N 0 completions on partial matrices


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1 N 0 completions on prtil mtrices C. Jordán C. Mendes Arújo Jun R. Torregros Instituto de Mtemátic Multidisciplinr / Centro de Mtemátic Universidd Politécnic de Vlenci / Universidde do Minho Cmino de Ver s/n Vlenci / Cmpus de Gultr 4710 Brg Spin / Portugl Abstrct An n n mtrix is clled n N 0 mtrix if ll its principl minors re nonpositive. In this pper we re interested in N 0 mtrix completion problems tht is when prtil N 0 mtrix hs n N 0 mtrix completion. In generl combintorilly or noncombintorilly symmetric prtil N 0 mtrix does not hve n N 0 mtrix completion. Here we prove tht combintorilly symmetric prtil N 0 mtrix with no null min digonl entries hs n N 0 mtrix completion if the grph of its specified entries is 1chordl grph or cycle. We lso nlyze the mentioned problem when the prtil mtrix hs some null min digonl entries. Key words: Prtil mtrix completion N 0 mtrix chordl grph cycle. 1 Introduction A prtil mtrix is mtrix in which some entries re specified nd others re not. We mke the ssumption throughout tht ll digonl entries re prescribed. A completion of prtil mtrix is the mtrix resulting from prticulr choice of vlues for the unspecified entries. The completion obtined by replcing ll the unspecified entries by zeros is clled the zero completion nd denoted A 0. A completion problem sks if we cn obtin completion of prtil mtrix with some prescribed properties. An n n prtil mtrix This reserch ws supported by the Portuguese Foundtion for Science nd Technology (FCT) through the reserch progrm POCTI nd by Spnish DGI grnt number MTM Correspondig uthor Emil ddresses: (C. Jordán) (C. Mendes Arújo) (Jun R. Torregros ). Preprint submitted to Elsevier 14 Jnury 2009
2 ( ij ) it sid to be combintorilly symmetric when ij is specified if nd only if ji is nd is sid to be wekly signsymmetric if ij ji 0 for ll i j { n} such tht both (i j) (j i) entries re specified. A nturl wy to described n n n prtil mtrix ( ij ) is vi grph G (V E) where the set of vertices V is { n} nd the edge or rc {i j} (i j) is in set E if nd only if position (i j) is specified; s ll min digonl entries re specified we omit loops. In generl directed grph is ssocited with noncombintorilly symmetric prtil mtrix nd when the prtil mtrix is combintorilly symmetric n undirected grph is used. A pth is sequence of edges (rcs) {i 1 i 2 } {i 2 i 3 }... {i k 1 i k } in which the vertices re distinct. A cycle is pth with the first vertex equl to the lst vertex. An undirected grph is chordl if it hs no induced cycles of length 4 or more [1]. The submtrix of mtrix A of size n n lying in rows α nd columns β α β N = { n} is denoted by A[α β] nd the principl submtrix A[α α] is bbrevited to A[α]. An n n rel mtrix ( ij ) is clled N 0 mtrix (Nmtrix) if ll its principl minors re nonpositive (negtive). These clsses of mtrices rise in multivrite nlysis [6] in liner complementry problems [45] nd in the theory of globl univlence of functions [2]. In the following proposition we give some properties tht re very useful in the study of N 0 mtrices. Proposition 1 Let ( ij ) be n n n N 0 mtrix. Then (1) If P is permuttion mtrix then P AP T is n N 0 mtrix. (2) If D is positive digonl mtrix then DA nd AD re N 0 mtrices. (3) If D is nonsingulr digonl mtrix then DAD 1 is n N 0 mtrix. (4) If ii 0 i = n then ij 0 i j { n}. (5) A is wekly signsymmetric (6) α { n} principl submtrix A[α] is n N 0 mtrix. From the bove properties it is esy to prove tht ny n n N 0 mtrix with no null digonl entries is digonlly similr to n N 0 mtrix in the set: S n = { ( ij ) : ij 0 nd sign( ij ) = ( 1) i+j+1 i j} On the other hnd the lst property of the previous proposition llows us to give the following definition. Definition 1 A prtil mtrix is sid to be prtil N 0 mtrix if every com 2
