On convexity of polynomial pats and generalized majorizations Marija Dodig Centro de Estruturas Lineares e Combinatórias, CELC, Universidade de Lisboa, Av. Prof. Gama Pinto 2, 1649-003 Lisboa, Portugal dodig@cii.fc.ul.pt Marko Stošić Instituto de Sistemas e Robótica and CAMGSD, Instituto Superior Técnico, Av. Rovisco Pais 1, 1049-001 Lisbon, Portugal mstosic@isr.ist.utl.pt Submitted: Nov 15, 2009; Accepted: Apr 5, 2010; Publised: Apr 19, 2010 Matematics Subject Classification: 05A17, 15A21 Abstract In tis paper we give some useful combinatorial properties of polynomial pats. We also introduce generalized majorization between tree sequences of integers and explore its combinatorics. In addition, we give a new, simple, purely polynomial proof of te convexity lemma of E. M. de Sá and R. C. Tompson. All tese results ave applications in matrix completion teory. 1 Introduction and notation In tis paper we prove some useful properties of polynomial pats and generalized majorization between tree sequences of integers. All proofs are purely combinatorial, and te presented results are used in matrix completion problems, see e.g. [2, 4, 7, 10, 11]. We study cains of monic polynomials and polynomial pats between tem. Polynomial pats are combinatorial objects tat are used in matrix completion problems, see [7, 9, 11]. Tere is a certain convexity property of polynomial pats appeared for te first time in [5]. In Lemma 2 we give a simple, direct polynomial proof of tat result. We also sow tat no additional divisibility relations are needed. Tis work was done witin te activities of CELC and was partially supported by FCT, project ISFL-1-1431, and by te Ministry of Science of Serbia, projects no. 144014 (M. D.) and 144032 (M. S.). Corresponding autor. te electronic journal of combinatorics 17 (2010), #R61 1
Also, we explore generalized majorization between tree sequences of integers. It presents a natural generalization of a classical majorization in Hardy-Littlewood-Pólya sense [6], and it appears frequently in matrix completion problems wen bot prescribed and te wole matrix are rectangular (see e.g [1, 4, 11]). We give some basic properties of generalized majorization, and we prove tat tere exists a certain pat of sequences, suc tat every two consecutive sequences of te pat are related by an elementary generalized majorization. E. Marques de Sá [7] and independently R. C. Tompson [10], gave a complete solution for te problem of completing a principal submatrix to a square one wit a prescribed similarity class. Te proof of tis famous classical result is based on induction on te number of added rows and columns, and one of te crucial steps is te convexity lemma. Te original proofs of te convexity lemma, wic are completely independent one from te anoter one, bot in [7] and [10] are rater long and involved. Later on, new combinatorial proof of tis lemma as appeared in [8]. In Teorem 1, we give simple and te first purely polynomial proof of tis result. 1.1 Notation All polynomials are considered to be monic. Let F be a field. Trougout te paper, F[λ] denotes te ring of polynomials over te field F wit variable λ. By f g, were f, g F[λ] we mean tat g is divisible by f. If ψ 1 ψ r is a polynomial cain, ten we make a convention tat ψ i = 1, for any i 0, and ψ i = 0, for any i r + 1. Also, for any sequence of integers satisfying c 1 c m, we assume c i = +, for i 0, and c i =, for i m + 1. 