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

Abstract

This paper considers a completion problem of a nonsingular 2×2 block matrix over the real quaternion algebra ℍ: Let 𝑚₁,𝑚₂,𝑛₁,𝑛₂ be nonnegative integers, 𝑚₁+𝑚₂=𝑛₁+𝑛₂=𝑛>0, and 𝐴₁₂∈ℍ^𝑚1×𝑛2,𝐴₂₁∈ℍ^𝑚2×𝑛1,𝐴₂₂∈ℍ^𝑚2×𝑛2,𝐵₁₁∈ℍ^𝑛1×𝑚1 be given. We determine necessary and sufficient conditions so that there exists a variant block entry matrix 𝐴₁₁∈ℍ^𝑚1×𝑛1 such that 𝐴=^𝐴11𝐴12_𝐴21𝐴22∈ℍ^𝑛×𝑛 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 ℝ. Suppose 𝑚₁,𝑚₂,𝑛₁,𝑛₂ are nonnegative integers, 𝑚₁+𝑚₂=𝑛₁+𝑛₂=𝑛>0,𝐴₁₁∈ℝ^𝑚1×𝑛1,𝐴₁₂∈ℝ^𝑚1×𝑛2,𝐴₂₁∈ℝ^𝑚2×𝑛1, and 𝐵₂₂∈ℝ^𝑛2×𝑚2. Determine a matrix 𝐴₂₂∈ℝ^𝑚2×𝑛2 such that 𝐴𝐴=¹¹𝐴₁₂𝐴₂₁𝐴₂₂ (1) is nonsingular and 𝐵₂₂ is the lower right block of a partitioning of 𝐴⁻¹. This problem has the form of 𝐴₁₁𝐴₁₂𝐴₂₁?⁻¹=???𝐵₂₂, (2) 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 ℝ.

Some other particular forms for 2×2 block matrices over ℝ have also been examined (see, e.g., [3]), such as 𝐴₁₁𝐴₁₂𝐴₂₁?⁻¹=𝐵₁₁?,𝐴??₁₁???⁻¹=???𝐵₂₂,𝐴₁₁??𝐴₂₂⁻¹=?𝐵₁₂𝐵₂₁?. (3)

The real quaternion matrices play a role in computer science, quantum physics, and so on (e.g., [4–6]). Quaternion matrices are receiving much attention as witnessed recently (e.g., [7–9]). 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: ℍ=𝑎₀+𝑎₁𝑖+𝑎₂𝑗+𝑎₃𝑖𝑘∣₂=𝑗²=𝑘²=𝑖𝑗𝑘=−1and𝑎₀,𝑎₁,𝑎₂,𝑎₃∈ℝ. (4)

Problem 1. Suppose 𝑚₁,𝑚₂,𝑛₁,𝑛₂ are nonnegative integers, 𝑚₁+𝑚₂=𝑛₁+𝑛₂=𝑛>0, and 𝐴₁₂∈ℍ^𝑚1×𝑛2,𝐴₂₁∈ℍ^𝑚2×𝑛1,𝐴₂₂∈𝑅^𝑚2×𝑛2,𝐵₁₁∈ℍ^𝑛1×𝑚1. Find a matrix 𝐴₁₁∈ℍ^𝑚1×𝑛1 such that 𝐴𝐴=¹¹𝐴₁₂𝐴₂₁𝐴₂₂∈ℍ^𝑛×𝑛 (5) is nonsingular, and 𝐵₁₁ is the upper left block of a partitioning of 𝐴⁻¹. That is ?𝐴₁₂𝐴₂₁𝐴₂₂⁻¹=𝐵₁₁?,?? (6) where ℍ^𝑚×𝑛 denotes the set of all 𝑚×𝑛 matrices over ℍ and 𝐴⁻¹ denotes the inverse matrix of 𝐴.

