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UNIVERSITY OF OSLO Department of Informatics

A note on nonnegative diagonally

dominant matrices

Geir Dahl

Report 269,

ISBN 82-7368-211-0

April 1999

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A note on nonnegative diagonally dominant matrices

Geir Dahl

April 1999

We make some observations concerning the setCnof real nonnegative, symmetric and diagonally dominant matrices of order n. This set is a convex cone and we determine its extreme rays. From this we derive dierent results, e.g., that the rank and the kernel of each matrixA∈ Cnis determined by a certain support graph ofA, and may be found explicitly.

Moreover, the set of doubly stochastic matrices inCnis studied.

Keywords: Diagonally dominant matrices, convex cones, graphs and ma- trices.

1 An observation

We recall that a real matrix A of order n is called diagonally dominant if

|ai,i| ≥ P

j6=i|ai,j| for i = 1, . . . , n. If all these inequalities are strict, A is strictly diagonally dominant. These matrices arise in many applications as e.g., discretization of partial dierential equations ([14]) and cubic spline interpola- tion ([10]), and a typical problem is to solve a linear system Ax = b where

A is (strictly) diagonally dominant, see also [13]. Strict diagonal dominance is a criterion (which is easy to check) for nonsingularity, and this is important for the estimation of eigenvalues (confer Ger²chgorin disks, see e.g. [7]). For more about diagonally dominant matrices, see [7] or [13]. A matrix is called nonnegative(positive) if all its elements are nonnegative (positive).

LetDn IRn,n denote the set of all matrices of ordernthat are nonnegative and diagonally dominant. The set of symmetric matrices inDn is denoted by

Dn. Both these sets are pointed polyhedral (convex) cones in the vector space

IRn,n of real matrices of ordernas we have

Dn ={AIRn,n: ai,j 0 for1i, jn;

ai,iP

j6=iai,j fori= 1, . . . , n},

Dn ={A∈ Dn: ai,j =aj,i for1i, jn}. (1)

University of Oslo, Dept. of Mathematics and Dept. of Informatics, P.O.Box 1080, Blindern, 0316 Oslo, Norway (Email:[email protected])

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Note that the set of diagonally dominant matrices in IRn,n is a nonconvex cone. The interior ofDnconsists of the positive and strictly diagonallydominant matrices. Similarly,the relative interior ofDnconsists of the symmetric, positive and strictly diagonally dominant matrices.

We mention an interesting result from [9] that is relevant to this note. It was shown that ifA∈ Dn, then Ais completely positive. This means thatA can be factored asA=BBT for some nonnegativen×m matrix. We return to this result in connection with Theorem 3 below.

LetS={v1, . . . ,vk}be a set ofk1vectors in a vector spaceV (over the reals). The nitely generated convex cone

cone(S) ={ Xk j=1

λjvj :λ1, . . . , λk 0}

is said to bespanned by S. If the vectors v1, . . . ,vk are linearly independent, cone(S)is called asimplex cone. We need a simple result on such cones.

Lemma 1

Letcone(S) be the convex cone spanned byS ={v1, . . . ,vk} ⊂ V. Then cone(S) is a simplex cone if and only if each point in cone(S) may be written uniquely as a conical (i.e., nonnegative linear) combination of the vectors

v1, . . . ,vk.

Proof.

Ifv1, . . . ,vkare linearly independent, then the representation is clearly unique. Conversely, assume that the uniqueness of such representations hold and that Pkj=1µjvj =0. Choose nonnegative numbers λj andλ0j forj = 1, . . . , k

such thatµj =λjλ0j for each j. Then 0=Pjµjvj =P

jλjvjP

jλ0jvj. ThereforePjλjvj =P

jλ0jvj so by assumption λj =λ0j andµj = 0 for allj. This shows thatv1, . . . ,vk are linearly independent.

We call the unique representation of a point v in a given simplex cone the conical representationofv.

