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

Marko Stoˇsi´c
Instituto de Sistemas e Rob´otica and CAMGSD, Instituto Superior T´ecnico,
Av. Rovisco Pais 1, 1049-001 Lisbon, Portugal

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
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 such that
(a
′
1
, a
′
2
, . . . , a
′
l
) ≺ (a
1
, a
2
, . . . , a
l
). (39)
If
(i) d
i
 g

i+1
, i = 1, . . . , ρ,
(ii) g
i
 f
i+l−1
, i = 1, . . . , ρ + 1,
(iii) d
i
= g
i+1
, i 
¯
h
1
, where
¯
h
1
= min{i|d
i
< g
i
},
(iv)
˜
h
q

i=1

f
i
−
˜
h
q
−q

i=1
g
i

q+1

i=2
a
′
i
, q = 1, . . . , l − 1, where
˜
h
q
= min{i|g
i−q+1
< f
i
},
(v)
ρ+l


i=1
f
i
=
ρ+1

i=1
g
i
+
l

i=2
a
′
i
=
ρ

i=1
d
i
+
l

i=1
a
′
i
,

then
h
q

i=1
f
i
−
h
q
−q

i=1
d
i

q

i=1
a
i
, q = 1, . . . , l, (40)
where h
q
= min{i|d
i−q+1
< f
i
}, q = 1, . . . , l.
Proof: From the definition of h

q
,
˜
h
q
and
¯
h
1
, we obtain the following inequalities
h
q
 max(
˜
h
q−1
, min(
¯
h
1
+ q − 1,
˜
h
q
)), q = 1, . . . , l − 1, (
˜
h
0
= 0), (41)
and

h
l
 max(
˜
h
l−1
,
¯
h
1
+ l − 1). (42)
This is true since for q = 1, . . . , l − 1, and j < min(
¯
h
1
+ q − 1,
˜
h
q
), we have that
d
j−q+1
 g
j−q+1
 f
j
.
Therefore, h
q
 min(

¯
h
1
+ q − 1,
˜
h
q
). Also, for every q = 1, . . . , l, and j <
˜
h
q−1
, we have
d
j−q+1
 g
j−q+2
 f
j
, which gives h
q

˜
h
q−1
. Furthermore, for every j <
¯
h
1
+ l − 1, we
have d

j−l+1
 g
j−l+1
 f
j
, and so h
l

¯
h
1
+ l − 1. Altogether, we have (41) and (42 ) .
Let q ∈ {1, . . . , l − 1}. From ( 41), we have the following three possibilities on h
q
:
a) h
q

˜
h
q
, in the case
˜
h
q

¯
h
1
+ q − 1,

b)
˜
h
q
> h
q
 max(
˜
h
q−1
,
¯
h
1
+ q − 1) if
˜
h
q
>
¯
h
1
+ q − 1,
c) h
q

˜
h
q
> max(

˜
h
q−1
,
¯
h
1
+ q − 1) if
˜
h
q
>
¯
h
1
+ q − 1.
Observe these cases separately:
the electronic journal of combinatorics 17 (2010), #R61 11
a) Let h
q

˜
h
q
(
˜
h
q

¯

h
1
+ q − 1), then by (iv) we have
h
q

i=1
f
i
=
˜
h
q

i=1
f
i
+
h
q

˜
h
q
+1
f
i

˜
h

q
−q

i=1
g
i
+
h
q

˜
h
q
+1
f
i
+
q+1

i=2
a
′
i

˜
h
q
−q

i=1

d
i
+
h
q
−q

˜
h
q
−q+1
d
i
+
q+1

i=2
a
′
i
=
h
q
−q

i=1
d
i
+
q+1


i=2
a
′
i
.
The second inequality is true since
˜
h
q
− q <
¯
h
1
. So, we have d
i
 g
i
for all i 
˜
h
q
− q.
Also, from h
q
< h
q+1
, we obtain f
i
 d

i−q
, for all i  h
q
< h
q+1
.
Finally, from (39), we have
q+1

i=2
a
′
i

q

i=1
a
i
,
and so
h
q

i=1
f
i

h
q

−q

i=1
d
i
+
q

i=1
a
i
,
which proves (40), as wanted.
b) Let
˜
h
q
> h
q
 max(
¯
h
1
+ q − 1,
˜
h
q−1
), then by (iv), we have
h
q


i=1
f
i
=
˜
h
q−1

i=1
f
i
+
h
q

˜
h
q−1
+1
f
i

˜
h
q−1
−q+1

i=1
g

i
+
h
q

˜
h
q−1
+1
f
i
+
q

i=2
a
′
i

˜
h
q−1
−q+1

i=1
g
i
+
h
q

−q+1

˜
h
q−1
−q+2
g
i
+
q

i=2
a
′
i
.
The second inequality is true, since h
q
<
˜
h
q
, and so, g
i−q+1
 f
i
, for all i  h
q
.
Moreover, since h

