Combinatorics of polynomial chains

Journal Pre-proof Combinatorics of polynomial chains

Marija Dodig, Marko Stoši´c

PII:

S0024-3795(19)30542-7

DOI:

https://doi.org/10.1016/j.laa.2019.12.028

Reference:

LAA 15226

To appear in:

Linear Algebra and its Applications

Received date:

6 December 2018

Accepted date:

17 December 2019

Please cite this article as: M. Dodig, M. Stoši´c, Combinatorics of polynomial chains, Linear Algebra Appl. (2020), doi: https://doi.org/10.1016/j.laa.2019.12.028.

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Combinatorics of polynomial chains Marija Dodig∗

Marko Stoˇsi´c†

Abstract In this paper we consider two polynomial chains and we study the existence of an intermediate polynomial chain satisfying interlacing inequality relations together with some convexity-like properties that arise from applications in completion problems. The presented result is a vast generalisation of existing results dealing with intermediate polynomial chains. The proof of the main result is completely novel to the field, different from all the existing methods used in similar problems. The obtained necessary and sufficient conditions are simple, explicit and constructive.

AMS classification: 05A20, 15A83 Keywords: Polynomial chains, convexity, completion of matrix pencils.

1

Introduction

Let F be a field. By a polynomial chain we mean a family of non-zero monic polynomials from F[λ] of the form α1 | · · · |αn . Let α : α1 | · · · |αn and γ : γ1 | · · · |γn+x+y be two polynomial chains satisfying γi |αi |γi+a+b ,

i = 1, . . . , n,

(1)

where a, b, x, y are nonnegative integers, a ≥ x and b ≥ y. The divisibility relations (1) are usually called interlacing inequalities. Polynomial chains satisfying interlacing inequalities are especially important in resolving completion problems, where polynomial chains appear as invariant factors of matrices and matrix pencils, see e.g. [4, 11, 12, 19, 21, 22, 23]. Although mainly motivated by applications in matrix theory, polynomial chains satisfying (1) have intrinsic combinatorial properties of independent interest. In this paper we solve the following general combinatorial problem: ∗

CEAFEL, Departamento de Mat´ematica, Universidade de Lisboa, Edificio C6, Campo Grande, 1749-016 Lisbon, Portugal, and Mathematical Institute SANU, Knez Mihajlova 36, 11000 Belgrade, Serbia. ([email protected]). Corresponding author. † CAMGSD, Departamento de Matem´ atica, Instituto Superior T´ecnico, Av. Rovisco Pais 1, 1049-001 Lisbon, Portugal, and Mathematical Institute SANU, Knez Mihajlova 36, 11000 Belgrade, Serbia.

1

Problem 1 Let a and b be positive integers. Let x, y, Z, Xi , i = 1, . . . , a−1 and Yj , j = 1, . . . , b − 1 be nonnegative integers such that a ≥ x and b ≥ y. Let α : α1 | · · · |αn and γ : γ1 | · · · |γn+x+y be polynomial chains satisfying γi |αi |γi+a+b ,

i = 1, . . . , n.

Find necessary and sufficient conditions for the existence of a polynomial chain β : β1 | · · · |βn+x , which satisfies the following: βi |αi |βi+a ,

n+min(a−j,x)

i = 1, . . . , n,

γi |βi |γi+b , i = 1, . . . , n + x, n+x i=1 d(βi ) = Z,

d(lcm(αi−a+j , βi )) ≤ Xj , j = 1, . . . , a − 1, i=1 n+x+min(k,y) d(lcm(βi−k , γi )) ≤ Yk , k = 1, . . . , b − 1. i=1

(2) (3) (4) (5) (6)

Although of independent combinatorial interest, Problem 1 is directly motivated by the general matrix pencil completion problem (GMPCP). Special cases of Problem 1 have appeared in solving classical particular cases of the GMPCP, see e.g. [1, 2, 4, 5, 6, 7, 19, 21, 23]. In particular, if a = b = x and y = 0, Problem 1 appears in the celebrated S´a-Thompson’s result describing the possible similarity class of a matrix with a prescribed principal submatrix [19, 21]. Also, if a = x and for specific values of Xj , j = 1, . . . , a−1 and Yk , k = 1, . . . , b−1, Problem 1 appears in the well-known completion problem for regular pencils [2, 4], where the possible similarity class of a matrix with a prescribed arbitrary submatrix is described (for details see Section 4). The main limitations of the existing approaches for solving Problem 1 are restrictions on the values of a and b, as well as restrictions on the forms and values of Xj and Yk appearing in (5) and (6). In this paper we overcome all these constraints, and give a complete, explicit and constructive solution to Problem 1 (Theorem 1 in Section 3). Hence, the combinatorial results on polynomial chains obtained in [1, 2, 4, 5, 6, 7, 19, 21, 23] can be derived from Theorem 1. As an illustration of this, in Section 4, we give a new, direct solution to the central combinatorial result from [2, 4]. Apart from being a vast generalisation of the existing results on polynomial combinatorics and interlacing inequalities, the proof of the main result itself is also completely novel to the field, and it is different from all the existing methods used in similar problems. Basically, by a convenient decomposition of the polynomials involved in suitable pieces we essentially reduce the problem to an interesting combinatorial question involving nonnegative integers which is then resolved in Lemma 2. Furthermore, the conditions appearing both in the main result (Theorem 1) and in the combinatorial 2

Lemma 2, are nice and concise. Finally, we note that the sufficiency part of the proof is constructive, i.e. we explicitly define a polynomial chain satisfying the wanted properties in Problem 1. Because of its generality many applications of Theorem 1 and Lemma 2 are expected, in particular in solving the general matrix pencil completion problem (for details see Section 4).

1.1

Notation

Although the necessity part of the proof is valid over an arbitrary field F, to prove the sufficiency of the conditions of the main result, we need an ¯ (the same type of restriction is also needed in algebraically closed field F some particular cases previously resolved e.g. [4, 5, 6, 7]). For a polynomial chain α : α1 | · · · |αn , we assume that αi ∈ F[λ] (or αi ∈ F[λ, μ] homogeneous) are all monic nonzero polynomials. For a homogeneous polynomial α(λ, μ) we say that it is monic if the monomial with the largest exponent of λ has coefficient 1. By convention we put αi = 1, for i ≤ 0, and αi = 0, for i ≥ n + 1. For a polynomial f , by d(f ) we denote its degree. Also, we assume d(1) = 0, and r d(0) = +∞. Whenever s and r are integers such r that s > r, we assume i=s φi = 1 for any polynomials φ1 , . . . , φn , and i=s ci = 0 for any integers c1 , . . . , cn . The following definition of the classical majorization will be used in Section 4: Definition 1 [15] Let a = (a1 , . . . , as ) and g = (g1 , . . . , gs ) be two sequences of integers. Let σ1 and σ2 be two permutations of the integers {1, . . . , s}  such that aσ1 (1) ≥ · · · ≥ aσ1 (s) and gσ2 (1) ≥ · · · ≥ gσ2 (s) . If  s s g = i i=1 i=1 ai and j 

gσ2 (i) ≤

i=1

j 

aσ1 (i) ,

j = 1, . . . , s,

i=1

then we say that g is majorized by a and write g ≺ a.

