Some New Algebraic and Topological Properties of the Minkowski Inverse in the Minkowski Space

Abstract

We introduce some new algebraic and topological properties of the Minkowski inverse 𝐴^⊕ of an arbitrary matrix 𝐴∈𝑀_𝑚,𝑛 (including singular and rectangular) in a Minkowski space 𝜇. Furthermore, we show that the Minkowski inverse 𝐴^⊕ in a Minkowski space and the Moore-Penrose inverse 𝐴⁺ in a Hilbert space are different in many properties such as the existence, continuity, norm, and SVD. New conditions of the Minkowski inverse are also given. These conditions are related to the existence, continuity, and reverse order law. Finally, a new representation of the Minkowski inverse 𝐴^⊕ is also derived.

1. Introduction and Preliminaries

In this work, we consider matrices over the field of complex numbers as ℂ and real numbers as ℝ. The set of 𝑚-by-𝑛 complex matrices is denoted by 𝑀_𝑚,𝑛(ℂ)=ℂ^𝑚×𝑛. For simplicity, we write 𝑀_𝑚,𝑛 instead of 𝑀_𝑚,𝑛(ℂ) or 𝑀_𝑚,𝑛(ℝ), and when 𝑚=𝑛, we write 𝑀_𝑛 instead of 𝑀_𝑛,𝑛. The notations 𝐴^𝑡, 𝐴^∗, 𝐴^∼,𝑟(𝐴), 𝑅(𝐴), 𝑁(𝐴), tr(𝐴), det(𝐴), ‖𝐴‖₂, 𝐴⁺, 𝐴^⊕, and 𝜎(𝐴) stand for the transpose, conjugate transpose, 𝜇-symmetric, rank, range, null space, trace, determinant, Frobenius norm, Moore-Penrose inverse, Minkowski inverse, and set of all eigenvalues of a matrix 𝐴, respectively.

The Moore-Penrose inverse is widely used in perturbation theory, singular systems, neural network problems, least-squares problems, optimization problems, and many other subjects [1–8]. The Moore-Penrose inverse of an arbitrary matrix 𝐴∈𝑀_𝑚,𝑛 is defined to be the unique solution of the following four matrix equations [3, 4, 8–10]: ()𝐴𝑋𝐴=𝐴,𝑋𝐴𝑋=𝑋,𝐴𝑋^∗=𝐴𝑋,(𝑋𝐴)^∗=𝑋𝐴, (1) and it is often denoted by 𝑋=𝐴⁺. Note that if we designate any matrix 𝑋 that satisfying the 𝑖th matrix equation (𝑖∈{1,2,3,4}) in (1) is called the 𝑖-inverse and denoted by 𝐴^(𝑖).

The Moore-Penrose inverse can be explicitly expressed by the singular value decomposition (SVD) due to van Loan [11]. For any matrix 𝐴∈𝑀_𝑚,𝑛 with 𝑟(𝐴)=𝑟, there exist unitary matrices 𝑈∈𝑀_𝑚 and 𝑉∈𝑀_𝑛 satisfying 𝑈^∗𝑈=𝐼_𝑚 and 𝑉^∗𝑉=𝐼_𝑛 such that 𝐴=𝑈𝑉𝐷000^∗, (2) where 𝐷=diag(𝛿₁,𝛿₂,…,𝛿_𝑟)∈𝑀_𝑟, 𝛿₁≥𝛿₂≥⋯≥𝛿_𝑟>0, and 𝛿²_𝑖 (𝑖=1,2,…,𝑟) are the nonzero eigenvalues of 𝐴^∗𝐴. Then, the Moore-Penrose inverse can be represented as 𝐴⁺𝐷=𝑉⁻¹0𝑈00^∗. (3)

Some algebraic properties concerning the null space, range, rank, continuity, and some representations of some types of the generalized inverses of a given matrix over complex and real fields are widely studied by many researchers [12–16]. The Minkowski inverse 𝐴^⊕ of an arbitrary matrix 𝐴∈𝑀_𝑚,𝑛 is one of the important generalized inverses for solving matrix equations in the Minkowski space 𝜇 with respect to the generalized reflection antisymmetric matrix 𝐴^∼ [17]. Some methods such as iterative, Borel summable, Euler-Knopp summable, Newton-Raphson, and Tikhonov’s methods are used for representation and computation of the Minkowski inverse 𝐴^⊕ in the Minkowski space 𝜇 [18, 19].