Throughout, over the real quaternion algebra ℍ, we denote the identity matrix with the appropriate size by 𝐼, the transpose of 𝐴 by 𝐴^𝑇, the rank of 𝐴 by 𝑟(𝐴), the conjugate transpose of 𝐴 by 𝐴^∗=(𝐴)^𝑇, a reflexive inverse of a matrix 𝐴 over ℍ by 𝐴⁺ which satisfies simultaneously 𝐴𝐴⁺𝐴=𝐴 and 𝐴⁺𝐴𝐴⁺=𝐴⁺. Moreover, 𝐿_𝐴=𝐼−𝐴⁺𝐴,𝑅_𝐴=𝐼−𝐴𝐴⁺, where 𝐴⁺ is an arbitrary but fixed reflexive inverse of 𝐴. Clearly, 𝐿_𝐴 and 𝑅_𝐴 are idempotent, and each is a reflexive inverse of itself. ℛ(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 ℍ, 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 𝐴∈ℍ^𝑚×𝑛 be of rank 𝑟. Then there exist unitary quaternion matrices 𝑈∈ℍ^𝑚×𝑚 and 𝑉∈ℍ^𝑛×𝑛 such that 𝐷𝑈𝐴𝑉=^𝑟0,00 (7) where 𝐷_𝑟=diag(𝑑₁,…,𝑑_𝑟) and the 𝑑_𝑗's are the positive singular values of 𝐴.

Let ℍ^𝑛_𝑐 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 ℍ^𝑛_𝑐 of dimension 𝑛(𝐴). We have the following lemma.

Lemma 2. Let 𝐴₁₁𝐴₁₂𝐴₂₁𝐴₂₂ (8) be a partitioning of a nonsingular matrix 𝐴∈ℍ^𝑛×𝑛, and let 𝐵₁₁𝐵₁₂𝐵₂₁𝐵₂₂ (9) be the corresponding (i.e., transpose) partitioning of 𝐴⁻¹. Then 𝑛(𝐴₁₁)=𝑛(𝐵₂₂).

Proof. It is readily seen that 𝐵₂₂𝐵₂₁𝐵₁₂𝐵₁₁,𝐴₂₂𝐴₂₁𝐴₁₂𝐴₁₁ (10) 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 𝐴₁₁𝐵₁₂+𝐴₁₂𝐵₂₂=0, (11) we have 𝐴₁₁𝐵₁₂𝐹=0. (12)
From 𝐴₂₁𝐵₁₂+𝐴₂₂𝐵₂₂=𝐼, (13) we have 𝐴₂₁𝐵₁₂𝐹=𝐹. (14) It follows that the rank 𝑟(𝐵₁₂𝐹)≥𝑐. In view of (12), this implies 𝑛𝐴₁₁≥𝑟𝐵₁₂𝐹≥𝑐=𝑛𝐵₂₂. (15)
Thus 𝑛𝐴₁₁=𝑛𝐵₂₂. (16)

Lemma 3 (see [10]). Let 𝐴∈ℍ^𝑚×𝑛, 𝐵∈ℍ^𝑝×𝑞, 𝐷∈ℍ^𝑚×𝑞 be known and 𝑋∈ℍ^𝑛×𝑝 unknown. Then the matrix equation 𝐴𝑋𝐵=𝐷 (17) is consistent if and only if 𝐴𝐴⁺𝐷𝐵⁺𝐵=𝐷. (18) In that case, the general solution is 𝑋=𝐴⁺𝐷𝐵⁺+𝐿_𝐴𝑌₁+𝑌₂𝑅_𝐵, (19) where 𝑌₁, 𝑌₂ are any matrices with compatible dimensions over ℍ.

By Lemma 1, let the singular value decomposition of the matrix 𝐴₂₂ and 𝐵₁₁ in Problem 1 be 𝐴₂₂=𝑄𝑅Λ000^∗,𝐵₁₁=𝑈𝑉Σ000^∗, (20)(21) where Λ=diag(𝜆₁,𝜆₂,…,𝜆_𝑠) is a positive diagonal matrix, 𝜆_𝑖≠0(𝑖=1,…,𝑠) are the singular values of 𝐴₂₂, 𝑠=𝑟(𝐴₂₂), Σ=diag(𝜎₁,𝜎₂,…,𝜎_𝑟) is a positive diagonal matrix, 𝜎_𝑖≠0(𝑖=1,…,𝑟) are the singular values of 𝐵₁₁ and 𝑟=𝑟(𝐵₁₁).