2 Convexity and polynomial pats Let α 1 α n and γ 1 γ n+m be two cains of monic polynomials. Let n+j π j := lcm(α i j, γ i ), j = 0,..., m. (1) We ave te following divisibility: Lemma 1 π j π j+1, j = 0,...,m 1 (i.e. π 0 π 1 π m ). Proof: By te definition of π j, j = 0,..., m, te statement of Lemma 1 is equivalent to i.e., n+j lcm(α i j, γ i ) n+j+1 lcm(α i j 1, γ i ), j = 0,...,m 1, n n lcm(α i, γ i+j ) γ j+1 lcm(α i, γ i+j+1 ), j = 0,...,m 1, (2) te electronic journal of combinatorics 17 (2010), #R61 2
wic is trivially satisfied. By Lemma 1 we can define te following polynomials σ j := π j π j 1, j = 1,...,m. (3) Ten, we ave te following convexity property of π i s: Lemma 2 σ j σ j+1, j = 1,...,m 1 (i.e. σ 1 σ 2 σ m ). Proof: By te definition of σ j, j = 1,..., m, te statement of Lemma 2 is equivalent to n+j lcm(α i j, γ i ) n+j 1 lcm(α i j+1, γ i ) n+j+1 i.e. for all j = 1,..., m 1, we ave to sow tat lcm(α i j 1, γ i ) n+j lcm(α, j = 1,...,m 1, i j, γ i ) γ j lcm(α 1, γ j+1 ) lcm(α 2, γ j+2 ) lcm(α n, γ j+n ) lcm(α 1, γ j ) lcm(α 2, γ j+1 ) lcm(α n, γ j+n 1 ) γ j+1 lcm(α 1, γ j+2 ) lcm(α 2, γ j+3 ) lcm(α n, γ j+n+1 ). (4) lcm(α 1, γ j+1 ) lcm(α 2, γ j+2 ) lcm(α n, γ j+n ) Before proceeding, note tat for every two polynomials ψ and φ we ave lcm (ψ, φ) = ψφ gcd(ψ, φ) (5) Tus, for every i and j, we ave lcm (α i, γ i+j ) = lcm(lcm(α i, γ i+j 1 ), γ i+j ) = By applying (6), equation (4) becomes equivalent to γ j γ i+j lcm(α i, γ i+j 1 ) gcd(lcm(α i, γ i+j 1 ), γ i+j ). (6) n n gcd(lcm(α i, γ i+j ), γ i+j+1 ) γ n+j+1 gcd(lcm(α i, γ i+j 1 ), γ i+j ). (7) By sifting indices, te rigt and side of (7) becomes n 1 γ n+j+1 gcd(lcm(α 1, γ j ), γ j+1 ) gcd(lcm(α i+1, γ i+j ), γ i+j+1 ). Tis, togeter wit obvious divisibilities γ j gcd(lcm(α 1, γ j ), γ j+1 ) and gcd(lcm(α n, γ n+j ), γ n+j+1 ) γ n+j+1, proves (7), as wanted. te electronic journal of combinatorics 17 (2010), #R61 3
Frequently wen dealing wit polynomial pats we ave te following additional assumptions and γ i α i, i = 1,...,n (8) α i γ i+m, i = 1,...,n. (9) Ten te following lemma follows trivially from te definition of π i s, for i = 0 and i = m: Lemma 3 π 0 = n α i and π m = n+m γ i. 2.1 Polynomial pats Let α = (α 1,..., α n ) and γ = (γ 1,...,γ n+m ) be two systems of nonzero monic polynomials suc tat α 1 α n and γ 1 γ n+m. A polynomial pat between α and γ as been defined in a following way in [7, 9], see also [11]: Definition 1 Let ǫ j = (ǫ j 1,...,ǫj n+j ), j = 0,...,m, be a system of nonzero monic polynomials. Let ǫ 0 := α and ǫ m := γ. Te sequence ǫ = (ǫ 0, ǫ 1,...,ǫ m ) is a pat from α to γ if te following is valid: ǫ j i ǫj i+1, i = 1,...,n + j 1, j = 0,..., m, (10) ǫ j i ǫj 1 i ǫ j i+1, i = 1,...,n + j 1, j = 1,...,m. (11) Consider te polynomials β j i := lcm(α i j, γ i ), i = 1,..., n + j, j = 0,...,m from (1). Let β j = (β1, j...,β j n+j ), j = 0,...,m. Ten te following proposition is valid (see Proposition 3.1 in [11] and Section 4 in [7]): Proposition 1 Tere exists a pat from α to γ, if and only if γ i α i γ i+m, i = 1,..., n. (12) Moreover, if (12) is valid, ten β = (β 0,...,β m ) is a polynomial pat between α and γ, and for every pat ǫ between α and γ old Hence, β is a minimal pat from α to γ. β j i ǫj i, i = 1,..., n + j, j = 0,...,m. Te polynomials π j from (1) are defined as π j = n+j βj i. Te polynomials σ i were used by Sá [7, 9] and by Zaballa [11], but te convexity of π j s, i.e. te result of Lemma 2, was obtained later by Goberg, Kaasoek and van Scagen [5]. We gave a direct polynomial proof of tis result and we ave sown tat it olds even witout te divisibility relations (12). te electronic journal of combinatorics 17 (2010), #R61 4