By letting ℂ^𝑛 be the space of complex 𝑛-tuples, we will index the components of a complex vector in ℂ^𝑛 from 0 to 𝑛−1; that is, 𝑢=(𝑢₀,𝑢₁,𝑢₂,…,𝑢_𝑛−1). In addition to that, let 𝐺 be the Minkowski metric tensor defined by 𝐺𝑢=(𝑢₀,−𝑢₁,−𝑢₂,…,−𝑢_𝑛−1). Clearly, the Minkowski metric matrix is defined by [18, 20] 𝐺=100−𝐼_𝑛−1∈𝑀_𝑛, (4) and 𝐺^∗=𝐺 and 𝐺²=𝐼_𝑛.

In [21, 22], the Minkowski inner product on ℂ^𝑛 is defined by (𝑢,𝑣)=[𝑢,𝐺𝑣], where [⋅,⋅] denotes the conventional Hilbert (unitary) space inner product. The space with the Minkowski inner product is called a Minkowski space and is denoted by 𝜇. For any square matrix 𝐴∈𝑀_𝑛 and vectors 𝑥 and 𝑦∈ℂ^𝑛, we have 𝐴𝑥,𝑦=𝐴𝑥,𝐺𝑦=𝑥,𝐴^∗𝐺𝑦=𝑥,𝐺𝐺𝐴^∗𝐺𝑦=𝑥,𝐺𝐴^∼𝑦=𝑥,𝐴^∼𝑦, (5) where 𝐴^∼=𝐺𝐴^∗𝐺 is called the Minkowski conjugate transpose of 𝐴 in the Minkowski space 𝜇. Naturally, the matrix 𝐴∈𝑀_𝑛 is called 𝜇-symmetric in the Minkowski space 𝜇 if 𝐴=𝐴^∼. Now, it is easy to show that 𝐴 is 𝜇-symmetric if and only if 𝐴𝐺 is Hermitian if and only if 𝐺𝐴 is Hermitian. Also, it is easy to verify that 𝐺⁻¹=𝐺 and 𝜎(𝐴^∼)=𝜎(𝐴). More generally, if 𝐴∈𝑀_𝑚,𝑛, then the Minkowski conjugate transpose of 𝐴 is defined by 𝐴^∼=𝐺₁𝐴^∗𝐺₂ (where𝐺₁ and 𝐺₂ are the Minkowski metric matrices of orders 𝑛×𝑛 and 𝑚×𝑚, resp.), and it satisfies the following algebraic properties as in the following result.

Lemma 1. Let 𝐴∈𝑀_𝑚,𝑛. Then, the following one given:(i)𝐴^∼ is unique,(ii)(𝐴^∼)^∼=𝐴,(iii)(𝐴𝐵)^∼=𝐵^∼𝐴^∼,(iv)~-cancellation rule 𝐴^∼𝐴𝑋=𝐴^∼𝐴𝑌⇒𝐴𝑋=𝐴𝑌,(v)𝑟(𝐴)=𝑟(𝐴^∼),(vi)𝑅(𝐴^∗)=𝑅(𝐴^∼),(vii)𝑁(𝐴^∗)=𝑁(𝐴^∼).

Finally, a matrix 𝐴∈𝑀_𝑚,𝑛 is said to be a range symmetric in unitary space (or equivalently 𝐴 is said to be EP) if 𝑁(𝐴^∗)=𝑁(𝐴). For further properties of EP matrices, one may refer to [3, 4, 10, 11].

In this paper, some algebraic properties concerning the rank, range, existence, uniqueness, continuity, and reverse order law of the Minkowski inverse 𝐴^⊕ are introduced. The relationships between 𝐴^⊕ and 𝐴^∼ are also discussed. Furthermore, a new representation of 𝐴^⊕ related to the full-rank factorization of the matrix 𝐴 is derived, and new conditions for the existence and continuity of 𝐴^⊕ are also given.

