View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Elsevier - Publisher Connector Linear Algebra and its Applications 317 (2000) 217–224 www.elsevier.com/locate/laa A note on diagonally dominant matrices Geir Dahl∗ Department of Mathematics and Department of Informatics, University of Oslo, P.O. Box 1080, Blindern, 0316 Oslo, Norway Received 29 April 1999; accepted 18 May 2000 Submitted by R.A. Brualdi Abstract The set Dn of diagonally dominant symmetric real matrices of order n with nonnegative diagonal elements is a polyhedral convex cone. Based on its extreme rays, we derive a de- scription of the kernel of each matrix in Dn in terms of a certain support graph. Moreover, the doubly stochastic matrices in Dn are studied. © 2000 Elsevier Science Inc. All rights reserved. Keywords: Diagonally dominant matrices; Convex cones; Graphs and matrices 1. Generators and kernels We recall (see,P e.g., [5]) that a real matrix A of order n is called diagonally dom- inant if |ai,i| > j6=i |ai,j | for i = 1,...,n. If all these inequalities are strict, A is strictly diagonally dominant. A matrix is called nonnegative if all its elements 0 are nonnegative. Let Dn denote the set of all diagonally dominant symmetric real matrices of order n and let 0 Dn ={A =[ai,j ]∈Dn: ai,i > 0fori = 1,...,n}. P Observe that A ∈ Dn iff ai,j = aj,i for each i, j and ai,i > j6=i i,j ai,j for i = 1,...,nand i,j ∈{−1, 1} for each i, j. It follows that Dn is a polyhedral (convex) cone in the vector space Rn,n of real matrices of order n. Moreover, this cone is pointed. (Similar results are true without the symmetry assumption, but we concen- 0 trate on the symmetric case below.) Note that Dn is also a cone, but it is nonconvex. ∗ Tel.: 47-22-85-2425; fax: 47-22-85-2401. E-mail address: geird@ifi.uio.no (G. Dahl). 0024-3795/00/$ - see front matter ( 2000 Elsevier Science Inc. All rights reserved. PII:S0024-3795(00)00178-6 218 G. Dahl / Linear Algebra and its Applications 317 (2000) 217–224 The relative interior of Dn consists of the strictly diagonally dominant symmetric matrices with positive elements. n Let ei denote the ith unit vector in R and define the following (−1, 0, 1)-matrices of order n: T (i) Di = eiei for i = 1,...,n; + T (ii) Di,j = (ei + ej )(ei + ej ) for 1 6 i<j6 n; (1) − T (iii) Di,j = (ei − ej )(ei − ej ) for 1 6 i<j6 n. Let Sn be the set of matrices in (1). All these matrices lie in Dn, are positive semi- definite and have rank 1. We let cone(Sn) denote the finitely generated convex cone spanned by Sn (consisting of all nonnegative linear combinations of matrices in Sn). The following result, and some related ones, may be found in [1]. Proposition 1. Dn = cone(Sn) and each matrix in Sn generates an extreme ray of Dn. Proof. As noted Sn ⊆ Dn and therefore cone(Sn) ⊆ Dn. To see the converse in- clusion, consider a matrix A ∈ Dn.DefineP ={(i, j): i<j,ai,j > 0} and N = {(i, j): i<j,ai,j < 0}. It is easy to check that Xn X X X + − A = ai,i − |ai,j | Di + ai,j Di,j + (−ai,j )Di,j . (2) i=1 j6=i (i,j)∈P (i,j)∈N Here all coefficients are nonnegative, so A ∈ cone(Sn) as desired. Moreover, one can check that no matrix in Sn is the sum of other matrices in Sn, so each matrix in Sn generates an extreme ray of Dn. A variation of this result concerns sign-restrictions on the matrix elements. Let T + and T − be a partition of the index set {(i, j):16 i<j6 n}. Consider the set + − + D(T ,T ) of matrices in Dn satisfying ai,j > 0forall(i, j) ∈ T and ai,j 6 0for all (i, j) ∈ T −. For instance, if T − is empty, we obtain the nonnegative matrices in + Dn,orifT is empty, we get the matrices in Dn with nonpositive off-diagonal ele- + − ments. As above, it is easy to see that D(T ,T ) is spanned by the matrices Di for + + − − + − i = 1,...,n, Di,j for (i, j) ∈ T and Di,j for (i, j) ∈ T . Moreover, D(T ,T ) is a simplex cone, meaning that the mentioned matrices (spanning the cone) are lin- early independent. We also remark that a