Leteidenote theith unit vector inIRn and dene the following matrices of ordern:

(i) i =eieTi fori= 1, . . . , n; (ii) i,j = (ei+ej)(ei+ej)T for1i < jn;

(iii) ¯i,j =ei(ei+ej)T fori6=j. (2) These are all(0,1)-matrices.ihas a single one which is in position(i, i). The four ones in the matrixi,j are in positions(i, i),(i, j),(j, i)and(j, j). Finally,

¯i,j has two ones, in positions(i, i)and(i, j). LetSn be the set of matrices in (2)(i) and (iii), and letSn be the set of matrices in (2)(i) and (ii). Note that all these matrices are nonnegative, diagonally dominant and have rank one.

Moreover, the matrices inSn are symmetric and positive semidenite.

Proposition 2

Dn = cone(Sn) and Dn =cone(Sn). Moreover, both Dn and

Dn are simplex cones.

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Proof.

LetAcone(Sn)so there are nonnegative numbersλi forinand

λi,j for1i < jnsuch that

() A= Xn i=1

λii+X

i<j

λi,ji,j.

From this it follows thatAis symmetric (as a linear combination of symmetric matrices) and nonnegative. Moreover, ()gives ai,j =λi,j for 1i < j n

as only the matrixi,j has a nonzero in position (i, j). Moreover, due to the structure of the matricesi,j we also get from()thatai,i=λi+Pj<iλj,i+ P

i<jλi,j=λi+P

j6=iai,jsoλi=ai,iP

j6=iai,j. From this we conclude that

Ais nonnegative and diagonally dominant and thereforeA∈ Dn. Conversely, eachA∈ Dnmay be written in the form()so we conclude thatDn=cone(Sn). Moreover, we see that eachA ∈ Dn has a unique representation as a conical combination of the matricesi and i,j. So, according to Lemma 1, Dn is a simplex cone. The proof of the results forDn is similar.

Related results on generators for certain cones are found in [1]. They study dierent convex cones associated with diagonally dominant matrices, e.g., the complex (or real) matrices of ordern satisfyingai,i P

j6=i|ai,j| (so the only nonnegativity requirements are on the diagonal elements). Note that the set of these matrices is a convex cone, but not a simplex cone.

We hereafter concentrate our study on the symmetric diagonally dominant matrices, i.e., the setDn.

From the previous proof we see that the conical representation of a symmet- ric, nonnegative and diagonally dominant matrixAis simply

A= Xn i=1

(ai,iX

j6=i

ai,j)∆i+X

i<j

ai,ji,j. (3)

We dene thesupport graphof a matrixA∈ Dn as the graphGA= (V, EA) with node set V = {v1, . . . , vn} and edges (i) [vi, vi] (a loop) when ai,i >

P

j6=iai,j for i = 1, . . . , n and (ii) [vi, vj] when ai,j > 0 for 1 i < j n. Thus, edges of GA correspond to the positive coecients in the conical repre- sentation ofA. This graph will be used below.

WhenEis some nite set,SE andxIRE we use the notationx(S) :=

P

eSxe. We also letx() := 0.

2 Some consequences

We now look at some consequences of our proposition.

Dimension and faces.

An immediate consequence of Proposition 2 is that

dim(Dn) =n2(so it is full-dimensional) anddim(Dn) =n(n1)/2.

The kernel.

In order to study the kernel of matrices inDn we need some graph notations. Consider a matrixA∈ Dn. Let C1, . . . , CtV be the con- nected components of the support graphGA. We may assume (after reordering)

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that for1jpthe subgraphGA[Cj]is bipartite and without any loop. Here

0pt andp= 0means that no component is bipartite and without loops.

For eachj p, let Cj+ andCj be the two color classes ofCj. Thus, Cj+,Cj

is a partition ofCj and each edge of GA[Cj] joins a node inCj+ and a node in

Cj. Let zCj IRn be a vector whose support is Cj, and ziCj = 1ifvi Cj+ andzCij =1ifvi Cj. (If the color classes change role, we obtain the nega- tive ofzCj, but this ambiguity will not matter below). Note that we allow the componentCj to be trivial, i.e., with a single nodevi (but no loop), and then

zCj =ei.

With this notation we have the following result on the kernel Ker(A) ={x

IRn:Ax=0}of a matrixA∈ Dn.

