sparsity patterns with high rank extremal positive …helton/billspapersscanned/hlw94.pdf · 300 j....

SIAM J. MATRIX ANAL. APPL.Vol. 15, No. I, pp. 299-312, January 1994

()1994 Society for Industrial and Applied Mathematics023

SPARSITY PATTERNS WITH HIGH RANK EXTREMAL POSITIVESEMIDEFINITE MATRICES*

J. WILLIAM HELTONi$, DANIEL LAMi, AND HUGO J. WOERDEMAN

Abstract. This article concerns the positive semidefinite matrices M+(G) with zero entries inprescribed locations; that is, matrices with given sparsity graph G. The issue here is the rank ofthe extremals of the cone M+(G). It was shown in [J. Agler, J. W. Helton, S. McCullough, andL. Rodman, Linear Algebra Appl., 107 (1988), pp. 101-149] that the key in constructing high rankextreme points resides in certain atomic graphs G called blocks and superblocks. The k-superblocksare defined to be sparsity graphs G that contain an extreme point of rank k while containing (in anextremely strong sense) no graph with the same property. The goal of this article is to write downall graphs that are superblocks. The article succeeds completely for k

_4 and it lists necessary

conditions in general as well as sufficient conditions. The subject is closely related to orthogonalrepresentations of graphs as studied earlier in [L. Lovsz, M. Saks, and A. Schrijver, Linear AlgebraAppl., 114/115 (1989), pp. 439-454] and in the previously mentioned paper by Alger et al. Indeed,the paper is an extension of the findings of Alger et al.

Key words, extremal matrix, order of a graph, superblocks, orthogonal representation

AMS subject classifications, primary 05C50; secondary 05B20, 15A57

Introduction. Let G be an undirected graph without multiple edges or loops.Let V(G) (= {1,..., n)) denote the set of vertices of G and E(G) c V(G) V(G) theset of edges. Note that the absence of loops means that (i, i) E(G), i 1,..., n.Define M+(G) to be the closed cone of all positive semidefinite n n real symmetricmatrices whose (i, j) entry is zero whenever (i, j) E E(G). (Note the difference indefinition compared to preceding papers on the subject ([AHMR], [HPR]); in thosepapers the zero entries would correspond to edges in the complementary graph.)

A matrix A in M+(G) is called an extremal when each additive decomposition ofA in M+(G) is a trivial one, i.e., A is an extremal when A B/C with B, C E M+(G)yields that B, C span {A}. We say that G has order k if k is the maximum of theranks of extremals in M+(G).

We are interested in determining the order of a given graph. The graphs of order 1have a very elegant characterization (see [AHMR], [PPS]) based on the main result in[GJSW]. The general case turns out to be very hard, and, therefore, some reductionsmust be made. Recall from [AHMR] the following definitions. A graph G is called ak-block if G has order k but no induced subgraph has order k. (The aph ( is aninduced subgraph of G if Y(() c V(G) and E(() E(G)f (Y() V(G)).) In termsof matrices this means that for a k-block G any rank k extremal in M+(G) does nothave zero rows or columns. A full description of k-blocks, k 1, 2,..., would give asolution to our problem, since the order of a graph equals the maximal k for whichthe graph has a k-block as an induced subgraph ([AHMR, Whm. 1.2]). Classifyingk-blocks is based on the study of much better behaved objects called k-superblocks.

*Received by the editors October 22, 1990; accepted for publication (in revised form) May 4,1992.

Department of Mathematics, University of California at San Diego, La Jolla, California 92093(heltonosiris.ucsd.edu, hugocs.wm.edu).

:Supported in part by the Air Force Office of Scientific Research and the National ScienceFoundation.

Supported by the Netherlands Organization for Scientific Research (NWO).

299

300 J. WILLIAM HELTON DANIEL LAM AND HUGO J. WOERDEMAN

A graph G is called a k-superblock when it is a k-block that does not properly containanother k-block. (In this paper "( is contained in G" always means Y() c V(G)and E() c E(G).) In terms of matrices, this means that as soon as you allow someof the zero entries prescribed by G to be nonzero there are no extremals of rank kanymore. It is true (see [AHMR]) that any k-block (or any order k graph, for thatmatter) contains a k-superblock. However, to obtain all k-blocks assuming one knowshow to characterize k-superblocks still requires work.