3 pletely specified principl submtrix is n N 0 mtrix. In [3] the uthors study the Nmtrix completion problem nd they close the problem for some types of prtil mtrices. Our interest here is in the N 0 mtrix completion problem tht is does prtil N 0 mtrix hve n N 0  mtrix completion? The study of this problem is different from the previous one since some minors or some entries of the prtil mtrix cn be zero. In Section 2 we nlyze this problem for combintorilly nd noncombintorilly symmetric prtil N 0 mtrices. The problem hs in generl negtive nswer. In Section 3 we show tht the 1chordl grphs gurntee the existence of n N 0 mtrix completion for prtil N 0 mtrix nd in Section 4 we complete the study of the mentioned completion problem for prtil N 0 mtrix whose ssocited grph is cycle. In both cses we consider prtil N 0 mtrix with no null min digonl entries. Finlly in Section 5 we nlyze the problems tht pper when some min digonl entries re zero. 2 Preliminry results The following exmple shows tht the N 0 mtrix completion problem hs in generl negtive nswer for combintorilly nd noncombintorilly symmetric prtil N 0 mtrices. Exmple 1 The prtil mtrices 1 1 x y x 3 nd B = re combintorilly nd noncombintorilly symmetric prtil N 0 mtrices respectively. However A hs no N 0 mtrix completion since det 1 x y. On the other hnd det B[{1 2}] 0 if nd only if x 1/2 but det B > 0 x 1/2. Therefore B hs no N 0 mtrix completion. We produce prtil N 0 mtrix of size n n n 4 which hs no N 0 mtrix completion by embedding the bove mtrix A (nlogously B) s principl submtrix nd putting 1 s on the min digonl nd unspecified entries on the remining positions tht is M = A X Y Ī 3
4 where Ī is prtil mtrix with ll entries unspecified except the entries of the min digonl tht re equl to 1 nd X nd Y re completely unspecified submtrices. Keeping Proposition 1 in mind when we suppose tht ll digonl entries re nonzero it would not mke sense to study the existence of N 0 mtrix completion of nonwekly signsymmetric prtil N 0 mtrices or of prtil N 0 mtrices which do not stisfy up to permuttion nd digonl similrity tht if (i j) entry is specified then sign( ij ) = ( 1) i+j+1. Therefore in this context (prtil mtrices with nonzero digonl entries) we ssume tht the prtil N 0 mtrix belongs to the set P S n of prtil mtrices such tht for ll (i j) specified entry sign( ij ) = ( 1) i+j+1. When restricting our study to prtil N 0 mtrices tht belong to P S n we re implicitly nlyzing the completion problem for prtil N 0 mtrices tht re permuttion or digonlly similr to prtil N 0 mtrix tht belongs to P S n. Proposition 2 Let A be prtil N 0 mtrix of size 3 3 with no null min digonl entries. Then there exists n N 0 mtrix completion A c of A. Proof: Since the clss of N 0 mtrices is invrint under left nd right positive digonl multipliction we my tke ll digonl entries of A to be 1. We denote by λ A the number of unspecified entries of A. If λ 0 A is not prtil mtrix nd if λ 6 the result is trivil. Let us first consider the cse in which A hs exctly one unspecified entry. By permuttion nd digonl similrities we cn ssume tht this entry is in position (13) nd tht ll upper digonl entries re equl to 1. Hence A hs the following form 1 1 x with nd 31 > 0. If 31 > 1 it suffices to choose x = 1 in order to obtin n N 0 mtrix completion of A. If 31 1 we consider the completion A c of A with x = 1/ The formultion of the problem in cse λ A > 1 reduces to tht of λ 1. In fct it is possible to complete some dequte unspecified entries in order to obtin prtil N 0 mtrix with single unspecified entry. Unfortuntely we cn not extend Proposition 2 for generl n s we see in the following exmple. 4
