Báo cáo toán học: "On convexity of polynomial paths and generalized majorizations" ppt

On convexity of polynomial paths and generalized majorizations ∗ Marija Dodig † Centro de Estruturas Lineares e Combinat´orias, CELC, Universidade de Lisboa, Av. Prof. Gama Pinto 2, 1649-003 Lisboa, Portugal dodig@cii.fc.ul.pt Marko Stoˇsi´c Instituto de Sistemas e Rob´otica and CAMGSD, Instituto Superior T´ecnico, Av. Rovisco Pais 1, 1049-001 Lisbon, Portugal mstosic@isr.ist.utl.pt Submitted: Nov 15, 2009; Accepted: Apr 5, 2010; Published: Apr 19, 2010 Mathematics Subject Classification: 05A17, 15A21 Abstract In this paper we give some useful combinatorial properties of polynomial paths. We also intro duce generalized majorization between thr ee sequences of integers and explore its combinatorics. In addition, we give a new, simple, p urely polynomial proof of the convexity lemma of E. M. de S´a and R. C. Thompson. All these results have applications in matrix completion theory. 1 Introduction and no tation In this paper we prove some useful properties of polynomial paths and generalized ma- jorization between three sequences of integers. All proo fs are purely combinatorial, and the presented results are used in matrix completion problems, see e.g. [2, 4, 7, 10, 11]. We study chains of monic polynomials and polynomial paths between them. Polyno- mial paths are combinatorial objects that are used in matrix completion problems, see [7, 9, 11]. There is a certain convexity property of polynomial paths appeared for the first time in [5]. In Lemma 2 we give a simple, direct polynomial proof of that result. We also show t hat no additional divisibility relations are needed. ∗ This work was done within the ac tivities of CE L C and was partially supported by FCT, project ISFL-1-1431, and by the Ministry of Science of Serbia , projects no. 144014 (M. D.) and 14403 2 (M. S.). † Corresponding author. the electronic journal of combinatorics 17 (2010), #R61 1 Also, we explore generalized majorization between three sequences of integers. It presents a natural generalization of a classical major ization in Hardy-Littlewood-P´olya sense [6], and it appears frequently in matrix completion problems when both prescribed and the whole matrix are rectangular (see e.g [1, 4, 11]). We give some basic properties of generalized major ization, and we prove that there exists a certain path of sequences, such that every two consecutive sequences of the path are related by an elementary generalized majorization. E. Marques de S´a [7] and independently R. C. Thompson [10], gave a complete solution for the problem of completing a principal submatrix to a square one with a prescribed similarity class. The proof of this famous classical result is based on induction on the number of added rows and columns, and one of the crucial steps is the convexity lemma. The original proofs of the convexity lemma, which are completely independent one from the ano ther one, both in [7] and [10] are rather long and involved. Later on, new combi- natorial proof of this lemma has appeared in [8]. In Theorem 1, we give simple and the first purely polynomial proof of this result. 