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Chromatic number and spectral radius
Linear Algebra and its Applications 426 (2007) 810–814 www.elsevier.com/locate/laa
Chromatic number and spectral radius Vladimir Nikiforov Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, USA Received 31 January 2007; accepted 8 June 2007 Available online 16 June 2007 Submitted by R.A. Brualdi
Abstract Write μ(A) = μ1 (A) · · · μmin (A) for the eigenvalues of a Hermitian matrix A. Our main result is: Let A be a Hermitian matrix partitioned into r × r blocks so that all diagonal blocks are zero. Then for every real diagonal matrix B of the same size as A 1 μ(B − A) μ B + A . r −1 Let G be a nonempty graph, χ(G) be its chromatic number, A be its adjacency matrix, and L be its Laplacian. The above inequality implies the well-known result of Hoffman χ (G) 1 +
μ(A) , −μmin (A)
and also, χ (G) 1 +
μ(A) . μ(L) − μ(A)
Equality holds in the latter inequality if and only if every two color classes of G induce a |μmin (A)|-regular subgraph. © 2007 Elsevier Inc. All rights reserved. AMS classification: 05C50 Keywords: Graph Laplacian; Largest eigenvalue; Least eigenvalue; k-Partite graph; Chromatic number
E-mail address: [email protected] 0024-3795/$ - see front matter ( 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.laa.2007.06.005
V. Nikiforov / Linear Algebra and its Applications 426 (2007) 810–814
811
Main results Write μ(A) = μ1 (A) · · · μmin (A) for the eigenvalues of a Hermitian matrix A. Given a graph G, let χ (G) be its chromatic number, A(G) be its adjacency matrix, and D(G) be the diagonal matrix of its degree sequence; set L(G) = D(G) − A(G). Letting G be a nonempty graph with L(G) = L and A(G) = A, we prove that μ(A) , μ(L) − μ(A) complementing the well-known inequality of Hoffman [1] χ (G) 1 +
(1)
μ(A) . (2) −μmin (A) Equality holds in (1) if and only if every two color classes of G induce a |μmin (A)|-regular subgraph. We deduce inequalities (1) and (2) from a theorem of its own interest. χ (G) 1 +
Theorem 1. Let A be a Hermitian matrix partitioned into r × r blocks so that all diagonal blocks are zero. Then for every real diagonal matrix B of the same size as A, 1 μ(B − A) μ B + A . (3) r −1 Proof of Theorem 1. Write n for the size of A, let [n] = ∪ri=1 Ni be the partition of its index set, and let b1 , . . . , bn be the diagonal entries of B. Set L = B − A, K = (r − 1)B + A, and select a unit eigenvector x = (x1 , . . . , xn ) to μ(K). Our proof strategy is simple: using x, we define specific n-vectors y1 , . . . , yr and show that r(r − 1)μ(L) μ(L) yi 2 Lyi , yi rKx, x = rμ(K). i∈[r]
i∈[r]
For i = 1, . . . , r define yi = (yi1 , . . . , yin ) as −(r − 1)xj if j ∈ Ni , yij = xj if j ∈ [n]\Ni . The Rayleigh principle implies that μ(L) yi 2 μ(L)yi 2 Lyi , yi . i∈[r]
Noting that yi 2 =
i∈[r]
we obtain,
yi 2 = r + r(r − 2)
i∈[r]
i∈[r]
|xj |2 + (r − 1)2
j ∈[n]\Ni
|xj |2 = 1 + r(r − 2)
j ∈Ni
On the other hand, we have Lyi , yi = bj |yij |2 − aj k yik yij . j,k∈[n]
|xj |2 ,
j ∈Ni
|xj |2 = r + r(r − 2) = r(r − 1).
