TOEPLITZ AND POSITIVE SEMIDEFINITE COMPLETION PROBLEM FOR CYCLE GRAPH

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1 English NUMERICAL MATHEMATICS Vol14, No1 Series A Journal of Chinese Universities Feb 2005 TOEPLITZ AND POSITIVE SEMIDEFINITE COMPLETION PROBLEM FOR CYCLE GRAPH He Ming( Λ) Michael K Ng(Ξ ) Abstract We present a sufficient an necessary conition for a so-calle Cn pattern to have positive semiefinite (PSD) completion Since the graph of the Cn pattern is compose by some simple cycles, our results exten those given in [1] for a simple cycle We also erive some results for a partial Toeplitz PSD matrix specifying the Cn pattern to have PSD completion an Toeplitz PSD completion Key wors Partial matrix, Toeplitz matrix, completion problem, Cn pattern AMS(2000)subject classifications 15A24, 65F10, 82C70 1 Introuction A partial matrix is a matrix in which some entries are specifie, while the remains entries are free to be chosen(from a certain set) A completion of a partial matrix is the conventional matrix resulting from a particular choice of values for the unspecifie entries For most of matrix completion problems there are some obvious conitions that must be satisfie for a completion of a certain class to exist For example, for real symmetric positive semiefinite (PSD) completions, all completely specifie submatrices must be symmetric PSD A partial matrix which satisfies such a conition is calle a partial PSD matrix In this paper another class of matrices which we concern is Toeplitz matrix It is nown that an n n matrix A =(a ij ) is calle a symmetric Toeplitz matrix if a i,j = r i j for all i, j =1, 2,,n So a partial symmetric Toeplitz matrix is a partial symmetric matrix an if an entry in position (i, j) is specifie then all entries in positions (i + l, j + l) (mo n) are also specifie an these (specifie) entries are equal Other Research supporte in part by National Natural Science Founation of China No Research supporte in part by RGC Grant Nos 7132/00P an 7130/02P, an HKU CRCG Grant Nos , an Receive: Sep 16, 2004

2 68 He Ming Michael K Ng types of partial matrices are efine similarly, see [4] A pattern for n n matrices is a list of positions of an n n matrix, that is, a subset of {1, 2,, n} {1, 2,, n} A partial matrix specifies the pattern if its specifie entries are exactly those liste in the pattern A pattern Q is calle symmetric if (i, j) Q implies (j, i) Q We assume throughout this paper that all the patterns we iscuss are symmetric an inclue all iagonal positions, an when we efine a pattern or calculate the number of the specifie entries of a pattern, we shall ignore iagonal an symmetric positions of the pattern (eg, Definition 1 below) an the entries in the positions For a certain class of matrices, one important area of research is to ecie for which patterns of specifie entries all partial matrices in the class are completable to the class of matrices Among the types of matrix completion problems that have been stuie are completion to positive efinite (semiefinite) matrices [2],[7], to M-matrices an inverse M-matrices [5], to P- matrices an P 0 matrices[6],[4] an to contractions [8], etc For the positive efinite completion problem, a solution was given by Grone, Johnson, Sa, an Wolowicz in [2] For the partial positive efinite Toeplitz completion problem, an interesting open problem was presente in [7] Now we give efinition of the Cn consier pattern an escribe the completion problems we shall Definition 1 Let n/2 > 1, the symmetric pattern Q = {(1,+1), (2,+2),,(n, n), (n +1, 1),,(n, )} (1) is calle a Cn pattern A Toeplitz C n pattern is a C n pattern with a restriction that the partial matrix specifying the pattern is a partial Toeplitz matrix Remar 1 In the efinition, replacing by n gives the same pattern So this is why we call it Cn pattern It is trivial to show that when n is even an = n, the pattern has PSD completion So in the following, we always assume that <n Remar 2 If we number the iagonals for an n n symmetric Toeplitz matrix in the following way: The main iagonal is given the number 0 an then the iagonals are numbere in increasing orer to the last, which becomes the (n 1)th, one Then the (Toeplitz) Cn pattern is, in fact, the th an (n )th iagonal In terms of the efinition, our problems are as follows Problem 1 Whether oes the C n pattern have a PSD completion? Without loss of generality, associate with a normalize of the ata (see [1]), we assume