5 Exmple 2 Let A be the prtil mtrix x It is not difficult to verify tht A is prtil N 0 mtrix nd A P S 4. However A hs no N 0 mtrix completion since det A[{1 2 4}] = x 0 nd det A[{1 3 4}] = 0.9x 9 0 if nd only if x 901/19 nd x 10 which is impossible. From this exmple we cn estblish de following result. Proposition 3 For every n 4 there is n n n noncombintorilly symmetric prtil N 0 mtrix belonging to P S n tht hs no N 0 mtrix completion. In Sections 3 nd 4 we re going to work with combintorilly symmetric prtil N 0 mtrices which belong to P S n nd with no null min digonl entries. 3 Chordl grphs We recll some very rich clique structure of chordl grphs (see [1] for further informtion). A clique in n undirected grph G is simply complete (ll possible edges) induced subgrph. We lso use clique to refer to complete grph nd use K p to indicte clique on p vertices. If G 1 is the clique K q nd G 2 is ny chordl grph contining the clique K p p < q then the clique sum of G 1 nd G 2 long K p is lso chordl. The cliques tht re used to build chordl grphs re the mximl cliques of the resulting chordl grph nd the cliques long which the summing tkes plce re the soclled miniml vertex seprtors of the resulting chordl grph. If the mximum number of vertices in miniml vertex seprtor is p then the chordl grph is clled pchordl. In this section we re interested in 1chordl grphs. Proposition 4 Let A be n n n prtil N 0 mtrix with no null min digonl entries the grph of whose specified entries is 1chordl with two mximl cliques one of them with two vertices. Then there exists n N 0 mtrix completion of A. 5
6 Proof: Since the clss of N 0 mtrices is invrint under left nd right positive digonl multipliction we my tke ll digonl entries of A to be 1. Moreover since there re no null min digonl entries A is permuttionlly similr to prtil N 0 mtrix in P S n. Therefore we my ssume without loss of generlity tht (ā ij ) hs the following form: 1 1 x 13 x 1n ( 1) n+1 2n x ( 1) n+2 3n.... x n1 ( 1) n+1 n2 ( 1) n+2 n3 1 Consider the completion A c of A obtined by replcing the unspecified entries in the following wy: x 1j = ā 2j j { n} x i1 = ā i2 i { n} To prove tht A c is n N 0 mtrix we only need to show tht det A c [{1} α] 0 for ll α {2... n}. In order to do so let α {2... n}. If 2 α det A c [{1} α] = ( 21 1) det A[α] 0. If 2 α A c [{1} α] is obtined from A c [{2} α] by multiplying the first row nd the first column by 1. Then det A c [{1} α] = det A c [{2} α] 0 Despite the preceding result being prticulr cse of the next proposition its proof will be very useful in the resolution of completion problems for certin prtil N 0 mtrices with specil types of ssocited grphs. Proposition 5 Let A be n n n prtil N 0 mtrix with no null min digonl entries the grph of whose specified entries is 1chordl with two mximl cliques. Then there exists n N 0 mtrix completion of A. Proof: Tking into ccount proposition 1 we my ssume without loss of generlity tht A hs the following form A X 21 T 1 T 23 Y 32 A 33 where X nd Y re completely unspecified mtrices nd the remining entries of A re prescribed nd where 12 is column vector with the sme number of 6