1.1 Notation All polynomials are considered to be monic. Let F b e a field. Throughout the paper, F[λ] denotes the ring of polynomials over the field F with variable λ. By f|g, where f, g ∈ F[λ] we mean that g is divisible by f. If ψ 1 | · · ·|ψ r is a polynomial chain, then we make a convention that ψ i = 1, for any i  0, and ψ i = 0, for a ny 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 paths Let α 1 | · · · |α n and γ 1 | · · · |γ n+m be two chains of monic polynomials. Let π j := n+j  i=1 lcm(α i−j , γ i ), j = 0, . . . , m. (1) We have the following divisibility: Lemma 1 π j | π j+1 , j = 0, . . . , m − 1 (i.e. π 0 |π 1 | · · · |π m ). Proof: By the definition of π j , j = 0, . . . , m, the statement of Lemma 1 is equivalent to n+j  i=1 lcm(α i−j , γ i ) | n+j+1  i=1 lcm(α i−j−1 , γ i ), j = 0, . . . , m − 1, i.e., n  i=1 lcm(α i , γ i+j ) | γ j+1 n  i=1 lcm(α i , γ i+j+1 ), j = 0, . . . , m − 1, (2) the electronic journal of combinatorics 17 (2010), #R61 2 which is trivially satisfied. By Lemma 1 we can define the following polynomials σ j := π j π j−1 , j = 1, . . . , m. (3) Then, we have the 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 the definition of σ j , j = 1, . . . , m, the statement of Lemma 2 is equivalent to  n+j i=1 lcm(α i−j , γ i )  n+j−1 i=1 lcm(α i−j+1 , γ i ) |  n+j+1 i=1 lcm(α i−j−1 , γ i )  n+j i=1 lcm(α i−j , γ i ) , j = 1, . . . , m − 1, i.e. for all j = 1, . . . , m − 1, we have to show that 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 ) lcm(α 1 , γ j+1 ) lcm(α 2 , γ j+2 ) · · · lcm(α n , γ j+n ) . (4) Before proceeding, note that for every two polynomials ψ and φ we have lcm (ψ, φ) = ψφ gcd(ψ, φ) (5) Thus, for every i and j, we have lcm (α i , γ i+j ) = lcm(lcm(α i , γ i+j−1 ), γ i+j ) = γ i+j lcm(α i , γ i+j−1 ) gcd(lcm(α i , γ i+j−1 ), γ i+j ) . (6) By applying (6), equation (4) becomes equivalent to γ j n  i=1 gcd(lcm(α i , γ i+j ), γ i+j+1 ) | γ n+j+1 n  i=1 gcd(lcm(α i , γ i+j−1 ), γ i+j ). (7) By shifting indices, the right hand side of (7) becomes γ n+j+1 gcd(lcm(α 1 , γ j ), γ j+1 ) n−1  i=1 gcd(lcm(α i+1 , γ i+j ), γ i+j+1 ). This, together with 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. the electronic journal of combinatorics 17 (2010), #R61 3 Frequently when dealing with polynomial paths we have the following additional as- sumptions γ i | α i , i = 1, . . . , n (8) and α i | γ i+m , i = 1, . . . , n. (9) Then the following lemma f ollows trivially from the definition of π i ’s, for i = 0 and i = m: Lemma 3 π 0 =  n i=1 α i and π m =  n+m i=1 γ i . 2.1 Polynomial paths Let α = (α 1 , . . . , α n ) and γ = (γ 1 , . . . , γ n+m ) be two systems of nonzero monic polynomials such that α 1 | · · · |α n and γ 1 | · · · |γ n+m . A polynomial path between α and γ has been defined in a following way in [7, 9], see also [1 1]: Definition 1 Let ǫ j = (ǫ j 1 , . . . , ǫ j n+j ), j = 0, . . . , m, be a system of non zero monic poly- nomials. Let ǫ 0 := α and ǫ m := γ. The sequence ǫ = (ǫ 0 , ǫ 1 , . . . , ǫ m ) is a path from α to γ if the following is valid: ǫ j i |ǫ j i+1 , i = 1, . . . , n + j − 1, j = 0, . . . , m, (10) ǫ j i |ǫ j−1 i |ǫ j i+1 , i = 1, . . . , n + j − 1, j = 1, . . . , m. (11) Consider the polynomials β j i := lcm(α i−j , γ i ), i = 1, . . . , n + j, j = 0, . . . , m from (1). Let β j = (β j 1 , . . . , β j n+j ), j = 0, . . . , m. Then the following proposition is valid (see Proposition 3.1 in [11] and Section 4 in [7]): Proposition 1 There exists a path from α to γ, if and only if γ i |α i |γ i+m , i = 1, . . . , n. (12) Moreo ver, if (12) is val i d, then β = (β 0 , . . . , β m ) is a pol yno mial path between α and γ, and for e very path ǫ between α and γ hold β j i | ǫ j i , i = 1, . . . , n + j, j = 0, . . . , m. Hence, β is a minimal path from α to γ. The polynomials π j from (1) are defined as π j =  n+j i=1 β j i . The polynomials σ i were used by S´a [7, 9] and by Zaballa [11], but the convexity of π j ’s, i.e. the result of Lemma 2, was obtained later by Gohberg, Kaashoek