Figure 4 describes Y 3 and Figure 5 shows Y 4. 9

10 Figure 4: Y 3 is 3-regular with λ (Y 3 ) = θ(3) = 5 The graph Y d is d-regular, has d + vertices and its edge-connectivity is. It has an equitable partition into four parts of sizes d 1,,,d 1 with the following quotient matrix: d 0 0 Ỹ d = d d 1 (14) 0 0 d The characteristic polynomial of this matrix equals P Yd (x) = (x d)(x + 1)(x + (3 d)x + (4 3d)) Thus, the eigenvalues of Ỹd are d, 1 and d 3± (d+3) 16. To simplify our notation, let θ(d) = d 3+ (d+3) 16. Note that θ(d) is the largest root of and T(x) = x + (3 d)x + (4 3d) (15) d 4 d + < θ(d) < d 4 d + 3. (16) Lemma 4.1. The second largest eigenvalue of Y d equals θ(d). Proof. Since the previous partition of V (Y d ) into four parts is equitable, it follows that the four eigenvalues of Ỹ d are also eigenvalues of Y d. Obviously, the second largest of the eigenvalues of Ỹ d is θ(d). By an argument similar to the one of Lemma 3., one can show that the other eigenvalues of Y d are ( 1) (d ) which implies the desired result. We are ready now to prove Theorem 1.5. Proof of Theorem 1.5. We will prove the contrapositive, namely we will show that among all d-regular graphs with edge-connectivity less than or equal to, the graph Y d has the smallest λ. Let G be a d-regular graph with κ (G). We will prove that λ (G) θ(d) with equality if and only if G = Y d. If κ (G) = 1, then Theorem 1.4, (8) and (16) imply that λ (G) ρ(d) d d 4 d+3 > θ(d). 10 (d 1) (d+1)(d+)

11 Figure 5: Y 4 is 4-regular with λ (Y 4 ) = θ(4) = If κ (G) =, then there exists a partition of V (G) into two parts V 1 and V such that e(v 1,V ) =. Let S 1 V 1 and S V denote the endpoints of the two edges between V 1 and V. We have that ( S 1, S ) {(1, ), (, ), (, 1)}. Let n i = V i for i {1, }. It is easy to see that n i d + 1 for each i {1, }. Note that n 1 = n = d + 1 is equivalent to G = Y d. Without loss of generality assume that n n 1 d + 1. If n 1 d + 3, consider the partition of V (G) into V 1 and V. The quotient matrix of this partition is [ d n 1 n 1 n d n and its eigenvalues are d and d n 1 n. Eigenvalue interlacing and n n 1 d+3 imply that λ (G) d n 1 n d 4. Using inequality (16), we obtain that λ d+3 (G) d 4 > θ(d) d+3 which finishes the proof of this case. If n 1 = d +, then we have a few cases to consider. If S 1 =, then e(s 1 ) = 0 or e(s 1 ) = 1. If e(s 1 ) = 0, consider the partition of V (G) into three parts: V 1 \ S 1,S 1,V. The quotient matrix of this partition is d d d 0 d d d n d n Its characteristic polynomial equals ( ) P 3 (x) = (x d) x d n + n dn d x (d n + n + dn d) dn dn If P (x) = x d n +n dn d dn ( where R(x) = dn +n d d n +d n 4 dn dn +n x d x (d n +n + dn d) dn, then ] P (x) = T(x) + R(x) ). The expression d n +d n 4 dn +n d is decreasing with n and thus it attains its maximum when n = d+. This maximum equals d3 +d 8 d +d+4 = 11

12 d 4(d+) < d 4. Thus, P ( d +d+4 d+ (θ(d)) = R(θ(d)) < dn +n d dn d 4 θ(d)) < 0 where d+ the last inequality follows from (16). This fact and eigenvalue interlacing imply that λ (G) > θ(d). If e(s 1 ) = 1, then we consider the same partition into three parts: V 1 \ S 1,S 1,V. The quotient matrix is the following d d 4 d 4 0 d d d n d n Its characteristic polynomial equals ( ) P 3 (x) = (x d) x d n + 4n dn d x d n + 4n + 8 6dn. dn dn If P (x) = x d n +4n dn d dn x d n +4n +8 6dn dn then P (x) = T(x) + R(x) ) ( where R(x) = (dn +n d) d n +dn n 4 dn dn +n x d. The expression d n +dn n 4 dn +n d decreases with n and thus, its maximum is attained at n = d +. This maximum equals d3 +3d 8 d +3d+4 = d 4(d+) < d 4. As before, the previous inequality and eigenvalue interlacing imply d +3d+4 d+ that λ (G) > θ(d). If S 1 = 1, then d must be even. Indeed, the subgraph induced by V 1 contains d + 1 vertices of degree d and one vertex of degree d. This cannot happen when d is odd. Thus, d is even and d 4. Consider the partition of G into three parts: V 1 \ S 1,S 1,V. The quotient matrix of this partition is d d d 0 d+1 d+1 d 0 0 n d n Its characteristic polynomial is ( ) P 3 (x) = (x d) x d n + n d x d n + d + 4n + 8 4dn (d + 1)n (d + 1)n If P (x) = x d n +n d (d+1)n x d n +d+4n +8 4dn, then where R(x) = dn +5n d d n +3dn 8n d 8 dn +5n d (d+1)n ( (d+1)n P (x) = T(x) + R(x) ) x. Because d 4, the expression d n +3dn 8n d 8 dn +5n d is decreasing with n and thus, its maximum is attained when n = d+. This maximum equals d3 +7d 4d 4 = d 1d+4 < d 4 d +7d+8 d +7d+8 interlacing imply λ (G) > θ(d). 1 d+. This fact and eigenvalue