The following theorem gives necessary conditions for a sparsity pattern to be ak-superblock.

THEOREM 0.1. Let G be a k-superblock. Then the following are true:(i) #E(G) 1/2 (k + 2)(k- 1);(ii) G contains no Kp,q, p + q > k + 1;(iii) For all il,...,im 6 V(G) with 1 <_ m < k we have that

(0.1) #((i,j) e E(G) i or j e {il,...,im}}1m(2k + 1 m)<1 (k + 2)(k 1) 1 (k m + 2)(k m 1)

Conversely, when k 1, 2, 3, or 4 these conditions imply that G is a k-superblock.Here Kp,q denotes the bipartite graph described by

V(Kp,q) {1,... ,p -F q}, E(Kp,q) ((i,j) 1 <_ i <_ p, p 4- 1 <_ j <_ p -F q}.

The necessary conditions (i) and (ii) were established earlier in [AHMR]. Condi-tion (iii) is implied by (i) and (ii) when k 1, 2, 3 but not when k _> 4. For k- 1, 2, 3the k-superblocks were described earlier in [AHMR], and indeed they are precisely thegraphs which satisfy the necessary conditions in Theorem 0.1. For k 4 this is alsotrue (as stated in Theorem 0.1). This follows from the description of 4-superblocksgiven in the next theorem, which is the second main result in this paper.

THEOREM 0.2. Let G be a graph with nine edges. The following are equivalent"(i) G is a 4-superblock;(ii) G cannot be obtained from

by identifying vertices;1(iii) G is a graph which, after identifying vertices, is one of the following 28

graphs:

A

See the definition of a collapse of a graph in 2.

HIGH RANK EXTREMAL POSITIVE MATRICES 301

GlO Gll G12=

G15

G22= G23

G28 -Note that (ii) in Theorem 0.2 is a restatement of the necessary conditions inTheorem 0.1 for k- 4.

In 2 we develop some results which give sufficient conditions for a graph to bea k-superblock. Unfortunately these conditions do not equal the necessary conditionsin Theorem 0.1. In the case when k 4, for instance, 17 of the 28 graphs in Theorem0.2(iii) meet the necessary conditions of Theorem 0.1, but not our general sufficientconditions. To prove that these 17 graphs are 4-superblocks, we used a computer pro-gram employing Mathematica (using integer arithmetic). It is natural to ask whetherthis gap in the theory can be dissolved in the following way.

SPECULATION 0.3. Let G be a graph satisfying (i), (ii), and (iii) in Theorem 0.1.Then G is a k-superblock.

We shall point out in the end of 1 what remains to be done in order to provethis speculation. We used our Mathematica program to check some likely candidatesfor counterexamples (with k 5 and 6), but so far we have been unsuccessful (partlybecause the program is very slow when k is large).

1. Making extremals in M+ (G). From [AHMR] one can deduce the followingrecipe for making all extremals in M+(G) of rank k.

Let G be a graph, and let k <_ #V(G).

302 J. WILLIAM HELTON, DANIEL LAM, AND HUGO J. WOERDEMAN

(Step I)

(Step II)

Find an assignment f: V(a)= {1,..., n} k such that(a) (f(i), f(j)) O, (i,j) e E(G);(b) span {f(j)[j e V(G)} k.Check if all M MT E ]kxk satisfying

(1.1) (Mr(i), f(j)) O, (i,j) e E(G)

(Step III)are multipliers of the k x k identity matrix Ik.If so, then

f(1)T(f(1) f(n))

T

is an extremal of rank k in M+(G).An assignment f: V(G) -- k such that (a) in Step I holds is called an orthogonal

representation of G. Such representations were introduced and studied independentlyin [LSS] and [AHMR] (for quite different reasons).

Example. Let

Then

1

G=6

5 4

1 1 00 0 10 (100110) 1 1 (11 1 -1 01 0 00 5 7

0 1 1 01 1 -1 0 7

are extremals in M+(G) of ranks 1 and 2, respectively. There are no extremals ofrank _> 3 because of dimension counting. Indeed, if k 3, either f(5) 0 or span{f(2), f(3), f(6) } has dimension of at most 2. This results in at most four constraints,appearing in (1.1), on M. This can never force M to be a scalar multiple of I3.Furthermore, when k > 3, the five constraints (1.1) on M are too few to force M tobe a scalar multiple of Ik.