Theorem 3

LetA∈ Dn. Then rank(A) =npand Ker(A) =span({zC1, . . . ,zCp}).

Proof.

Consider the conical representationA= Pnj=1λii+P

i<jλi,ji,j

where λi = ai,iP

j6=iai,j 0and λi,j = ai,j 0. Let x IRn. From the simple structure of the matricesi andi,j we obtain the following identities

(i) Ax = Pni=1λixiei+P

i<jλi,j(xi+xj)(ei+ej);

(ii) xTAx = Pni=1λix2i +P

i<jλi,j(xi+xj)2. (4) Moreover, in (4) it suces to sum over those i for which [vi, vi] EA (i.e.,

λi > 0) and those i < j for which [vi, vj] EA (i.e., λi,j > 0). Note that

xTAx0(soAis positive semidenite).

Let nowxKer(A). Then Ax=0and therefore xTAx= 0. Thus, from (4)(ii) we see that

(a) xi = 0 whenever[vi, vi]EA, and (b) xi =xj whenever[vi, vj]EA (andi6=j).

From (a) and (b) it easily follows that xi = 0 for each node vi that lies in a componentCj (of GA) which contains an odd cycle or a loop. Consider a componentCj where 1j p (so GA[Cj] is a bipartite component with no loop). Then it follows from (b) that, for some real numberα, xi =αfor each

vi Cj+ and xi = α for each vi Cj. Thus, the restriction of x to the nodes inCj lies in span(zCj). This holds for everyj pand we conclude that

xspan({zC1, . . . ,zCp}).

Conversely, assume that x span({zC1, . . . ,zCp}). Thenxi = 0 whenever

vilies in one of the componentsCp+1, . . . , Ct. Moreover, for each edge[vi, vj]

EA that belongs to one of the components C1, . . . , Cp we have xi = xj. If

Cj = {vi} is a trivial component (with no loop), then λi = λk,i = λi,j = 0 for 1 k < i and i < j n, so both the ith row and the ith column of A are the zero vector. From these observations and (4)(i) it follows thatAx=0 so x Ker(A). This proves the description of the kernel. Finally, we note that all the vectors spanning the kernel are nonzero and have disjoint supports,

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so they are linearly independent. Therefore, the kernel has dimensionp and rank(A) =np.

From this result we see the interesting fact that the kernel and the rank of a symmetric, nonnegative diagonally dominant matrix A depends only on the support graph. In other words, the kernel and the rank are determined by which coecients in the conical representation (3) that are positive; otherwise the magnitudes of these numbers are irrelevant. The reduced row-echelon form ofAalso has this feature; it only depends onGA. It is also interesting to note that the kernel has a basis consisting of orthogonal(1,0,1)-vectors. Finally, we see that the calculation of rank(A)and Ker(A)is easily done by a breadth- rst-search in the support graphGA (so no numerical calculation is required).

Remark.

Theorem 3 and its proof is related to the already mentioned result of [9] saying that each matrixA∈ Dnis completely positive. In the proof of this result [9] considered the graph GA and dened its weighted incidence matrixB as follows. B has a row for each node ofGA and a column for each edge in EA, and bvi,[vi,vj] = bvj,[vi,vj] = a1/2i,j when [vi, vj] EA, bvi,[vi,vi] = (ai,iP

j6=iai,j)1/2 while all other entries are zero. Then one can check that

A=BBT. In connection with the proof of Theorem 3 we note that Ker(BT) = Ker(BBT) = Ker(A), and thatBTx=0is just conditions (a) and (b) in our proof.

Range and linear systems.

LetA∈ Dn. SinceAis symmetric, we have Ran(A) =Ker(A) where Ran(A) ={Ax:xIRn}is the range ofA. Thus, Ran(A)consists of the vectorsxIRn satisfying

x(Cj+) =x(Cj) forj= 1, . . . , p. (5) (IfCjconsists of a single node, the equation says that the corresponding variable

xi is zero). Choose, for each j p, an index k(j) such that vk(j) Cj (if

Cj = {vi}, let k(j) = i). Then, a basis of Ran(A) consists of the vectors

ei+ek(j),iCj+ andeiek(j),iCj forj= 1, . . . , p. Let nowbIRn and consider the linear system of equationsAx=b. This system has a solution if and only ifbsatises (5). Moreover, if this condition holds the solution set of

Ax=bis the ane setx0+span({zC1, . . . ,zCp})wherex0is some solution of

Ax=b.