3 Toeplitz an Positive Semiefinite Completion Problem For Cycle Graph 69 that the partial PSD matrix specifying the Cn pattern is of the form: 1 cosθ 1 cos θ n +1? 1 cosθ 2? cos θ 1 1? cos θn C = cos θ 2? 1 cos θ n +1? cos θn? cosθ n cos θ n 1, (2) where 0 θ i π, the?s inicate unspecifie entries Thus Problem 1 can be restate as: Problem 1 When oes the partial PSD matrix (2) amit a PSD completion? Next problem is about Toeplitz C n pattern an so more special than the previous one Problem 2 Whether oes a Toeplitz C n pattern have a Toeplitz PSD completion? Lie Problem 1, Problem 2 can be restate as: Problem 2 When oes the partial Toeplitz PSD matrix 1 cosθ cos φ? 1 cosθ? cos θ 1? cos φ T = cos θ? 1 cos φ? cos θ? cosφ cos θ 1 (3) amit a Toeplitz PSD completion? Here we assume that in the partial matrix T, 0 θ π, 0 φ π an cos θ, cosφ lies in the th, the (n )th iagonal, respectively In the case = 1, Problem 1 or Problem 1 has been resolve by W Barrett, C R Johnson an P Tarazaga in [1] (see Theorem 2) A partial answer to Problem 2 or 2 in the case =1 was also given by them (see Theorem 9 below) Our main contribution in this note is to exten their results to arbitary (n/2 > 1) for Problem 1 (see next section) an give some results for Problem 2 (see section 3) 2 Completion of Cn pattern Graphs an igraphs are very effective to stuy matrix completion problems [3] Since the Cn pattern is a symmetric pattern, we shall use unirecte graph to stuy its completion problem Let Q be a symmetric pattern for n n matrices, the graph G=(V,E) of Q is efine to have noes set V = N {1, 2,,n} an ege (i, j) E if an only if (i, j) Q

4 70 He Ming Michael K Ng A path in a graph G =(V,E) is a sequence of noes v 1,v 2,,v,v +1 in V such that for i =1, 2,,,(v i,v i+1 ) E an all noes are istinct except possibly v 1 = v +1 Ifv 1 = v +1 in the path v 1,v 2,,v,v +1, then the path forms a simple circle, which will be enote by {v 1,v 2,,v +1 (= v 1 )}, its length is Accoring to the efinition, the graph of the Cn 1 pattern is a simple circle of length n, for whose PSD completion problem the following result was given in [1] Theorem 2 Let n 4anN = {1, 2,, n}, 0 θ 1,θ 2,, θ n π, then the matrix 1 cosθ 1 cos θ n cos θ 1 1 cosθ 2? C = cos θ 2 1 (4)? cos θn 1 cos θ n cos θ n 1 1 has a PSD completion if an only if for each S N with S o, θ i ( S 1)π + θ i (5) i S i N\S A graph is choral if it has no minimal simple cycle of length four or more The following result is well-nown[2] Theorem 3 Every partial PSD matrix specifying a pattern Q has a PSD completion if an only if the graph of Q is choral A graph is connecte if there is a path from any noe to any other noe; otherwise it is isconnecte A subgraph of the graph G =(V G,E G ) is a graph H =(V H,E H ), where V H V G an E H E G an that (u, v) E H requires u, v V H since H is a graph A component of a graph is maximal connecte subgraph The graph G =(V G,E G ) is isomorphic to the graph H =(V H,E H ) by isomorphism φ if φ is a one-to-one map from V G onto V H an (u, v) E G if an only if (φ(u),φ(v)) E H Relabeling the noes of a graph iagram correspons to perform a graph isomorphism Since the class of PSD matrices is close uner permutation similarity, we are free to relabel graphs as esire an we have Lemma 4 Let Q be a pattern an G its graph If every component of G is isomorphic to a graph of a pattern an the pattern has PSD completion, then Q has PSD completion Proof The proof of the lemma is similar to that of Lemma 34 of [4] an is omitte here In the following, we let gc(n, ) enote the greatest common ivisor of the integers n an Let x = x (mo n), =gc(n, ) ant = n Now we escribe the graph of the C n pattern Lemam 5 Every component of the graph of Cn pattern is isomorphic to a simple circle