3 Generalized majorization Let d 1 d ρ, f 1 f ρ+l and a 1 a l, be nonincreasing sequences of integers. Definition 2 We say tat f (d, a), i.e., we ave a generalized majorization between te partitions d = (d 1,...,d ρ ), a = (a 1,..., a l ) and f = (f 1,...,f ρ+l ), if and only if d i f i+l, i = 1,..., ρ, (13) ρ+l f i = ρ d i + l a i, (14) q f i d i q a i, q = 1,...,l, (15) were q = min{i d i q+1 < f i }, q = 1,..., l. Remark 1 Recall tat in Section 1.1 we ave made a convention tat f i = + and d i = +, for i 0, and tat f i =, for i > ρ + l, and d i =, for i > ρ. Tus, q s are well-defined. In particular, for every q = 1,..., l, we ave q q q + l, and 1 < 2 <... < l. Note tat if ρ = 0, ten te generalized majorization reduces to a classical majorization (in Hardy-Littlewood-Pólya sense [6]) between te partitions f and a (f a). If l = 1, (13) (15) are equivalent to Indeed, for l = 1, (15) becomes d i f i+1, i = 1,...,ρ, (16) ρ+1 f i = ρ d i + a 1, (17) d i = f i+1, i 1. (18) 1 f i Te last inequality togeter wit (14), gives ρ+1 i= 1 +1 1 1 f i d i + a 1. ρ i= 1 d i. (19) Finally, from (13), we obtain tat (19) is equivalent to (18), as wanted. Generalized majorization for te case l = 1 will be called elementary generalized majorization, and will be denoted by f 1 (d, a). te electronic journal of combinatorics 17 (2010), #R61 5
In particular, if l = 1, and f, d and a satisfy d i f i, i = 1,...,ρ and (17), ten 1 = ρ + 1, and so f 1 (d, a). Note tat if f (d, a), ten in te same way as in te proof of te equivalence of (15) and (18), we ave d i = f i+l, i l l + 1. (20) Te aim of tis section is to sow tat tere is a generalized majorization between te partitions d, a and f if and only if tere are elementary majorizations between tem, i.e. if and only if tere exist intermediate sequences tat satisfy (16) (18). In certain sense, we sow tat tere exists a pat of sequences between d and f suc tat every neigbouring two satisfy te elementary generalized majorization (see Teorems 5 and 7 below). More precisely, we sall sow tat f (d, a) if and only if tere exist sequences g i = (g i 1,...,g i ρ+i), i = 1,..., l 1, wit g i 1 g i ρ+i, and wit te convention g 0 := d and g l := f, suc tat g i 1 (gi 1, a i ), i = 1,...,l. Lemma 4 Let f, d and a be te sequences from Definition 1. If f (d, a), ten tere exist integers g 1 g ρ+l 1, suc tat (i) g i f i+1, i = 1,...,ρ + l 1, (ii) d i g i+l 1, i = 1,...,ρ, (iii) g i = f i+1, i, were := min{i g i < f i }, (iv) (v) q g i ρ+l f i = ρ+l 1 d i q a i, q = 1,...,l 1, were q = min{i d i q+1 < g i }, a l. Proof: Let H 1,...,H l 1 be integers defined as and H q := q q a i f i + H 0 := 0. d i, q = 1,...,l 1, te electronic journal of combinatorics 17 (2010), #R61 6