2

Auxiliary results

We recall a simple combinatorial result from [4]: Lemma 1 [4, Lemma 3] Let α, β, γ ∈ F[λ] \ {0}. Then lcm(α, γ) |

lcm(α, β) lcm(γ, β) . β

3

(7)

The following combinatorial lemma will be essential in the proof of the main result. Lemma 2 Let a and b be positive integers. Let z, x1 , . . . , xa−1 , y1 , . . . , yb−1 be nonnegative integers. Let xa = 0, yb = 0. Let A¯u,v , u = 1, . . . , a, v = 1, . . . , b be nonnegative integers (or +∞). Let z ≤ xj + yk +

j  k 

A¯u,v ,

for all

j = 1, . . . , a,

k = 1, . . . , b.

(8)

u=1 v=1

Then there exist nonnegative integers Au,v , u = 1, . . . , a, v = 1, . . . , b, satisfying Au,v ≤ A¯u,v , z =

xj

≥

u = 1, . . . , a,

v = 1, . . . , b,

a  b 

Au,v u=1 v=1 b a  

(9) (10)

Au,v ,

j = 1, . . . , a − 1,

(11)

Au,v ,

k = 1, . . . , b − 1,

(12)

u=j+1 v=1

yk ≥

a b   u=1 v=k+1

and such that for all u ∈ {1, . . . , a − 1}, v ∈ {1, . . . , b − 1} we have Au+1,v+1 > 0 ⇒ Au,v = A¯u,v .

(13)

Proof: Before proceeding, let x ˜j := min{z, x1 , x2 , . . . , xj },

j = 1, . . . , a,

y˜k := min{z, y1 , y2 , . . . , yk },

k = 1, . . . , b.

Then xj ≥ x ˜j , for j = 1, . . . , a,

and

yk ≥ y˜k , for k = 1, . . . , b,

(14)

z ≥ y˜1 ≥ y˜2 ≥ · · · ≥ y˜b = 0.

(15)

as well as z≥x ˜1 ≥ x ˜2 ≥ · · · ≥ x ˜a = 0,

Let j ∈ {1, . . . , a} and k ∈ {1, . . . , b} be fixed. From (8) for 1 ≤ j  ≤ j and 1 ≤ k  ≤ k we have: 

z ≤ xj  + yk +



j  k 

A¯u,v ≤ xj  + yk +

u=1 v=1

j  k  u=1 v=1

4

A¯u,v .

(16)

Together with the obvious z ≤ z, (16) gives z≤x ˜j + y˜k +

j  k 

A¯u,v ,

for

j = 1, . . . , a,

k = 1, . . . , b.

(17)

u=1 v=1

Now, in order to prove the lemma, we are going to prove that if (17) holds, one can define nonnegative integers Au,v , u = 1, . . . , a, v = 1, . . . , b, which satisfy (9), (10), (13) and x ˜j

≥

a b  

Au,v ,

j = 1, . . . , a − 1,

(18)

Au,v ,

k = 1, . . . , b − 1.

(19)

u=j+1 v=1

y˜k ≥

a b   u=1 v=k+1

By (14) such defined Au,v , u = 1, . . . , a, v = 1, . . . , b, will also satisfy (11) and (12), as wanted. The proof of the existence of such nonnegative integers Au,v , u = 1, . . . , a, v = 1, . . . , b, is done by induction on a + b. Let a + b = 2, i.e. let a = b = 1. Then (17) is equivalent to z ≤ A¯1,1 , so we set A1,1 := z, which clearly satisfies (9) and (10), and trivially (18), (19) and (13). Now let a + b > 2, a, b ≥ 1. We set ˜a−1 ) Aa,1 := min(A¯a,1 , x

(20) a 

Aj,1 := min(A¯j,1 , x ˜j−1 −

Ai,1 ),

j = a − 1, . . . , 2,

(21)

i=j+1

A1,1 := min(A¯1,1 , z −

a 

Ai,1 ).

i=2

The rest of the proof is split into two cases: Case I

Au,1 = A¯u,1

for all

u = 1, . . . , a.

5

(22)

In this case definitions (20)–(22) give x ˜j

a 

≥

z ≥

A¯u,1 ,

j = 1, . . . , a − 1,

u=j+1 a 

A¯u,1 .

(23)

(24)

u=1

If b = 1, we set Au,1 := A¯u,1 ,

u = 1, . . . , a.

Such defined Au,1 , u = 1, . . . , a, trivially satisfy (9), (10), (13), (18) and (19). Indeed, condition (17) for j = a and k = b = 1 together with (24) gives (10); (23) is equal to (18). Finally, (19) and (13) are void for b = 1 and so trivially satisfied. If b > 1, we set z  := z −

a 

A¯u,1

(25)

u=1

x ˜j

a 

:= x ˜j −

A¯u,1 ,

j = 1, . . . , a,

(26)

u=j+1

y˜k := y˜k ,

k = 1, . . . , b.

(27)

By (23) and (24) we have that x ˜j ≥ 0, for j = 1, . . . , a and z  ≥ 0. By definition y˜k ≥ 0, for k = 1, . . . , b. Also, by (25)-(27) and (15), inequalities (17) become: z ≤ x ˜j + y˜k + z  ≤ y˜1 .

j  k 

A¯u,v ,

j = 1, . . . , a,

k = 2, . . . , b,

(28)

u=1 v=2

(29)

˜1 , . . . , x ˜a , From (28), by induction hypothesis for a = a, b = b − 1, z  , x y˜2 , . . . , y˜b , and A¯u,v , u = 1, . . . , a, v = 2, . . . , b, there exist Au,v , u = 1, . . . , a and v = 2, . . . , b, satisfying 0 ≤ Au,v ≤ A¯u,v , z = x ˜j y˜k

≥

u = 1, . . . , a,

v = 2, . . . , b,

a  b 

Au,v u=1 v=2 b a  

(30) (31)

Au,v ,

j = 1, . . . , a − 1,

(32)

Au,v ,

k = 2, . . . , b − 1,

(33)

u=j+1 v=2

≥

a b   u=1 v=k+1

6

and for all u = 1, . . . , a − 1, v = 2, . . . , b − 1 Au+1,v+1 > 0 ⇒ Au,v = A¯u,v .

(34)

Au,1 := A¯u,1 ,

(35)

Finally, we set u = 1, . . . , a.

Such obtained Au,v , u = 1, . . . , a and v = 1, . . . , b satisfy (9), (10), (13), (18) and (19). Indeed, (30) and (35) give (9); (25), (31) and (35) give (10); (26) and (32) give (18); (27) and (33) give (19) for k = 2, . . . , b − 1. (19) for k = 1 follows from (29). As for (13), for u = 1, . . . , a − 1 and v = 2, . . . , b − 1 it follows from (34), and for u = 1, . . . , a − 1 and v = 1 it follows from (35), as wanted. Case II

There exists i ∈ {1, . . . , a},

for which

Ai,1 < A¯i,1 .

Let r ∈ {1, . . . , a} be the largest index i ∈ {1, . . . , a} for which Ai,1 < ¯ Ai,1 . If r = 1, by (20)–(22) it follows that x ˜j

≥

z ≤

a 

A¯u,1 ,

j = 1, . . . , a − 1,

(36)

u=j+1 a 

A¯u,1 .