Positive (semi)denite.

It is well-known that each symmetric diagonally dominant matrix is positive semidenite. The fact that this is true for nonneg- ative matrices is an immediate consequence of Proposition 2 as we have noted that each matrix inSn is positive semidenite (or it was observed in the proof of Theorem 3). The set of (symmetric) positive semidenite matrices of order

nis a (nonpolyhedral) convex cone PSDn which containsDn as a subcone. See [5] for a discussion of many aspects of PSDn, related cones and convex sets.

The positive denite matrices in Dn may be characterized in terms of the support graph in the following way.

Corollary 4

LetA∈ Dn. Then the following three statements are equivalent:

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(i) Ais positive denite.

(ii) Ais nonsingular.

(iii) Each component of GA contains a loop or an odd cycle.

Proof.

The equivalence of (i) and (ii) follows from the fact thatAis positive semidenite. Moreover,A is nonsingular i rank(A) = n which, by Theorem 3 means that p = 0, i.e., each component of GA contains a loop or an odd cycle.

For instance, consider the matrices

A=

2 1 1

1 2 1

1 1 2

, B=

1 1 0

1 2 1

0 1 1

.

Ais positive denite becauseGAis an odd cycle (a triangle) whileBis singular (GB is a path). Note that these matrices are not strictly diagonally dominant, in fact,ai,i=Pj6=iai,j for eachi(and similar equations hold forB).

LetA∈ Dn be tridiagonal, i.e.,ai,j = 0when|ij|>1, and assume that

ai,i+1 = ai+1,i > 0for i = 1, . . . , n. Then GA contains the path [vi, vi+1] for

i= 1, . . . , n1, soGAis connected. Thus, by Corollary 4,Ais nonsingular (and positive denite) if and only ifai,i > ai,i+1+ai1,i for somei. Such matrices are of interest in connection with cubic splines, see [10]. More generally, assume thatA∈ Dn is not decomposable, i.e., there is no permutation matrixPsuch that PTAP = A1A2 where A1 and A2 are square, nonvacuous matrices.

This implies (in fact, it is equivalent to) thatGAis connected. Thus, Corollary 4 gives thatAis nonsingular (and positive denite) if and only ifGA contains a loop (ai,i>Pj6=iai,j) or an odd cycle.

We refer to [6], [7] and [15] for other criteria for a diagonallydominant matrix to be nonsingular.

Faces.

Recall that a (nontrivial) face of a convex setC is the intersection between C and one of its supporting hyperplanes. Consider A ∈ Dn and let

F(A)denote the smallest face ofDn that containsA. Thendim(F(A)) =|EA|

andF(A)is the simplex cone spanned by the matricesi for[vi, vi]EAand

i,j with [vi, vj] EA. It follows from Theorem 3 that the maximum rank among the matrices inF(A)is npwhere pis the number of components of

GA that are bipartite and without any loop. This maximum rank is achieved for all matrices in the relative interior ofF(A)(i.e., the matrices having conical representation with positive coecient for each edge inEA).

Related polytopes.

Letα >0and consider the set of matrices inDnwith traceα, i.e.,Dn(α) ={A∈ Dn:Pni=1ai,i=α}. ThenDn(α)is a simplex and its vertices are the zero matrix and the points in the intersection between the extreme rays ofDnand the hyperplanePni=1ai,i=α. Thus the nonzero vertices ofDn(α)are the matricesα∆i fori= 1, . . . , nand(α/2)∆i,jfor1i < jn.

Concave optimization.

Let f real-valued convex function dened on

IRn,n, i.e., f((1λ)A+λB) (1λ)f(A) +λf(B) for each A,B IRn,n

and0λ1. An example isf(A) =kAkwhere k · k is an arbitrary matrix norm. Another example is the linear function f(A) = hC,Ai =P

i,jci,jai,j.