7 rows s A is lso column vector with the sme number of rows s A 33 T 21 is row vector with the sme number of columns s A 11 nd T 23 is lso row vector with the sme number of columns s A 33. Consider the completion of A A T 23 A c = T 21 1 T T A 33 We re going to see tht A c is n N 0 mtrix. Let α nd β be the subsets of N = { n} such tht A c [α] = A nd A c [β] = T T A 33 nd ssume α = k (thus k is the index of the overlpping entry). Let γ N. Then there re two cses to consider: () k γ. In this cse det A c [γ] = ( 1) det A c [γ α] det A c [γ β] 0. (b) k γ. We consider γ = N \ {k}. For nother γ we proceed in nlogous wy. We re going to distinguish two cses. (b1) A 11 is nonsingulr. From det A c [α] 0 we obtin λ = T 21A nd since det A c [β] 0 we hve det (A T 23) 0. Now by pplying Guss elimintion method we obtin det A c [γ] = det A 11 det (A 33 + λ 32 T 23). To prove tht det A c [γ] 0 we need to show tht det (A 33 + λ 32 T 23) 0. By simplicity we denote A 33 = (c ij ) p ij=1 nd 32 T 23 = (b ij ) p ij=1 where p = n k. It is esy to prove tht given ρ R with det (A 33 + ρ 32 T 23) = det A 33 + ρm M = det M 1 + det M p nd where b 11 c 12 c 1p c 11 c 1p 1 b 1p b M 1 = 21 c 22 c 2p c M p = 21 c 2p 1 b 2p b p1 c p2 c pp c p1 c pp 1 b pp 7
8 For ρ = 1 we hve det (A 33 + ρ 32 T 23) = det A 33 + ρm = det (A T 23) 0 then M det A For ρ = λ det (A 33 + λ 32 T 23) 0 λm det A 33. Since λ 1 nd M 0 we hve λm M det A 33. Therefore det (A 33 + λ 32 T 23) 0 nd then det A c [γ] 0. (b2) A 11 is singulr. In this cse it is esy to see tht rnk A 11 < k 1 32 T 21 nd then rnk A T 23 < k 1 + n k = n T 21 A 33 Therefore det A c [γ] = 0. We cn extend this result in the following wy: Theorem 1 Let G be n undirected connected 1chordl grph. Then ny prtil N 0 mtrix with no null min digonl entries the grph of whose specified entries is G hs n N 0 mtrix completion. Proof: Let A be prtil N 0 mtrix the grph of whose specified entries is G. The proof is by induction on the number p of mximl cliques in G. For p = 2 we obtin the desired completion by pplying Proposition 5. Suppose tht the result is true for 1chordl grph with p 1 mximl cliques nd we re going to prove it for p mximl cliques. Let G 1 be the subgrph induced by two mximl cliques with common vertex. By pplying Proposition 5 to the submtrix A 1 of A the grph of whose specified entries is G 1 nd by replcing the obtined completion A 1c in A we obtin prtil N 0 mtrix such tht whose ssocited grph is 1 chordl with p 1 mximl cliques. The induction hypothesis llows us to obtin the result. 8
9 A prtil mtrix A is sid to be block digonl if A cn be prtitioned s A 1 X 12 X 1k X 21 A 2 X 2k... X k1 X k2 A k where X ij re completely unspecified rectngulr mtrices nd ech A i is prtil mtrix i = 1... k. Let A be prtil N 0 mtrix whose ssocited grph G is nonconnected. It is not difficult to prove the following theorem which estblishes tht if ech submtrix ssocited with ech connected component of G hs n N 0 mtrix completion then so does the whole mtrix A. Theorem 2 If prtil N 0 mtrix A is permuttion similr to block digonl prtil mtrix in which ech digonl block hs n N 0 mtrix completion then A dmits n N 0 mtrix completion. From this result nd tking into ccount tht prtil mtrix whose grph is nonconnected is permuttion similr to block digonl mtrix we cn ssume without loss of generlity tht the ssocited grph of prtil N 0  mtrix is connected grph. The completion problem for prtil N 0 mtrices whose ssocited grph is p chordl p > 1 is still unresolved. We note here tht ny pchordl grph p > 1 contins s n induced subgrph 2chordl grph with four vertices. By left nd right positive digonl multipliction nd by permuttion similrity we cn ssume keeping in mind proposition 1 tht 4 4 prtil N 0 mtrix the grph of whose prescribed entries is 2chordl grph hs the form x y with nd It is esy to prove tht det ( 32 1)xy x det A 0 [{2 3 4} {1 2 3}] y det A 0 [{1 2 3} {2 3 4}]+det A 0 (1) 9