The type of arguments used in the example led in [AHMR] to the following result.Let G be a graph. If order G k, then #E(G) > 1/2 (k + 2)(k 1). Furthermore,

if order G k and #E(G) 1/2 (k + 2)(k- 1), then G contains no Kp,q, p+q > k + 1.Let us now prove that a k-superblock must satisfy the conditions (i), (ii), and

(iii) in Theorem 0.1.Proof of the necessary part of Theorem 0.1. Let G be a k-superblock. In par-

ticular, order G k, and thus the recipe works for a representation f, say. Letil,...,ik V(G) be such that f(ij), j 1,... ,ik, span ]1(k. Furthermore, choose


distinct ik+l,...,i such that each edge in E(G) has an endpoint in (i,...,i} andwrite

k

f(ip) ).P) f(ij) p k + 1,... ,1.j--1

Let M be a k k matrix, and put wj Mf(ij), j 1 k. Let uj) u(j) denote... ..., pj

the vertices adjacent to ij that are not in the set i1,..., ij_. Put

Uj [f(uj)) ()f(Upj )], j= 1,...,/.

Put

and

"--f(i2) --f(i3) 0 f(ik 0 0f(il) 0 --f(i3) 0 --f(ik) 00 f(il) f(i2) 0 0 0

0 0 0 0 0 f(ik)0 0 0 f(i) f(i2) f(ik-1)

That is to say, if 1/2 j(j- 1) < p < 1/2 (j + 1)j then column p in has on the p- 1/2 j(j- 1)row the entry -f(ij+l) and on the (j+l) row the entry f(ip-j(_)/2) (j 1,..., k-l).Note that W is of size k #E(G) and of size k 1/2 k(k- 1). The equations (1.1)are equivalent to

col (wi)k= E cokernel W

and the symmetry of M is ensured by

col (wi)ik= E cokernel .Checking Step II in the recipe now comes down to checking that

(1.2) cokernel [W, ] span {col (f(ij))=l)

Note that the inclusion D in (1.2) is always fulfilled since f is an orthogonal repre-sentation. We assume that the recipe works, and therefore (1.2) holds. Suppose that#E(G) > 1/2 (k + 2)(k 1), then the number of columns in [W, ] is _> k2. Since thecolumns in are linearly independent, and because of (1.2), we can remove a columnin W without changing the cokernel of [W, ]. But removing a column in W corre-sponds to removing an edge in G, yielding that G properly contains a k-block. Thus#E(G) 1/2 (k + 2)(k- 1).

Now it follows from the quoted result before the proof that (ii) holds. It remainsto prove (iii). First note that if the recipe works for f, it also works for an orthogonalrepresentation f with the property that Ill(i)- f(i)ll is small enough. Indeed, such aperturbation will not destroy the invertibility of a (k2-1) (k2-1) invertible submatrix


of [W, ]. Since the graph does not contain any Kp,q’s, p / q > k + 1, we know from[LSS] that in any neighborhood of f we can find an orthogonal representation ] thathas the property that_ any set of k representing vectors are linearly independent (inthe terms of [LSS]: f is in general position). Thus without loss of generality wemay assume that f has the latter property. Choose now il,..., ira E V(G), m < k,arbitrary. Then f(il),..., f(ira) are linearly independent. Choosing the i,..., ira asthe first m vertices in il,... ,ik,ik+,it} we can set up the matrix W and as before.After permutation of columns of the matrix [W, ] we obtain the matrix

(1.3) [All A12]0 A:

where

I(i ) o oI(i ) o o0 0 0 0

0 0 0 --f(ira_)0 f(il) f(i2) f(ira-)

Note thatcol (f(ij))r=i coker All.

But then, since the cokernel of (1.3) should have dimension exactly equal to 1, thematrix A22 should have at least as many columns as rows. Since A22 is of size

(k2-mk) X (k2-1)- #{(i,j) eE(G)liorje{il,...,ira})+-

this inequality precisely yields (0.1). r]

In order to prove Speculation 0.3, it remains to prove that for a graph satisfying(i), (ii), and (iii) there is an orthogonal representation f such that the matrix [W, ],constructed in the proof of Theorem 0.1, has a one-dimensional cokernel.