𝑄=(𝑄₁𝑄₂)∈ℍ^𝑚2×𝑚2, 𝑅=(𝑅₁𝑅₂)∈ℍ^𝑛2×𝑛2, 𝑈=(𝑈₁𝑈₂)∈ℍ^𝑛1×𝑛1, 𝑉=(𝑉₁𝑉₂)∈ℍ^𝑚1×𝑚1 are unitary quaternion matrices, where 𝑄₁∈ℍ^𝑚2×𝑠, 𝑅₁∈ℍ^𝑛2×𝑠, 𝑈₁∈ℍ^𝑛1×𝑟,  and  𝑉₁∈ℍ^𝑚1×𝑟.

Theorem 4. Problem 1 has a solution if and only if the following conditions are satisfied:(a)𝑟^𝐴12_𝐴22=𝑛₂,(b)𝑛₂−𝑟(𝐴₂₂)=𝑚₁−𝑟(𝐵₁₁),  that is 𝑛₂−𝑠=𝑚₁−𝑟,(c)ℛ(𝐴₂₁𝐵₁₁)⊂ℛ(𝐴₂₂),(d)ℛ(𝐴^∗₁₂𝐵^∗₁₁)⊂ℛ(𝐴^∗₂₂).
In that case, the general solution has the form of 𝐴₁₁=𝐵⁺₁₁+𝐴₁₂𝑅Λ⁻¹𝑄^∗₁𝐴₂₁𝑈₁𝑉Σ0𝐻−^∗₂𝐴₁₂𝑅₂⁻¹×𝑉^∗𝐵⁺₁₁+𝑌−𝑌𝐵₁₁𝐵⁺₁₁, (22) where 𝐻 is an arbitrary matrix in ℍ^{(𝑛2−𝑠)×𝑟} and 𝑌 is an arbitrary matrix in ℍ^𝑚1×𝑛1.