and van Schagen [5]. We gave a direct poly- nomial proof of this result and we have shown that it holds even without the divisibility relations (12). the 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 that f ≺ ′ (d, a), i.e., we have a generalized majorization between the partitions d = (d 1 , . . . , d ρ ), a = (a 1 , . . . , a l ) and f = (f 1 , . . . , f ρ+l ), if and only i f d i  f i+l , i = 1, . . . , ρ, (13)  ρ+l i=1 f i =  ρ i=1 d i +  l i=1 a i , (14)  h q i=1 f i −  h q −q i=1 d i   q i=1 a i , q = 1, . . . , l, (15) where h q = min{i|d i−q+1 < f i }, q = 1, . . . , l. Remark 1 Recall that in Section 1.1 we have made a convention that f i = +∞ and d i = +∞, for i  0, and that f i = −∞, for i > ρ + l, and d i = −∞, for i > ρ. Thus, h q ’s are well-defined. In particular, for every q = 1, . . . , l, we have q  h q  q + l, and h 1 < h 2 < . . . < h l . Note that if ρ = 0, then the generalized majorization reduces to a classical majorization (in Hardy-Littlewood-P´olya sense [6]) between the partitions f and a (f ≺ a). If l = 1, (13)–(15) are equivalent to d i  f i+1 , i = 1, . . . , ρ, (16)  ρ+1 i=1 f i =  ρ i=1 d i + a 1 , (17) d i = f i+1 , i  h 1 . (18) Indeed, for l = 1, (15) becomes h 1  i=1 f i  h 1 −1  i=1 d i + a 1 . The last inequality together with (14), gives ρ+1  i=h 1 +1 f i  ρ  i=h 1 d i . (19) Finally, from (13), we o bta in that (19) is equivalent to (18), as wanted. Generalized majorization for the case l = 1 will be called elementary generalized majorization, and will b e denoted by f ≺ ′ 1 (d, a). the 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), then h 1 = ρ + 1, and so f ≺ ′ 1 (d, a). Note that if f ≺ ′ (d, a), then in the same way as in the proof of the equivalence of (15) and ( 18), we have d i = f i+l , i  h l − l + 1. (20) The aim of this section is to show that there is a generalized majorization between the partitions d, a and f if and only if there are elementary majorizations b etween them, i.e. if and only if there exist intermediate sequences that satisfy (16)–(18). In certain sense, we show that there exists a path of sequences b etween d and f such that every neighbouring two satisfy the elementary generalized majorization (see Theorems 5 and 7 below). More precisely, we shall show that f ≺ ′ (d, a) if and only if there exist sequences g i = (g i 1 , . . . , g i ρ+i ), i = 1, . . . , l−1, with g i 1  · · ·  g i ρ+i , and with the convention g 0 := d and g l := f, such that g i ≺ ′ 1 (g i−1 , a i ), i = 1, . . . , l. Lemma 4 Let f , d and a be the sequences from Defin i tion 1. If f ≺ ′ (d, a), then there exist integers g 1  · · ·  g ρ+l−1 , such that (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  h, where h := min{i|g i < f i }, (iv) ˜ h q  i=1 g i − ˜ h q −q  i=1 d i  q  i=1 a i , q = 1, . . . , l − 1, where ˜ h q = min{i|d i−q+1 < g i }, (v) ρ+l  i=1 f i = ρ+l−1  i=1 g i + a l . Proof: Let H 1 , . . . , H l−1 be integers defined as H q := q  i=1 a i − h q  i=1 f i + h q −q  i=1 d i , q = 1, . . . , l − 1, and H 0 := 0. the electronic journal of combinatorics 17 (2010), #R61 6 Note that fro m ( 15), we have that H q  0, q = 1, . . . , l − 1. Let S q := h q −q  i=h q−1 −q+2 d i − h q  i=h q−1 +1 f i , q = 1, . . . , l − 1. Thus H q − H q−1 = S q + a q , q = 1, . . . , l − 1. Since a 1  · · ·  a l−1 , we have H 1 − S 1  H 2 − H 1 − S 2  · · ·  H l−1 − H l−2 − S l−1 . (21) Now, define the numbers H ′ i := min(H i , H i+1 , . . . , H l−1 ), i = 0, . . . , l − 1. (22) Thus, we have 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 . The wanted g 1  · · ·  g ρ+l−1 will be defined