(37)

u=1

Then we set: u = 2, . . . , a, Au,1 := A¯u,1 , a  A1,1 := z − A¯u,1 ,

(38) (39)

u=2

Au,v := 0,

u = 1, . . . , a,

v = 2, . . . , b.

a

(40) A¯u,1 ,

˜1 ≥ u=2 Equation (36) for j = 1 and the definition of x ˜1 give z ≥ x and so A1,1 ≥ 0. Such defined Au,v , u = 1, . . . , a, v = 1, . . . , b satisfy (9), (10), (13), and (19). Indeed, (37) and (39) imply A1,1 ≤ A¯1,1 . This, together with and (40) give (9). By summing (38)-(40) we obtain (10), while (36), and (40) give (18). From (40) we obtain (19), as wanted. Finally, by we have that (13) trivially holds.

(18) (38) (38) (40)

If r > 1, by (20)–(22) it follows that x ˜j

≥

x ˜r−1 ≤

a 

A¯u,1 ,

u=j+1 a 

A¯u,1 .

j = r, . . . , a − 1,

(41)

(42)

u=r

7

Let z  = z − x ˜r−1 and x ˜j = x ˜j − x ˜r−1 , j = 1, . . . , r − 1. By (15) the   ˜j , j = 1, . . . , r − 1, are nonnegative. From (17) it follows numbers z and x that z  ≤ x ˜j + y˜k +

j  k 

A¯u,v ,

j = 1, . . . , r − 1,

k = 1, . . . , b,

u=1 v=1

and so by the induction hypothesis for a = r − 1 (1 ≤ a < a), b, z  , ˜r−1 , y˜1 , . . . , y˜b , and A¯u,v , u = 1, . . . , a , v = 1, . . . , b, there exist x ˜1 , . . . , x nonnegative integers Au,v , u = 1, . . . , a , v = 1, . . . , b, satisfying Au,v ≤ A¯u,v , z  = x ˜j ≥ y˜k ≥

(43)

r−1  b 

Au,v u=1 v=1 b r−1  

(44)

Au,v ,

j = 1, . . . , r − 2,

(45)

Au,v ,

k = 1, . . . , b − 1,

(46)

u=j+1 v=1 r−1  b  u=1 v=k+1

and for all u = 1, . . . , r − 2, v = 1, . . . , b − 1 Au+1,v+1 > 0 ⇒ Au,v = A¯u,v .

(47)

The remaining Au,v are defined in the following way: Au,1 := A¯u,1 , ˜r−1 − Ar,1 := x

u = r + 1, . . . , a, a 

A¯i,1 ,

(48) (49)

i=r+1

Au,v := 0,

if u ≥ r and v ≥ 2.

(50)

Such obtained Au,v , u = 1, . . . , a, v = 1, . . . , b satisfy (9), (10), (18) and (19). Indeed, (41) for j = r and (42) give that 0 ≤ Ar,1 ≤ A¯r,1 , which together with (43), (48) and (50) prove (9); (10) follows from (44), (48), (49) and the definition of z  ; (46) and (50) give (19); (45), (48), (49), (50) and the definition of xj , for j = 1, . . . , r − 2, give (18) for j = 1, . . . , r − 2; (48),(49) and (50) give (18) for j = r − 1. Formulas (41), (48) and (50) give that (18) holds for r ≤ j ≤ a − 1, as wanted. Finally, by (50) Au+1,v+1 can be strictly positive only for 1 ≤ u < r − 1 and v ≥ 1. Thus (13) follows from (47). This finishes the proof.

8

3

Main result

In the following theorem we give a complete, explicit and constructive solution to Problem 1. Theorem 1 Let a and b be positive integers and let x and y be nonnegative integers such that a ≥ x, b ≥ y. Let α : α1 | · · · |αn and γ : γ1 | · · · |γn+x+y be polynomial chains. Let X1 , . . . , Xa−1 , Y1 , . . . , Yb−1 and Z be nonnegative integers. If there exists a polynomial chain β : β1 | · · · |βn+x satisfying βi |αi |βi+a ,

(i) (ii) (iii) (iv) (v)

n+min(a−j,x)

i = 1, . . . , n,

γi |βi |γi+b , i = 1, . . . , n + x, n+x i=1 d(βi ) = Z,

d(lcm(αi−a+j , βi )) ≤ Xj , j = 1, . . . , a − 1, i=1 n+x+min(k,y) d(lcm(βi−k , γi )) ≤ Yk , k = 1, . . . , b − 1, i=1

then γi |αi |γi+a+b , n+min(x+y,a−j+k,x+k)  i=1

i = 1, . . . , n,

(51)

d(lcm(αi−a+j−k , γi )) ≤ Xj + Yk − Z,

(52)

j = 0, . . . , a, k = 0, . . . , b, such that a − j ≥ x or k ≤ y. n n+x+y Here Xa = i=1 d(αi ), X0 = Z, Yb = i=1 d(γi ) and Y0 = Z. for

Conversely, when F is an algebraically closed field, then conditions (51) and (52) are also sufficient for the existence of a polynomial chain β : β1 | · · · |βn+x satisfying (i) − (v). Proof: Necessity: Let there exists a polynomial chain β : β1 | · · · |βn+x satisfying (i) − (v). We shall prove that conditions (51) and (52) hold. Conditions (i) and (ii) give (51). Before proceeding with the proof of (52), we note that from the definition of X0 , Y0 , Xa , Yb , and by conditions (i) and (ii), condition (iv) holds for all j = 0, . . . , a, and condition (v) holds for all k = 0, . . . , b. Let j ∈ {0, . . . , a} and k ∈ {0, . . . , b}. By Lemma 1 (applied on the polynomials αi−a+j−k , βi−k , γi ), we have that n+min(x+y,a−j+k,x+k)



n+min(x+y,a−j+k,x+k)

d(lcm(αi−a+j−k , γi )) ≤

i=1

 i=1

9

d(lcm(βi−k , γi ))+ (53)

n+min(x+y−k,a−j,x)



+

n+min(x+y−k,a−j,x)



d(lcm(αi−a+j , βi )) −

i=1

d(βi ).

i=1

Now we consider two cases: a−j ≥x

k ≤ y.

or

If a − j ≥ x, then by applying (iii), (iv) and (v), the right hand side of (53) becomes n+x+min(y,k)



n+x+min(y−k,0)



d(lcm(βi−k , γi ))+

i=1

n+x+min(y−k,0)



d(lcm(αi−a+j , βi ))−

i=1

≤ Yk +

n+x 

d(βi ) ≤

i=1 n+x 

d(lcm(αi−a+j , βi )) −

i=1

d(lcm(αi−a+j , βi ))−

i=n+x+min(y−k,0)+1

−

n+x 

d(βi ) +

i=1

d(βi ) ≤

i=n+x+min(y−k,0)+1

n+x 

≤ Yk +Xj −Z −

n+x 

n+x 

d(lcm(αi−a+j , βi ))+

i=n+x+min(y−k,0)+1

d(βi ) ≤

i=n+x+min(y−k,0)+1

≤ Yk + Xj − Z, which proves (52), as wanted. If k ≤ y, then by applying (iii), (iv) and (v), the right hand side of (53) becomes n+k+min(a−j,x)



n+min(a−j,x)

d(lcm(βi−k , γi )) +

i=1



d(lcm(αi−a+j , βi )) −

i=1

≤ Xj + Yk − Z −

n+x+k 

n+min(a−j,x)



d(βi ) ≤

i=1

d(lcm(βi−k , γi )) +

i=n+k+min(a−j,x)+1

n+x 

d(βi ) ≤

i=n+min(a−j,x)+1

≤ Xj + Yk − Z, which again proves (52), as wanted. Sufficiency: Let F be an algebraically closed field, and let conditions (51) and (52) be valid. Our aim is to define a polynomial chain β : β1 | · · · |βn+x which satisfies conditions (i) − (v). The definition of the polynomials βi , i = 1, . . . , n + x is given in the formula (88) below, by using two different sets of auxiliary polynomials which will be defined in two steps: 10