similar result to Proposition 1 holds for the possibly nonsymmetric matrices that are diagonally dominant and have nonnegative diagonal. The dimension of Dn is n(n + 1)/2. Let A ∈ Dn. We shall determine the kernel of A and to this end some graph nota- tion is introduced. Define the node set V ={v1,...,vP n} and the following three edge sets. L consists of the loops [vi,vi ] when ai,i > j6=i |ai,j | for i = 1,...,n,andP (resp. N) consists of the edges [vi,vj ] when ai,j > 0 (resp. ai,j < 0) for 1 6 i<j6 n (this notation is consistent with P and N defined before (2)). Let V1,...,Vt ⊆ V be the connected components of the graph (V, N).LetH denote the graph with G. Dahl / Linear Algebra and its Applications 317 (2000) 217–224 219 node set {V1,...,Vt } and edges [Vi,Vj ] whenever [vi ,vj ]∈P for some vi ∈ Vi and vj ∈ Vj (when i = j this is a loop). Moreover, H contains a loop [Vi,Vi] whenever [vi,vi ]∈L for some vi ∈ Vi. We call H the support graph of A.LetC1,...,Cp denote the components of H that are bipartite and without any loop (p = 0 means that no such component exists.) Note that C1,...,Cp correspond to a coarser partition of V than V1,...,Vt does. Thus, we may write vi ∈ Cj (meaning that vi ∈ Vi and + − Vi ∈ Cj for some Vi). For each j 6 p,letCj and Cj be the two color classes of + − + Cj . Thus, Cj , Cj is a partition of Cj and each edge of H joins a node in Cj and − j n j + a node in Cj .Letz ∈ R be a vector whose support is Cj ,andzi = 1ifvi ∈ Cj j − and zi =−1ifvi ∈ Cj . (If the color classes change role, we obtain the negative of j z , but this ambiguity will not matter below.) Note that we allow the component Cj j to be trivial, i.e., with a single node Vj (but no loop), and then zi = 1ifvi ∈ Vj and j zi = 0otherwise. With this notation, we have the following result on the kernel Ker(A) of a matrix A ∈ Dn. p Theorem 2. Let A ∈ Dn. Then Ker(A) = span({z1,...,z }) and therefore rank(A) = n − p. Proof. Consider the conical representation of A given in (2) and define λi = ai,i − P n j6=i |ai,j | for i = 1,...,n.Letx ∈ R . From the simple structure of the matrices + − Di, Di,j and Di,j , we obtain the following identities: Xn X (i) Ax = λi xiei + ai,j (xi + xj )(ei + ej ) i=1 X (i,j)∈P + (−ai,j )(xi − xj )(ei − ej ); (i,j)∈N Xn X (3) T 2 2 (ii) x Ax = λixi + ai,j (xi + xj ) i=1 X (i,j)∈P 2 + (−ai,j )(xi − xj ) . (i,j)∈N Moreover, in (3) it suffices to sum over those i for which [vi ,vi ]∈L (i.e., λi > 0). Note that xTAx > 0(A is positive semidefinite). Let now x ∈ Ker(A).ThenAx = 0 and therefore xTAx = 0. Thus, (3)(ii) implies that (a) xi = 0 whenever [vi ,vi ]∈L, and (b) xi =−xj whenever [vi ,vj ]∈P and (c) xi = xj whenever [vi ,vj ]∈N. From (c), we see that for each component Vj of (V, N) (i.e., each node of H)there is a number αj such that xi = αj for all vi ∈ Vj . If there is a loop [vi,vi ]∈L for some vi ∈ Vj or if [vi,vk]∈P for some vi ,vk ∈ Vj , then it follows that αj = 0. Moreover, if H contains an edge between Vi and Vj , then we conclude from (b) that αi =−αj . It follows that αi = 0 for all nodes Vi in a component of H that contains a loop or an odd cycle. Consider one of the components Cj , j 6 p (which 220 G. Dahl / Linear Algebra and its Applications 317 (2000) 217–224 has no loop and is bipartite). Then there is some number α (depending on j) such that + − xi = α for each vi ∈ Cj and xi =−α for each vi ∈ Cj . Thus, the restriction of x j to the nodes in Cj lies in span(z ). This holds for every j 6 p and we conclude that x ∈ span({z1,...,zp}). Conversely, assume that x ∈ span({z1,...,zp}). To prove that Ax = 0,itissuf- k k ficient to prove that Az = 0 for each k 6 p. First, observe that (Az )i = 0 for each k k + vi 6∈ Ck as z has its support in Ck (confer (3)(i)). Recall that zi = 1ifvi ∈ Ck k − and zi =−1ifvi ∈ Ck . It follows that λi = 0 for each vi ∈ Ck (as Ck has no loop), k k k k zi =−zj for each [vi ,vj ]∈P (with vi,vj ∈ Ck), and zi = zj for each [vi,vj ]∈N k (with vi ,vj ∈ Ck).