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 𝐴₂₁𝐵₁₁+𝐴₂₂𝐵₂₁𝐵=0,₁₁𝐴₁₂+𝐵₁₂𝐴₂₂=0, (23) so that (c) and (d) are satisfied.
By (11), we have 𝑟𝐴₂₂+𝑛𝐴₂₂=𝑛₂,𝑟𝐵₁₁+𝑛𝐵₁₁=𝑚₁. (24) From Lemma 2, Notice that 𝐴₁₁𝐴₁₂𝐴₂₁𝐴₂₂ is the corresponding partitioning of 𝐵⁻¹, we have 𝑛𝐵₁₁=𝑛𝐴₂₂, (25) implying that (b) is satisfied.
Conversely, from (c), we know that there exists a matrix 𝐾∈ℍ^𝑛2×𝑚1 such that 𝐴₂₁𝐵₁₁=𝐴₂₂𝐾. (26) Let 𝐵₂₁=−𝐾. (27) From (20), (21), and (26), we have 𝐴₂₁𝑈𝑉Σ000^∗=𝑄𝑅Λ000^∗𝐾. (28)
It follows that 𝑄^∗𝐴₂₁𝑈𝑉Σ000^∗𝑉=𝑄^∗𝑄𝑅Λ000^∗𝐾𝑉. (29)
This implies that 𝑄^∗₁𝐴₂₁𝑈₁𝑄^∗₁𝐴₂₁𝑈₂𝑄^∗₂𝐴₂₁𝑈₁𝑄^∗₂𝐴₂₁𝑈₂=𝑅Σ000Λ000^∗1𝐾𝑉₁𝑅^∗₁𝐾𝑉₂𝑅^∗₂𝐾𝑉₁𝑅^∗₂𝐾𝑉₂. (30)
Comparing corresponding blocks in (30), we obtain 𝑄^∗₂𝐴₂₁𝑈₁=0. (31)
Let 𝑅^∗𝐾𝑉=𝐾. From (29), (30), we have Λ𝐾=⁻¹𝑄^∗₁𝐴₂₁𝑈₁Σ0𝐻𝐾₂₂,𝐻∈ℍ^{(𝑛2−𝑠)×𝑟},𝐾₂₂∈ℍ^{(𝑛2−𝑠)×(𝑚1−𝑟)}. (32)
In the same way, from (d), we can obtain 𝑉^∗₁𝐴₁₂𝑅₂=0. (33) Notice that 𝐴₁₂𝐴₂₂ in (a) is a full column rank matrix. By (20), (21), and (33), we have 0𝑄^∗𝑉^∗0𝐴₁₂𝐴₂₂𝑉𝑅=Λ000_∗1𝐴₁₂𝑅₁𝑉^∗₁𝐴₁₂𝑅₂𝑉^∗₂𝐴₁₂𝑅₁𝑉^∗₂𝐴₁₂𝑅₂, (34) so that 𝑛₂𝐴=𝑟¹²𝐴₂₂=𝑟0𝑄^∗𝑉^∗0𝐴₁₂𝐴₂₂𝑅𝑉=𝑟Λ000_∗1𝐴₁₂𝑅₁𝑉^∗₁𝐴₁₂𝑅₂𝑉^∗₂𝐴₁₂𝑅₁𝑉^∗₂𝐴₁₂𝑅₂=𝑟(Λ)+𝑟𝑉^∗₂𝐴₁₂𝑅₂=𝑠+𝑟𝑉^∗₂𝐴₁₂𝑅₂. (35) It follows from (b) and (35) that 𝑉^𝑇₂𝐴₁₂𝑅₂ is a full column rank matrix, so it is nonsingular.
From 𝐴𝐵=𝐼, we have the following matrix equation: 𝐴₁₁𝐵₁₁+𝐴₁₂𝐵₂₁=𝐼, (36) that is 𝐴₁₁𝐵₁₁=𝐼−𝐴₁₂𝐵₂₁,𝐼∈ℍ^𝑚1×𝑚1, (37) where 𝐵₁₁, 𝐴₁₂ were given, 𝐵₂₁=−𝐾 (from (27)). By Lemma 3, the matrix equation (37) has a solution if and only if 𝐼−𝐴₁₂𝐵₂₁𝐵⁺₁₁𝐵₁₁=𝐼−𝐴₁₂𝐵₂₁. (38) By (21), (27), (32), and (33), we have that (38) is equivalent to: 𝐼+𝐴₁₂𝐾𝑉Σ⁻¹0𝑈00^∗𝑈𝑉Σ000^∗=𝐼+𝐴₁₂𝐾. (39) We simplify the equation above. The left hand side reduces to (𝐼+𝐴₁₂𝐾)𝑉₁𝑉^∗₁ and so we have 𝐴₁₂𝐾𝑉₁𝑉^∗₁−𝐴₁₂𝐾=𝐼−𝑉₁𝑉^∗₁. (40) So, 𝐴₁₂𝑅𝐾𝑉^∗𝑉₁𝑉^∗₁−𝐴₁₂𝑅𝐾𝑉^∗=𝑉₁𝑉₂𝑉^∗₁𝑉^∗₂−𝑉₁𝑉^∗₁. (41) This implies that 𝐴₁₂𝑅𝐾𝑉^∗₁𝑉₁𝑉^∗₂𝑉₁𝑉^∗₁−𝐴₁₂𝑅𝐾𝑉^∗₁𝑉^∗₂=𝑉₂𝑉^∗₂, (42) so that 𝐴₁₂𝑅𝐾𝐼0𝑉^∗₁−𝐴₁₂𝑅𝐾𝑉^∗₁𝑉^∗₂=𝑉₂𝑉^∗₂. (43) So, −𝐴₁₂𝑅𝐾0𝑉^∗₂=𝑉₂𝑉^∗₂, (44) and hence, −𝐴₁₂𝑅₁𝐴₁₂𝑅₂Λ⁻¹𝑄^∗₁𝐴₂₁𝑈₁Σ0𝐻𝐾₂₂0𝑉^∗₂=𝑉₂𝑉^∗₂. (45) Finally, we obtain 𝐴₁₂𝑅₂𝐾₂₂𝑉^∗₂=−𝑉₂𝑉^∗₂. (46) Multiplying both sides of (46) by 𝑉^∗ from the left, considering (33) and the fact that 𝑉^∗₂𝐴₁₂𝑅₂ is nonsingular, we have 𝐾₂₂=−𝑉^∗₂𝐴₁₂𝑅₂⁻¹. (47) From Lemma 3, (38), (47), Problem 1 has a solution and the general solution is 𝐴₁₁=𝐵⁺₁₁+𝐴₁₂𝑅Λ⁻¹𝑄^∗₁𝐴₂₁𝑈₁𝑉Σ0𝐻−^∗₂𝐴₁₂𝑅₂⁻¹×𝑉^∗𝐵⁺₁₁+𝑌−𝑌𝐵₁₁𝐵⁺₁₁, (48) where 𝐻 is an arbitrary matrix in ℍ^{(𝑛2−𝑠)×𝑟} and 𝑌 is an arbitrary matrix in ℍ^𝑚1×𝑛1.