q
− q + 1 
¯
h
1
, by conditions (iii) and (v), we have
h
q
−q+1

i=1
g
i
=
ρ+1

i=1
g
i
−
ρ+1

i=h
q
−q+2
g
i
=
ρ


i=1
d
i
+ a
′
1
−
ρ

i=h
q
−q+1
d
i
=
h
q
−q

i=1
d
i
+ a
′
1
,
and so
h
q
−q+1


i=1
g
i
+
q

i=2
a
′
i
=
h
q
−q

i=1
d
i
+
q

i=1
a
′
i
.
Last equality together with (39 ) gives
h
q


i=1
f
i

h
q
−q

i=1
d
i
+
q

i=1
a
i
,
the electronic journal of combinatorics 17 (2010), #R61 12
which proves (40), as wanted.
c) Let h
q

˜
h
q
> max(
¯
h

1
+ q − 1,
˜
h
q−1
), then by (iv), we have
h
q

i=1
f
i
=
˜
h
q−1

i=1
f
i
+
h
q

˜
h
q−1
+1
f
i


˜
h
q−1
−q+1

i=1
g
i
+
h
q

˜
h
q−1
+1
f
i
+
q

i=2
a
′
i
=
˜
h
q−1

−q+1

i=1
g
i
+
˜
h
q
−1

˜
h
q−1
+1
f
i
+
h
q

˜
h
q
f
i
+
q

i=2

a
′
i

˜
h
q−1
−q+1

i=1
g
i
+
˜
h
q
−q

˜
h
q−1
−q+2
g
i
+
h
q
−q

˜

h
q
−q
d
i
+
q

i=2
a
′
i
=
h
q
−q

i=1
d
i
+ a
′
1
+
q

i=2
a
′
i

.
The second inequality follows from the definition of
˜
h
q
and the fact that h
q
< h
q+1
, while
the last equality is true since
˜
h
q
−q 
¯
h
1
. Now, we finish the proof as in the previous case.
The only remaining case is q = l. Let i > h
l
. Since h
l
 max(
˜
h
l−1
,
¯
h

1
+ l −1), we have
i >
˜
h
l−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 f
i
= g
i−l+1
. Also, since i >
¯
h
1

+ l − 1, from (iii) we have
g
i−l+1
= d
i−l
, and thus
f
i
= d
i−l
, i > h
l
. (43)
Now, by (v), condition (40) for q = l is equivalent to
ρ+l

i=h
l
+1
f
i

ρ

i=h
l
−l+1
d
i
.

Finally, from (i), we have that d
i
 f
i+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 d
1
 · · ·  d
ρ
, f
1
 · · ·  f
ρ+l
, a
1
 · · ·  a
l
and a
′
1
, . . . , a
′
l
be
integers, such that
(a
′
1

, . . . , a
′
l
) ≺ (a
1
, . . . , a
l
).
Moreo ver, for every j = 1, . . . , l − 1, let g
j
= (g
j
1
, . . . , g
j
ρ+j
) be s uch that g
j
1
 · · ·  g
j
ρ+j
.
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 13
Thus, Theorems 5 and 7 prove the existence of a path of sequences, as announced
before Lemma 4. In part icular, we have
Corollary 8 Let l  2, d
1
 · · ·  d
ρ
, f
1
 · · ·  f
ρ+l
, a
1
 · · ·  a
l
be integers.
Then
f ≺
′

(d, a)
if and only if there exists g = (g
1
, . . . , g
ρ+s
), for some 0 < s < l, such that g
1
 · · ·  g
ρ+s
and
f ≺
′
(g, a
′
)
g ≺
′
(d, a
′′
)
where a
′
= (a
1
, . . . , a
l−s
) and a
′′
= (a
l−s+1

, . . . , a
l
).
4 Convexity lemma
In this section we give a short polynomial proof of the convexity lemma, which is the
crucial step in S´a-Thompson theorem [7, 10]. The original proofs of S´a and Thompson
were long and complicated, and relied on very involved techniques. The proof in [7]
(Proposition 4.1 and Lemma 4.2) uses nonelementary analytical tools, while the 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 α
1
| · · · |α
n
and γ
1
| · · · |γ
n+m
be two polynomial chains.
For every j = 0, . . . , m, let
δ
j
i
:= lcm(α
i−2j
, γ
i
), i = 1, . . . , n + j,
δ

j
:=
n+j

i=1
δ
j
i
.
The difference between the convexity in this case and the result f r om Lemma 2 is in
a different shift in the definition of δ
j
comparing to π
j
. 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
), f or j = 1, . . . , m − 1.
Before going to the proof we give one simple lemma:
the electronic journal of combinatorics 17 (2010), #R61 14
Lemma 10 Let φ
1
, φ
2
, ψ
1
and ψ
2
be polynomials such that φ
1
|φ
2
and ψ
1
|ψ
2
. Then
lcm(φ
1
, ψ
1
) lcm(φ
2
, ψ
2