5 Toeplitz an Positive Semiefinite Completion Problem For Cycle Graph 71 of length t Proof Let Q be the Cn pattern efine in (1) an G =(N,E) its graph Using the notation of x, the ege set can be written as E = {(i +(j 1), i + j) 1 i ;1 j t} By noting that i + t = i, it is not ifficult to prove that the graph G is compose by the following inepenent simple cycles of length t: {1, 1+, 1+2, 1+(t 1), 1+t} {2, 2+, 2+2, 2+(t 1), 2+t} {3, 3+, 3+2, 3+(t 1), 3+t},,,, {, +, +2, +(t 1), + t} The proof is complete Remar 3 The lemma shows that the graph of the Cn pattern is a cycle graph[9], that is, it is compose by some simple cycles But only one simple cycle also is possible even for >1 if n an have no common ivisor larger than an equal to 2 In this case, t = n an (2) is permutation similar to (4) For example, let n =5, = 2, then it is easy to verify that the following two partial PSD matrices are permutation similar: 1? c 1 c 4?? 1? c 2 c 5 c 1? 1? c 3 c 4 c 2? 1?? c 5 c 3? 1 1 c 1?? c 4 c 1 1 c 3??? c 3 1 c 5??? c 5 1 c 2 c 4?? c 2 1, here we enote cosθ i by c i 1 By using Lemma 4 an 5, we easily erive the following results for Problem 1 an Problem Theorem 6 PSD completion Theorem 7 The C n pattern has a PSD completion if an only if the C 1 n pattern has a The partial PSD matrix (2) is permutation similiar to the partial PSD matrix:

6 72 He Ming Michael K Ng C = C 1 C 2 C,where C i = 1 cosθ i cos θ i+(t 1) cos θ i 1 cosθ i+? cos θ i+ 1? cos θi+(t 2) cos θ i+(t 1) cos θ i+(t 2) 1 (6) an all entries outsie the iagonal blocs of C are unspecifie Thus the partial PSD matrix C of (2) can be complete to a PSD matrix if an only if the partial PSD matrix C can It is easy to complete C ( then complete C of (2) to a PSD martrix: complete every C i to a PSD matrix by using the metho given in [1], an then set all (unspecifie) entries outsie the iagonal blocs of C to zero On the other han, if the partial PSD matrix C of (2) has PSD completion, then C an so every C i of(6)musthaveapsd completion too So for Problem 1, the following solution is given by applying Theorem 7 an Theorem 2 Theorem 8 Let n 4, 1 <n/2, = gc(n, ) ant = n/ Thenwehave a) If t = 3, then the partial PSD matrix (2) has a PSD completion b) If t 4, then the partial PSD matrix (2) has a PSD completion if an only if for each S i N i {i, i +, i +2,, i +(t 1)} with S i o, θ j ( S i 1)π + θ j, (7) j S i j N i\s i i =1, 2,, Proof to 3 From the assumption of < n/2, we now that t is an integer larger than or equal When t = 3, Lemma 5 shows that the graph of the Cn pattern is choral ( simple cycles of length 3) So C has a PSD completion by Theorem 3, a) is prove When t 4, applying Theorem 2 to each C i of (6), b) is obtaine 3 Completion of Toeplitz Cn Pattern Now we consier Problem 2 or 2 : Toeplitz PSD completion problem for the Toeplitz Cn pattern, which is more ifficult than non-toeplitz completion problem We shall see that Lemma 5 is not irectly applicable to the problem [1] has not solve Problem 2 for the case =1an only presente a result for non-toeplitz PSD completion as follows