Note tat from (15), we ave tat H q 0, q = 1,...,l 1. Let Tus S q := Since a 1 a l 1, we ave Now, define te numbers i= q 1 q+2 d i q i= q 1 +1 f i, q = 1,..., l 1. H q H q 1 = S q + a q, q = 1,..., l 1. H 1 S 1 H 2 H 1 S 2 H l 1 H l 2 S l 1. (21) Tus, we ave H i := min(h i, H i+1,..., H l 1 ), i = 0,..., l 1. (22) H 1 H l 1, (23) H l 1 = H l 1 and H i H i, i = 1,..., l 2. (24) We are going to define certain integers g 1,...,g ρ+l 1. Te wanted g 1 g ρ+l 1 will be defined as te nonincreasing ordering of g 1,...,g ρ+l 1. Let g i := d i l+1, i > l 1. (25) We sall split te definition of g 1,...,g l 1 into l 1 groups. For arbitrary j = 1,...,l 1, we define g i, i = j 1 + 1,..., j, (wit convention 0 := 0) in a following way: If f j H j H j 1 S j, (26) ten we define g j 1 +1 g j 1 as a nonincreasing sequence of integers suc tat and j 1 i= j 1 +1 g i d i j+1 g i f i j 1 i= j 1 +1 f i = H j H j 1 (tis is obviously possible because of (26)). Also, in tis case, we define g j := f j. If f j < H j H j 1 S j, (27) te electronic journal of combinatorics 17 (2010), #R61 7
ten we define and g i := d i j+1, i = j 1 + 1,..., j 1, g j := H j H j 1 S j. Note tat in bot of te previous cases, (26) and (27), we ave j i= j 1 +1 g i j i= j 1 +1 f i = H j H j 1, j = 1,..., l 1. (28) and g i = max(f i, H i H i 1 S i), i = 1,..., l 1. Now, let i {1,..., l 2}. If g i+1 = f i+1, ten g i+1 f i g i. If g i+1 = H i+1 H i S i+1 > f i+1, ten, from (28), we ave tat H i+1 > H i, and so H i = H i. However, tis togeter wit (21), gives Hence, we ave g i+1 = H i+1 H i S i+1 H i+1 H i S i+1 = H i+1 H i S i+1 H i H i 1 S i = H i H i 1 S i H i H i 1 S i g i. g 1 g 2 g l 1. (29) Also, from te definition of i, i = 1,...,l 1, te subsequence of g i s for i {1,..., ρ + l 1} \ { 1,..., l 1 } is in nonincreasing order, and satisfies: For i l, from (20), we ave d i j+1 g i f i, j 1 < i < j, j = 1,...,l. (30) d i l+1 = g i = f i+1, i l. (31) Now, since g i f i+1 for all i = 1,..., ρ + l 1, and since g i s are te nonincreasing ordering of g i s, we ave (i). Moreover, since g l 1 f l 1 > d l 1 l+2 = g l 1 +1, we ave tat g i = g i, for i > l 1. Ten, from (30), we ave g i f i, for i < l, wic togeter wit g l = g l = d l l+1 < f l, implies = l. Tus, (31) implies (iii). If we denote by ν 1 ν ρ te subsequence of g i s for i {1,...,ρ + l 1} \ { 1,..., l 1 }, ten from (30) and (31) we ave wic implies (ii). d i ν i, i = 1,..., ρ, (32) te electronic journal of combinatorics 17 (2010), #R61 8
Also, by summing all inequalities from (28), for j = 1,..., l 1, we ave l 1 l 1 g i f i = H l 1, wic togeter wit (24) and te definition of H l 1, gives l 1 l 1 l+1 g i l 1 d i = a i. Te last equation, togeter wit te definition of te remaining g i s (25), te fact tat ρ+l 1 g i = ρ+l 1 g i, and (14), gives (v). Before going to te proof of (iv), we sall establis some relations between q s and q s. So, let q {1,...,l 1}. Te sequence of g i s is defined as te nonincreasing ordering of g i s. As we ave sown, te sequence of g i s is te union of two nonincreasing sequences: g 1 g 2... g l 1 and ν 1 ν 2... ν ρ. Let r q be te index suc tat ν rq g q > ν rq+1. First of all, from te definition of g q and q, we ave tat g q f q > d +1 ν +1, and so r q q q. (33) Furtermore, te subsequence g 1 g 2... g rq+q is te nonincreasing ordering of te union of sequences g 1 g 2... g q and ν 1 ν 2... ν rq, wit g q being te smallest among tem, i.e. g rq+q = g q. Tus, ν i g i+q 1, for i = 1,...,r q, and so from (32), for every i r q we ave tat d i ν i g i+q 1, i.e. q r q + q. (34) By (33), we ave two possibilities for r q : If r q = q q, as proved above, we ave g q = g q, wic ten implies g q f q > d +1 ν +1, and so q q, wic togeter wit (34) in tis case gives q = q = r q + q. If r q < q q, ten g q > ν f q, and so from te definition of g i s, we ave tat ν i = d i, for i = r q + 1,..., q q. Tus g rq+q = g q > ν rq+1 = d rq+1, and so q r q + q, wic togeter wit (34) gives q = r q + q. Tus, altogeter we ave tat q q, and g 1 g 2... g q is te nonincreasing ordering of te union of sequences g 1 g 2... g q and ν 1 ν 2... ν, wit g q = g q, and tat q < q implies ν i = d i, for i = q q + 1,..., q q. te electronic journal of combinatorics 17 (2010), #R61 9