First step: For every i = 1, . . . , n + x, we start by defining ab polynomials i , u = 1, . . . , a, v = 1, . . . , b, in the following way: ψ¯u,v •

If i − a + u ≤ n and i + v ≤ n + x + y, then lcm(αi−a+u−1 , γi+v ) lcm(αi−a+u , γi+v−1 ) i := . ψ¯u,v lcm(αi−a+u−1 , γi+v−1 ) lcm(αi−a+u , γi+v )

•

If i − a + u = n + 1 and i + v ≤ n + x + y, then lcm(αn , γi+v ) i ψ¯u,v . := lcm(αn , γi+v−1 )

•

(55)

If i − a + u ≤ n and i + v = n + x + y + 1, then lcm(αi−a+u , γn+x+y ) i . := ψ¯u,v lcm(αi−a+u−1 , γn+x+y )

•

(54)

(56)

If i − a + u = n + 1 and i + v = n + x + y + 1, then i ψ¯u,v := 0.

(57)

• In all other cases, i.e. if i − a + u > n + 1 or i + v > n + x + y + 1, we put i := 1. ψ¯u,v

(58)

Since (54)–(58) depend only on i + u and i + v, we trivially have that i+1 i ψ¯u,v = ψ¯u+1,v+1 ,

u = 1, . . . , a−1, v = 1, . . . , b−1, i = 1, . . . , n+x−1. (59)

Next, we set i ), A¯iu,v := d(ψ¯u,v

and A¯u,v :=

i = 1, . . . , n + x, n+x 

A¯iu,v ,

u = 1, . . . , a,

u = 1, . . . , a,

v = 1, . . . , b,

v = 1, . . . , b.

i=1

¯ u,v , u = 1, . . . , a, v = 1, . . . , b: Rewriting condition (52) in terms of A   Before proceeding, we calculate the value of ju=1 kv=1 A¯u,v , for the different values of j = 0, . . . , a and k = 0, . . . , b: •

Case I:

Let

a−x
and

y < k ≤ b,

A¯u,v = +∞.

u=1 v=1

11

then (60)

Indeed, let u0 be such that a − x + 1 ≤ u0 ≤ min(j, a − x + k − y). Since a − x + 1 ≤ j and k > y, such u0 exists. Let v0 := x + y + u0 − a and i0 := n + 1 + a − u0 . Then 1 ≤ i0 ≤ n + x, 1 ≤ u0 ≤ j, 1 ≤ v0 ≤ k and i0 − a + u0 = n + 1, i0 + v0 = n + x + y + 1. Therefore by (57) ψ¯ui00 ,v0 = 0 and so j  k 

A¯u,v =

u=1 v=1

j  n+x k  i=1 u=1 v=1

i d(ψ¯u,v ) ≥ d(ψ¯ui00 ,v0 ) = +∞,

which gives (60). •

Case II: j  k 

0≤k≤y

Let

and

0 ≤ j ≤ a, then

n+min(x,a−j)



A¯u,v =

u=1 v=1

−

d(lcm(αi−a+j , γi )) +

i=1 n+x 

n+x+k 

d(lcm(αi−a−k , γi ))−

i=1

(61)

n+k+min(x,a−j)

d(lcm(αi−a , γi )) −

i=1



d(lcm(αi−a+j−k , γi )).

i=1

Indeed, we have j  k  u=1 v=1

A¯u,v =

j  n+x k  i=1 u=1 v=1

i d(ψ¯u,v ).

(62)

The values of the indices i, u and v in the summation of (62) satisfy i + v ≤ i appearing in (62) are defined by (54), n + x + y, and so the polynomials ψ¯u,v (55) or (58). Therefore for i, u and v such that 1 ≤ ii ≤ n + x, 1 ≤ u ≤ j : and 1 ≤ v ≤ k, we have three possible value of kv=1 ψ¯u,v i ’s from (62) are defined by (54) and so: • If i − a + u ≤ n, then ψ¯u,v k  v=1

i = ψ¯u,v

=

k  lcm(αi−a+u , γi+v−1 ) lcm(αi−a+u−1 , γi+v ) lcm(αi−a+u−1 , γi+v−1 ) lcm(αi−a+u , γi+v )

v=1

lcm(αi−a+u , γi ) lcm(αi−a+u−1 , γi+k ) . lcm(αi−a+u−1 , γi ) lcm(αi−a+u , γi+k )

12

(63)

i ’s from (62) are defined by (55) and so: • If i − a + u = n + 1, then ψ¯u,v k  v=1

i = ψ¯u,v

k  lcm(αn , γi+v ) lcm(αn , γi+k ) = . lcm(αn , γi+v−1 ) lcm(αn , γi )

(64)

v=1

i ’s from are defined by (58) and so: • If i − a + u > n + 1, then ψ¯u,v k  v=1

Now we have j n+x 

k 

i=1 u=1 v=1

i = ψ¯u,v

⎛

n+min(x,a−j) j  

i =⎝ ψ¯u,v

k 

k 

u=1 v=1

i=1

1 = 1.

(65)

v=1

⎞⎛ i ⎠⎝ ψ¯u,v

n+x 

j  k 

i=n+min(x,a−j)+1 u=1 v=1

⎞ i ⎠ . ψ¯u,v

(66) The indices i and u from the first parenthesis on the right-hand side of (66) i satisfy i − a + u ≤ n and therefore all ψ¯u,v appearing in that product are defined by (54) and so by (63) we have n+min(x,a−j) j   i=1

k 

u=1 v=1

n+min(x,a−j) j  

i = ψ¯u,v

u=1

i=1 n+min(x,a−j)



=

i=1

lcm(αi−a+u , γi ) lcm(αi−a+u−1 , γi+k ) lcm(αi−a+u−1 , γi ) lcm(αi−a+u , γi+k )

lcm(αi−a , γi+k ) lcm(αi−a+j , γi ) . lcm(αi−a , γi ) lcm(αi−a+j , γi+k )

(67)

The second parenthesis in (66) is clearly equal to 1 if a − j ≥ x. If a − j < x, then for every i with n + a − j + 1 ≤ i ≤ n + x, let ui := n + 1 + a − i. Then i one has 1 ≤ ui ≤ j. Moreover, for u < ui the polynomials ψ¯u,v are defined by (54), for u = ui they are defined by (55) and for u > ui are defined by (58) and so from (63), (64) and (65) one gets: n+x 

j  k 

i=n+min(x,a−j)+1 u=1 v=1

=

n+x 



u −1 i  lcm(αi−a+u , γi ) lcm(αi−a+u−1 , γi+k ) lcm(αn , γi+k )

i=n+min(x,a−j)+1

=

n+x  i=n+min(x,a−j)+1

=

n+x  i=n+min(x,a−j)+1

=

i = ψ¯u,v

n+x  i=n+min(x,a−j)+1

u=1

lcm(αi−a+u−1 , γi ) lcm(αi−a+u , γi+k )

lcm(αn , γi )

lcm(αi−a+ui −1 , γi ) lcm(αi−a , γi+k ) lcm(αn , γi+k ) = lcm(αi−a , γi ) lcm(αi−a+ui −1 , γi+k ) lcm(αn , γi ) lcm(αn , γi ) lcm(αi−a , γi+k ) lcm(αn , γi+k ) = lcm(αi−a , γi ) lcm(αn , γi+k ) lcm(αn , γi ) lcm(αi−a , γi+k ) . lcm(αi−a , γi ) 13