7 Toeplitz an Positive Semiefinite Completion Problem For Cycle Graph 73 Theorem 9 [1] Let C be an n n matrix, if n 4anθ, φ [0,π], then C = 1 cosθ cos φ cos θ 1 cosθ? cos θ 1? cos θ cos φ cos θ 1 (8) has PSD completion if an only if φ (n 1)θ (n 2)π + φ, n is even; (9) φ (n 1)θ (n 1)π φ, n is o (10) The following result generalizes the theorem Theorem 10 Let n 4, 1 <n/2, = gc(n, ) ant = n/ Thenwehave a) If t = 3, then the partial Toeplitz PSD matrix (3) has a PSD completion b) If t 4, then the partial Toeplitz PSD matrix (3) has a PSD completion if an only if : 1) when n is o an is even, (n )θ (n )π + φ (2n 2)π (n )θ (11) 2) when n is even an is o, φ ( )π +(n )θ (n + 2)π φ (12) 3) when both n an are o, (n )θ (n )π + φ (n 2)π +(n )θ (13) Proof a) is obvious from the proof of Theorem 8 We prove b) by applications of Theorem 7anTheorem8 First, note that it is impossible that both n an are even, since =gc(n, ) Seconly, we now from Theorem 7 that T is permutation similar to T = C 1 C 2 C where C i is efine in (6), but θ j is θ or φ an each C i contains n cos θ an cos φ Thus C i is generally not a partial Toeplitz matrix, so it is Theorem 8 other than Theorem 9 that is applie in the following proof T has a PSD completion if an only if every C i has In fact, all C i (i =1, 2,,)havethe same form So we only nee to consier the PSD completion of one C i

8 74 He Ming Michael K Ng It is obvious that the inequality (7) hols automatically if θ or φ occurs on both sies of (7), since θ π ( φ π ) So we only nee to consier the following two cases: one case is that t is o an all θ an φ appear on the left sie of the inequality (7) The another case is that all n θ just appears on one sie of (7), all φ appears in the other sie of (7) Thus we have 1) When n is o, is even, we only nee to consier S i, in Theorem 8, such that S i = n an S i = t The require inequalities are (n ) (n ) θ ( 1)π + (n ) φ an θ + φ (n 1)π (14) 2) When n is even, is o, we only nee to consier S i such that S i = an S i = t The require inequalities are ( n )φ (n 1)π + n θ an (n ) θ + φ (n 1)π (15) 3) When both n S i = The require inequalities are an are o, we have to consier S i such that S i = n an (n ) (n ) θ ( 1)π + φ an (n )φ (n 1)π + n θ (16) We get (11), (12) an (13) from (14), (15) an (16), respectively The theorem is prove Corollary 11 When θ = φ, ift is three or even, then the partial Toeplitz PSD matrix (3) has a PSD completion; if t is o larger than three, then the partial Toeplitz PSD matrix (3) has a PSD completion if an only if θ n n π or cos θ cos π (17) n Proof We o not nee to prove the case t 3 When t is even, both n an are o n By applying 3) of Theorem 10 an noting that, the inequalities (13) hol automatically When t is o larger than three, applying 1) an 2) of Theorem 10, we fin the require inequality only is (17) By using a similar proof, we get another interesting result Corollary 12 If t 4, θ = π φ or cos θ = cos φ, then the partial Toeplitz PSD matrix (3) has a PSD completion if an only if 1) when n is o an is even, θ n n π or cos θ cos n π