i∈[r] j ∈Ni
j ∈[n]
(4)
(5)
812
V. Nikiforov / Linear Algebra and its Applications 426 (2007) 810–814
For every i ∈ [n], we see that bj |yij |2 = bj |xj |2 + r(r − 2) bj |xj |2 , j ∈[n]
j ∈[n]
j ∈Ni
and, likewise, aj k yik yij = aj k xk xj − r j,k∈[n]
j,k∈[n]
j ∈Ni ,k∈[n]
i∈[r],j ∈[n]
−r
aj k xk xj + r
j,k∈[n]
= r(r − 1)
i∈[r]
bj |xj |2 − r
j ∈[n]
= r(r − 1)
bj |xj |2
i∈[r],j ∈Ni
bj |xj | + r
j ∈[n]
aj k xk xj + r
j ∈Ni ,k∈[n]
aj k xk xj + 2r
j,k∈[n] 2
aj k xk xj .
k∈Ni ,j ∈[n]
Summing these equalities for all i ∈ [r], we find that Lyi , yi = bj |xj |2 + r(r − 2) i∈[r]
aj k xk xj − r
aj k xk xj
k∈Ni ,j ∈[n]
aj k x k x j
j,k∈[n]
aj k xk xj = rKx, x = rμ(K).
j,k∈[n]
Hence, in view of (4) and (5), we obtain (r − 1)μ(B − A) μ(K), completing the proof.
Lemma 2. Let A be an irreducible nonnegative symmetric matrix and R be the diagonal matrix of its rowsums. Then 1 r μ R+ A μ(A) (6) r −1 r −1 with equality holding if and only if all rowsums of A are equal. Proof. Let A = (aij ) and n be its size. Note first that for any vector x = (x1 , . . . , xn ) aij (xi − xj )2 0. (R − A)x, x = 1i
Hence, R − A is positive semidefinite; since A is irreducible, if (R − A)x, x = 0, then all entries of x are equal. 1 Let x = (x1 , . . . , xn ) be an eigenvector to μ = μ(D + r−1 A). We have μ= R +
n n n n 1 1 xi2 aij + aij xi xj A x, x = r −1 r −1 i=1
=
aij (xi − xj )2 +
1i
r r −1 n
= (R − A)x, x +
i=1 j =1
aij xi xj
i=1 j =1
n
r r μ(A), aij xi xj r −1 r −1 i=1 j =1
proving (6).
j =1 n n
(7)
V. Nikiforov / Linear Algebra and its Applications 426 (2007) 810–814
813
Let now equality holds in (6). Then equality holds in (7), and so (R − A)x, x = 0 and x is an eigenvector of A to μ(A). Therefore x1 = · · · = xn and the rowsums of A are equal. If the rowsums of A are equal, the vector j = (1, . . . , 1) is an eigenvector of A to μ(A) and of R 1 to μ(R); therefore j is an eigenvector of R + r−1 A to μ, and so equality holds in (6), completing the proof. Proof of (1) and (2). Let G be a graph with chromatic number χ = r. Coloring the vertices of G into r colors defines a partition of its adjacency matrix A = A(G) with zero diagonal blocks. Letting B be the zero matrix, Theorem 1 implies inequality (2). Letting now B = D = D(G), Lemma 2 implies that 1 r μ D+ A μ(A), r −1 r −1 and inequality (1) follows. The following argument for equality in (1) was kindly suggested by the referee. If equality holds in (1), by Lemma 2, G is regular; hence equality holds also in (2). Setting μ(G) = k, |μmin (G)| = τ and writing α(G) for the independence number of G, let us recall Hoffman’s bound on α(G): for every k-regular graph G, α(G)
nτ . k+τ
(8)
On the other hand, we have n n nτ α(G) = = . χ (G) 1 + k/τ k+τ and thus, equality holds in (8). It is known (see, e.g., [2], Lemma 9.6.2) that this is only possible if χ(G) = n/α(G) and every two color classes of G induce a τ -regular bipartite subgraph. Concluding remarks For the complete graph of order n without an edge, inequality (1) gives χ = n − 1, while (2) gives only χ n/2 + 2. By contrast, for a sufficiently large wheel W1,n , i.e., a vertex joined to all vertices of a cycle of length n, we see that (1) gives χ 2, while (2) gives χ 3. A natural question is to determine when equality holds in (3). A particular answer, building upon [4], can be found in [5]: if G is a connected graph, then μ(D(G) − A(G)) = μ(D(G) + A(G)) if and only if G is bipartite. Problem 3. Determine when equality holds in (3). Finally, any lower bound on μ(A(G)), together with (1), gives a lower bound on μ(L(G)). This approach helps deduce some inequalities for bipartite graphs given in [3] and [6]. Acknowledgments Thanks to Peter Rowlinson, Sebi Cioab˘a and Cecil Rousseau for useful suggestions. The author is most indebted to the referee for the exceptionally thorough, helpful and kind report.