10 From (1) we obtin sufficient conditions for the existence of the desired completion. Specificlly if det A 0 [{1 2 3} {2 3 4}] > 0 we re going to see tht A dmits n N 0 mtrix completion. We consider the following cses: (i) Submtrix A[{2 3}] is singulr. Then A hs the form x y Let Ā the prtil N 0mtrix obtined by replcing entry x by c such tht 0 < c < min { }. Now the determinnt of ny principl submtrix contining position (4 1) is polynomil in y with negtive leding coefficient. Therefore there exists M R M > 0 such tht by completing position (4 1) by every d > M we obtin N 0 mtrix completion of A. (ii) Submtrix A[{2 3}] is nonsingulr. Now consider the completion c A c = d where c d R such tht d > 0 nd 0 < c < min { det A 0[{1 2 3} {2 3 4}] } The determinnt of ny principl submtrix contining position (4 1) is polinomil in d with negtive leding coefficient. Therefore s in the previous cse there exists M R M > 0 such tht A c is n N 0 mtrix for d > M. 4 Cycles In this section we re going to prove the existence of n N 0 completion for prtil N 0 mtrix with no null min digonl entries whose ssocited grph is cycle. Lemm 1 Let A be n 4 4 prtil N 0 mtrix with no null min digonl entries whose ssocited grph is cycle. Then there exists n N 0 mtrix completion. 10
11 Proof: Since the clss of N 0 mtrices is invrint under left nd right positive digonl multipliction we my tke ll digonl entries of A to be 1. We lso know tht A is permuttionlly similr to prtil N 0 mtrix in P S n. Hence we my ssume tht A hs the following form: 1 1 x x 24 x x where nd Let Ā the prtil N 0mtrix obtined from A by replcing x 31 = 32 nd x 42 = 41. It is esy to prove tht there exists x 13 1/ 32 such tht det Ā[{1 3 4}] 0 nd there exists x 24 1/ 41 such tht det Ā[{2 3 4}] 0. Since det Ā = ( 21 1) det Ā[{1 3 4}] we cn conclude tht there exists n N 0 mtrix completion of A. We extend this result in the following wy. Theorem 3 Let A be n n n n 4 prtil N 0 mtrix with no null min digonl entries whose ssocited grph is cycle. Then there exists n N 0  mtrix completion. Proof: By left nd right positive digonl multipliction nd by permuttion similrity we my ssume tht A hs the following form: 1 1 x 13 x 1n 1 ( 1) n 1n x 2n 1 x 2n x x 3n 1 x 3n..... x n 11 x n 12 x n ( 1) n n1 x n2 x n3 nn 1 1 where 1n n1 > 0 nd ii 1 1 i = n. 11
12 The proof is by induction on n. For n = 4 see Lemm 1. Now let A be n n n mtrix n > 4. Consider the following prtil N 0 mtrix: 1 1 x 13 ( 1) n 1 1n x 1n x 2n 1 x 2n x x 3n 1 x 3n Ā = ( 1) n 1 n1 x n 12 x n x n1 x n2 x n3 nn 1 1 Ā[{ n 1}] is prtil N 0 mtrix such tht its ssocited grph is n (n 1)cycle. By induction hypothesis there exists n N 0 mtrix completion Ā[{ n 1}] c. Let Â be the prtil N 0mtrix obtined by replcing in Ā the completion Ā[{ n 1}] c. Now Â is prtil N 0 mtrix whose ssocited grph is 1chordl with two mximl cliques one of them with two vertices. By pplying Proposition 4 to mtrix Â we obtin n N 0mtrix completion A c of A. In the lst section we re going to consider the possibility tht some min digonl entries re zero. 5 Zero entries in the min digonl The results of Sections 3 nd 4 do not hold if some min digonl entries re zero s we cn see in the following exmple. Exmple 3 () Consider the prtil N 0 mtrix 1 12 x 13 x x x 41 with > 0 whose ssocited grph is 1chordl with two mximl cliques. Mtrix A hs no N 0 mtrix completion since det A[{1 2 4}] = > 0 for ll x 14 nd x
