# Extending an Almost Complete Pair of Matrices to a Complete Triple

Authors: Sheng Chen; Chao Xia
Journal: The Scientific World Journal
 (2014)

---

## Abstract

Motivated by the concept of complete pairs, which was introduced by Krein and Langer, we present the concepts of an almost complete pair of matrices and a complete triple of matrices. It is proved that an almost complete pair of matrices can be extended to a complete triple. An application of the result to differential equations is also given.


---

## Body

## 1. Introduction

Let(1)
L
(
λ
)
=
I
λ
l
+
A
l
-
1
λ
l
-
1
+
⋯
+
A
0
be a monic matrix polynomial, where A
i
∈
M
n
(
ℂ
) for i
=
0,1
,
…
,
l
-
1. If there exist a matrix S and a polynomial L
1
(
λ
) of degree l
-
1 such that L
(
λ
)
=
L
1
(
λ
)
(
I
λ
-
S
), then I
λ
-
S is said to be a monic right divisor of L
(
λ
). As is well known, I
λ
-
S is a monic right divisor of L
(
λ
) if and only if S is a root of L
(
λ
); that is,
(2)
S
l
+
A
l
-
1
S
l
-
1
+
⋯
+
A
0
=
0
.
Let C be the companion matrix of L
(
λ
); that is,
(3)
C
=
(
0
I
0
⋯
0
0
0
I
⋯
0
⋮
⋮
⋮
⋮
0
0
0
⋯
I
-
A
0
-
A
1
-
A
2
⋯
-
A
l
-
1
)
.
Then we have det
⁡
(
I
λ
-
C
)
=
det
⁡
(
L
(
λ
)
). If S
1
,
S
2
,
…
,
S
l
∈
M
n
(
ℂ
) are the different l roots of L
(
λ
), it is clear that the Vandermonde matrix
(4)
W
(
S
1
,
S
2
,
…
,
S
l
)
=
(
I
I
I
⋯
I
S
1
S
2
S
3
⋯
S
l
S
1
2
S
2
2
S
3
2
⋯
S
l
2
⋮
⋮
⋮
⋮
S
1
l
-
1
S
2
l
-
1
S
3
l
-
1
⋯
S
l
l
-
1
)
plays a role in the investigation of the spectral analysis of L
(
λ
) [1].Definition 1.
We say that thel-tuple (
S
1
,
S
2
,
…
,
S
l
) of complex square matrices of order n is complete if the Vandermonde matrix W
(
S
1
,
S
2
,
…
,
S
l
) is invertible. When l
=
2 and l
=
3, we call a complete l-tuple of matrices a complete pair and a complete triple, respectively.Suppose that(
S
1
,
S
2
) is complete. Take P
=
S
2
-
S
1. Then P is invertible. If we take
(5)
A
1
=
-
P
S
2
P
-
1
-
S
1
,
A
2
=
P
S
2
P
-
1
S
1
,
L
(
λ
)
=
I
λ
2
+
A
1
λ
+
A
2
,
then I
λ
-
S
1, I
λ
-
S
2, form a complete pair of right divisors of L
(
λ