814
V. Nikiforov / Linear Algebra and its Applications 426 (2007) 810–814
References [1] A.J. Hoffman, On eigenvalues and colorings of graphs, Graph Theory and its Applications, Academic Press, New York, 1970, pp. 79–91. [2] C. Godsil, G. Royle, Algebraic Graph Theory, Graduate Texts in Mathematics, vol. 207, Springer-Verlag, New York, 2001, xx+439pp. [3] Y. Hong, X.-D. Zhang, Sharp upper and lower bounds for largest eigenvalue of the Laplacian matrices of trees, Discrete Math. 296 (2005) 187–197. [4] R. Merris, Laplacian matrices of graphs: a survey, Linear Algebra Appl. 197–198 (1994) 143–176. [5] J.-L. Shu, Y. Hong, K. Wen-Ren, A sharp upper bound on the largest eigenvalue of the Laplacian matrix of a graph, Linear Algebra Appl. 347 (2002) 123–129. [6] A.M. Yu, M. Lu, F. Tian, On the spectral radius of graphs, Linear Algebra Appl. 387 (2004) 41–49.
Chromatic number and spectral radius Vladimir Nikiforov Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, USA Received 31 January 2007; accepted 8 June 2007 Available online 16 June 2007 Submitted by R.A. Brualdi
Abstract Write μ(A) = μ1 (A) · · · μmin (A) for the eigenvalues of a Hermitian matrix A. Our main result is: Let A be a Hermitian matrix partitioned into r × r blocks so that all diagonal blocks are zero. Then for every real diagonal matrix B of the same size as A 1 μ(B − A) μ B + A . r −1 Let G be a nonempty graph, χ(G) be its chromatic number, A be its adjacency matrix, and L be its Laplacian. The above inequality implies the well-known result of Hoffman χ (G) 1 +
μ(A) , −μmin (A)
and also, χ (G) 1 +
μ(A) . μ(L) − μ(A)
Equality holds in the latter inequality if and only if every two color classes of G induce a |μmin (A)|-regular subgraph. © 2007 Elsevier Inc. All rights reserved. AMS classification: 05C50 Keywords: Graph Laplacian; Largest eigenvalue; Least eigenvalue; k-Partite graph; Chromatic number
E-mail address: [email protected] 0024-3795/$ - see front matter ( 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.laa.2007.06.005
V. Nikiforov / Linear Algebra and its Applications 426 (2007) 810–814
811
Main results Write μ(A) = μ1 (A) · · · μmin (A) for the eigenvalues of a Hermitian matrix A. Given a graph G, let χ (G) be its chromatic number, A(G) be its adjacency matrix, and D(G) be the diagonal matrix of its degree sequence; set L(G) = D(G) − A(G). Letting G be a nonempty graph with L(G) = L and A(G) = A, we prove that μ(A) , μ(L) − μ(A) complementing the well-known inequality of Hoffman [1] χ (G) 1 +
(1)