13 Assume now that n 1 = d + 1. This implies that V 1 induces a subgraph isomorphic to K d 1 K and consequently, S 1 = and e(s 1 ) = 0. If n d + 3, then consider the partition of G into three parts: V 1 \ S 1,S 1,V. The quotient matrix of this partition is d 0 d n d n Its characteristic polynomial is If P (x) = x which implies that P 3 (x) = (x d) (d n ) x + + n d, then ( ) x (d n x + + ) d. n P (x) = T(x) + dn + n (n )x n P (θ(d)) = dn + n (n )θ(d) n The expression dn + n (n )θ(d) n is decreasing with n and therefore, its maximum is attained when n = d + 3. Thus, P (θ(d)) d + d 4 (d + 1)θ(d) d + 3 = (d + 1)( d 4 θ(d)) d+1 < 0. d + 3 This inequality and eigenvalue interlacing imply that λ (G) > θ(d). Thus, the remaining case is n 1 = d + 1 and n d +. If d is odd, then n d +. Otherwise, the sum of the degrees of the graph induced by V would equal d V = d(d + ) which is an odd number. This is impossible and thus, n = d + 1. This implies G = Y d. If n = d + and d is even, then we have a few cases to consider. If S = 1, then both vertices in S 1 are adjacent to the vertex a of S. Thus, a has exactly d neighbours in V which means there are 3 vertices of V which are not adjacent to a. Each of these three vertices has degree d and the only way this can happen is they form a clique and each of them is adjacent to the d neighbours of a in V. Using a degree argument, it also follows that the d neighbours of a in V induce a subgraph isomorphic to the complement of a perfect matching on d vertices M d. Using the notation from the previous section, it follows that G = K d 1 K K 1 M d K 3. The graph G has an obvious equitable partition into five parts whose quotient matrix is d d d d d 13

14 Its characteristic polynomial equals P 5 (x) = (x d)p 4 (x) where P 4 (x) = x 4 + ( d + 4)x 3 + ( 4d+4)x 8x+6d 1. Dividing P 4 (x) by T(x) we get that P 4 (x) = T(x)Q 4 (x)+r 4 (x) where Q 4 (x) = x + x 3 and R 4 (x) = 3x 3d. It follows that P 4 (θ(d)) = R 4 (θ(d)) = 3θ(d) 3d < 0. This fact and eigenvalue interlacing imply that λ (G) > θ(d). The case remaining is S =. If e(s ) = 0, then each vertex of S has exactly d 1 neighbours in V. Let S = {x,y }. If x and y are adjacent to the same d 1 vertices of V, then because V = n = d +, there is exactly one vertex of V outside the vertices of S and their d 1 common neighbours. This vertex cannot have degree d. Thus, this case cannot happen and therefore, both x and y have exactly d common neighbours in V. We call this set U. Let {u,w } = V \ ({x,y } U). Because x and y have d 1 neighbours in V, we may assume that x is adjacent to u and y is adjacent to w. A degree argument implies that u and w must be adjacent and both of them are adjacent to each vertex of U. Finally, the subgraph induced by U must be (d 4)-regular. The following is a five-part equitable partition of G: V 1 \ S 1,S 1,S,U, {u,w }. The quotient matrix of this partition is the following: d d d d d 1 Its characteristic polynomial equals P 5 (x) = (x d)p 4 (x) where P 4 (x) = x 4 +(5 d)x 3 +(10 5d)x +( 6d+7)x d. Dividing P 4 (x) by T(x) we get that P 4 (x) = T(x)Q 4 (x)+r 4 (x) where Q 4 (x) = x + x and R 4 (x) = x d. It follows that P 4 (θ(d)) = R 4 (θ(d)) = θ(d) d < 0. This and eigenvalue interlacing show that λ (G) > θ(d). The final case of our proof is e(s ) = 1. Because V \ S = d and both x and y have d neighbours in V \ S, it follows that x and y have at least d 4 common neighbours in V \ S. If x and y have at least d 3 common neighbours in V \ S, we deduce that there exists at least one vertex z of V \ S that is not adjacent to x nor y. The vertex z cannot have degree d since its only possible neighbours are in V \ (S {z}) which has size d 1. We conclude that x and y must have precisely d 4 common neighbours in V \ S. There are 4 remaining vertices a,b,c,d in V \ S and without loss of generality, assume that x is adjacent to both a and b and y is adjacent to both c and d. Each of the vertices a,b,c,d will be adjacent to every other vertex of V \ S. The degree constraint implies that the remaining d 4 vertices of V \ S induce a (d 6)-regular subgraph of G. Obviously, this argument shows this case is only possible when d 6. The following is a five-part equitable partition of G: V 1 \ S 1,S 1,S, {a,b,c,d },V \ (S {a,b,c,d } ). Its quotient matrix is the following: d d d d d 6 14