Finally, we can pass to te proof of (iv). Let q {1,..., l 1}. We sall prove (iv) for tis q in te following equivalent form q g i If q = q, (35) is equivalent to q q d i H q + f i d i. (35) (g i f i) H q, (36) wic follows from (24) and (28). If q < q, we ave tat ν i = d i, for i = q q + 1,..., q q. Hence, te condition (35) is again equivalent to (36), wic concludes our proof. By iterating te previous result, we obtain te following Teorem 5 Let f, d and a be te sequences from Definition 1. If f (d, a), ten tere exist sequences of integers g j = (g j 1,...,g j ρ+j ), j = 1,...,l 1, wit gj 1 g j ρ+j, suc tat g j 1 (gj 1, a j ), j = 1,...,l, were g 0 = d and g l = f. Proof: For l = 1, te claim of teorem follows trivially. Let l > 1, and suppose tat teorem olds for l 1. By Lemma 4, tere exists a sequence g = (g 1,..., g ρ+l 1 ), suc tat g 1 g ρ+l 1 and suc tat tey satisfy conditions (i) (v) from Lemma 4. Set g l 1 := g. From (i), (iii) and (v) we ave From (ii), (iv) and (v), we ave f 1 (gl 1, a l ). (37) g l 1 (d, a ), (38) were a = (a 1,...,a l 1 ). By induction ypotesis tere exist sequences g 1,...,g l 2, suc tat Tis togeter wit (37) finises our proof. g j 1 (gj 1, a j ), j = 1,..., l 1. Te following two results give converse of Lemma 4 and Teorem 5: te electronic journal of combinatorics 17 (2010), #R61 10
Lemma 6 Let d 1 d ρ, f 1 f ρ+l and g 1 g ρ+1 be integers. Let a 1 and a 2 a l be integers. Let a 1 a l be integers suc tat If ten (i) d i g i+1, i = 1,..., ρ, (ii) g i f i+l 1, i = 1,..., ρ + 1, (a 1, a 2,...,a l ) (a 1, a 2,...,a l ). (39) (iii) d i = g i+1, i 1, were 1 = min{i d i < g i }, (iv) (v) q f i ρ+l ρ+1 f i = q+1 g i a i, q = 1,..., l 1, were q = min{i g i q+1 < f i }, l ρ a i = d i + q f i d i were q = min{i d i q+1 < f i }, q = 1,..., l. l a i, q a i, q = 1,...,l, (40) Proof: From te definition of q, q and 1, we obtain te following inequalities and q max( q 1, min( 1 + q 1, q )), q = 1,...,l 1, ( 0 = 0), (41) l max( l 1, 1 + l 1). (42) Tis is true since for q = 1,...,l 1, and j < min( 1 + q 1, q ), we ave tat d j q+1 g j q+1 f j. Terefore, q min( 1 + q 1, q ). Also, for every q = 1,..., l, and j < q 1, we ave d j q+1 g j q+2 f j, wic gives q q 1. Furtermore, for every j < 1 + l 1, we ave d j l+1 g j l+1 f j, and so l 1 + l 1. Altogeter, we ave (41) and (42). Let q {1,..., l 1}. From (41), we ave te following tree possibilities on q : Observe tese cases separately: a) q q, in te case q 1 + q 1, b) q > q max( q 1, 1 + q 1) if q > 1 + q 1, c) q q > max( q 1, 1 + q 1) if q > 1 + q 1. te electronic journal of combinatorics 17 (2010), #R61 11