2. A sufficiency result. For a vertex v V(G) the degree is defined to be thenumber of adjacent vertices.

THEOREM 2.1. Let P be a graph with an induced subgraph G that satisfies(i) #E(G) > 1/2 (k + 2)(k- 1);(ii) G contains no Kp,q ’s, p + q >_ k + 1;(iii) for any {i,..., ira} C V(G) with 1 <_ m <_ k- 1,

#((i,j) e E(G) i orj e (i,...,ira)}1(k- m + 2)(k- m- 1)< #E(G)- -(iv) G has k- 1 vertices {Ul,..., Uk-1} Of degree k- 1 such that one of the

vertices uj (1 _< j _< k 1) is not adjacent to any ui, j 7 i {1,..., k- 1}.Then order P > k.The conclusion remains true when (iv) is replaced by(v) G has at least k vertices of degree k-1, and, after deleting m of these vertices,

the remaining graph always contains a Kp,q, p + q > k m + 1, m 1,..., k 2.


It should be noted that it follows from the last part of the proof of Theorem 0.1that (iii) is a necessary condition for G to be of order > k.

Before proving the theorem, let us make some remarks. Condition (iv) is a verystringent one since it requires the subgraph G to be fairly condensed. However, com-bined with the next result from [AHMR] the theorem shows that a substantial numberof graphs have order _> k.

We introduce~ the following partial ord.ering on graphs. We say that G _<c G (G isa collapse of G) if G can be obtained from G by identifying vertices without identifyingedges. For example,

10 8=9

1=122" 2=-5 3

1 5 9

4 7 8 11 12

PROPOSITION 2.2. Let G and G be graphs satisfying G <_e G. Then

order G < order (.

This is a restatement of Theorem 4.7 in [AHMR].Example. Theorem 2.1 and Proposition 2.2 together prove that the order of

is _> 4. Indeed, the graph

has order > 4 by Theorem 2.1, and (2.4) is a collapse of (2.3). Therefore, (2.3) hasorder > 4. In fact, (2.3) has precisely order 4 since, by Theorem 0.1, a graph of order>_ 5 should at least have 14 edges, the number of edges of a 5-superblock.

In order to prove Theorem 2.1, we need some attxiliaxy results. Recall that anorthogonal representation f V(G) I:tk is in general position if any set of k repre-senting vectors is linearly independent.

LEMMA 2.3. Let G be a graph that does not contain Kp,q’s p + q > k + 1, andlet f: V(G) -. Rk be an orthogonal representation in general position. IfM MT

xk satisfies (1.1), then for any v V(G) with degree equal to k- 1 the vector f(v)is an eigenvector of M.

Proof. Both f(v) and Mr(v) belong to the orthogonal complement of span {f(u)(u, v) e E(G)}. Since deg v k- 1, the vectors Mr(v) and f(v) both belong to a

one-dimensional space.


Recall from [LSS] that an orthogonal representation f of a graph G is calledfaithful if (f(u), f(v)) 0 if and only if (u, v) E E(G).

LEMMA 2.4. Let G be a graph that contains no Kp,q ’s, p + q >_ k + 1. Further,suppose that u is a vertex of degree k- 1 that has r nonadjacent vertices of degreek- 1. Then for any faithful orthogonal representation f ofG in Rk in general positionthe set of symmetric matrices M satisfying (1.1) is either span (Ik or contains anelement of rank < k r.

Proof. Let M be symmetric such that (1.1) holds. Since f(u) is an eigenvector ofM (at/0, say), which is not orthogonal to r other eigenvectors, the dimension of theeigenspace of M at/0 is at least the dimension of the span of f(u) and these other reigenvectors. Since f is in general position we obtain that

rank (M-/0I) _< max {0, k- (r + 1)}.

Since M-/0I is symmetric and satisfies (1.1), the lemma is proved.PROPOSITION 2.5. Let G be a graph that contains no Kp,q ’s, p+q >_ k + 1. Then

.for every orthogonal representation f: V(G) -- Rk there exists a symmetric matrix Mof rank 1 satisfying (1.1) if and only if there is a set V of at most k 1 vertices in Gsuch that any edge in G has an endpoint in V.

Proof. Suppose such a set Y exists. Choose 0 w e Rk such that (w, f(v)) 0for any v V. It is easy to check that M:-- wwT satisfies (1.1).