13 (b) Now consider the prtil N 0 mtrix 0 1 x x B = 24 x x whose ssocited grph is cycle. Mtrix B hs no N 0 mtrix completion since det B[{1 2 4}] = 1 > 0 for ll x 24 nd x 42. On the other hnd the completion problem for prtil N 0 mtrices whose ssocited grph is pchordl p > 1 with some null min digonl entries hs lso negtive nswer. Exmple 4 The prtil N 0 mtrix x x the grph of whose specified entries is 2chordl hs no N 0 mtrix completion since det A[{1 2 4}] = 1 for ll x 14 nd x 41. Proposition 2 itself does not hold when some min digonl entries re zero. Exmple 5 The prtil N 0 mtrix 1 0 x hs no N 0 mtrix completion since det 1 for ll x 13. In this context the role of set S n is tken by ws n = { ( ij ) : ij = 0 or sign( ij ) = ( 1) i+j+1 i j}. Unfortuntely n n n N 0 mtrix with some null digonl entries is not in generl digonlly similr to n N 0 mtrix tht belongs to ws n s the following exmple shows. 13
14 Exmple 6 Consider the N 0 mtrix It is esy to observe tht there is not digonl mtrix D such tht DAD 1 is n element of ws n. In order to obtin n N 0 mtrix completion when there re null elements in the min digonl of the prtil N 0 mtrix we introduce the following condition. Definition 2 Let A be prtil N 0 mtrix. We sy tht A stisfies condition (ND) if ii jj = 0 = ij ji = 0 when ij nd ji re specified. Proposition 6 Let A be combintorilly symmetric prtil N 0 mtrix of size 3 3 with some null min digonl entries. If A stisfies condition (ND) then there exists n N 0 mtrix completion A c of A. Proof: Since the clss of N 0 mtrices is invrint under left nd right positive digonl multipliction nd under permuttion similrity we my ssume without loss of generlity tht A hs the form x y where ii 0 i = Consider the following cses: () 22 = 0. From condition (ND) we hve = 0 nd = 0. Therefore 12 = 0 or 21 = 0 nd 23 = 0 or 32 = 0. We cn obtin in ll cses vlues for x nd y such tht xy nd det x y 0. (b) Then 11 = 0 or 33 = 0. In both cses the completion A 0 is n N 0 mtrix. The formultion of the problem in cse of mtrix A hs more thn two unspecified entries is reduced to this one. 14
15 However condition (N D) is not necessry condition since mtrix 1 0 x y 1 1 is prtil N 0 mtrix which does not stisfy condition (ND) but by tking x = y = 1 we obtin n N 0 mtrix completion. On the other hnd condition (ND) is not sufficient condition for noncombintorilly symmetric prtil N 0 mtrix s we see in the following exmple. Exmple 7 Consider the prtil N 0 mtrix 1 1 x which stisfies condition (ND) nd since det 1 x it hs no n N 0  mtrix completion. Motivted by the results obtined in sections 2 4 we dd new restriction to our study: we will consider prtil N 0 mtrices belonging to the set P ws n of n n prtil mtrices ( ij ) such tht for ll specified entry (i j) ij = 0 or sign( ij ) = ( 1) i+j+1. Note tht we will implicitly nlyze the completion problem for prtil N 0 mtrices tht re permuttion or digonlly similr to prtil N 0 mtrix tht belongs to P ws n. It is not difficult to prove the following result. Proposition 7 Let A be n 3 3 noncombintorilly symmetric prtil N 0  mtrix with some null min digonl entries such tht A P ws 3. If A stisfies condition (ND) then there exists n N 0 mtrix completion of A. The following exmple shows tht there exists prtil N 0 mtrix tht belong to P ws 4 which stisfies condition (ND) but hs no n N 0 mtrix completion. Therefore we cn not extend Proposition 6 nd Proposition 7 for generl n. 15