(68)

=

Therefore, from (67) and (68), the expression (66) becomes: j n+x 

k 

i=1 u=1 v=1

i = ψ¯u,v

n+min(x,a−j)

 i=1

n+x lcm(αi−a+j , γi )  lcm(αi−a , γi+k ) , lcm(αi−a+j , γi+k ) lcm(αi−a , γi ) i=1

and so j  k 

n+min(x,a−j)



A¯u,v =

u=1 v=1

d(lcm(αi−a+j , γi )) +

i=1

−

n+x 

n+x 

d(lcm(αi−a , γi+k ))−

i=1

(69)

n+min(x,a−j)



d(lcm(αi−a , γi )) −

i=1

d(lcm(αi−a+j , γi+k )).

i=1

Since n+x+k 

n+k+min(x,a−j)



d(lcm(αi−a−k , γi )) −

i=1

=

n+x+k 

d(lcm(αi−a+j−k , γi )) =

i=1

d(lcm(αi−a−k , γi ))+

k 

n+k+min(x,a−j)



d(γi )−

i=1

i=1+k

=

n+x 

d(lcm(αi−a+j−k , γi ))−

k  i=1

i=1+k n+min(x,a−j)



d(lcm(αi−a , γi+k )) −

i=1

d(lcm(αi−a+j , γi+k )),

i=1

(69) is equal to (61). •

Case III: j  k 

Let

A¯u,v =

u=1 v=1

−

n+x 

0≤j ≤a−x

and

n+x+min(y,k)

d(lcm(αi−a+j , γi )) +

i=1 n+x  i=1

0 ≤ k ≤ b, then 

d(lcm(αi−a−k , γi ))−

i=1

(70)

n+x+min(y,k)

d(lcm(αi−a , γi )) −



d(lcm(αi−a+j−k , γi )).

i=1

Indeed, in this case the summation indices from (62) satisfy i + u ≤ n + a, i appearing in (62) are defined by (54), (56) or and so all polynomials ψ¯u,v (58). Completely analogously as in Case II we obtain (70). We note that if k ≤ y and j ≤ a − x, i.e. when Cases II and III overlap, the righthand sides of (61) and (70) coincide.

14

d(γi ) =

Now, let z :=

Z−

n+x 

d(lcm(αi−a , γi ))

i=1 n+min(x,a−j)



xj := Xj −

d(lcm(αi−a+j , γi )),

j = 1, . . . , a − 1,

d(lcm(αi−a−k , γi )),

k = 1, . . . , b − 1.

i=1 n+x+min(y,k)



yk := Yk −

i=1

Also, we set xa := 0,

yb := 0,

x0 = y0 := z.

(71)

By condition (52) for k = 0 and j = 0 we obtain z ≥ 0, for k = 0 and by (71) we obtain xj ≥ 0,

j = 0, . . . , a,

and for j = 0 and by (71) we obtain yk ≥ 0,

k = 0, . . . , b.

For j = 0, . . . , a and k = 0, . . . , b, denote by n+min(x,a−j)

M j,k :=



n+x+min(y,k)

d(lcm(αi−a+j , γi )) +

i=1

−

n+x 



d(lcm(αi−a−k , γi ))−

i=1 n+min(x+y,a−j+k,x+k)

d(lcm(αi−a , γi )) −

i=1



d(lcm(αi−a+j−k , γi )).

i=1

Now, condition (52) becomes: z ≤ xj + yk + M j,k

(72)

for all j = 0, . . . , a and k = 0, . . . , b such that a − j ≥ x or k ≤ y. As seen in (61) and (70) in the cases k ≤ y or a − j ≥ x, respectively, we have j  k  M j,k = A¯u,v . u=1 v=1

15

Also, as shown in (60) if k > y and a − j < x, then j  k 

A¯u,v = +∞.

u=1 v=1

Therefore by (72), we have that z ≤ x j + yk +

j  k 

A¯u,v ,

for all

j = 0, . . . , a,

k = 0, . . . , b.

(73)

u=1 v=1

Application of Lemma 2: Now, from condition (73) by Lemma 2 there exist nonnegative integers Au,v , u = 1, . . . , a, v = 1, . . . , b, satisfying Au,v ≤ A¯u,v , a  b z = u=1 v=1 Au,v ,  a xj ≥ u=j+1 bv=1 Au,v , j = 1, . . . , a − 1,   yk ≥ au=1 bv=k+1 Au,v , k = 1, . . . , b − 1.

(74) (75) (76) (77)

In addition, for all u = 1, . . . , a − 1, v = 1, . . . , b − 1, Au+1,v+1 > 0 implies that Au,v = A¯u,v . Our next goal is to define nonnegative integers Aiu,v , i = 1, . . . , n + x, u = 1, . . . , a, v = 1, . . . , b, which satisfy Aiu,v ≤ A¯iu,v Au,v =

n+x 

(78)

Aiu,v .

(79)

i=1

This will be done in the following way: fix u ∈ {1, . . . , a} and v ∈ {1, . . . , b}. By (74) we have n+x 

A¯iu,v = 0 ≤ Au,v ≤

n+x 

i=n+x+1

A¯iu,v .

i=1

Let i0 ∈ {1, . . . , n + x} be such that n+x 

A¯iu,v ≤ Au,v ≤

i=i0 +1

n+x 

A¯iu,v .

i=i0

Let Aiu,v := A¯iu,v , 0 Aiu,v := Au,v −

i ≥ i0 + 1 n+x 

A¯iu,v ,

(80) (81)

i=i0 +1

Aiu,v := 0,

i < i0 . 16

(82)

Such defined Aiu,v , i = 1, . . . , n + x, u = 1, . . . , a, v = 1, . . . , b, satisfy (78) and (79), as wanted. Second step: Next, for every i = 1, . . . , n + x we shall define ab monic i , u = 1, . . . , a, v = 1, . . . , b, in the following way: polynomials ψu,v • If i−a+u ≤ n+1, i+v ≤ n+x+y +1, are such that i−a+u = n+1 i is defined as a polynomial satisfying or i + v = n + x + y + 1, then ψu,v i i |ψ¯u,v ψu,v

i d(ψu,v ) = Aiu,v .

and

(83)

i • If i − a + u = n + 1 and i + v = n + x + y + 1, then ψu,v is defined as a polynomial satisfying i gcd(ψu,v , lcm(αn , γn+x+y )) = 1

and

i d(ψu,v ) = Aiu,v .

(84)

• If i − a + u > n + 1 or i + v > n + x + y + 1, then we put i ψu,v = 1.

(85)

i = 1, and thus 0 ≤ Ai ¯i We also note that in this case ψ¯u,v u,v ≤ Au,v .