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From convexity we know that a convex function dened on a polytope achieves its maximum in one of the vertices, so max{f(A) : A ∈ Dn(α)} equals the maximum of the numbers f(0), f(α∆i) for i = 1, . . . , n and f((α/2)∆i,j)for

1i < jn. (Whenf is positively homogeneous,αmay be moved out of the maximization.) As an example, letf be the spectral norm sof(A) =kAk2 is the largest eigenvalue ofA(asAis symmetric). The characteristic polynomial ofi is 1)λn1 and the characteristic polynomial of i,j is 2)λn1. This givesf(α∆i) =α·λmax(∆i) =αand f(α∆i,j) = (α/2)·λmax(∆i,j) =α. Therefore

max{kAk2:A∈ Dn(α)}=α

and the maximum is attained for all the matricesi andi,j.

Matrices with nonpositive o-diagonal elements.

In the discretization of certain partial dierential equations one is interested in symmetric diagonally dominant matrices with nonnegative diagonal elements, but nonpositive o- diagonal elements. LetMn denote the set of such matrices of ordern. Using similar proof techniques as above one may show the following results. Mn is a simplex cone spanned by the matricesi (as before) and(eiej)(eiej)T for1 i < j n. Dene the support graph GA = (V, EA) nearly as before:

V ={v1, . . . , vn}and edges (i) [vi, vi](a loop) when ai,i > Pj6=i|ai,j| fori =

1, . . . , nand (ii)[vi, vj]whenai,j<0for1i < jn. Again, the edges ofGA correspond to the positive coecients in the conical representation of A. We then have forA∈ Mn that

Ker(A) =span({χC1, . . . , χCq})

whereC1, . . . , Cqare the components ofGAwithout a loop andχCj is the(0,1)- incidence vector ofCj (i.e., χCij is 1 if vi Cj and 0 otherwise). Moreover, rank(A) =nq. Note that, in contrast to the case of nonnegative matrices, whether the components ofGAare bipartite or not plays no role for the kernel or the rank. But again we have the interesting fact that the kernel and the rank depends only on the support graph.

We also see that A ∈ Mn is nonsingular if and only if each component of

GA contains a loop. To recognize this condition, we see that after simultaneous permutations of rows and columns ofAit may be written as the direct sum of smaller matrices, sayA1, . . . ,Ar, each corresponding to a component of GA. Clearly, eachAilies inMk for somek. Now,Aiis irreducible as it corresponds to a component of GA (and Ai is symmetric). Moreover, the statement that this component has a loop just means thatai,i>Pj6=i|ai,j|for somei(where nodevi lies in that component). Thus, Ai is irreducibly diagonally dominant, a known criterion forAi to be nonsingular (see [7]).

Note that if A∈ Mn is nonsingular and ai,i>0for each i (meaning that

Ahas no zero row), then Ais a Stieltjes matrix, i.e., a symmetricM-matrix.

ThusA10.

Eigenvalues in a restricted case.

Consider a matrixA∈ Dn such that

ai,j∈ {0,1}wheni6=jandai,i=Pj6=iai,jfor eachi. Thus, the support graph

GAhas no loop and it determinesAuniquely. We note that by changing sign on

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all o-diagonal elements of Awe obtain the Laplacian matrix ofGA. Assume that GA is bipartite, say with color classes I and J. Thus, A is singular (by Corollary 4) and 0 is the smallest eigenvalue of A. LetµA denote the second smallest eigenvalue ofA. ThenµA is related to connectivity properties of the support graphGA. To clarify this, note rst that

() µA= min{ X

[i,j]EA

(xi+xj)2:kxk= 1, x(I) =x(J)}

asz=χIχJis an eigenvector corresponding to the eigenvalue 0 (zKer(A));

confer the Courant-Fischer minmax theorem, see [7]. Introducing the change of variablesyi=xi foriIandyi=xi foriJ we see from () that

µA= min{ X

[i,j]EA

(yiyj)2:yU}

whereU consists of the vectorsyIRn with kyk= 1andeTy=Pni=1yi = 1. This means thatµA is equal to the so-calledalgebraic connectivity of the graph

GA. We refer to [2] for a discussion of algebraic connectivity (and the Laplacian matrix of a graph). Several properties ofµAis known, but here we just mention that (i)µA is positive if and only ifGA is connected, and (ii)µA is no greater than the node connectivity ofGA.