as the nonincreasing ordering of g ′ 1 , . . . , g ′ ρ+l−1 . Let g ′ i := d i−l+1 , i > h l−1 . (25) We shall split the definition of g ′ 1 , . . . , g ′ h l−1 into l − 1 groups. For arbitrary j = 1, . . . , l − 1, we define g ′ i , i = h j−1 + 1, . . . , h j , (with convention h 0 := 0) in a following way: If f h j  H ′ j − H ′ j−1 − S j , (26) then we define g ′ h j−1 +1  · · ·  g ′ h j −1 as a nonincreasing sequence of integers such that d i−j+1  g ′ i  f i and h j −1  i=h j−1 +1 g ′ i − h j −1  i=h j−1 +1 f i = H ′ j − H ′ j−1 (this is obviously possible because of (26)). Also, in this case, we define g ′ h j := f h j . If f h j < H ′ j − H ′ j−1 − S j , (27) the electronic journal of combinatorics 17 (2010), #R61 7 then we define g ′ i := d i−j+1 , i = h j−1 + 1, . . . , h j − 1, and g ′ h j := H ′ j − H ′ j−1 − S j . Note that in both of the previous cases, (26) and (27), we have h j  i=h j−1 +1 g ′ i − h j  i=h j−1 +1 f i = H ′ j − H ′ j−1 , j = 1, . . . , l − 1. (28) and g ′ h i = max(f h i , H ′ i − H ′ i−1 − S i ), i = 1, . . . , l − 1. Now, let i ∈ {1, . . . , l − 2}. If g ′ h i+1 = f h i+1 , then g ′ h i+1  f h i  g ′ h i . If g ′ h i+1 = H ′ i+1 − H ′ i − S i+1 > f h i+1 , then, from (28), we have that H ′ i+1 > H ′ i , and so H ′ i = H i . However, this together with (21), gives g ′ h 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 ′ h i . Hence, we have g ′ h 1  g ′ h 2  · · ·  g ′ h l−1 . (29) Also, from the definition of h i , i = 1, . . . , l − 1 , the subsequence of g ′ i ’s for i ∈ {1, . . . , ρ + l − 1} \ {h 1 , . . . , h l−1 } is in nonincreasing order, and satisfies: d i−j+1  g ′ i  f i , h j−1 < i < h j , j = 1, . . . , l. (30) For i  h l , from (20), we have d i−l+1 = g ′ i = f i+1 , i  h l . (31) Now, since g ′ i  f i+1 for all i = 1, . . . , ρ + l − 1, and since g i ’s are the nonincreasing ordering of g ′ i ’s, we have (i). Moreover, since g ′ h l−1  f h l−1 > d h l−1 −l+2 = g ′ h l−1 +1 , we have that g i = g ′ i , for i > h l−1 . Then, from (30), we have g i  f i , for i < h l , which together with g h l = g ′ h l = d h l −l+1 < f h l , implies h = h l . Thus, (31) implies (iii). If we denote by ν 1  · · ·  ν ρ the subsequence of g ′ i ’s for i ∈ {1, . . . , ρ + l − 1} \ {h 1 , . . . , h l−1 }, then from (30) and (31) we have d i  ν i , i = 1, . . . , ρ, (32) which implies (ii). the electronic journal of combinatorics 17 (2010), #R61 8 Also, by summing all inequalities from (28), for j = 1, . . . , l − 1, we have h l−1  i=1 g ′ i − h l−1  i=1 f i = H ′ l−1 , which together with (24) and the definition of H l−1 , gives h l−1  i=1 g ′ i − h l−1 −l+1  i=1 d i = l−1  i=1 a i . The last equation, together with the definition of the remaining g ′ i ’s (25), the fact that  ρ+l−1 i=1 g i =  ρ+l−1 i=1 g ′ i , and (14), gives (v). Before going to the proof of (iv), we shall establish some relations between h q ’s a nd ˜ h q ’s. So, let q ∈ {1, . . . , l − 1}. The sequence of g i ’s is defined as the nonincreasing ordering of g ′ i ’s. As we have shown, the sequence of g ′ i ’s is the union of two nonincreasing sequences: g ′ h 1  g ′ h 2  . . .  g ′ h l−1 and ν 1  ν 2  . . .  ν ρ . Let r q be the index such that ν r q  g ′ h q > ν r q +1 . First of all, from the definition of g ′ h q and h q , we have that g ′ h q  f h q > d h q −q+1  ν h q −q+1 , and so r q  h q − q. (33) Furthermore, the subsequence g 1  g 2  . . .  g r q +q is the nonincreasing ordering of the union of sequences g ′ h 1  g ′ h 2  . . .  g ′ h q and ν 1  ν 2  . . .  