Linea Algeba and its Applications 426 2007) 810 814 www.elsevie.com/locate/laa Chomatic numbe and spectal adius Vladimi Nikifoov Depatment of Mathematical Sciences, Univesity of Memphis, Memphis, TN 38152, USA Received 31 Januay 2007; accepted 8 June 2007 Available online 16 June 2007 Submitted by R.A. Bualdi Abstact Wite μa) = μ 1 A) μ min A) fo the eigenvalues of a Hemitian matix A. Ou main esult is: Let A be a Hemitian matix patitioned into blocks so that all diagonal blocks ae zeo. Then fo evey eal diagonal matix B of the same size as A μb A) μ B + 1 ) A. Let G be a nonempty gaph, χg) be its chomatic numbe, A be its adjacency matix, and L be its Laplacian. The above inequality implies the well-known esult of Hoffman and also, χg) 1 + μa) μ min A), μa) χg) 1 + μl) μa). Equality holds in the latte inequality if and only if evey two colo classes of G induce a μ min A) -egula subgaph. 2007 Elsevie Inc. All ights eseved. AMS classification: 05C50 Keywods: Gaph Laplacian; Lagest eigenvalue; Least eigenvalue; k-patite gaph; Chomatic numbe E-mail addess: vnikifv@memphis.edu 0024-3795/$ - see font matte 2007 Elsevie Inc. All ights eseved. doi:10.1016/j.laa.2007.06.005
V. Nikifoov / Linea Algeba and its Applications 426 2007) 810 814 811 Main esults Wite μa) = μ 1 A) μ min A) fo the eigenvalues of a Hemitian matix A. Given a gaph G, let χg) be its chomatic numbe, AG) be its adjacency matix, and DG) be the diagonal matix of its degee sequence; set LG) = DG) AG). Letting G be a nonempty gaph with LG) = L and AG) = A, we pove that μa) χg) 1 + μl) μa), 1) complementing the well-known inequality of Hoffman [1] χg) 1 + μa) μ min A). 2) Equality holds in 1) if and only if evey two colo classes of G induce a μ min A) -egula subgaph. We deduce inequalities 1) and 2) fom a theoem of its own inteest. Theoem 1. Let A be a Hemitian matix patitioned into blocks so that all diagonal blocks ae zeo. Then fo evey eal diagonal matix B of the same size as A, μb A) μ B + 1 ) A. 3) Poof of Theoem 1. Wite n fo the size of A, let [n] = i=1 N i be the patition of its index set, and let b 1,...,b n be the diagonal enties of B. Set L = B A, K = )B + A, and select a unit eigenvecto x = x 1,...,x n ) to μk). Ou poof stategy is simple: using x, we define specific n-vectos y 1,...,y and show that )μl) μl) y i 2 i, y i Kx, x =μk). i [] i [] Ly Fo i = 1,..., define y i = y i1,...,y in ) as { )xj if j N y ij = i, x j if j [n]\n i. The Rayleigh pinciple implies that μl) y i 2 μl) y i 2 i, y i. 4) i [] i [] i [] Ly Noting that y i 2 = x j 2 + ) 2 x j 2 = 1 + 2) x j 2, \N i j N i j N i we obtain, y i 2 = + 2) x j 2 = + 2) = ). 5) i [] i [] j N i On the othe hand, we have Ly i, y i = b j y ij 2 a jk y ik y ij.
812 V. Nikifoov / Linea Algeba and its Applications 426 2007) 810 814 Fo evey i [n], we see that b j y ij 2 = b j x j 2 + 2) b j x j 2, j N i and, likewise, a jk y ik y ij = a jk x k x j a jk x k x j a jk x k x j. j N i,k [n] k N i, Summing these equalities fo all i [], we find that Ly i, y i = b j x j 2 + 2) b j x j 2 i [] i [], i [],j N i ) a jk x k x j + a jk x k x j a jk x k x j + i [] = ) b j x j 2 = ) b j x j 2 + j N i,k [n] a jk x k x j + 2 k N i, a jk x k x j a jk x k x j = Kx, x =μk). Hence, in view of 4) and 5), we obtain )μb A) μk), completing the poof. Lemma 2. Let A be an ieducible nonnegative symmetic matix and R be the diagonal matix of its owsums. Then μ R + 1 ) A μa) 6) with equality holding if and only if all owsums of A ae equal. Poof. Let A = a ij ) and n be its size. Note fist that fo any vecto x = x 1,...,x n ) R A)x, x = a ij x i x j ) 2 0. 1 i<j n Hence, R A is positive semidefinite; since A is ieducible, if R A)x, x =0, then all enties of x ae equal. Let x = x 1,...,x n ) be an eigenvecto to μ = μd + 1 1 A). Wehave μ= R + 1 ) n n A x, x = xi 2 a ij + 1 n n a ij x i x j = poving 6). 1 i<j n a ij x i x j ) 2 + = R A)x, x + i=1 n j=1 n n a ij x i x j n a ij x i x j μa), 7)
V. Nikifoov / Linea Algeba 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 eigenvecto of A to μa). Theefoe x 1 = =x n and the owsums of A ae equal. If the owsums of A ae equal, the vecto j = 1,...,1) is an eigenvecto of A to μa) and of R to μr); theefoe j is an eigenvecto of R + 1 1 A to μ, and so equality holds in 6), completing the poof. Poof of 1) and 2). Let G be a gaph with chomatic numbe χ =. Coloing the vetices of G into colos defines a patition of its adjacency matix A = AG) with zeo diagonal blocks. Letting B be the zeo matix, Theoem 1 implies inequality 2). Letting now B = D = DG), Lemma 2 implies that μ D + 1 ) A μa), and inequality 1) follows. The following agument fo equality in 1) was kindly suggested by the efeee. If equality holds in 1), by Lemma 2, G is egula; hence equality holds also in 2). Setting μg) = k, μ min G) =τ and witing αg) fo the independence numbe of G, let us ecall Hoffman s bound on αg): fo evey k-egula gaph G, αg) nτ k + τ. 8) On the othe hand, we have αg) n χg) = n 1 + k/τ = nτ 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 χg) = n/αg) and evey two colo classes of G induce a τ-egula bipatite subgaph. Concluding emaks Fo the complete gaph of ode n without an edge, inequality 1) givesχ = n 1, while 2) gives only χ n/2 + 2. By contast, fo a sufficiently lage wheel W 1,n, i.e., a vetex joined to all vetices of a cycle of length n, we see that 1)givesχ 2, while 2) givesχ 3. A natual question is to detemine when equality holds in 3). A paticula answe, building upon [4], can be found in [5]: if G is a connected gaph, then μdg) AG)) = μdg) + AG)) if and only if G is bipatite. Poblem 3. Detemine when equality holds in 3). Finally, any lowe bound on μag)), togethe with 1), gives a lowe bound on μlg)). This appoach helps deduce some inequalities fo bipatite gaphs given in [3] and [6]. Acknowledgments Thanks to Pete Rowlinson, Sebi Cioabă and Cecil Rousseau fo useful suggestions. The autho is most indebted to the efeee fo the exceptionally thoough, helpful and kind epot.
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