9 Toeplitz an Positive Semiefinite Completion Problem For Cycle Graph 75 2) when n 3) when both n is even an is o, θ n π or cos θ cos n π an are o, n π θ n n π or cos n cos θ cos n π In terms of the two corollaries, the two partial Toeplitz PSD matrices 1? 1 1? 1? 1 1?? 1? 1 1? 1? 1 1 1? 1? 1 an 1? 1? 1 1 1? 1? 1 1? 1?? 1 1? 1? 1 1? 1 have PSD completions, but the two partial Toeplitz PSD matrices 1? 1 1?? 1? 1 1 1? 1? 1 an 1 1? 1?? 1 1? 1 have no PSD completions 1? 1 1?? 1? 1 1 1? 1? 1 1 1? 1?? 1 1? 1 Now we turn to Problem 2 an iscuss the case = 1 first For notation convenience an obvious reason, in the following by To(a 1,a 2,,a n ) we shall enote a symmetric Toeplitz matrix whose first row is (a 1,a 2,,a n ) The partial Toeplitz PSD matrix (8) can be written as To(1,cosθ,?,,?,cosφ) Theorem 13 The partial Toeplitz PSD matrix (8) has a Toeplitz PSD completion if 1 cos φ cos(n 1)θ cos θ an 0 1 (18) n 1 cos θ cos(n 1)θ or if or if cos θ 1 n 1 an 0 cos φ cos(n 1)θ ( 1) n 1 (19) cos θ cos(n 1)θ 1 n 1 cos θ 1 n 1 an { 0 cos θ cos φ 2cosθ 1, if n is o, cos φ =cosθ, if n is even (20) Proof Define three symmetric Toeplitz matrices: B = To(1, cos θ, cos 2θ,,cos(n 1)θ), (21) E = To(1, cos θ, cos θ,,cos θ), (22) an F = To(1, cos θ, cos θ,,( 1) n cos θ) (23) We first prove that B is a PSD matrix, E is positive semiefinite if cos θ 1 n 1,anF is positive semiefinite if cos θ 1 n 1

10 76 He Ming Michael K Ng In fact, B is positive semiefinite, since 1 0 ( cos θ sin θ 1 cosθ cos 2θ cos(n 1)θ B = cos 2θ sin 2θ 0 sinθ sin 2θ sin(n 1)θ cos(n 1)θ sin(n 1)θ E is positive semiefinite if cos θ 1 n 1, since it can be written as E =(1 cos θ)i n + (cos θ)ee T, where I n is n n ientity matrix an e =(1, 1,,1) T Thus it is easy to verify that E is symmetric an has n-1 eigenvalues 1 cosθ an one eigenvalue 1 + (n 1)cosθ all eigenvalues of E are nonnegative uner the assumption Similarly, F is positive semiefinite if cosθ 1, since it has n-1 eigenvalues 1 + cosθ an one eigenvalue 1 (n 1)cosθ n 1 Now we prove that (8) has a Toeplitz PSD completion if (18) hols Define T t =(1 t)b+te, then T t is a Toeplitz PSD matrix for any t [0, 1] if cos θ 1, since it is the sum of two n 1 Toeplitz PSD matrices: (1 t)b an te cos φ cos(n 1)θ Let t 0 = cos θ cos(n 1)θ,then0 t 0 1 by the assumption (18) an cos φ =(1 t 0 )cos(n 1)θ + t 0 cos θ Thus T t0 = To(1, cos θ, (1 t 0 )cos2θ + t 0 cos θ,,(1 t 0 )cos(n 2)θ + t 0 cos θ, cos φ) is a Toeplitz PSD completion of (8) ) Similarly, if we efine T t completion if (19) hols = (1 t)b + tf, we can prove that (8) has a Toeplitz PSD Finally, we prove that (8) has a Toeplitz PSD completion if (20) hols Now we efine T t =(1 t)e + tf,thent t is a Toeplitz PSD matrix for any t [0, 1] if 1 n 1 cos θ 1 n 1 Note now that T t = To(1, cos θ, (1 2t)cosθ,,[1 (1 ( 1) n )t]cosθ) When n is even, it is necessary that cos φ =cosθ in orer that T t is a Toeplitz PSD completion cos θ cos φ of (8) When n is o, let t 0 =,then0 t 0 1 by the assumption (20) an 2cosθ cos φ =(1 2t 0 )cosθ Therefore T t0 is a Toeplitz PSD completion of (8) Now we give a result for Problem 2 : Theorem 14 The partial Toeplitz PSD matrix (3) has a Toeplitz PSD completion if 1 n 1 n cos φ cos cos θ an 0 θ cos θ cos n 1 (24) θ or if when is o, cos θ 1 n 1 an 0 cos φ cos n θ ( 1) n +1 cos θ cos n 1 (25) θ