In order to prove the only i.:part, let f be an orthogonal representation in generalposition (such an f exists: Theorem 1.1 in [LSS]). Also let M wwT with w 0satisfy (1.1). Then for all edges (i,j) e E(G)

l(J)) 0,

thus w is orthogonal to one of the endpoints of each edge in G. Since w can beorthogonal to at most k- 1 linearly independent vectors we obtain the propositionabove. D

We are now ready to prove Theorem 2.1.Proof of Theorem 2.1. Suppose (i)-(iv) hold. We have to prove that order G _> k.

Let f: V(G) -- Rk be in general position and faithful (existence is assured by Theorem1.1 and (the proof of) Corollary 1.4 in [LSS]). Lemma 2.4 yields, because G satisfies(iv), that either the set of symmetric matrices M satisfying (1.1) is span {Ik} or has anelement of rank 1. Since G satisfies condition (iii) in the theorem, the latter possibilityis ruled out. (Since, if M is symmetric of rank 1 satisfying (1.1), then Proposition 2.5yields that (2.1) is violated (for m- k- 1).)

Now suppose that (i), (ii), (iii), and (v) hold. Let f: V(G) --. Rk be an orthogonalrepresentation in general position. Further suppose that M is positive semidefinite,satisfies (1.1) and has rank d, with 1 < d < k. (We can always assume that M ispositive semidefinite, since if M satisfies (1.1) then any linear combination of M andIk satisfies (1.1) also.) Since G has k vertices of degree k- 1, the representing vectorscorresponding to these vertices are a basis of eigenvectors of M. But then k d ofthese vertices represent the kernel. Delete those vertices. Then from (1.1) it easilyfollows that M1/2f, defined by

(M1/2f)(v) M/2(f(v)),

is an orthogonal representation in general position of the remaining graph. Since (v)holds, this is impossible by Theorem 1.1 in [LSS]. As before, a symmetric M satisfying


(1.1) cannot have rank 1. But then all symmetric M satisfying (1.1) must be in spanD

3. 4-superblocks. In this section we prove Theorem 0.2.Proof of Theorem 0.2. The implication (i) = (ii) is merely a restatement of the

necessary conditions in Theorem 0.1 in this special case.For the proof of (ii) (iii) we determine the _<c-minimal elements in the set of

graphs described under (ii) in Theorem 0.1, i.e., is the set of graphs with nineedges that are not a collapse of one of the three graphs under (ii). The result is givenin the following proposition.

PROPOSITION 3.1. The <_c-minimal elements in q are the graphs Gi, i 1,..., 28,defined in Theorem 0.1 (iii).

The proof requires a lemma.LEMMA 3.2. Let G E 5 be <_c-minimal. Then G does not have a vertex u of

degree 1 and a vertex v of degree <_ 2 such that the distance d(u, v) between u and vis larger than 2.

Proof. Suppose that G has vertices u and v with deg u 1, deg v _< 2 andd(u, v) > 2. By identifying u and v we do not create a degree-4 node. Suppose wecreate a K2,3. Then we must have had

(e)

with three other edges. In case (1) these three other edges do not form a subgraphK1,3; otherwise G _<c 3 K1,3, which contradicts G E {5. Here 3 gl,3 denotes thegraph on the right-hand side of (2.2). But then G must be disconnected. This yieldsthat we can identify one of the vertices in a connected component of G not containingu with u and obtain a graph in G that is _<c-smaller than G. For the second possibility(2), the reasoning is similar.

Suppose that by identifying u and v we create a graph that is <_c-smaller than3 K1,3. Since G has no vertices of degree 4, G must, in this case, have vertices ofdegree <_ 2 besides u and v. But then u may be identified with one of these othervertices and stay in the class . D

Proof of Proposition 3.1. In the reasoning to follow we shall quite frequently usethe fact that a nine-edge graph satisfies

(3.1) Z deg (u) 18.ev(a)

Let us now determine the <_c-minimal elements G in {5.