) (cf. Lemma 2.4 in [2]). The concept of complete pair was introduced in [3, 4]. In [5], Lancaster gave some important applications of complete pairs to solutions of differential equations. Motivated by the results in [5], we are interested in the following question.Question 1.
Suppose that(
S
1
,
S
2
) is not a complete pair. Is it possible for us to find S
3 and a cubic matrix polynomial L
(
λ
) such that (
S
1
,
S
2
,
S
3
) is a complete triple and
(6)
I
λ
-
S
1
,
I
λ
-
S
2
,
I
λ
-
S
3
are right divisors of L
(
λ
)?If(
S
1
,
S
2
,
S
3
) is a complete triple, then
(7)
(
I
I
S
1
S
2
S
1
2
S
2
2
)
is of full column rank. Note that if S
1
,
S
2
∈
M
n
(
ℂ
) and the rank of S
1
-
S
2 is less than n
-
1, we cannot always find S
3 such that (
S
1
,
S
2
,
S
3
) is a complete triple. For example, let S
1
,
S
2
∈
M
n
(
ℂ
) be defined as follows:
(8)
S
1
=
(
A
1
α
β
A
2
)
,
where A
1, A
2 are square matrices of orders n
-
2 and 2, respectively, and S
2
-
S
1 is in the form
(9)
(
I
n
-
2
0
0
0
0
0
0
0
0
)
.
Then
(10)
S
2
2
-
S
1
2
=
(
2
A
1
+
I
n
-
2
α
β
0
)
.Ifrank
⁡
(
α
)
=
1, then the last two columns of (
S
2
-
S
1
S
2
2
-
S
1
2
) are linearly dependent. For any S
3
∈
M
n
(
ℂ
),
(11)
(
I
I
I
S
1
S
2
S
3
S
1
2
S
2
2
S
3
2
)
cannot be invertible.Ifrank
⁡
(
α
)
=
2, then
(12)
(
I
I
S
1
S
2
S
1
2
S
2
2
)
is of full column rank. From the condition that
(13)
(
I
I
S
1
S
2
S
1
2
S
2
2
)
is of full column rank, we cannot conclude that rank
⁡
(
S
2
-
S
1
)
≥
n
-
1.After the above discussion, we give the following definition.Definition 2.
For two matricesS
1
,
S
2
∈
M
n
(
ℂ
), one says they form an almost complete pair if the matrix
(14)
(
I
I
S
1
S
2
S
1
2
S
2
2
)
is of full column rank and the rank of S
1
-
S
2 is n
-
1. For simplification, one says (
S
1
,
S
2
) is an almost complete pair of matrices of order n.The following is the main result of this paper.Theorem 3.
Suppose thatS
1
,
S
2
∈
M
n
(
ℂ
), n
≥
2, and (
S
1
,
S
2
) is an almost complete pair. Then one can find a matrix S
3
∈
M
n
(
ℂ
) such that (
S
1
,
S
2
,
S
3
) is a complete triple; that is,
(15)
(
I
I
I
S
1
S
2
S
3
S
1
2
S
2
2
S
3
2
)
is invertible.We prove this theorem in Section2. In Section 3, we partially answer the question mentioned before and give an application of the main result to differential equations.