2. Some Algebraic Properties of the Minkowski Inverse

In this section, we derive some attractive algebraic properties and the reverse order law property of the Minkowski inverse in a Minkowski space.

The Minkowski inverse of an arbitrary matrix 𝐴∈𝑀_𝑚,𝑛 (including singular and rectangular), analogous to the Moore-Penrose inverse, is defined as follows.

Definition 2. Let 𝐴∈𝑀_𝑚,𝑛 be any matrix in the Minkowski space 𝜇. Then, the Minkowski inverse of 𝐴 is the matrix 𝐴^⊕∈𝑀_𝑛,𝑚 which satisfies the following four matrix equations: 𝐴𝐴^⊕𝐴=𝐴,𝐴^⊕𝐴𝐴^⊕=𝐴^⊕,𝐴𝐴^⊕∼=𝐴𝐴^⊕,𝐴^⊕𝐴^∼=𝐴^⊕𝐴. (6)

Theorem 3. Let 𝐴∈𝑀_𝑚,𝑛 be any matrix in the Minkowski space 𝜇. Then, the Minkowski inverse 𝐴^⊕ satisfies the following properties:(i)(𝐴^⊕)^∼=(𝐴^∼)^⊕,(ii)𝐴^⊕ is a unique matrix,(iii)(𝐴^⊕)^⊕=𝐴,(iv)𝐴𝐴^⊕ and 𝐴^⊕𝐴 are idempotents (i.e., 𝐴𝐴^⊕ and 𝐴^⊕𝐴 are projectors on 𝑅(𝐴) and 𝑅(𝐴^⊕), resp.),(v)𝐴^⊕𝐴+𝛼𝐼_𝑛 and 𝐴𝐴^⊕+𝛼𝐼_𝑚 are invertible matrices, where 𝛼>0,(vi)(𝐴^∼𝐴)^⊕𝐴^∼=𝐴^⊕=𝐴^∼(𝐴𝐴^∼)^⊕,(vii)𝑅(𝐴^⊕)=𝑅(𝐴^∼).