3 Doubly stochastic diagonally dominant matri- ces

A matrix is doubly stochastic if it is nonnegative and each row and column sum is 1. We let Bn denote the set of doubly stochastic matrices of order n. The set Bn is a convex polytope in IRn,n, often called the Birkho polytope. The classical Birkho-von Neumann theorem states thatBn is the convex hull of all permutation matrices of ordern. For more information about this theorem and doubly stochastic matrices, we refer to [2] and [7]. We are here concerned with the setDBn of symmetric, diagonally dominant and doubly stochastic matrices (of ordern), i.e.,

DBn =Dn∩ Bn.

Note that DBn is a (convex) polytope and that the only integral matrix in

DBn is the identity matrix. We shall give dierent representations ofDBn. Let

Bn be the set of symmetric matrices inBn. Let δ(v) denote the set of edges incident to a nodevin a a graph (including, possibly, the loop[v, v]). We need the following lemma concerning the fractional perfect matching polytope in graphs with loops. It may be proved using techniques explained in [12] (see also [3]).

Lemma 5

LetG= (V, E)be a connected graph, possibly with loops and dene the polytope F M(G) ={xIRE :x0, x(δ(v)) = 1 for allvV}.

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Then xIRE is a vertex ofF M(G) if and only ifxe∈ {0,1/2,1}for each

eE and the edges ewith xe= 1/2form node disjoint odd cycles.

This result may be reformulated in terms of matrices. A symmetric matrix

A IRn,n may be represented by a weighted graph G = (V, E) with nodes

v1, . . . , vn and edges [vi, vj] with associated weight xi,j := ai,j = aj,i for 1 i j n (when i = j we have a loop [vi, vi]). We see that A is symmetric and doubly stochastic i x IRE is nonnegative and x(δ(vi)) = 1 for each

i n. Thus, Bn and F M(G) are anely isomorphic. Let A be a vertex of

Bn. Consider the corresponding vertex x of F M(G) and choose an ordering of the vertices so that (i) the vertices of each fractional cycle (having edges with xe = 1/2) occur consecutively, and (ii) the endnodes of each edge with

xe= 1occur consecutively. The node ordering corresponds to simultaneous line permutations inAand we see from Lemma 5 that the resulting matrixQTAQ may be written as the direct sum of the matrices

[1],

0 1

1 0

, and C(p)

whereC(p)= [ci,j]IRp,p is dened byci,i+1=ci1,i= 1/2for2ip1,

c1,2 =c1,p = cp,1 = cp,p1 = 1/2and ci,j = 0 otherwise. Here the rst and the second matrix corresponds to a loop and an edge withxe= 1, respectively, while C(p) corresponds to a fractional cycle of length pwhere p is odd. This also shows the following result due to [8] (see also [4]).

Proposition 6

The setBn of symmetric doubly stochastic matrices is the con- vex hull of matrices of the form(1/2)(P+PT)wherePis a permutation matrix of order n.

Observe that a matrixA with Pjai,j = 1satises ai,i Pj6=iai,j if and only if it satises ai,i 1/2. It follows that DBn consists of the matricesA satisfying the following linear system

Pn

j=1ai,j = 1 fori= 1, . . . , n;

ai,j =aj,i for1i < jn;

ai,i 1/2 fori= 1, . . . , n;

ai,j 0 for1i < jn.

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Other descriptions ofDBn are contained in the following proposition. We letIn denote the identity matrix of ordern.

Corollary 7

(i)DBn= (1/2)·In+ (1/2)· Bn.

(ii)DBn is the convex hull of the matrices(1/2)·In+ (1/4)·(P+PT)where

Pis a permutation matrix of ordern.

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