## 2. Proof of the Main Theorem

In this section, we prove Theorem3. Note that there is an invertible matrix P such that P
-
1
(
S
2
-
S
1
)
P is the Jordan normal form of S
2
-
S
1. Let Q
=
P
⊕
P
⊕
P. Then the condition that
(16)
X
=
(
I
I
I
S
1
S
2
S
3
S
1
2
S
2
2
S
3
2
)
is invertible is equivalent to that
(17)
Q
-
1
X
Q
is invertible. So from the beginning, we can assume that S
2
-
S
1 is in Jordan normal form.First we prove a lemma.Lemma 4.
AssumeS
1
,
S
2
∈
M
n
(
ℂ
) and the pair (
S
1
,
S
2
) is an almost complete pair with S
2
-
S
1
=
I
n
-
1
⊕
0. Write
(18)
S
1
=
(
A
α
β
s
)
,
where A
∈
M
n
-
1
(
ℂ
). Then α is not a zero vector.Proof.
By some computations, we have(19)
S
2
2
-
S
1
2
=
(
2
A
+
I
n
-
1
α
β
0
)
.
The last column of
(20)
(
S
2
-
S
1
S
2
2
-
S
1
2
)
is (
0,0
,
…
,
0
,
α
,
0
)
T, which is a 2
n-dimensional column vector. By the definition of almost complete pair, α is not a zero vector.Now we prove the main theorem.Proof.
From now on, we writeS
1
=
(
a
11
⋯
a
1
n
⋮
⋱
⋮
a
n
1
⋯
a
n
n
). We denote by P
(
i
,
j
)
∈
M
n
-
1
(
ℝ
) the matrix which is obtained by exchanging the columns of i and j of (
n
-
1
)
×
(
n
-
1
) identity matrix.
Without loss of generality, we can assumeS
2
-
S
1 is in the Jordan normal form.
Case  1. Firstly, we consider the case that
(21)
S
2
-
S
1
=
(
I
n
-
1
0
0
0
)
.
If a
1
n
≠
0, then we take
(22)
S
3
=
S
1
+
(
0
0
t
I
n
-
1
0
)
.
By some computation, the last columns of S
2
2
-
S
1
2 and S
3
2
-
S
1
2 are (
a
1
n
,
a
2
n
,
…
,
a
n
-
1
n
,
0
)
T and (
0
,
t
a
1
n
,
t
a
2
n
,
…
,
t
a
n
-
1
n
)
T, respectively, and the ith (1
≤
i
≤
n
-
2) column of S
3
2
-
S
1
2 is
(23)
(
f
i
,
1
(
t
)
,
f
i
,
2
(
t
)
,
…
,
f
i
,
i
+
1
(
t
)
,
h
h
t
2
+
f
i
,
i
+
2
(
t
)
,
f
i
,
i
+
3
(
t
)
,
…
,
f
i
,
n
(
t
)
)
T
,
and the (
n
-
1
)th column of S
3
2
-
S
1
2 is (
f
n
-
1,1
(
t
)
,
f
n
-
1,2
(
t
)
,
…
,
f
n
-
1
,
n
(
t
)
)
T, where f
i
,
1
=
t
a
1
i
+
1 and f
i
,
j
=
t
(
a
j
i
+
1
+
a
j
-
1
i
)
(
1
≤
i
≤
n
-
1
,
2
≤
j
≤
n
). It is clear that
(24)
det
⁡
(
S
2
-
S
1
S
3
-
S
1
S
2
2
-
S
1
2
S
3
2
-
S
1
2
)
is a polynomial in t. If a
1
n
≠
0, then the degree of this polynomial is 2
n
-
2 and the leading term of this polynomial is a
1
n
2
t
2
n
-
2. If |
t
| is sufficiently large, we can get that
(25)