μ(A) . (2) −μmin (A) Equality holds in (1) if and only if every two color classes of G induce a |μmin (A)|-regular subgraph. We deduce inequalities (1) and (2) from a theorem of its own interest. χ (G) 1 +
Theorem 1. Let A be a Hermitian matrix partitioned into r × r blocks so that all diagonal blocks are zero. Then for every real diagonal matrix B of the same size as A, 1 μ(B − A) μ B + A . (3) r −1 Proof of Theorem 1. Write n for the size of A, let [n] = ∪ri=1 Ni be the partition of its index set, and let b1 , . . . , bn be the diagonal entries of B. Set L = B − A, K = (r − 1)B + A, and select a unit eigenvector x = (x1 , . . . , xn ) to μ(K). Our proof strategy is simple: using x, we define specific n-vectors y1 , . . . , yr and show that r(r − 1)μ(L) μ(L) yi 2 Lyi , yi rKx, x = rμ(K). i∈[r]
i∈[r]
For i = 1, . . . , r define yi = (yi1 , . . . , yin ) as −(r − 1)xj if j ∈ Ni , yij = xj if j ∈ [n]\Ni . The Rayleigh principle implies that μ(L) yi 2 μ(L)yi 2 Lyi , yi . i∈[r]
Noting that yi 2 =
i∈[r]
we obtain,
yi 2 = r + r(r − 2)
i∈[r]
i∈[r]
|xj |2 + (r − 1)2
j ∈[n]\Ni
|xj |2 = 1 + r(r − 2)
j ∈Ni
On the other hand, we have Lyi , yi = bj |yij |2 − aj k yik yij . j,k∈[n]
|xj |2 ,
j ∈Ni
|xj |2 = r + r(r − 2) = r(r − 1).
i∈[r] j ∈Ni
j ∈[n]
(4)
(5)
812
V. Nikiforov / Linear Algebra and its Applications 426 (2007) 810–814
For every i ∈ [n], we see that bj |yij |2 = bj |xj |2 + r(r − 2) bj |xj |2 , j ∈[n]
j ∈[n]
j ∈Ni
and, likewise, aj k yik yij = aj k xk xj − r j,k∈[n]
j,k∈[n]
j ∈Ni ,k∈[n]
i∈[r],j ∈[n]
−r
aj k xk xj + r
j,k∈[n]
= r(r − 1)
i∈[r]
bj |xj |2 − r
j ∈[n]
= r(r − 1)
bj |xj |2
i∈[r],j ∈Ni
bj |xj | + r
j ∈[n]
aj k xk xj + r
j ∈Ni ,k∈[n]
aj k xk xj + 2r
j,k∈[n] 2
aj k xk xj .
k∈Ni ,j ∈[n]
Summing these equalities for all i ∈ [r], we find that Lyi , yi = bj |xj |2 + r(r − 2) i∈[r]
aj k xk xj − r
aj k xk xj
k∈Ni ,j ∈[n]
aj k x k x j
j,k∈[n]
aj k xk xj = rKx, x = rμ(K).
j,k∈[n]
Hence, in view of (4) and (5), we obtain (r − 1)μ(B − A) μ(K), completing the proof.
Lemma 2. Let A be an irreducible nonnegative symmetric matrix and R be the diagonal matrix of its rowsums. Then 1 r μ R+ A μ(A) (6) r −1 r −1 with equality holding if and only if all rowsums of A are equal. Proof. Let A = (aij ) and n be its size. Note first that for any vector x = (x1 , . . . , xn ) aij (xi − xj )2 0. (R − A)x, x = 1i
Hence, R − A is positive semidefinite; since A is irreducible, if (R − A)x, x = 0, then all entries of x are equal. 1 Let x = (x1 , . . . , xn ) be an eigenvector to μ = μ(D + r−1 A). We have μ= R +
n n n n 1 1 xi2 aij + aij xi xj A x, x = r −1 r −1 i=1
=
aij (xi − xj )2 +
1i
r r −1 n
= (R − A)x, x +
i=1 j =1
aij xi xj
i=1 j =1
n
r r μ(A), aij xi xj r −1 r −1 i=1 j =1
proving (6).