Proof. (i)Since the following four matrix equations are satisfied: 𝐴^∼𝐴^⊕∼𝐴^∼=𝐴𝐴^⊕𝐴^∼=𝐴^∼,𝐴^⊕∼𝐴^∼𝐴^⊕∼=𝐴^⊕𝐴𝐴^⊕∼=𝐴^⊕∼,𝐴^∼𝐴^⊕∼∼=𝐴^⊕∼∼(𝐴^∼)^∼=𝐴^⊕𝐴=𝐴^⊕𝐴^∼=𝐴^∼𝐴^⊕∼,(𝐴^⊕)^∼𝐴^∼∼=((𝐴^∼)^∼)𝐴^⊕∼∼=𝐴𝐴^⊕=𝐴𝐴^⊕∼=𝐴^⊕∼𝐴^∼, (7) then, by (6), we get the result.
(ii)Let 𝐺₁ and 𝐺₂ be two Minkowski metric tensors such that 𝐴^⊕₁ and 𝐴^⊕₂ are two Minkowski inverses of a matrix 𝐴; then, by using Lemma 1 and Theorem 3(i), we have 𝐴^⊕₁=𝐴^⊕₁𝐴𝐴^⊕₁=𝐴^⊕₁𝐴𝐴^⊕₁^∼=𝐴^⊕₁𝐴^⊕₁^∼𝐴^∼=𝐴^⊕₁𝐴^⊕₁^∼𝐴^∼𝐴^⊕₂^∼𝐴^∼=𝐴^⊕₁𝐴𝐴^⊕₁^∼𝐴𝐴^⊕₂^∼=𝐴^⊕₁𝐴𝐴^⊕₁𝐴𝐴^⊕₂=𝐴^⊕₁𝐴𝐴^⊕₂=𝐴^⊕₁𝐴𝐴^⊕₂𝐴𝐴^⊕₂=𝐴^⊕₁𝐴^∼𝐴^⊕₂𝐴^∼𝐴^⊕₂=𝐴^∼𝐴^⊕₁^∼𝐴^∼𝐴^⊕₂^∼𝐴^⊕₂=𝐴^∼𝐴^⊕₂^∼𝐴^⊕₂=𝐴^⊕₂𝐴^∼𝐴^⊕₂=𝐴^⊕₂𝐴𝐴^⊕₂=𝐴^⊕₂. (8)
This means that 𝐴^⊕ is a unique matrix.
(iii) It follows by applying the four matrix equations in (6).
(iv) By using the matrix equations in (6), we have (𝐴𝐴^⊕)²=𝐴(𝐴^⊕𝐴𝐴^⊕)=𝐴𝐴^⊕ and (𝐴^⊕𝐴)²=𝐴^⊕(𝐴𝐴^⊕𝐴)=𝐴^⊕𝐴.
(v) Since 𝐴𝐴^⊕ is an idempotent matrix, then eigenvalues of 𝐴𝐴^⊕ are 0 or 1. That is, det(𝐴𝐴^⊕+𝛼𝐼_𝑚)=0⇔𝛼=0 or 𝛼=−1. So, for all 𝛼>0, we have det(𝐴𝐴^⊕+𝛼𝐼_𝑚)≠0 (i.e., 𝐴^⊕𝐴+𝛼𝐼_𝑛is an invertible matrix). Similarly, we can prove that 𝐴𝐴^⊕+𝛼𝐼_𝑚 is also an invertible matrix.
(vi) Since 𝐴^∼𝐴(𝐴^∼𝐴)^⊕𝐴^∼𝐴=𝐴^∼𝐴, then, from ~-cancellation, we have 𝐴(𝐴^∼𝐴)^⊕𝐴^∼𝐴=𝐴. (9) Now, by using the four matrix equations in (6), Theorem 3, and Lemma 1, we have (𝐴^∼𝐴)^⊕𝐴^∼𝐴(𝐴^∼𝐴)^⊕𝐴^∼=(𝐴^∼𝐴)^⊕(𝐴^∼𝐴)(𝐴^∼𝐴)^⊕𝐴^∼=(𝐴^∼𝐴)^⊕𝐴^∼,𝐴(𝐴^∼𝐴)^⊕𝐴^∼∼=𝐴(𝐴^∼𝐴)^⊕∼𝐴^∼=𝐴((𝐴^∼𝐴)^∼)^⊕𝐴^∼=𝐴(𝐴^∼𝐴)^⊕𝐴^∼,(𝐴^∼𝐴)^⊕𝐴^∼𝐴^∼=𝐴^∼𝐴(𝐴^∼𝐴)^⊕=𝐴^∼𝐴(𝐴^∼𝐴)^⊕∼=(𝐴^∼𝐴)^⊕𝐴^∼𝐴. (10)(11)(12) Consequently, (9), (10), and (11) show that (𝐴^∼𝐴)^⊕𝐴^∼=𝐴^⊕.
(vii) Equations (9) and (10) show that 𝑟(𝐴^⊕)=𝑟(𝐴)=𝑟(𝐴^∼). Now, by applying Theorem 3(vi), we have 𝑅(𝐴^⊕)⊂𝑅(𝐴^∼); then, the equality holds.

The reverse order law property for the Moore-Penrose inverse of the product of two matrices is investigated by many researchers; one may refer to [23]. Analogous to Greville’s conditions that were stated in [6], we reached the following result.

Theorem 4. Let 𝐴∈𝑀_𝑚,𝑛 and 𝐵∈𝑀_𝑛.𝑝 be two matrices in the Minkowski space 𝜇 such that the Minkowski inverses 𝐴^⊕, 𝐵^⊕, and (𝐴𝐵)^⊕ exist. Then, (𝐴𝐵)^⊕=𝐵^⊕𝐴^⊕ if and only if 𝑅(𝐴^∼𝐴𝐵)⊂𝑅(𝐵) and 𝑅(𝐵𝐵^∼𝐴^∼)⊂𝑅(𝐴^∼).