a) Let q q ( q 1 + q 1), ten by (iv) we ave q q f i = f i + q q+1 f i d i + q q+1 +1 q+1 f i + a i q+1 d i + a i = q+1 d i + a i. Te second inequality is true since q q < 1. So, we ave d i g i for all i q q. Also, from q < q+1, we obtain f i d i q, for all i q < q+1. Finally, from (39), we ave q+1 q a i a i, and so wic proves (40), as wanted. q f i d i + q a i, b) Let q > q max( 1 + q 1, q 1 ), ten by (iv), we ave q f i = q 1 f i + q q 1 +1 f i q 1 q+1 q q 1 +1 f i + q a i q 1 q+1 +1 q 1 q+2 q a i. Te second inequality is true, since q < q, and so, g i q+1 f i, for all i q. Moreover, since q q + 1 1, by conditions (iii) and (v), we ave +1 ρ+1 g i = g i ρ+1 i= +2 g i = ρ d i + a 1 ρ i= +1 d i = d i + a 1, and so +1 q Last equality togeter wit (39) gives a i = d i + q a i. q f i d i + q a i, te electronic journal of combinatorics 17 (2010), #R61 12
wic proves (40), as wanted. c) Let q q > max( 1 + q 1, q 1 ), ten by (iv), we ave q f i = q 1 f i + q q 1 +1 f i q 1 q+1 q q 1 +1 f i + q a i = q 1 q+1 q 1 q 1 +1 f i + q q f i + q a i = q 1 q+1 q 1 q+2 q d i + a 1 + a i. d i + q a i Te second inequality follows from te definition of q and te fact tat q < q+1, wile te last equality is true since q q 1. Now, we finis te proof as in te previous case. Te only remaining case is q = l. Let i > l. Since l max( l 1, 1 +l 1), we ave i > l 1. From (ii), (iv) and (v) we ave tat f (g, a ), were a = (a 2, a 3,...,a l ), and so (see (20)) we ave f i = g i l+1. Also, since i > 1 + l 1, from (iii) we ave g i l+1 = d i l, and tus f i = d i l, i > l. (43) Now, by (v), condition (40) for q = l is equivalent to ρ+l i= l +1 f i ρ i= l l+1 Finally, from (i), we ave tat d i f i+l, i = 1,...,ρ, and so condition (40) for q = l is equivalent to (43), wic concludes our proof. d i. By iterating te previous result, we obtain te following one: Teorem 7 Let d 1 d ρ, f 1 f ρ+l, a 1 a l and a 1,...,a l be integers, suc tat (a 1,...,a l ) (a 1,...,a l ). Moreover, for every j = 1,...,l 1, let g j = (g j 1,...,g j ρ+j ) be suc tat gj 1 g j ρ+j. Also, let g 0 := d, and g l := f. If g j 1 (gj 1, a j ) for j = 1,..., l, ten f (d, a). te electronic journal of combinatorics 17 (2010), #R61 13
Tus, Teorems 5 and 7 prove te existence of a pat of sequences, as announced before Lemma 4. In particular, we ave Corollary 8 Let l 2, d 1 d ρ, f 1 f ρ+l, a 1 a l be integers. Ten f (d, a) if and only if tere exists g = (g 1,...,g ρ+s ), for some 0 < s < l, suc tat g 1 g ρ+s and f (g, a ) g (d, a ) were a = (a 1,..., a l s ) and a = (a l s+1,..., a l ). 4 Convexity lemma In tis section we give a sort polynomial proof of te convexity lemma, wic is te crucial step in Sá-Tompson teorem [7, 10]. Te original proofs of Sá and Tompson were long and complicated, and relied on very involved tecniques. Te proof in [7] (Proposition 4.1 and Lemma 4.2) uses nonelementary analytical tools, wile te proof in [10] is elementary but very long and does not involve te concept of convexity. Later on sorter, combinatorial proof was given in [8]. Here we give te first purely polynomial proof of te convexity lemma. Let α 1 α n and γ 1 γ n+m be two polynomial cains. For every j = 0,..., m, let δ j i := lcm(α i 2j, γ i ), i = 1,..., n + j, n+j δ j := δ j i. Te difference between te convexity in tis case and te result from Lemma 2 is in a different sift in te definition of δ j comparing to π j. Tis makes te problem muc more difficult, and in particular ere we do not ave tat δ j 1 δ j. However, te convexity of te degrees of δ j olds: Teorem 9 (Convexity Lemma) d(δ j ) d(δ j 1 ) d(δ j+1 ) d(δ j ), for j = 1,...,m 1. Before going to te proof we give one simple lemma: te electronic journal of combinatorics 17 (2010), #R61 14