j =1 n n
(7)
V. Nikiforov / Linear Algebra and its Applications 426 (2007) 810–814
813
Let now equality holds in (6). Then equality holds in (7), and so (R − A)x, x = 0 and x is an eigenvector of A to μ(A). Therefore x1 = · · · = xn and the rowsums of A are equal. If the rowsums of A are equal, the vector j = (1, . . . , 1) is an eigenvector of A to μ(A) and of R 1 to μ(R); therefore j is an eigenvector of R + r−1 A to μ, and so equality holds in (6), completing the proof. Proof of (1) and (2). Let G be a graph with chromatic number χ = r. Coloring the vertices of G into r colors defines a partition of its adjacency matrix A = A(G) with zero diagonal blocks. Letting B be the zero matrix, Theorem 1 implies inequality (2). Letting now B = D = D(G), Lemma 2 implies that 1 r μ D+ A μ(A), r −1 r −1 and inequality (1) follows. The following argument for equality in (1) was kindly suggested by the referee. If equality holds in (1), by Lemma 2, G is regular; hence equality holds also in (2). Setting μ(G) = k, |μmin (G)| = τ and writing α(G) for the independence number of G, let us recall Hoffman’s bound on α(G): for every k-regular graph G, α(G)
nτ . k+τ
(8)
On the other hand, we have n n nτ α(G) = = . χ (G) 1 + k/τ k+τ and thus, equality holds in (8). It is known (see, e.g., [2], Lemma 9.6.2) that this is only possible if χ(G) = n/α(G) and every two color classes of G induce a τ -regular bipartite subgraph. Concluding remarks For the complete graph of order n without an edge, inequality (1) gives χ = n − 1, while (2) gives only χ n/2 + 2. By contrast, for a sufficiently large wheel W1,n , i.e., a vertex joined to all vertices of a cycle of length n, we see that (1) gives χ 2, while (2) gives χ 3. A natural question is to determine when equality holds in (3). A particular answer, building upon [4], can be found in [5]: if G is a connected graph, then μ(D(G) − A(G)) = μ(D(G) + A(G)) if and only if G is bipartite. Problem 3. Determine when equality holds in (3). Finally, any lower bound on μ(A(G)), together with (1), gives a lower bound on μ(L(G)). This approach helps deduce some inequalities for bipartite graphs given in [3] and [6]. Acknowledgments Thanks to Peter Rowlinson, Sebi Cioab˘a and Cecil Rousseau for useful suggestions. The author is most indebted to the referee for the exceptionally thorough, helpful and kind report.
814
V. Nikiforov / Linear Algebra and its Applications 426 (2007) 810–814
References [1] A.J. Hoffman, On eigenvalues and colorings of graphs, Graph Theory and its Applications, Academic Press, New York, 1970, pp. 79–91. [2] C. Godsil, G. Royle, Algebraic Graph Theory, Graduate Texts in Mathematics, vol. 207, Springer-Verlag, New York, 2001, xx+439pp. [3] Y. Hong, X.-D. Zhang, Sharp upper and lower bounds for largest eigenvalue of the Laplacian matrices of trees, Discrete Math. 296 (2005) 187–197. [4] R. Merris, Laplacian matrices of graphs: a survey, Linear Algebra Appl. 197–198 (1994) 143–176. [5] J.-L. Shu, Y. Hong, K. Wen-Ren, A sharp upper bound on the largest eigenvalue of the Laplacian matrix of a graph, Linear Algebra Appl. 347 (2002) 123–129. [6] A.M. Yu, M. Lu, F. Tian, On the spectral radius of graphs, Linear Algebra Appl. 387 (2004) 41–49.