X
=
(
I
I
I
S
1
S
2
S
3
S
1
2
S
2
2
S
3
2
)
is invertible. If a
1
n
=
0, by Lemma 4, then there must be some
(26)
a
i
n
≠
0
,
where 2
≤
i
≤
n
-
1. So
(27)
(
P
(
1
,
i
)
0
0
1
)
(
A
α
β
0
)
(
P
(
1
,
i
)
0
0
1
)
=
(
(
P
(
1
,
i
)
)
A
P
(
1
,
i
)
P
(
1
,
i
)
α
β
P
0
)
,
where the first entry of the vector P
(
1
,
i
)
α is a
i
n. Take S
3 such that
(28)
(
P
(
1
,
i
)
⊕
1
)
(
S
3
-
S
1
)
(
P
(
1
,
i
)
⊕
1
)
=
(
0
0
t
I
n
-
1
0
)
.
Then
(29)
X
=
(
I
I
I
S
1
S
2
S
3
S
1
2
S
2
2
S
3
2
)
is invertible.
Case  2. Now we consider the case that S
2
-
S
1 has the Jordan normal form
(30)
J
(
λ
1
)
⊕
J
(
λ
2
)
⊕
⋯
⊕
J
(
λ
k
)
⊕
0
,
where λ
i
≠
0, 1
≤
i
≤
k. If a
1
n
≠
0, then we take
(31)
S
3
=
S
1
+
(
0
0
t
I
n
-
1
0
)
.
If a
1
n
=
0 and a
i
n
≠
0
(
1
<
i
<
n
), then we take S
3 such that
(32)
(
P
(
1
,
i
)
⊕
1
)
(
S
3
-
S
1
)
(
P
(
1
,
i
)
⊕
1
)
=
(
0
0
t
I
n
-
1
0
)
.
In any case when |
t
| is sufficiently large, we can get that
(33)
X
=
(
I
I
I
S
1
S
2
S
3
S
1
2
S
2
2
S
3
2
)
is invertible.
Case  3. Now we consider the case that 0 is not a single root of characteristic polynomial of S
2
-
S
1; that is, S
2
-
S
1 has the Jordan normal form:
(34)
J
(
λ
1
)
⊕
J
(
λ
2
)
⊕
⋯
⊕
J
(
λ
s
)
⊕
J
(
0
)
.
If n
=
2, then
(35)
S
2
-
S
1
=
(
0
0
1
0
)
,
By Lemma 4, a
12
≠
0. Take
(36)
S
3
=
S
1
+
(
1
0
0
0
)
,
Then
(37)
det
⁡
(
I
I
I
S
1
S
2
S
3
S
1
2
S
2
2
S
3
2
)
=
det
⁡
(
S
2
-
S
1
S
3
-
S
1
S
2
2
-
S
1
2
S
3
2
-
S
1
2
)
=
-
t
2
a
12
2
≠
0
.
We can conclude that
(38)
X
=
(
I
I
I
S
1
S
2
S
3
S
1
2
S
2
2
S
3
2
)
is invertible.
Firstly, we prove the case that the normal form itself is(39)
(
0
0
t
I
n
-
1
0
)
.
If a
1
n
≠
0, then we take
(40)
S
3
=
S
1
+
(
t
0
0
0
⋯
0
t
0
0
0
0
⋯
0
0
t
t
0
0
⋯
0
0
0
0
t
0
⋯
0
0
⋮
⋮
⋮
⋱
⋮
⋮
0
0
0
0
⋯
t
0
)
.
So
(41)
(
S
3
-
S
1
)
2
=
(
t
2
0
0
⋯
0
t
2
0
0
0
0
⋯
0
0
0
t
2
0
0
⋯
0
0
0
0
t
2
0
⋯
0
0
0
⋮
⋮
⋱
⋮
⋮
⋮
0
0
0
⋯
t
2
0
0
)
.
By some computation,
(42)
det
⁡
X
=
det
⁡
(
I
I
I
S
1
S
2
S
3
S
1
2
S
2
2
S
3
2
)
is a polynomial in t, where the leading term of it is a
1
n
t
2
n
-
1. So if a
1
n
≠
0, then X is invertible when |
t
| is sufficiently large. If a
1
n
=
0, then there must be an i
∈
{
2,3
,
…
,
n
-
1
} such that a
i
n
≠
0. If i
∈
{
2,3
,
…
,
n
-
2
}, take S
3 such that
(43)
(
P
(
1
,
i
-
1
)
⊕
1
)
(
S
3
-
S
1