16 Exmple 8 Let A be the prtil mtrix x y (We cn tke y = 1 for n exmple of noncombintorilly symmetric prtil mtrix) It is not difficult to verify tht A is prtil N 0 mtrix A P ws 4 nd it stisfies condition (ND). But A hs no n N 0 mtrix completion since det 2 for ll x nd y. Tking into ccount this exmple nturl question rises: given n n n prtil N 0 mtrix A whose ssocited grph is n 1chordl grph or cycle re conditions (ND) nd A P ws n sufficient conditions in order to obtin the desired completion? Proposition 8 Let A be 4 4 prtil N 0 mtrix belonging to P ws 4 with some null digonl entries the grph of whose specified entries is cycle. If A stisfies condition (ND) then there exists n N 0 mtrix completion of A. Proof: Let N d be the number of null min digonl entries. Consider the following cses: () N d = 1. (1) 11 = 0. Since the clss of N 0 mtrices is invrint under left nd right positive digonl multipliction nd under permuttion similrity we my ssume without loss of generlity tht mtrix A hs the form: 0 12 x x 24 x x
17 with nd > 0. We cn prove tht mtrix A c = is n N 0 mtrix completion of A. (2) 22 = 0. In this cse mtrix A hs the form 1 12 x x 24 x x with nd > 0. Mtrix A c = is n N 0 mtrix completion of A. (3) 33 = 0. In this cse mtrix A is permuttion similr to mtrix of type (2). (4) 44 = 0. Anlogously to cse (3) now mtrix A is permuttion similr to mtrix of type (1). (b) N d > 1. Mtrix A 0 is lwys n N 0 mtrix completion of A. If in the previous proposition we leve out condition (ND) the result does not hold s we cn see in Exmple 3 mtrix B. In ddition if mtrix A does not belong to P ws 4 in generl it hs no n N 0 mtrix completion. For exmple 0 1 x x 24 x x
18 is prtil N 0 mtrix tht does not belong to P ws 4 stisfies condition (ND) nd hs no N 0 mtrix completion since det A[{1 2 4}] 0 nd det A[{2 3 4}] 0 if nd only if x 24 < 0 nd x 24 > 0 which is impossible. Proposition 9 Let A be 4 4 prtil N 0 mtrix belonging to P ws 4 with some null min digonl entries the grph of whose specified entries is 1 chordl with two mximl cliques nd null vertex seprtor. If A stisfies condition (ND) then there exists n N 0 mtrix completion of A. Proof: Since the clss of N 0 mtrices is invrint under permuttion similrity we cn suppose tht the vertex seprtor is in position (2 2). We consider the following cses: () ii 0 i = We my ssume using right nd left positive digonl multipliction nd permuttion similrity tht mtrix A hs the form 1 12 x 13 x x x By replcing x 13 = x 31 = 1 x 14 = 34 nd x 41 = 43 we obtin n N 0 mtrix completion of A. (b) 11 = 0. Completion A 0 is n N 0 mtrix. (c) Now we distinguish three possibilities: (c1) 33 = 44 = 0 (c2) 33 = 0 nd 44 0 nd (c3) 33 0 nd 44 = 0. We obtin n N 0 mtrix completion of A in ech of them by nlyzing lot of cses depending on ech of the offdigonl specified entries re or not zero. Our study bout the N 0 mtrix completion problem when the n n prtil N 0 mtrix hs some null digonl entries llows us to conjecture tht if we dd condition (ND) s n hypothesis the results showed in Sections 3 nd 4 hold for prtil N 0 mtrices with some null digonl entries. In ddition the mentioned condition cn be left out in the cse of 1chordl grphs when the vertex seprtor is nonzero. References [1] J.R.S. Blir B. Peyton An introduction to chordl grphs nd clique trees The IMA volumes in Mthemtics nd its Applictions Vol. 56 Springer New 18
19 York 1993 pp [2] K. Ind The production coefficient mtrix nd the StolperSmuelson condition Econometric 39 (1971) [3] C. Mendes Jun R. Torregros An M. Urbno Nmtrix completion problem Liner Algebr nd its Applictions 372 (2003) [4] S.R. Mohn R. Sridhr On chrcterizing Nmtrices using liner complementrity Liner Algebr nd its Applictions 160 (1992) [5] S.R. Mohn Degenercy in liner complementrity problems: survey Annls of Opertions Reserch (1993) [6] S.R. Prnjpe Simple proofs for the infinite divisibility of multivrite gmm distributions Snkhy Ser. A 40 (1978)
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