Hence i d(ψu,v ) = Aiu,v , for i = 1, . . . , n + x,

and

i i ψu,v |ψ¯u,v ,

i = 1, . . . , n + x,

u = 1, . . . , a, u = 1, . . . , a,

Definition of the polynomials β1 | · · · |βn+x : nomial chain β : β1 | · · · |βn+x is defined by βi := lcm(αi−a , γi )

a  b  u=1 v=1

i ψu,v ,

v = 1, . . . , b, v = 1, . . . , b.

(86) (87)

Finally, the wanted poly-

i = 1, . . . , n + x.

(88)

We are left with proving that such defined polynomials satisfy conditions (i)−(v), and that they indeed form a polynomial chain, i.e. that β1 | · · · |βn+x . By (79), (86) and the definition of z, (75) gives (iii). Next we shall prove that β1 , . . . , βn+x satisfy condition (iv). Fix j ∈ {1, . . . , a − 1}. In order to prove (iv) we shall first show that d(lcm(αi−a+j , βi )) ≤ d(lcm(αi−a+j , γi )) +

a b   u=j+1 v=1

17

i d(ψu,v ),

(89)

for i = 1, . . . , n + min(a − j, x). From (88) follows ⎛ ⎛ d(lcm(αi−a+j , βi )) = d ⎝lcm ⎝αi−a+j , lcm(αi−a , γi )

j  b  u=1 v=1

≤ d lcm(αi−a+j , lcm(αi−a , γi )

j  b  u=1 v=1



⎛

i ψu,v ) +d ⎝

i ψu,v

b a  

u=j+1 v=1

a b   u=j+1 v=1

⎞⎞ i ⎠⎠ ψu,v ≤

⎞ i ⎠ ψu,v , i = 1, . . . , n+min(a−j, x).

Thus, in order to prove (89) we shall prove that

 j  b  i lcm αi−a+j , lcm(αi−a , γi ) ψu,v = lcm(αi−a+j , γi ),

(90)

u=1 v=1

for i = 1, . . . , n + min(a − j, x). Obviously the righthand side in (90) divides the lefthand side, and therefore it is enough proving that lcm(αi−a , γi )

j  b  u=1 v=1

i | lcm(αi−a+j , γi ), ψu,v

i = 1, . . . , n + min(a − j, x). (91)

Since (87) holds, we shall prove (91) by proving lcm(αi−a , γi )

j  b  u=1 v=1

i | lcm(αi−a+j , γi ), ψ¯u,v

i = 1, . . . , n + min(a − j, x).

(92) To that end, let us fix i ∈ {1, . . . , n + min(a − j, x)}. By definitions (54)-(58) it follows that: •

i are defined by (54), and thus If v ≤ n + x + y − i, ψ¯u,v j  u=1

•

lcm(αi−a , γi+v ) lcm(αi−a+j , γi+v−1 ) i = . ψ¯u,v lcm(αi−a , γi+v−1 ) lcm(αi−a+j , γi+v )

i are defined by (56), and thus If v = n + x + y − i + 1, ψ¯u,v j  u=1

•

lcm(αi−a+j , γn+x+y ) i . = ψ¯u,v lcm(αi−a , γn+x+y )

i are defined by (58), and thus If v > n + x + y − i + 1, ψ¯u,v j  u=1

i = 1. ψ¯u,v

18

Finally, in order to prove (92), we consider two cases depending on the value of n + x + y − i: −

If n + x + y − i < b, then

lcm(αi−a , γi )

j  b  u=1 v=1

= lcm(αi−a , γi )

i = lcm(αi−a , γi ) ψ¯u,v

n+x+y−i  v=1

= lcm(αi−a , γi )

n+x+y−i  v=1

(

j  u=1

i ) ψ¯u,v

j  u=1

lcm(αi−a , γi+v ) lcm(αi−a+j , γi+v−1 ) lcm(αi−a , γi+v−1 ) lcm(αi−a+j , γi+v )

i = ψ¯u,n+x+y−i+1

lcm(αi−a+j , γn+x+y ) = lcm(αi−a , γn+x+y )

lcm(αi−a+j , γi ) lcm(αi−a , γn+x+y ) lcm(αi−a+j , γn+x+y ) = lcm(αi−a , γi ) lcm(αi−a+j , γn+x+y ) lcm(αi−a , γn+x+y ) = lcm(αi−a+j , γi ).

−

If n + x + y − i ≥ b, then lcm(αi−a , γi )

j  b  u=1 v=1

i = lcm(αi−a , γi ) ψ¯u,v

j b   i ( )= ψ¯u,v

v=1 u=1

b  lcm(αi−a , γi+v ) lcm(αi−a+j , γi+v−1 )

= lcm(αi−a , γi )

v=1

lcm(αi−a , γi+v−1 ) lcm(αi−a+j , γi+v )

=

lcm(αi−a+j , γi ) lcm(αi−a , γi+b ) = lcm(αi−a , γi ) lcm(αi−a+j , γi+b ) γi+b | lcm(αi−a+j , γi ), = lcm(αi−a+j , γi ) lcm(αi−a+j , γi+b ) = lcm(αi−a , γi )

as wanted. This proves (92) and therefore (89). From (89) we have n+min(x,a−j)



n+min(x,a−j)

d(lcm(αi−a+j , βi )) ≤

i=1



a b n+min(x,a−j)   

+

u=j+1 v=1



u=j+1 v=1 i=1

n+min(x,a−j)

=

(93)

b n+x a   

d(lcm(αi−a+j , γi )) +

i=1



i d(ψu,v )≤

i=1

n+min(x,a−j)

≤

d(lcm(αi−a+j , γi ))+

i=1

d(lcm(αi−a+j , γi )) +

a b   u=j+1 v=1

i=1

19

i d(ψu,v )=

Au,v ≤ Xj .

Here the last inequality follows from (76) and the definition of xj . This proves (iv). Now we shall prove that β1 , . . . , βn+x satisfy condition (v). Let us fix k ∈ {1, . . . , b − 1}. In order to prove (v) we first show that d(lcm(βi−k , γi )) ≤ d(lcm(αi−a−k , γi )) +

a b  

i−k d(ψu,v ),

u=1 v=k+1

(94)

for i = k + 1, . . . , n + x + min(k, y). Equation (94) is equivalent to d(lcm(βi , γi+k )) ≤ d(lcm(αi−a , γi+k )) +

a b   u=1 v=k+1

i d(ψu,v ),

(95)

i = 1, . . . , n + x + min(0, y − k). Analogously as in the proof of (89), from (88) we have 



a a  k b    i i d(lcm(βi , γi+k )) = d lcm γi+k , lcm(αi−a , γi ) ≤ ψu,v ψu,v u=1 v=1





≤ d lcm γi+k , lcm(αi−a , γi )

a  k  u=1 v=1

 i ψu,v

u=1 v=k+1

+d

a b   u=1 v=k+1

 i ψu,v

,

= 1, . . . , n + x + min(0, y − k). Thus, in order to prove (94) we shall prove the following lcm(γi+k , lcm(αi−a , γi )

a  k  u=1 v=1

i ψu,v ) = lcm(αi−a , γi+k ),

(96)

for i = 1, . . . , n + x + min(0, y − k). Obviously the righthand side divides the lefthand side, and therefore we are left with proving lcm(αi−a , γi )

a  k  u=1 v=1

i ψu,v | lcm(αi−a , γi+k ), i = 1, . . . , n + x + min(0, y − k).