Lemma 10 Let φ 1, φ 2, ψ 1 and ψ 2 be polynomials suc tat φ 1 φ 2 and ψ 1 ψ 2. Ten Proof: For i = 1, 2, we ave lcm(φ 1, ψ 1 ) lcm(φ 2, ψ 2 ) lcm(φ 2, ψ 1 ) lcm(φ 1, ψ 2 ). (44) lcm(φ i, ψ 2 ) = lcm(φ i, ψ 1, ψ 2 ) = lcm(lcm(φ i, ψ 1 ), ψ 2 ) = lcm(φ i, ψ 1 )ψ 2 gcd(lcm(φ i, ψ 1 ), ψ 2 ). Now, by replacing tis expression for i = 1 and i = 2 into (44), it becomes equivalent to te following obvious divisibility relation: gcd(lcm(φ 1, ψ 1 ), ψ 2 ) gcd(lcm(φ 2, ψ 1 ), ψ 2 ). Proof of Teorem 9: In order to prove te convexity, it is enoug to prove tat By definition, we ave Since for all i and j we ave δ j δ j δ j 1 δ j+1, j = 1,..., m 1. (45) n+j δ j = lcm(α i 2j, γ i ), j = 0,...,m. (46) lcm(α i 2j, γ i ) = lcm(α i 2j, lcm(α i 2j 2, γ i )) = we can rewrite (46) as δ j = n+j α i 2j lcm(α i 2j 2, γ i ) gcd(α i 2j, lcm(α i 2j 2, γ i )) = n j α i α i 2j lcm(α i 2j 2, γ i ) gcd(α i 2j, lcm(α i 2j 2, γ i )), n+j lcm(α i 2j 2, γ i ) n j gcd(α i, lcm(α i 2, γ i+2j )). (47) We replace one δ j on te left and side and δ j+1 on te rigt and side of (45) by te expression (46), wile we replace te oter δ j and δ j 1 by te expression (47). Ten (45) becomes equivalent to n+j lcm(α i 2j, γ i ) n j α n+j i lcm(α i 2j 2, γ i ) n j gcd(α i, lcm(α i 2, γ i+2j )) n+j+1 lcm(α i 2j 2, γ i ) n j+1 α n+j 1 i lcm(α i 2j, γ i ) n j+1. gcd(α i, lcm(α i 2, γ i+2j 2 )) te electronic journal of combinatorics 17 (2010), #R61 15
After cancellations, te last divisibility becomes equivalent to n j+1 lcm(α n j, γ n+j ) gcd(α i, lcm(α i 2, γ i+2j 2 )) By using te obvious divisibility relation we are left wit proving tat n j lcm(α n j 1, γ n+j+1 )α n j+1 gcd(α i, lcm(α i 2, γ i+2j )). gcd(α i, lcm(α i 2, γ i+2j 2 )) gcd(α i, lcm(α i 2, γ i+2j )), lcm(α n j, γ n+j ) gcd(α n j+1, lcm(α n j 1, γ n+j 1 )) α n j+1 lcm(α n j 1, γ n+j+1 ). (48) However, since gcd(α n j+1, lcm(α n j 1, γ n+j 1 )) = α n j+1 lcm(α n j 1, γ n+j 1 ), lcm(α n j+1, γ n+j 1 ) (48) becomes equivalent to te following lcm(α n j 1, γ n+j 1 ) lcm(α n j, γ n+j ) lcm(α n j 1, γ n+j+1 ) lcm(α n j+1, γ n+j 1 ), wic follows directly from Lemma 10. References [1] I. Baragaña, I. Zaballa, Column completion of a pair of matrices, Linear and Multilinear Algebra, 27 (1990) 243-273. [2] M. Dodig, Matrix pencils completion problems, Linear Algebra Appl. 428 (2008), no. 1, 259-304. [3] M.Dodig, M. Stošić, Similarity class of a matrix wit prescribed submatrix, Linear and Multilinear Algebra, 57 (2009) 217-245. [4] M. Dodig, Feedback invariants of matrices wit prescribed rows, Linear Algebra Appl. 405 (2005) 121-154. [5] I. Goberg, M. A. Kaasoek, F. van Scagen, Eigenvalues of completions of submatrices, Linear and Multilinear Algebra, 25 (1989) 55-70. [6] G. Hardy, J.E. Littlewood, G. Pólya, Inequalities, Cambridge University Press, 1991. [7] E. M. Sá, Imbeding conditions for λ -matrices, Linear Algebra Appl. 24 (1979) 33-50. [8] E. M. Sá, A convexity lemma on te interlacing inequalities for invariant factors, Linear Algebra Appl. 109 (1988) 107-113. [9] E. M. Sá, Imersão de matrizes e entrelaçamento de factores invariantes, PD Tesis, Univ. of Coimbra, 1979. [10] R. C. Tompson, Interlacing inequalities for invariant factors, Linear Algebra Appl. 24 (1979) 1-31. [11] I. Zaballa, Matrices wit prescribed rows and invariant factors, Linear Algebra Appl. 87 (1987) 113-146. te electronic journal of combinatorics 17 (2010), #R61 16