)
(
P
(
1
,
i
-
1
)
⊕
1
)
=
(
t
0
0
0
⋯
0
t
0
0
0
0
⋯
0
0
t
t
0
0
⋯
0
0
0
0
t
0
⋯
0
0
⋮
⋮
⋮
⋱
⋮
⋮
0
0
0
0
⋯
t
0
)
.
When |
t
| is sufficiently large, we have X being invertible. If i
=
n
-
1, that is to say, the last column of S
2
2
-
S
1
2 is (
0,0
,
…
,
0
,
a
n
-
1
n
)
T, then we take
(44)
S
3
=
S
1
+
(
t
0
0
0
⋯
1
0
0
t
0
0
⋯
0
0
0
0
t
0
⋯
0
0
0
0
0
t
⋯
0
0
⋮
⋮
⋮
⋮
⋱
⋮
⋮
0
0
0
0
⋯
t
0
0
0
0
0
⋯
0
0
)
.
We have
(45)
(
S
3
-
S
1
)
2
=
(
t
2
0
0
0
⋯
2
t
0
0
t
2
0
0
⋯
0
0
0
0
t
2
0
⋯
0
0
0
0
0
t
2
⋯
0
0
⋮
⋮
⋮
⋮
⋱
⋮
⋮
0
0
0
0
⋯
t
2
0
0
0
0
0
⋯
0
0
)
,
where the first entry of the last column of S
3
2
-
S
1
2 is a
n
-
1
n. So the leading term of the polynomial
(46)
det
⁡
X
=
det
⁡
(
I
I
I
S
1
S
2
S
3
S
1
2
S
2
2
S
3
2
)
is a
n
-
1
n
2
t
2
n
-
3. Thus
(47)
X
=
(
I
I
I
S
1
S
2
S
3
S
1
2
S
2
2
S
3
2
)
is invertible when |
t
| is sufficiently large.
Now we prove the general case thatS
2
-
S
1 has the Jordan normal form:
(48)
J
(
λ
1
)
⊕
J
(
λ
2
)
⊕
⋯
⊕
J
(
λ
s
)
⊕
J
(
0
)
,
where λ
i
≠
0, 1
≤
i
≤
s. If there is some a
n
i
≠
0, m
+
1
≤
i
≤
n, we assume that J
(
0
) is a matrix of order n
-
m. For any square matrix B of order n
-
m, we can construct a square matrix A of order n
-
m such that
(49)
(
J
(
0
)
A
(
J
(
0
)
+
B
)
2
-
J
2
(
0
)
(
A
+
B
)
2
-
A
2
)
is invertible. Take
(50)
S
3
=
S
1
+
(
t
I
n
-
m
⊕
A
)
.
If a
i
n
=
0, for all m
+
1
≤
i
≤
n, there must be some j such that 1
≤
j
≤
m, a
j
n
≠
0. Let Y
=
X
P, where P is the matrix which is obtained by exchanging the columns m
+
1 and n of n
×
n identity matrix. Then det
⁡
X
=
-
det
⁡
Y and
(51)
Y
=
J
(
λ
1
)
⊕
J
(
λ
2
)
⊕
⋯
⊕
J
(
λ
i
)
⊕
0
⊕
C
,
where the square matrix C of order n
-
m
+
1 has the form
(52)
(
0
1
I
n
-
m
0
)
.
Let T be a matrix such that T
-
1
C
T has the Jordan normal form J. Then we can consider the matrix Z
=
(
I
⊕
T
-
1
)
Y
(
I
⊕
T
). Note that Z has the form
(53)
Z
=
J
(
λ
1
)
⊕
J
(
λ
2
)
⊕
⋯
⊕
J
(
λ
i
)
⊕
0
⊕
J
.
Denote S
=
J
(
λ
1
)
⊕
J
(
λ
2
)
⊕
⋯
⊕
J
(
λ
i
)
⊕
0. For any square matrix B, we can construct a square matrix A of order m
+
1 such that
(54)
(
S
A
(
S
+
B
)
2
-
S
2
(
A
+
B
)
2
-
A
2
)
is invertible. Take S
3
=
S
1
+
A
⊕
t
I
m
-
1. Then we can conclude that
(55)
(
I
I
I
S
1
S
2
S
3
S
1
2
S
2
2
S
3
2
)
is invertible. The proof is complete.