(97)

Expression (97) will be proved by showing that lcm(αi−a , γi )

a  k  u=1 v=1

i | lcm(αi−a , γi+k ), i = 1, . . . , n + x + min(0, y − k). ψ¯u,v

(98)

20

Equation (98) follows completely analogously as (91) does. Thus we obtain (94). From (94), we have n+x+min(k,y)



n+x+min(k,y)



d(lcm(βi−k , γi )) ≤

i=k+1

+

a b n+x+min(0,y−k)    i d(ψu,v )≤ u=1 v=k+1

i=1

n+x+min(k,y)



≤

d(lcm(αi−a−k , γi ))+

i=k+1

d(lcm(αi−a−k , γi )) +

a n+x b    u=1 v=k+1 i=1

i=k+1 n+x+min(k,y)



=

d(lcm(αi−a−k , γi )) +



Au,v ≤

u=1 v=k+1

i=k+1 n+x+min(k,y)

≤

a b  

i d(ψu,v )=

n+x+min(k,y)

d(lcm(αi−a−k , γi )) + Yk −



d(lcm(αi−a−k , γi )) =

i=1

i=k+1

= Yk −

k 

d(γi ).

i=1

Here the last inequality follows from (77) and the definition of yk . So, n+x+min(k,y)



n+x+min(k,y)

d(lcm(βi−k , γi )) =

i=1



d(lcm(βi−k , γi )) +

k 

d(γi ) ≤ Yk ,

i=1

i=k+1

which proves (v). Next, we prove that β1 , . . . , βn+x satisfy conditions (i) and (ii). From (88) follows that αi |βi+a , for i = 1, . . . , n + x − a, as well as γi |βi for i = 1, . . . , n + x. Also, from (51), (88) and (91) for j = a, follows βi |αi for i = 1, . . . , n. Analogously from (51), (88) and (97) for k = b, follows that βi |γi+b for i = 1, . . . , n + x + y − b. Since γi = 0 for i > n + x + y, we trivially obtain βi |γi+b for i = n + x + y − b + 1, . . . , n + x. Also, since βi = 0 for i > n + x we trivially obtain αi |βi+a for i = n + x − a + 1, . . . , n + x. Altogether, we have proved (i) and (ii). We are left with proving that β is a polynomial chain, i.e. that βi |βi+1 , i = 1, . . . , n + x − 1. Let us fix i ∈ {1, . . . , n + x − 1}. Then i i i i βi = lcm(αi−a , γi )ψ1,1 · · · ψ1,b ψ2,1 · · · ψa,1

21

a  b  u=2 v=2

i ψu,v ,

and βi+1 = lcm(αi+1−a , γi+1 )

a  b  u=1 v=1

i+1 ψu,v .

We shall prove that i i i i lcm(αi−a , γi )ψ1,1 · · · ψ1,b ψ2,1 · · · ψa,1 | lcm(αi−a+1 , γi+1 ),

(99)

and a−1  b−1  u=1 v=1

i ψu+1,v+1 |

a  b  u=1 v=1

i+1 ψu,v ,

(100)

thus obtaining βi |βi+1 , as wanted. Since

i i i i i i ¯i i ψ1,1 · · · ψ1,b ψ2,1 · · · ψa,1 |ψ¯1,1 · · · ψ¯1,b , ψ2,1 · · · ψ¯a,1

in order to prove (99) it is enough to prove i i ¯i i lcm(αi−a , γi )ψ¯1,1 · · · ψ¯1,b | lcm(αi−a+1 , γi+1 ). ψ2,1 · · · ψ¯a,1

(101)

By (92) for j = 1, we get lcm(αi−a+1 , γi ) i i ψ¯1,1 . · · · ψ¯1,b | lcm(αi−a , γi )

(102)

Since by (54) lcm(αi−a , γi+1 ) lcm(αi−a+1 , γi ) i ψ¯1,1 = , lcm(αi−a , γi ) lcm(αi−a+1 , γi+1 ) from (98) for k = 1 we get lcm(αi−a+1 , γi+1 ) i i ψ¯2,1 . · · · ψ¯a,1 | lcm(αi−a+1 , γi )

(103)

Relations (102) and (103) give (101), and thus (99) follows. In order to prove (100) it is enough to prove i i+1 ψu+1,v+1 | ψu,v ,

u = 1, . . . , a − 1,

v = 1, . . . , b − 1.

(104)

i Let us fix u ∈ {1, . . . , a − 1} and v ∈ {1, . . . , b − 1}. If ψu+1,v+1 = 1, then i i i (104) is trivially satisfied. If ψu+1,v+1 = 1, then Au+1,v+1 = d(ψu+1,v+1 ) > 0, ¯ and so Au+1,v+1 > 0, which by Lemma 2 gives Au,v = Au,v . Thus Ai+1 u,v = i+1 ¯i+1 A¯i+1 u,v , which further implies ψu,v = ψu,v (recall that i ∈ {1, . . . , n + x − 1}). Then (59) gives: i i i+1 i+1 ψu+1,v+1 | ψ¯u+1,v+1 = ψ¯u,v = ψu,v ,

which proves (104), and thus (100). Hence, β is a polynomial chain, as wanted. This finishes our proof. 22

Remark 2 We note that (83) is the only place in the paper where algebraically closed field is needed. Remark 3 In equations (89) and (94) the equality holds. This can be obtained directly from the definitions (83)-(85). However the equality is not important for the main result, and so we don’t prove it here. Remark 4 The second inequality in (93) is equality. Indeed, for i > n+a−j i = 1, for all and u ≥ j + 1, we have that i + u > n + a + 1 and so ψ¯u,v i v = 1, . . . , b. Hence Au,v = 0.

4

Applications

In this section we illustrate the importance of Problem 1 by using Theorem 1 in solving one of the most notable results in completion problems: the GMPCP in the regular case. It is a classical, hard, well studied problem where one needs to characterize the possible Kronecker invariants of a regular pencil with a prescribed subpencil: Problem 2 (The GMPCP in the regular case) Let A(λ) ∈ F[λ](n+p)×(n+m) be a matrix pencil, having α1 | · · · |αn , c1 ≥ · · · ≥ cm and r1 ≥ · · · ≥ rp as Kronecker invariants. Let D(λ) ∈ F[λ](n+p+m+s)×(n+p+m+s) be a regular matrix pencil having γ1 | · · · |γn+m+p+s as homogeneous invariant factors. When there exist pencils X(λ), Y (λ) and Z(λ) such that the pencil   A(λ) X(λ) (105) Y (λ) Z(λ) is strictly equivalent to D(λ)? Necessary conditions to Problem 2 have been obtained by I. Gohberg, M. A. Kaashoek, and F. van Schagen in [14]. A proof of the sufficiency of the conditions from [14] was much more involved. In fact, I. Cabral and F. C. Silva in [2] have obtained an implicit solution to Problem 2 by reducing it to a combinatorial problem which is a special case of Problem 1. In [4, Theorem 1] we have proved sufficiency of the conditions from [14]. However, the proof is very technical, involved and long. It includes a combinatorial structure involving a solution to the Carlson problem [3, 12], the LittlewoodRichardson coefficients and Young diagrams (for details and explanations see [4]). The solution to Problem 1 given in Theorem 1 allows skipping technical parts of the proof in [4], and reducing once long and involved proof to a one page (see Corollary 6). In an analogous way, by applying Theorem 1, one can directly obtain Theorem 4.1 in [5] and Theorem 3 in [7], as well as the combinatorial results involving polynomial chains in the main results from [1, 5, 6, 19, 21, 23]. 23