)| lcm(φ
2
, ψ
1
) lcm(φ
1
, ψ
2
). (44)
Proof: For i = 1, 2, we have
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 this expression for i = 1 and i = 2 into (44), it becomes equivalent to
the following obvious divisibility relation:
gcd(lcm(φ
1
, ψ
1
), ψ
2
) | gcd(lcm(φ
2
, ψ
1
), ψ
2
).
Proof of Theorem 9:
In order to prove the convexity, it is enough to prove that
δ
j

δ
j
| δ
j−1
δ
j+1
, j = 1, . . . , m − 1 . (45)
By definition, we have
δ
j
=
n+j

i=1
lcm(α
i−2j
, γ
i
), j = 0, . . . , m. (46)
Since for a ll i and j we have
lcm(α
i−2j
, γ
i
) = lcm(α
i−2j
, lcm(α
i−2j−2
, γ
i

)) =
α
i−2j
lcm(α
i−2j−2
, γ
i
)
gcd(α
i−2j
, lcm(α
i−2j−2
, γ
i
))
,
we can rewrite (46) as
δ
j
=
n+j

i=1
α
i−2j
lcm(α
i−2j−2
, γ
i
)

gcd(α
i−2j
, lcm(α
i−2j−2
, γ
i
))
=

n−j
i=1
α
i

n+j
i=1
lcm(α
i−2j−2
, γ
i
)

n−j
i=1
gcd(α
i
, lcm(α
i−2
, γ
i+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 o ther δ
j
and δ
j−1
by the expression (47). Then (45)
becomes equivalent to

n+j
i=1
lcm(α
i−2j
, γ
i
)

n−j
i=1
α
i

n+j
i=1
lcm(α

i−2j−2
, γ
i
)

n−j
i=1
gcd(α
i
, lcm(α
i−2
, γ
i+2j
))
|

n+j+1
i=1
lcm(α
i−2j−2
, γ
i
)

n−j+1
i=1
α
i

n+j−1

i=1
lcm(α
i−2j
, γ
i
)

n−j+1
i=1
gcd(α
i
, lcm(α
i−2
, γ
i+2j−2
))
.
the electronic journal of combinatorics 17 (2010), #R61 15
After cancellations, the last divisibility becomes equivalent to
lcm(α
n−j
, γ
n+j
)
n−j+1

i=1
gcd(α
i
, lcm(α

i−2
, γ
i+2j−2
))
| lcm(α
n−j−1
, γ
n+j+1
)α
n−j+1
n−j

i=1
gcd(α
i
, lcm(α
i−2
, γ
i+2j
)).
By using the obvious divisibility relation
gcd(α
i
, lcm(α
i−2
, γ
i+2j−2
)) | gcd(α
i
, lcm(α

i−2
, γ
i+2j
)),
we are left with proving that
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 the 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

),
which fo llows directly from Lemma 10.
References
[1] I. Baraga˜na, I. Zaballa, Column completion of a pair of matrices, Linear a nd Multilinear Algebr a,
27 (1990) 243-273.
[2] M. Dodig, Matrix pencils completion problems, Linear Algebra Appl. 428 (2008), no. 1, 259-304.
[3] M.Dodig, M. Stoˇsi´c, Similarity class of a matrix with prescribed submatrix, Linear and Multilinear
Algebra, 57 (2009) 217-245.
[4] M. Dodig, Feedback invariants of matrices with prescribed rows, Linear Algebra Appl. 405 (2005)
121-154.
[5] I. Gohberg, M. A. Kaashoek, F. van Schagen, Eigenva lues of completions of submatrices, Linear
and Multilinear Algebra, 25 (1989) 55-70.
[6] G. Hardy, J.E. Littlewood, G. P´o lya, Inequalities, Cambridge University Press, 1991.
[7] E. M. S´a, Imbeding conditions for λ -matrices, Linear Algebr a Appl. 24 (1979) 33-50.
[8] E. M. S´a, A convexity lemma on the interlacing inequalities for invariant fac tors, Linear Alg e bra
Appl. 109 (1988) 107-113.
[9] E. M. S´a, Imers˜ao de matrizes e entrela¸camento de factores invariantes, PhD Thesis, Univ. of
Coimbra, 1979.
[10] R. C. Thompson, Interlacing inequalities for invariant factors, Linea r Algebra Appl. 24 (1979) 1-31.
[11] I. Zaballa, Matrices with prescribed rows and invariant factors, Linear Algebra Appl. 87 (1987)
113-146.
the electronic journal of combinatorics 17 (2010), #R61 16