15 Its characteristic polynomial equals P 5 (x) = (x d)p 4 (x) where P 4 (x) = x 4 + ( d + 4)x 3 + ( 4d+6)x +( d 1)x+3d 6. Dividing P 4 (x) by T(x) we get that P 4 (x) = T(x)Q 4 (x)+ R 4 (x) where Q 4 (x) = x +x 1 and R 4 (x) = x. It follows that P 4 (θ(d)) = R 4 (θ(d)) = θ(d) < 0. This and eigenvalue interlacing imply that λ (G) > θ(d) which finishes our proof. 5 Some Remarks Any strongly regular graph of degree d 3 satisfies the condition λ d and thus, is d-edge-connected. The fact that the edge-connectivity of a strongly regular graph equals its degree, was observed by Plesńik in 1975 (cf. []). As mentioned in the introduction, much more is true, namely the vertex-connectivity of any distance-regular graph equals its degree (see [5]). It is known that any vertex transitive d-regular graph whose second largest eigenvalue is simple has λ (G) d and consequently, is d-edge-connected. In fact, any vertex transitive d-regular graph is d-edge-connected as shown by Mader in 1971 (see [14] or Chapter 3 of [8]). I expect that Theorem 1.4 and Theorem 1.5 can be extended to other values of edgeconnectivity and vertex-connectivity. For example, it seems that λ (G) d 1 implies κ(g). Note however that in many cases, Fiedler s bound κ(g) d λ cannot be improved. When k d and dk is even, consider the graph K d k+1 H where H is a (k d)-regular graph on k vertices. This graph is d-regular, has vertex connectivity k and its second largest eigenvalue equals d k. 6 Acknowledgments I am grateful to David Gregory, Edwin van Dam, Willem Haemers and an anonymous referee for many suggestions that have greatly improved the original manuscript. I thank Professor Nair Abreu, the organizers and the participants in the Spectral Graph Theory in Rio Workshop for their support and enthusiasm. References [1] N.M.M. de Abreu, Old and new results on algebraic connectivity of graphs, Linear Algebra and its Applications 43 (007), [] A.E. Brouwer, Spectrum and connectivity of a graph, CWI Quarterly 9 (1996), [3] A.E. Brouwer, A.M. Cohen and A. Neumaier, Distance Regular Graphs, Springer (1989). [4] A.E. Brouwer and W.H. Haemers, Eigenvalues and perfect matchings, Linear Algebra and its Applications, 395 (005), [5] A.E. Brouwer and J. Koolen, The vertex-connectivity of a distance-regular graph, European Journal of Combinatorics 30 (009),

16 [6] S.L. Chandran, Minimum cuts, girth and spectral threshold, Inform. Process. Lett. 89 (004), no. 3, [7] M. Fiedler, Algebraic connectivity of graphs, Czech. Math. J. 3 (1973), [8] C. Godsil and G. Royle, Algebraic Graph Theory, Springer 001. [9] W.H. Haemers, Eigenvalue techniques in design and graph theory, Ph.D. Thesis (1979), Eindhoven University of Technology. [10] W. Haemers, Interlacing eigenvalues and graphs, Linear Algebra and its Applications (1995) 6-8, [11] S. Hoory, N. Linial and A. Wigderson, Expander graphs and their applications, Bull. Amer. Math. Soc. (006) 46, [1] S. Kirkland, J.J. Molitierno, M. Neumann and B.L. Shader, On graphs with equal algebraic and vertex connectivity, Linear Algebra and its Applications 341 (00), [13] M. Krivelevich and B. Sudakov, Pseudo-random graphs, in More Sets, Graphs and Numbers, Bolyai Society Mathematical Studies 15 (006), [14] W. Mader, Minimale n-fach kantenzusammenhängende Graphen, Math. Ann. 191 (1971), 1-8. [15] B. Mohar, Some applications of Laplace eigenvalues of graphs, Graph Symmetry: Algebraic Methods and Applications, Eds. G. Hahn and G. Sabidussi, NATO ASI Ser. C 487, Kluwer, (1997),