Volume 2013, Issue 1 271978
Research Article
Open Access

Completing a 2 × 2 Block Matrix of Real Quaternions with a Partial Specified Inverse

Yong Lin

Yong Lin

Department of Mathematics, Shanghai University, Shanghai 200444, China shu.edu.cn

School of Mathematics and Statistics, Suzhou University, Suzhou 234000, China usts.edu.cn

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Qing-Wen Wang

Corresponding Author

Qing-Wen Wang

Department of Mathematics, Shanghai University, Shanghai 200444, China shu.edu.cn

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First published: 10 April 2013
Citations: 4
Academic Editor: K. Sivakumar

Abstract

This paper considers a completion problem of a nonsingular 2 × 2 block matrix over the real quaternion algebra : Let m1,   m2,   n1,   n2 be nonnegative integers, m1 + m2 = n1 + n2 = n > 0, and be given. We determine necessary and sufficient conditions so that there exists a variant block entry matrix such that is nonsingular, and B11 is the upper left block of a partitioning of A−1. The general expression for A11 is also obtained. Finally, a numerical example is presented to verify the theoretical findings.

1. Introduction

The problem of completing a block-partitioned matrix of a specified type with some of its blocks given has been studied by many authors. Fiedler and Markham [1] considered the following completion problem over the real number field . Suppose m1,   m2,   n1,   n2 are nonnegative integers, , and . Determine a matrix such that
()
is nonsingular and B22 is the lower right block of a partitioning of A−1. This problem has the form of
()
and the solution and the expression for A22 were obtained in [1]. Dai [2] considered this form of completion problems with symmetric and symmetric positive definite matrices over .
Some other particular forms for 2 × 2 block matrices over have also been examined (see, e.g., [3]), such as
()
The real quaternion matrices play a role in computer science, quantum physics, and so on (e.g., [46]). Quaternion matrices are receiving much attention as witnessed recently (e.g., [79]). Motivated by the work of [1, 10] and keeping such applications of quaternion matrices in view, in this paper we consider the following completion problem over the real quaternion algebra:
()

Problem 1. Suppose m1,   m2,   n1,   n2 are nonnegative integers, m1 + m2 = n1 + n2 = n > 0, and . Find a matrix such that

()
is nonsingular, and B11 is the upper left block of a partitioning of A−1. That is
()
where m×n denotes the set of all m × n matrices over and A−1 denotes the inverse matrix of A.

Throughout, over the real quaternion algebra , we denote the identity matrix with the appropriate size by I, the transpose of A by AT, the rank of A by r(A), the conjugate transpose of A by , a reflexive inverse of a matrix A over by A+ which satisfies simultaneously AA+A = A and A+AA+ = A+. Moreover, LA = IA+A,   RA = IAA+, where A+ is an arbitrary but fixed reflexive inverse of A. Clearly, LA and RA are idempotent, and each is a reflexive inverse of itself. (A) denotes the right column space of the matrix A.

The rest of this paper is organized as follows. In Section 2, we establish some necessary and sufficient conditions to solve Problem 1 over , and the general expression for A11 is also obtained. In Section 3, we present a numerical example to illustrate the developed theory.

2. Main Results

In this section, we begin with the following lemmas.

Lemma 1 (singular-value decomposition [9]). Let Am×n be of rank r. Then there exist unitary quaternion matrices Um×m and Vn×n such that

()
where Dr = diag (d1, …, dr) and the dj′s are the positive singular values of A.

Let denote the collection of column vectors with n components of quaternions and A be an m × n quaternion matrix. Then the solutions of Ax = 0 form a subspace of of dimension n(A). We have the following lemma.

Lemma 2. Let

()
be a partitioning of a nonsingular matrix An×n, and let
()
be the corresponding (i.e., transpose) partitioning of A−1. Then n(A11) = n(B22).

Proof. It is readily seen that

()
are inverse to each other, so we may suppose that n(A11) < n(B22).