Proof. Since 𝐵𝐵^⊕ is a projector on 𝑅(𝐵) as in Theorem 3(iv), then 𝐵𝐵^⊕𝐴^∼𝐴𝐵=𝐴^∼𝐴𝐵. (13) Now, by Definition 2 and Theorem 3, we have 𝐴^⊕𝐴𝐵𝐵^∼𝐴^∼=𝐵𝐵^∼𝐴^∼. (14) Taking the Minkowski conjugate transpose of the two sides of (13), we have 𝐵^∼𝐴^∼𝐴𝐵𝐵^⊕=𝐵^∼𝐴^∼𝐴. (15) Multiplying the right side and the left side of (15) by 𝐴^⊕ and ((𝐴𝐵)^∼)^⊕, respectively, we have ((𝐴𝐵)^∼)^⊕(𝐴𝐵)^∼(𝐴𝐵)𝐵^⊕𝐴^⊕=((𝐴𝐵)^∼)^⊕(𝐴𝐵)^∼𝐴𝐴^⊕. (16) Since 𝑅(𝐴𝐵)⊂𝑅(𝐴)=𝑅(𝐴𝐴^⊕), then we have 𝐴𝐵𝐵^⊕𝐴^⊕=(𝐴𝐵)(𝐴𝐵)^⊕∼(𝐴𝐵)𝐵^⊕𝐴^⊕=(𝐴𝐵)(𝐴𝐵)^⊕∼𝐴𝐴^⊕=(𝐴𝐵)(𝐴𝐵)^⊕. (17) Also, multiplying the right side and the left side of (14) by ((𝐴𝐵)^∼)^⊕ and 𝐵^⊕, respectively, and applying Theorem 3 and Definition 2 for (𝐴𝐵)^⊕, we have 𝐵^⊕𝐴^⊕𝐴𝐵(𝐴𝐵)^⊕(𝐴𝐵)=𝐵^⊕𝐵𝐵^∼𝐴^∼((𝐴𝐵)^∼)^⊕. (18) Since 𝐵^⊕𝐵 is a projector on 𝑅(𝐵^∼), we have 𝐵^⊕𝐴^⊕𝐴𝐵=𝐵^∼𝐴^∼((𝐴𝐵)^∼)^⊕=((𝐴𝐵))^⊕(𝐴𝐵). (19) Equations (17) and (19) imply that 𝐵^⊕𝐴^⊕ satisfies the first, third, and fourth equations in (6). Finally, by taking the Minkowski conjugate transpose of the two sides of the first and the second equations in (6) for matrices 𝐴 and 𝐵 and by using Theorem 3(vi), we have 𝐵^∼𝐴^∼=𝐵^∼𝐵𝐵^⊕𝐴^⊕𝐴𝐴^∼,𝐵^⊕𝐴^⊕=𝐵^⊕𝐵^⊕∼𝐵^∼𝐴^∼(𝐴𝐴^∼)^⊕. (20) This equation shows that 𝑟𝐵^⊕𝐴^⊕=𝑟(𝐵^∼𝐴^∼)=𝑟(𝐴𝐵)^∼=𝑟(𝐴𝐵). (21) Consequently, 𝐵^⊕𝐴^⊕ satisfies the second equation in (6).

3. Existence of the Minkowski Inverse

The Minkowski inverse of a matrix 𝐴 exists if and only if 𝑟(𝐴𝐴^∼)=𝑟(𝐴^∼𝐴)=𝑟(𝐴) [12]. In this section, we give some equivalent conditions for the existence and derive a new representation of the Minkowski inverse. If 𝐴∈𝑀_𝑚,𝑛 is a matrix of full row rank (column rank), then 𝐴𝐴^∗ and 𝐴^∗𝐴 are invertible matrices of orders 𝑚×𝑚 and 𝑛×𝑛, respectively, in a Hilbert (Euclidian) space. Here, in a Minkowski space, if we define ‖𝐴‖₀₂=(tr(𝐴𝐴^∼))^1/2, then the following example shows that 𝐴^∼𝐴 and 𝐴𝐴^∼ are, in general, not invertible matrices and also ‖𝐴‖₀₂≠‖𝐴‖₂.