4.1

A new solution to the regular case GMPCP

By using the notation from Problem 2 we cite here the main result from [2]:  Theorem 5 [2, Theorem 1] Let = m i=1 ci . Problem 2 has a solution if and only if there exists a polynomial chain β : β1 | · · · |βn+p satisfying the following conditions: βi |αi |βi+p ,

i = 1, . . . , n,

γi |βi |γi+m+2s , (r1 + 1, . . . , rp + 1) ≺ where

ξj

=

ξ1j

j · · · ξn+j− ,

ξij

j , η j = η1j · · · ηp+n+j

i = 1, . . . , n + p, −

d(ξ p−1 ), . . . , d(ξ 1 )

(107) −

d(ξ 0 )),

(108)

= lcm(αi−j+ , βi+ ), j = 0, . . . , p, i = 1, . . . , n + j − , n + p + m ≥ d(η m ),

(c1 + 1, . . . , cm + 1) ≺ (n + p + m − where

(d(ξ p )

(106)

d(η m−1 ), d(η m−1 )

ηij = lcm(βi−j , γi ),

−

(109) d(η m−2 ), . . . , d(η 1 )

−

j = 0, . . . , m, i = 1, . . . , n + p + j.

Now we give an explicit solution to Problem 2 as a direct corollary of Theorem 1: Corollary 6 [4, Theorem 1] Let F be an algebraically closed field. Necessary and sufficient conditions for Problem 2 are

θ=

γi | αi | γi+m+p+2s , i = 1, . . . , n, p m+p i=1 (ci + 1) + i=1 (ri + 1) − i=1 d(σi ) ≥ 0 and

m

(111) (112)

(113) (c1 +1, . . . , cm +1)∪(r1 +1, . . . , rp +1) ≺ (θ+d(σm+p ), d(σm+p−1 ), . . . , d(σ1 )),  i , πi = n+i where σi = ππi−1 j=1 lcm(αj−i , γj ), i = 1, . . . , m + p. Proof: By Theorem 5, Problem 2 has a solution if and only if there exists a polynomial chain β : β1 | · · · |βn+p satisfying (106)-(110). Since d(ξ p ) − d(ξ p−1 ) ≥ · · · ≥ d(ξ 1 ) − d(ξ 0 ) and n + p + m − d(η m−1 ) ≥ m−1 d(η ) − d(η m−2 ) ≥ · · · ≥ d(η 1 ) − d(η 0 ) (see [11] for details), by using the definition of majorization the conditions (106)-(110) can be written as βi |αi |βi+p ,

i = 1, . . . , n,

γi |βi |γi+m+2s , i = 1, . . . , n + p, n+p n   (ri + 1) = d(βi ) − d(αi ),

p 

i=1 n+p−j  i=1+

i=1

d(lcm(αi−p+j , βi )) ≤ m  i=1

n+p+k  i=1

n+p  i=1

(114) (115) (116)

i=1+

d(βi ) −

j

i=1 (ri

(ci + 1) = n + p + m −

d(lcm(βi−k , γi )) ≤ n + p + m − 24

+ 1), j = 1, . . . , p − 1,(117)

n+p  i=1

m−k  i=1

d(η 0 )), (110)

d(βi ),

(118)

(ci + 1), k = 1, . . . , m. (119)

 Before proceeding, since = m i=1 ci , from the Kronecker canonical form of the pencil A(λ) (see e.g. [13]) we have that αi = 1 for i = 1, . . . , . Hence, from (114) it follows that βi = 1 for i = 1, . . . , , as well as n 

d(αi ) =

i=1+

n 

n+p−j 

d(αi ) and

i=1

d(lcm(αi−p+j , βi )) =

n+p−j 

i=1+

d(lcm(αi−p+j , βi )).

i=1

Let us introduce additional notation in order to state conditions (114)-(119) in the form of conditions appearing in Problem 1. Let a := p, x := p, b := m + 2s, p n   (ri + 1) + d(αi ), Z := i=1

Xj

:= Z −

y := m + s,

(120) (121)

i=1 j 

(ri + 1),

j = 1, . . . , a − 1,

(122)

i=1

Yk := n + m + p −

m−k 

(ci + 1),

k = 1, . . . , m.

(123)

i=1

Yk := +∞ X0 := Z,

for

k = m + 1, . . . , b − 1,

Y0 := Z,

Xa :=

n 

d(αi ),

(124) Yb :=

i=1

n+m+p+s 

d(γi ). (125)

i=1

Note that if m = b, then s = 0 and the two definitions of Ym and Yb from (123) and (125) coincide since the pencil D(λ) is regular, therefore n+m+p d(γi ) = n + m + p. i=1 In this notation conditions (114)-(119) become βi |αi |βi+a ,

i = 1, . . . , n,

γi |βi |γi+b , i = 1, . . . , n + x, n+x i=1 d(βi ) = Z,

n+a−j

d(lcm(αi−a+j , βi )) ≤ Xj , j = 1, . . . , a − 1, i=1 n+x+k d(lcm(βi−k , γi )) ≤ Yk , k = 1, . . . , b − 1, n p m i=1 i=1 (ci + 1) + i=1 d(αi ) = (n + p + m) − i=1 (ri + 1).

(126) (127) (128) (129) (130) (131)

By Theorem 1 there exists a polynomial chain β : β1 | · · · |βn+x satisfying (126)-(130) if and only if γi |αi |γi+m+p+2s , n+p+k−j  i=1

i = 1, . . . , n,

d(lcm(αi−p+j−k , γi )) ≤ Xj + Yk − Z,

for

j = 0, . . . , p, 25

k = 0, . . . , m.

(132) (133)

Finally, by the definitions (120)-(125) conditions (131), (132) and (133) can we written as (111)-(113), which finishes our proof.

Acknowledgements: The authors would like to thank the referee for valuable comments and suggestions that have significantly improved the presentation of the paper. This work was done within the activities of CEAFEL and was partially supported by FCT, project ISFL-1-1431, and by the Ministry of Science of Serbia, project no. 174020 (M.D.) and no. 174012 (M.S.).

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[18] J. Loiseau, S. Mondi´e, I. Zaballa, P. Zagalak, Assigning the Kronecker invariants of a matrix pencil by row or column [19] E. M. S´ a, Imbedding conditions for λ-matrices, Linear Algebra Appl. 24 (1979) 33-50. [20] E. M. S´ a, Imers˜ ao de Matrizes e Entrela¸camento de Factores Invariantes, Ph.D. Thesis, Univ. Coimbra (1979). [21] R. C. Thompson, Interlacing inequalities for invariant factors, Linear Algebra Appl. 24 (1979) 1-31. [22] I. Zaballa, Matrices with prescribed rows and invariant factors, Linear Algebra Appl. 87 (1987) 113-146. [23] I. Zaballa, Interlacing inequalities and control theory, Linear Algebra Appl. 101 (1988), 9-31.

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