Case 1. G has a vertex u of degree 1.Let v denote the vertex adjacent to u. When deg v 1 the remaining vertices

should have degree 3 (Lemma 3.2). This is impossible by (3.1). Consider now thecase when deg v 2, and let w u be the other neighbor of v. The cases, deg w 1and deg w 2, are quickly disregarded again by using Lemma 3.2 and (3.1). Whendeg w 3, the only graph one obtains by requiring that all other vertices have degree3 (which must be the case because of Lemma 3.2) contains a K2,3 and is thereforenot in . Consequently, we are left with the case that deg v 3. Let {u, wl, w2}denote the adjacency set of v. When deg wl deg w2 1 we obtain G2 as theonly possibility. The cases {deg w 1, deg w2 2}, {deg w deg w2 2}, and


{deg wl deg w2 3} are quickly discarded, leaving the case {deg wl 2, deg w2

3}. From this we obtain G as the only possibility.Case 2. All vertices of G have degree 2.Then G must have nine vertices and consist of circuits. The only possibilities are

G3, G4, Gh, and G6.We are left with the cases that G has some vertices of degree 3 and some of degree

2. Because of (3.1), the number of degree-3 vertices must be even.Case 3. G has two vertices of degree 3 and six of degree 2.First we consider the case when the vertices of degree 3 are adjacent. Disconnected

graphs with these requirements are easily recognized to be G7 and Gs. Consideringthe possible paths between the vertices of degree-3 nodes, one obtains the graphs Gg,G0, and G in case there are three different paths; when there is only one path, thegraphs G12 and G13 are obtained (note that two paths between the degree-3 nodesis impossible). In a similar way, one obtains the graphs G4-Gs for the case whenvertices of the degree 3 have distance 2 or 3.

Case 4. G has four vertices of degree 3 and three of degree 2.Since G c 3 gl,3, there should be no three vertices of degree 3 that are all

nonadjacent. Let ul, u2, u3, and u4 denote the vertices of degree 3. When u,..., u4are all adjacent to one another, one obtains the only possibility, G26. When they form

Ul

u3

u2

u4

one obtains G25. When Ul,..., ua form a square (K2,2), one obtains G23. When?1 U4 form

3

u u2

one obtains G9 and G2. The case when u,..., u4 form a connected line givespossibilities G20, G22, and G28. The last case is when ul,..., u4 form a line of threevertices and an isolated vertex. This gives as the only possibility, G24.

Case 5. G only has vertices of degree 3.Here G27 is the only possibility.This proves Proposition 3.1.This concludes the proof of (ii) (iii).To prove (iii) (i) we need to show that for the graphs G1-G28 we can find a

rank-4 extremal in M+(G). Then Proposition 2.2 yields that all graphs under (iii)have a rank-4 extremal. The graphs G, G2, G9, G20, G2, G22, G24, G27, andG28 have a vertex of degree 3 that is not adjacent to two other vertices of degree 3.Therefore, by Theorem 2.1 (i)-(iv), the order of these graphs is > 4. But then, sincethe numbers of edges is smaller than 14 we obtain that the order is at most 4, givingequality. The graph G26 has as its complement the graph K3,4. Consequently, G26has order 4, by Theorem 6.1 in [HPR]. The graphs G23 and G28 are recognized tohave order > 4 by Theorem 2.1 using (i)-(iii) and (v).


The remaining graphs G3-Gls and G25 are dealt with by "brute force." A pro-gram using Mathematica (using integer arithmetic) produced for us the followingrank-4 extremals in M+(G) for G in G3-Gls, G25. For each Gi this rank-4 extremalis given by ATAi, where Ai, i- 3,..., 18, 25, is given by

1 0 0 2 -2 -2 _9 3_7 3222 4 45

0 2 -2 5 9 -3 s -3 3s6

0 -3 3 -2 -1 -3 1 1 2

0 1 2 -I -3 -2 -3 -2 2

1

A4-0

0

0

1

0A=

0

0

1

0A6=

0

0

1

0AT=

0

0

1

0As=

0

0

1

0Ag:

0

0

1

0AlO

0

0

2 -3 0 0 -3 -5 -6

1 3 2 4 3 -5 -1

3 2 2 -1 3 3 -3

2 -3 -3 2 -3 3 32 --12 0 0 63 -1 2 -I 2 -2 1/2 3

-2 3 -2 3 3 2 -3

1 3 -1 -3 1 1 -1

0 -2 1 1/2 0-3-6

-2 -I -3 1 1 -9 -2 --3 3 -I 2 2 5 -i 3

3 2 1 -2 1 -1 -I -i

2 3 0 0 3_ -I2 2

-I 1 3 2_ -i i3 2

1 -i -2 2 -2 -2

1 -I i 2 -2 1

0 0 -i 2 -14 16

-3 -i 3 _8_ -9 23

-3 -I 2 2 3 3

-2 3 2 1 -2 2

0 0 1 1 15 _2_2 5

2 6 11 5 -i 32 2

-3 3 -3 -3 -2 1

-i 3 -2 -2 2 1

0 0 2 1 2 7

-1 -3 -2 0 -3

1 1 -I 1 1 -i

-I 3 1 1 -3 -2

1 0 0 -2 59 io 6__6 3 59

0 -3 1 _5 7 -2 13

0 -2 -I 1 1 -I -3

0 1 -2 -3 3 3 -I

-1911

1

lO150

1

2

8

2

2

-110622772324277

-2

-2


A12

1 0 1 -3 0 0 9

0 -1 5 2 -2 4 -2 1/40 -3 -3 -2 2 1 1 -3

0 2 -2 -2 -2 -3 -2 2

1

0

0

0

180 0 0 -1 -2 -1 Yi2 753 3 1 4 2-

32 -3 3 -3 1 --2 3 -I 3 2 -1 -i

1

0

0

0

0 0 -1 3 2 7

2 -3 -4 0 4_ 13

2 -1 3 -I -2 3

-2 1 -1 -i 2 -2

16

1

0

0

0

0 0 0 1 27 24

2 2 2 __3 11 52 2 2

-3 -1 -2 3 3 2

1 3 -3 -3 -2 -1

48223140223

2

2

1

0

0

0

0 2 0 0 25 4 82 2 11

-1 12 -3 -5 -3 -3 48

-2 -3 3 -3 -3 3 -2

3 2 -2 3 1 -1 -2

1

0

0

0

0 -2 -- 0 0 _6_2 5

2 -3 3 -3 _5_ -i3

-3 -3 -2 2 2 -3

3 -i 2 -3 3 -3

8061

96

2

-2

1

0

0

0

0 0 0 -3 -1 3

-1 -3 1 12 10 --33 1 3 -2 2 1

-2 -3 2 -3 -2 2

29_415199302117302

-3

1 0 -2 0 _2 3{}3 7

0 3 -- -i 2 273 14

0 -1 2 -3 3 1

0 3 2 3 -1 2

8120

-9

2

-1


Here the numbering of the nodes is basically from top to bottom and from left toright, or, more explicitly, given by the following.

2

54 5 6 7 $ 9 6 7 $ 9

2 3 )t 2 3

G6= 5 G7 4,"/ ’ 5 ] G8 2L3= A46 7 8 9 8 6 7 5 678

2 2 2 3

G9 5 G10 4 Gll 4 5

67 87 8

2

G12 d7 G13 3 6

5 6 87 8

2 2 3

G15 6 G16 6 74 5

7 88

8

G14=46

7 8

2 4 7

G18 6G25

7 8 5 6

It is easy to check by hand that the representations of G3-Gls and G25, indicated byA3-Als and A25, respectively, satisfy the conditions in Step I of the recipe. In orderto check that Step II is satisfied, one has to go through more elaborate calculations.These checks were made using the Mathematica program.

Acknowledgments. We wish to thank Leiba Rodman for attracting our atten-tion to the reference [LSS] and Neola Crimmins for her masterful production of thismanuscript.

REFERENCES

[AHMR] J. AGLER, J. W. HELTON, S. MCCULLOUGH, AND L. RODMAN, Positive semidefinitematrices with a given sparsity pattern, Linear Algebra Appl., 107 (1988), pp. 101-149.

[GJSW] R. GRONE, C. R. JOHNSON, E. MAIQUES DE SA, AND H. WOLKOWICZ, Positive definitecompletions of partial Hermitian matrices, Linear Algebra Appl., 58 (1984), pp. 109-124.

[HPR] J.W. HELTON, S. PIERCE, AND L. I:tODMAN, The ranks of extremal positive semideflnitematrices with a given sparsity pattern, SIAM J. Matrix Anal. Appl., 10 (1989), pp. 407-423.


[LSS] L. Lovsz, M. SAKS, AND A. SCHRIJVER, Orthogonal representations and connectivity ofgraphs, Linear Algebra Appl., 114/115 (1989), pp. 439-454.

[PPS] V. I. PAULSEN, S. C. POWER, AND R. R. SMITH, Schur products and matrix completions,J. Funct. Anal., 85 (1989), pp. 151-178.

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