Example 5. Let 𝐴=¹⁻¹¹₁₁₁. Then, 𝑟(𝐴)=2, and 𝐴^∼=𝐺₁𝐴^∗𝐺₂==,1000−1000−111−1111100−11−111−11𝐴𝐴^∼==.1−111111−111−11−1−111 (22) Note that 𝑟(𝐴𝐴^∼)=1 (i.e., det(𝐴𝐴^∼)=0), and hence 𝐴𝐴^∼ is not invertible. Also, ‖𝐴‖₂=tr(𝐴𝐴^∗)=√6, and ‖𝐴‖₀₂=0, which are not equal.

Lemma 6. Let 𝐴∈𝑀_𝑚,𝑛 and 𝐵∈𝑀_𝑛.𝑝 be two matrices. Then, the following are considered.(i)If 𝑟(𝐴)=𝑛, then 𝑟(𝐴𝐵)=𝑟(𝐵).(ii)If 𝑟(𝐵)=𝑛, then 𝑟(𝐴𝐵)=𝑟(𝐴).

Proof. (i) Since 𝑟(𝐴)=𝑛, then 𝐴 is a left invertible; thus, there exists a matrix 𝑋∈𝑀_𝑛,𝑚 such that 𝑋𝐴=𝐼_𝑛. Hence, 𝑟(𝐵)=𝑟(𝑋𝐴𝐵)≤𝑟(𝐴𝐵)≤𝑟(𝐵), which implies that 𝑟(𝐴𝐵)=𝑟(𝐵).Similarly, we can prove (ii).

Theorem 7. Let 𝐴=𝐵𝐶∈𝑀_𝑚,𝑛 be a rank factorization of rank 𝑟. Then, (𝐶𝐶^∼)⁻¹ and (𝐵^∼𝐵)⁻¹ exist if and only if 𝑟(𝐴𝐴^∼)=𝑟(𝐴^∼𝐴)=𝑟(𝐴).

Proof. Since 𝐴=𝐵𝐶, then 𝐵 and 𝐶 are of orders 𝑚×𝑟 and 𝑟×𝑛, respectively, and 𝑟(𝐴)=𝑟(𝐵)=𝑟(𝐶)=𝑟. Hence, 𝐴𝐴^∼=𝐵𝐶𝐺₁𝐶^∗𝐺𝐺𝐵^∗𝐺₂=𝐵𝐶𝐶^∼𝐵^∼, (23) where 𝐺, 𝐺₁, and 𝐺₂ are the Minkowski metric matrices of orders 𝑟  ×  𝑟, 𝑛  ×  𝑛, and 𝑚×𝑚, respectively. Since 𝐵 and 𝐵^∼ are matrices of full ranks (i.e., 𝑟(𝐵)=𝑟(𝐵^∼)=𝑟), then, by Lemma 6, we have 𝑟(𝐴𝐴^∼)=𝑟(𝐶𝐶^∼). Similarly, we can prove that 𝑟(𝐴^∼𝐴)=𝑟(𝐵^∼𝐵).
Now, since 𝐵^∼𝐵 and 𝐶𝐶^∼ are square matrices of order 𝑟  ×  𝑟, then they are invertible if and only if they are of rank 𝑟.

By applying the four matrix equations in (6), we can get a new representation of the Minkowski inverse as shown in the following result.

Theorem 8. Let 𝐴=𝐵𝐶∈𝑀_𝑚,𝑛 be a rank factorization of rank 𝑟. Then, 𝐴^⊕=𝐶^∼(𝐶𝐶^∼)⁻¹(𝐵^∼𝐵)⁻¹𝐵^∼. (24)

4. Some Topological Properties of the Minkowski Inverse

In this section, we establish some attractive topological properties and new conditions for the continuity of the Minkowski inverse in a Minkowski space.

It is known that, in normed algebra of bounded linear operators, the map of linear invertible operators associated with its usual inverse is continuous. The following example shows that this property is not valid in the Minkowski space.