## 3. Applications

For given monic matrix polynomialsL
1
(
λ
)
,
L
2
(
λ
)
,
…
,
L
n
(
λ
), the construction of a common multiple of them was presented in [2].Suppose that(
S
1
,
S
2
) is an almost complete pair. By Theorem 3 in Section 1, we can find S
3
∈
M
n
(
ℂ
) such that (
S
1
,
S
2
,
S
3
) is a complete triple; that is,
(56)
(
I
I
I
S
1
S
2
S
3
S
1
2
S
2
2
S
3
2
)
is invertible. Let
(57)
W
m
(
S
1
,
S
2
,
S
3
)
=
(
I
I
I
S
1
S
2
S
3
⋮
⋮
⋮
S
1
m
-
1
S
2
m
-
1
S
3
m
-
1
)
.
Then we can conclude that
(58)
min
⁡
{
m
≥
1
∣
Ker
⁡
W
m
=
Ker
⁡
W
m
+
1
}
=
3
.By Theorem 9.11 of [2], there exists a monic matrix polynomial L
(
λ
) with degree 3 which is a common multiple of I
λ
-
S
1, I
λ
-
S
2, I
λ
-
S
3.As is well known, spectral method for the analysis of monic matrix polynomials was important (cf. [6–13]). The spectral method basically uses the construction of standard pairs and standard triples (e.g., [14, 15]) to solve differential equations.Now we recall definitions of standard pair and standard triple for monic polynomials.Definition 5 (see [2]).
A pair of matrices(
X
,
T
), where X is n
×
n
l and T is n
l
×
n
l, is called a standard pair for the monic matrix polynomial L
(
λ
) if the following conditions are satisfied:(i)
col
(
X
T
i
)
i
=
0
l
-
1 is nonsingular, where
(59)
col
(
X
T
i
)
i
=
0
l
-
1
=
(
X
X
T
⋮
X
T
l
-
1
)
;
(ii)
Σ
i
=
0
l
-
1
A
i
X
T
i
+
X
T
l
=
0.
IfT is in Jordan normal form, we call (
X
,
T
) a Jordan pair for the monic matrix polynomial L
(
λ
).Definition 6 (see [2]).
A triple of matrices(
X
,
T
,
Y
), where X is n
×
n
l, T is n
l
×
n
l, and Y is n
l
×
n, is called a standard triple for the monic matrix polynomial L
(
λ
) if (
X
,
T
) is a standard pair for L
(
λ
), and Y
=
(
X
X
T
⋮
X
T
l
-
1
)
-
1
(
0
⋮
0
I
).Now we recall the explicit formulas for solutions of the following differential equation:(60)
L
(
d
d
t
)
x
(
t
)
=
d
l
x
(
t
)
d
t
l
+
Σ
j
=
0
l
-
1
A
j
d
j
x
(
t
)
d
t
j
=
f
(
t
)
.Lemma 7 (Theorem 2.9 in [2]).
The general solution of (60) is given by the formula
(61)
x
(
t
)
=
X
e
t
T
c
+
X
∫
a
t
‍
e
(
t
-
s
)
T
Y
f
(
s
)
d
s
,
t
∈
[
a
,
b
]
,
where (
X
,
T
,
Y
) is a standard triple of L
(
λ
) and c
∈
ℂ
n
l is arbitrary. In particular, the general solution of the homogeneous equation
(62)
L
(
d
d
t
)
x
(
t
)
=
0
is given by the formula
(63)
x
(
t
)
=
X
e
t
T
c
,
c
∈
ℂ
n
l
.Before we give the main result of this section, we introduce some notations. Suppose(64)
J
1
=
X
1
-
1
S
1
X
1
,
J
2
=
X
2
-
1
S
2
X
2
,
J
3
=
X
3
-
1
S
3
X
3
are the Jordan forms of S
1, S
2, S
3, respectively. Write
(65)
X
=
(
X
1
X
2
X
3
)
,
J
=
J
1
⊕
J
2
⊕
J
3
.Theorem 8.
Let(
S
1
,
S
2
,
S
3
) be a complete triple of L
(
λ
)
=
0, where L
(
λ
)
=
I
λ
3
+
A
2
λ
2
+
A
1
λ
+
A
0. Then every solution of
(66)
L
(
d
d
t
)
u
=
f
is given by
(67)
u
(
t
)
=
e
S
1
t
c
1
+
e
S
2
t
c
2
+
e
S
3
t
c
3
+
X
∫
a
t
‍
e
(
t
-
s
)
J
(
Y
1
Y
2
Y
3
)
f
(
s
)
d
s
,
t
∈
[
a
,
b
]
,
for some c
1
,
c
2
,
c
3
∈
ℂ
n, where
(68)
(
Y
2
Y
3
)
=
(
S
2
-
S
1
S
3
-
S
1
S
2
2
-
S
1
2
S
3
2
-
S
1
2
)
-
1
(
0
X
3
-
1
)
,
Y
1
=
-
X
1
-
1
X
2
Y
2
-
X
1
-
1
X
3
Y
3
.
In particular, every solution of
(69)
L
(
d
d
t
)
u
=
0
has the form
(70)
u
(
t
)
=
e
S
1
t
c
1
+
e
S
2
t
c
2
+
e
S
3
t
c
3
,
where c
1
,
c
2
,
c
3
∈
ℂ
n.Proof.
Using Lemma7, we take
(71)
Y
=
(
Y
1
Y
2
Y
3
)
=
X
-
1
(
I
I
I
S
1
S
2
S
3
S
1
2
S
2
2
S
3
2
)
-
1
(
0
0
I
)
.
Then (
X
,
J
,
Y
) is a standard triple; it is easy to get the conclusion.Remark 9.
The above theorem and its proof are motivated by those in [5] (cf. also Sections 2.4 and 2.5 in [2]).



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*Source: 493606-2014-04-27.xml*