Extending an Almost Complete Pair of Matrices to a Complete Triple

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.


Definition 1.
We say that the -tuple ( 1 , 2 , . . . , ) of complex square matrices of order is complete if the Vandermonde matrix ( 1 , 2 , . . . , ) is invertible. When = 2 and = 3, we call a complete -tuple of matrices a complete pair and a complete triple, respectively.
Suppose that ( 1 , 2 ) is complete. Take = 2 − 1 . Then is invertible. If we take then − 1 , − 2 , form a complete pair of right divisors of ( ) (cf. Lemma 2.4 in [2]). The concept of complete pair was 2 The Scientific World Journal 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.
is not a complete pair. Is it possible for us to find 3 and a cubic matrix polynomial ( ) such that ( 1 , 2 , 3 ) is a complete triple and are right divisors of ( )?
is a complete triple, then is of full column rank. Note that if 1 , 2 ∈ (C) and the rank of 1 − 2 is less than − 1, we cannot always find 3 such that ( 1 , 2 , 3 ) is a complete triple. For example, let 1 , 2 ∈ (C) be defined as follows: where 1 , 2 are square matrices of orders − 2 and 2, respectively, and 2 − 1 is in the form If rank( ) = 1, then the last two columns of ( 2 − 1 ) are linearly dependent. For any 3 ∈ (C), cannot be invertible. If rank( ) = 2, then is of full column rank. From the condition that is of full column rank, we cannot conclude that rank( 2 − 1 ) ≥ − 1.
After the above discussion, we give the following definition.
Definition 2. For two matrices 1 , 2 ∈ (C), one says they form an almost complete pair if the matrix is of full column rank and the rank of 1 − 2 is − 1. For simplification, one says ( 1 , 2 ) is an almost complete pair of matrices of order .
The following is the main result of this paper.
We prove this theorem in Section 2. In Section 3, we partially answer the question mentioned before and give an application of the main result to differential equations.

Proof of the Main Theorem
In this section, we prove Theorem 3. Note that there is an invertible matrix such that −1 ( 2 − 1 ) is the Jordan normal form of 2 − 1 . Let = ⊕ ⊕ . Then the condition that is invertible. So from the beginning, we can assume that 2 − 1 is in Jordan normal form. First we prove a lemma.
The Scientific World Journal 3 Proof. By some computations, we have The last column of is (0, 0, . . . , 0, , 0) , which is a 2 -dimensional column vector. By the definition of almost complete pair, is not a zero vector.
Now we prove the main theorem.
Proof. From now on, we write 1 = (
Without loss of generality, we can assume 2 − 1 is in the Jordan normal form.
Case 2. Now we consider the case that 2 − 1 has the Jordan normal form where ̸ = 0, 1 ≤ ≤ . If 1 ̸ = 0, then we take If 1 = 0 and ̸ = 0 (1 < < ), then we take 3 such that In any case when | | is sufficiently large, we can get that is invertible.
Case 3. Now we consider the case that 0 is not a single root of characteristic polynomial of 2 − 1 ; that is, 2 − 1 has the Jordan normal form: The Scientific World Journal By Lemma 4, 12 ̸ = 0. Take We can conclude that is invertible. Firstly, we prove the case that the normal form itself is If 1 ̸ = 0, then we take ) .

(43)
When | | is sufficiently large, we have being invertible. If = − 1, that is to say, the last column of 2 2 − 2 1 is (0, 0, . . . , 0, −1 ) , then we take ) . (44) We have where the first entry of the last column of is invertible when | | is sufficiently large. Now we prove the general case that 2 − 1 has the Jordan normal form: where ̸ = 0, 1 ≤ ≤ . If there is some ̸ = 0, + 1 ≤ ≤ , we assume that (0) is a matrix of order − . For any square matrix of order − , we can construct a square matrix of order − such that is invertible. Take If = 0, for all + 1 ≤ ≤ , there must be some such that 1 ≤ ≤ , ̸ = 0. Let = , where is the matrix which is obtained by exchanging the columns + 1 and of × identity matrix. Then det = − det and The Scientific World Journal 5 where the square matrix of order − + 1 has the form Let be a matrix such that −1 has the Jordan normal form . Then we can consider the matrix = ( ⊕ −1 ) ( ⊕ ). Note that has the form For any square matrix , we can construct a square matrix of order + 1 such that is invertible. Take 3 = 1 + ⊕ −1 . Then we can conclude that is invertible. The proof is complete.

Applications
For given monic matrix polynomials 1 ( ), 2 ( ), . . . , ( ), the construction of a common multiple of them was presented in [2]. Suppose that ( 1 , 2 ) is an almost complete pair. By Theorem 3 in Section 1, we can find 3 ∈ (C) such that ( 1 , 2 , 3 ) is a complete triple; that is, is invertible. Let Then we can conclude that By Theorem 9.11 of [2], there exists a monic matrix polynomial ( ) with degree 3 which is a common multiple of − 1 , − 2 , − 3 .
Now we recall definitions of standard pair and standard triple for monic polynomials.
Definition 5 (see [2]). A pair of matrices ( , ), where is × and is × , is called a standard pair for the monic matrix polynomial ( ) if the following conditions are satisfied: If is in Jordan normal form, we call ( , ) a Jordan pair for the monic matrix polynomial ( ).
Definition 6 (see [2]). A triple of matrices ( , , ), where is × , is × , and is × , is called a standard triple for the monic matrix polynomial ( ) if ( , ) is a standard pair for ( ), and = ( . . . Now we recall the explicit formulas for solutions of the following differential equation: Lemma 7 (Theorem 2.9 in [2]). The general solution of (60) is given by the formula where ( , , ) is a standard triple of ( ) and ∈ C is arbitrary. In particular, the general solution of the homogeneous equation is given by the formula Before we give the main result of this section, we introduce some notations. Suppose In particular, every solution of ( ) = 0 (69) has the form where 1 , 2 , 3 ∈ C .
Proof. Using Lemma 7, we take = ( Then ( , , ) 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]).