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: 𝐴₁₂=12+𝑗2𝑘1−𝑘1+2𝑗,𝐴₂₁=32+121𝑖−21𝑗−2𝑘121𝑗+2𝑘32+12𝑖,𝐴₂₂=2𝑖2𝑗𝑘,𝐵₁₁=.1𝑖𝑗𝑘 (49)
It is easy to show that (c), (d) are satisfied, and that 𝑛₂𝐴=𝑟¹²𝐴₂₂𝑛=2,₂−𝑟𝐴₂₂=𝑚₁−𝑟𝐵₁₁=0, (50) so (a), (b) are satisfied too. Therefore, we have 𝐵⁺₁₁=12−12𝑗−121𝑖−2𝑘,𝐴₂₂=𝑄𝑅Λ000^∗,𝐵₁₁=𝑈𝑉Σ000^∗, (51) where 1𝑄=√21𝑖2√𝑗𝑘,Λ=0√202,1𝑅=1001,𝑈=√2,√1𝑖𝑗𝑘Σ=0√202,𝑉=.1001 (52) We also have 𝑄₁=1√21𝑖𝑗𝑘,𝑅₁=,𝑈1001₁=1√21𝑖𝑗𝑘,𝑉₁=.1001 (53)
By Theorem 4, for an arbitrary matrices 𝑌∈ℍ^2×2, we have𝐴₁₁=𝐵⁺₁₁+𝐴₁₂𝑅Λ⁻¹𝑄^∗₁𝐴₂₁𝑈₁Σ𝑉^∗𝐵⁺₁₁+𝑌−𝑌𝐵₁₁𝐵⁺₁₁=32+141𝑗+4𝑘34+143𝑖−2𝑗121−𝑖+41𝑗−4𝑘14−341𝑖−2,𝑗−𝑘 (54) it follows that 3𝐴=2+141𝑗+4𝑘34+143𝑖−21𝑗2+𝑗2𝑘121−𝑖+41𝑗−4𝑘14−341𝑖−21𝑗−𝑘−𝑘1+2𝑗32+121𝑖−21𝑗−21𝑘2𝑖21𝑗+2𝑘32+12,𝐴𝑖2𝑗𝑘₋₁=31𝑖−1−1𝑗𝑘0−1−10412−34𝑗1−1−121−𝑖2−121𝑖−2.𝑗−𝑘 (55) 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).