ν r q , with g ′ h q being the smallest among them, i.e. g r q +q = g ′ h q . Thus, ν i  g i+q−1 , for i = 1, . . . , r q , and so from (32), for every i  r q we have that d i  ν i  g i+q−1 , i.e. ˜ h q  r q + q. (34) By (33), we have two possibilities for r q : If r q = h q − q, as proved above, we have g h q = g ′ h q , which then implies g h q  f h q > d h q −q+1  ν h q −q+1 , and so ˜ h q  h q , which t ogether with (34) in this case gives ˜ h q = h q = r q + q. If r q < h q − q, then g ′ h q > ν h q −q  f h q , and so from the definition of g ′ i ’s, we have that ν i = d i , for i = r q + 1, . . . , h q − q. Thus g r q +q = g ′ h q > ν r q +1 = d r q +1 , and so ˜ h q  r q + q, which together with (34) gives ˜ h q = r q + q. Thus, altogether we have that ˜ h q  h q , and g 1  g 2  . . .  g ˜ h q is the nonincreasing ordering of the union of sequences g ′ h 1  g ′ h 2  . . .  g ′ h q and ν 1  ν 2  . . .  ν ˜ h q −q , with g ˜ h q = g ′ h q , and that ˜ h q < h q implies ν i = d i , for i = ˜ h q − q + 1, . . . , h q − q. the electronic journal of combinatorics 17 (2010), #R61 9 Finally, we can pass to the proof of (iv). Let q ∈ {1, . . . , l − 1}. We shall prove (iv) for this q in the following equivalent form ˜ h q  i=1 ˜g i − ˜ h q −q  i=1 d i  H q + h q  i=1 f i − h q −q  i=1 d i . (35) If ˜ h q = h q , (35) is equiva lent to h q  i=1 (g ′ i − f i )  H q , (36) which fo llows from (24) a nd (28). If ˜ h q < h q , we have that ν i = d i , for i = ˜ h q − q + 1, . . . , h q − q. Hence, the condition (35) is ag ain equivalent to (36), which concludes our proof. By iterating the previous result, we obtain the following Theorem 5 Let f, d and a be the sequences from Definition 1. If f ≺ ′ (d, a), then there exis t sequences of integers g j = (g j 1 , . . . , g j ρ+j ), j = 1, . . . , l − 1, with g j 1  · · ·  g j ρ+j , such that g j ≺ ′ 1 (g j−1 , a j ), j = 1, . . . , l, where g 0 = d and g l = f. Proof: For l = 1, the claim of theorem follows trivially. Let l > 1, and suppose that theorem holds for l − 1. By Lemma 4, there exists a sequence g = (g 1 , . . . , g ρ+l−1 ), such that g 1  · · ·  g ρ+l−1 and such that they satisfy conditions (i) − (v) from Lemma 4. Set g l−1 := g. From (i), (iii) a nd (v) we have f ≺ ′ 1 (g l−1 , a l ). (37) From (ii), (iv) and (v), we have g l−1 ≺ ′ (d, a ′ ), (38) where a ′ = (a 1 , . . . , a l−1 ). By induction hypothesis there exist sequences g 1 , . . . , g l−2 , such that g j ≺ ′ 1 (g j−1 , a j ), j = 1, . . . , l − 1. This together with (37) finishes our proof. The following two r esults give converse of Lemma 4 and Theorem 5 : the electronic journal of combinatorics 17 (2010), #R61 10 [...]... that g1 and f ≺′ (g, a′ ) ··· gρ+s g ≺′ (d, a′′ ) where a′ = (a1 , , al−s ) and a′′ = (al−s+1 , , al ) 4 Convexity lemma In this section we give a short polynomial proof of the convexity lemma, which is the crucial step in S´-Thompson theorem [7, 10] The original proofs of S´ and Thompson a a were long and complicated, and relied on very involved techniques The proof in [7] (Proposition 4.1 and Lemma... proof in [10] is elementary but very long and does not involve the concept of convexity Later on shorter, combinatorial proof was given in [8] Here we give the first purely polynomial proof of the convexity lemma Let α1 | · · · |αn and γ1 | · · · |γn+m be two polynomial chains For every j = 0, , m, let j δi := lcm(αi−2j , γi ), i = 1, , n + j, n+j j δi j δ := i=1 The difference between the convexity. .. the convexity in this case and the result from Lemma 2 is in a different shift in the definition of δ j comparing to πj This makes the problem much more difficult, and in particular here we do