If n(B22) = 0, necessarily n(A11) = 0 and we are finished. Let n(B22) = c > 0, then there exists a matrix F with c right linearly independent columns, such that B22F = 0. Then, using

()
we have
()

From

()
we have
()
It follows that the rank r(B12F) ≥ c. In view of (12), this implies
()

Thus

()

Lemma 3 (see [10].)Let Am×n, Bp×q, Dm×q be known and Xn×p unknown. Then the matrix equation

()
is consistent if and only if
()
In that case, the general solution is
()
where Y1, Y2 are any matrices with compatible dimensions over .

By Lemma 1, let the singular value decomposition of the matrix A22 and B11 in Problem 1 be
()
()
where Λ = diag (λ1, λ2, …, λs) is a positive diagonal matrix, λi ≠ 0  (i = 1, …, s) are the singular values of A22, s = r(A22), Σ = diag (σ1, σ2, …, σr) is a positive diagonal matrix, σi ≠ 0  (i = 1, …, r) are the singular values of B11 and r = r(B11).

, , , are unitary quaternion matrices, where , , ,  and  .

Theorem 4. Problem 1 has a solution if and only if the following conditions are satisfied:

  • (a)

    ,

  • (b)

    n2r(A22) = m1r(B11),  that is n2s = m1r,

  • (c)

    (A21B11) ⊂ (A22),

  • (d)

    .

In that case, the general solution has the form of

()
where H is an arbitrary matrix in and Y is an arbitrary matrix in .

Proof. If there exists an m1 × n1 matrix A11 such that A is nonsingular and B11 is the corresponding block of A−1, then (a) is satisfied. From AB = BA = I, we have that

()
so that (c) and (d) are satisfied.

By (11), we have

()
From Lemma 2, Notice that is the corresponding partitioning of B−1, we have
()
implying that (b) is satisfied.

Conversely, from (c), we know that there exists a matrix such that

()
Let
()
From (20), (21), and (26), we have
()

It follows that

()

This implies that

()

Comparing corresponding blocks in (30), we obtain

()

Let . From (29), (30), we have

()

In the same way, from (d), we can obtain

()
Notice that in (a) is a full column rank matrix. By (20), (21), and (33), we have
()
so that
()
It follows from (b) and (35) that is a full column rank matrix, so it is nonsingular.

From AB = I, we have the following matrix equation:

()
that is
()
where B11, A12 were given, B21 = −K (from (27)). By Lemma 3, the matrix equation (37) has a solution if and only if
()
By (21), (27), (32), and (33), we have that (38) is equivalent to:
()
We simplify the equation above. The left hand side reduces to and so we have
()
So,
()
This implies that
()
so that
()
So,
()
and hence,
()
Finally, we obtain
()
Multiplying both sides of (46) by V* from the left, considering (33) and the fact that is nonsingular, we have
()
From Lemma 3, (38), (47), Problem 1 has a solution and the general solution is
()
where H is an arbitrary matrix in and Y is an arbitrary matrix in .

3. An Example

In this section, we give a numerical example to illustrate the theoretical results.

Example 5. Consider Problem 1 with the parameter matrices as follows:

()

It is easy to show that (c), (d) are satisfied, and that

()
so (a), (b) are satisfied too. Therefore, we have
()
where
()
We also have
()

By Theorem 4, for an arbitrary matrices Y2×2, we have

()
it follows that
()
The results verify the theoretical findings of Theorem 4.

Acknowledgments

The authors would like to give many thanks to the referees and Professor K. C. Sivakumar for their valuable suggestions and comments, which resulted in a great improvement of the paper. This research was supported by Grants from the Key Project of Scientific Research Innovation Foundation of Shanghai Municipal Education Commission (13ZZ080), the National Natural Science Foundation of China (11171205), the Natural Science Foundation of Shanghai (11ZR1412500), the Discipline Project at the corresponding level of Shanghai (A. 13-0101-12-005), and Shanghai Leading Academic Discipline Project (J50101).

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