11 Toeplitz an Positive Semiefinite Completion Problem For Cycle Graph 77 when is even, or if when is o, 1 n 1 cos θ an 0 cos φ cos n θ ( 1) n cos θ cos n 1 (26) θ when is even, 1 n 1 cos θ 1 n 1 an { 0 cos θ cos φ 2cosθ 1, if n is even, cos φ = cosθ, if n is o (27) 1 n 1 cos θ an { 0 cos θ cos φ 2cosθ 1, if n is o, cos φ =cosθ, if n is even (28) Proof To prove the theorem, we nee to efine two more symmetric Toeplitz matrices except E, F in (22),(23) B = To(1, cos α, cos 2α,,cos(n 1)α), (29) here α = θ an F = To(1, cos θ, cos θ,,( 1) n 1 cos θ) (30) 1 B is clearly positive semiefinite F is positive semiefinite if cos θ, since it has n-1 n 1 eigenvalues 1 cos θ an one eigenvalue 1 + (n 1) cos θ Similar to the proof of Theorem 13, efine T t to be one of the following matrices: 1) (1 t) B + te, 2) (1 t) B + tf when is o, (1 t) B + t F when is even, 3) (1 t)e + tf when is o, (1 t)e + t F when is even We can complete the proof of the theorem From Theorem 14, we easily get Corollary 15 When θ = φ, the partial Toeplitz PSD matrix (3) has a Toeplitz PSD completion if cos θ 1 1 or if n is even, is o an cos θ n 1 n 1 When θ = π φ an n is o, the partial Toeplitz PSD matrix (3) has a Toeplitz PSD completion if is o an cosθ 1 1,orif is even an cos θ n 1 n 1 From the proof of Theorem 14, we can see that it is not necessary to connect (n )th iagonal to th iagonal, so the result of Theorem 14 can easily be extene to the following more general pattern: Q1 ={(1,+1), (2,+2),,(n, n), (l +1, 1),,(n, n l)},

12 78 He Ming Michael K Ng where 1 <l n 1 In fact, the pattern just is two ifferent iagonals: the th an the lth iagonal A final remar We have trie many numerical examples but faile to fin a partial Toeplitz PSD matrix specifying the C n pattern which has a PSD completion but no Toeplitz PSD completion This mae us to believe the following is true: Conjecture The partial Toeplitz PSD matrix (3) that has a PSD completion has a Toeplitz PSD completion References 1 Barrett W, Johnson C R, Tarazaga P The real positive efinite completion problem for a simple cycle Linear Algebra Appl, 1993,192: Grone R, Johnson C R, Sa E, Wolowicz H Positive efinite completions of partial Hermitian matrices Linear Algebra Appl, 1984,24: Hogben L Graph theoretic mathos for matrix completion problems Linear Algebra Appl, 2001,328: Hogben L Matrix completion problem for pairs of relate classes of matrices Linear Algebra Appl, 2003,373: Johnson C R, Smith R L The completion problem for M-matrices an inverse M-matrices Linear Algebra Appl, 1996, : Johnson C R, Kroschel B K The combinatorially symmetric P -matrix completion problem Electronic J Linear Algebra, 1996, 1: Johnson C R Matrix completions problem: A survey, in Matrix Theory an Applications Proceeing of Symposia of Applie Mathematics, Vol 40, C R Johnson, e, American Mathematical Society, Provience, RI, 1990, 40: Neval, G On the completion of partial given triangular Toeplitz matrices to contractions SIAM J Matrix Anal Appl, 1993,14: Yong X R, Acenjian T The numbers of spanning trees of cubic cycle CN 3 an quaruple cycle CN 4 Discrete Math, 1997, 169: He Ming Department of Information an Computational Mathematics, Xiamen University, Xiamen , PRC MichaelKNg Hongong Department of Mathematics, University of Hong Kong, Pofulam Roa,