not have that δ j−1 |δ j However, the convexity of the degrees of δ j holds: Theorem 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 the proof we give one simple lemma:... that g1 Also, let g 0 := d, and g l := f If g j ≺′1 (g j−1, a′j ) for j = 1, , l, then f ≺′ (d, a) the electronic journal of combinatorics 17 (2010), #R61 ··· j gρ+j 13 Thus, Theorems 5 and 7 prove the existence of a path of sequences, as announced before Lemma 4 In particular, we have Corollary 8 Let l Then 2, d1 ··· dρ , f1 ··· fρ+l , a1 ··· al be integers f ≺′ (d, a) if and only if there exists... , l Proof: ˜ From the definition of hq , hq and ¯ 1 , we obtain the following inequalities h hq ˜ ¯ ˜ max(hq−1 , min(h1 + q − 1, hq )), ˜ q = 1, , l − 1, (h0 = 0), (41) and ˜ ¯ max(hl−1 , h1 + l − 1) (42) ¯ This is true since for q = 1, , l − 1, and j < min(h1 + q − 1, ˜ q ), we have that h hl dj−q+1 gj−q+1 fj ¯ ˜ Therefore, hq min(h1 + q − 1, ˜ q ) Also, for every q = 1, , l, and j < hq−1... the electronic journal of combinatorics 17 (2010), #R61 14 Lemma 10 Let φ1 , φ2 , ψ1 and ψ2 be polynomials such that φ1 |φ2 and ψ1 |ψ2 Then lcm(φ1 , ψ1 ) lcm(φ2 , ψ2 )| lcm(φ2 , ψ1 ) lcm(φ1 , ψ2 ) (44) Proof: For i = 1, 2, we have lcm(φi , ψ2 ) = lcm(φi , ψ1 , ψ2 ) = lcm(lcm(φi , ψ1 ), ψ2 ) = lcm(φi , ψ1 )ψ2 gcd(lcm(φi , ψ1 ), ψ2 ) Now, by replacing this expression for i = 1 and i = 2 into (44), it... Baraga˜ a, I Zaballa, Column completion of a pair of matrices, Linear and Multilinear Algebra, n 27 (1990) 243-273 [2] M Dodig, Matrix pencils completion problems, Linear Algebra Appl 428 (2008), no 1, 259-304 [3] M.Dodig, M Stoˇi´, Similarity class of a matrix with prescribed submatrix, Linear and Multilinear sc Algebra, 57 (2009) 217-245 [4] M Dodig, Feedback invariants of matrices with prescribed rows,... γi ) = gcd(αi−2j , lcm(αi−2j−2 , γi)) n−j i=1 n−j i=1 αi n+j i=1 lcm(αi−2j−2 , γi) gcd(αi , lcm(αi−2 , γi+2j )) (47) We replace one δ j on the left hand side and δ j+1 on the right hand side of (45) by the expression (46), while we replace the other δ j and δ j−1 by the expression (47) Then (45) becomes equivalent to n+j i=1 lcm(αi−2j , γi) n−j i=1 | n−j i=1 αi n+j i=1 lcm(αi−2j−2 , γi ) gcd(αi , lcm(αi−2... inequality follows from the definition of hq and the fact that hq < hq+1 , while ˜ q −q h1 Now, we finish the proof as in the previous case ¯ the last equality is true since h ˜ ¯ The only remaining case is q = l Let i > hl Since hl max(hl−1 , h1 + l − 1), we have ˜ i > hl−1 From (ii), (iv) and (v) we have that f ≺′ (g, a′′), where a′′ = (a′2 , a′3 , , a′l ), ¯ and so (see (20)) we have fi = gi−l+1... di−l , and thus fi = di−l , i > hl (43) Now, by (v), condition (40) for q = l is equivalent to ρ ρ+l di fi i=hl +1 i=hl −l+1 Finally, from (i), we have that di fi+l , i = 1, , ρ, and so condition (40) for q = l is equivalent to (43), which concludes our proof By iterating the previous result, we obtain the following one: Theorem 7 Let d1 integers, such that ··· dρ , f1 ··· fρ+l , a1 ··· al and a′1 . result is based on induction on the number of added rows and columns, and one of the crucial steps is the convexity lemma. The original proofs of the convexity lemma, which are completely independent. in [7] and [10] are rather long and involved. Later on, new combi- natorial proof of this lemma has appeared in [8]. In Theorem 1, we give simple and the first purely polynomial proof of this. we assume c i = +∞, for i  0, and c i = −∞, for i  m + 1. 2 Convexity and polynomial paths Let α 1 | · · · |α n and γ 1 | · · · |γ n+m be two chains of monic polynomials. Let π j := n+j  i=1 lcm(α i−j ,