Example 9. Let 𝐴_𝑛=¹⁰_01/𝑛 be a sequence of matrices for 𝑛∈ℕ; then, 𝐴^⊕_𝑛=¹⁰_0𝑛, lim_𝑛→∞𝐴_𝑛=¹⁰₀₀, and (lim_𝑛→∞𝐴_𝑛)^⊕=¹⁰₀₀. Note that lim_𝑛→∞𝐴^⊕_𝑛 does not exist. That is, for the map 𝑇, we have 𝑇lim_𝑛→∞𝐴_𝑛≠lim_𝑛→∞𝑇𝐴_𝑛. (25)
For ‖𝐴^⊕‖₀₂≠0, the following results are very important for finding the new conditions for the continuity of the Minkowski inverse of rectangular matrices in a Minkowski space.

Lemma 10. Let 𝐴∈𝑀_𝑚,𝑛. Then, for any 𝑥∈𝑅(𝐴^∼), ‖𝐴𝑥‖₀₂≥‖𝑥‖₀₂𝐴^⊕₀₂. (26)

Proof. By using Theorem 3((iv) and (vii)), then, for any 𝑥∈𝑅(𝐴^∼), we have 𝑥=𝐴^⊕𝐴𝑥. Thus, ‖𝑥‖₀₂=‖𝐴^⊕𝐴𝑥‖₀₂≤‖𝐴^⊕‖₀₂‖𝐴𝑥‖₀₂, and then we get the result.

Lemma 11. Let 𝐴 and 𝐸∈𝑀_𝑚,𝑛 such that‖𝐴^⊕‖₀₂≠0 and ‖𝐸‖₀₂<1/‖𝐴^⊕‖₀₂. Then, 𝑟(𝐴+𝐸)≥𝑟(𝐴). (27)

Proof. Suppose that 𝑟(𝐴)=𝑟 and {𝑣₁,…,𝑣_𝑟} are the basis of 𝑅(𝐴^∼). Then, the set {(𝐴+𝐸)(𝑣₁),…,(𝐴+𝐸)(𝑣_𝑟)} is a subset of 𝑅(𝐴+𝐸). Now, suppose that ∑^𝑟_𝑖=1𝛼_𝑖(𝐴+𝐸)(𝑣_𝑖)=0, for 𝛼_𝑖∈ℂ; then, 𝑥=∑^𝑟_𝑖=1𝛼_𝑖𝑣_𝑖≠0, and we have 0=^𝑟_𝑖=1𝛼_𝑖(𝐴+𝐸)𝑣_𝑖02=(𝐴+𝐸)^𝑟_𝑖=1𝛼_𝑖𝑣_𝑖02=‖𝐴𝑥+𝐸𝑥‖₀₂≥‖𝐴𝑥‖₀₂−‖𝐸‖₀₂‖𝑥‖₀₂>‖𝐴𝑥‖₀₂−‖𝑥‖₀₂𝐴^⊕₀₂. (28) Now, by using Lemma 10, we have 0>(‖𝑥‖₀₂/‖𝐴^⊕‖₀₂)−(‖𝑥‖₀₂/‖𝐴^⊕‖₀₂)=0, which is impossible, and thus 𝑥=0. As {𝑣₁,…,𝑣_𝑟} is linearly independent, it follows that 𝛼_𝑖=0 (for all 𝑖=1,…,𝑟). Consequently, 𝑟(𝐴+𝐸)≥𝑟=𝑟(𝐴).

Corollary 12. Let 𝐴 and 𝐵∈𝑀_𝑚,𝑛 such that ‖𝐴^⊕‖₀₂≠0,‖𝐵^⊕‖₀₂≠0, and ‖𝐴−𝐵‖₀₂<1/max{‖𝐴^⊕‖₀₂,‖𝐵^⊕‖₀₂}. Then, 𝑟(𝐴)=𝑟(𝐵). (29)

Proof. Set 𝐸=𝐴−𝐵, and since ‖𝐵^⊕‖₀₂≠0, then, by using Lemma 11, we have ‖𝐸‖₀₂<1/‖𝐵^⊕‖₀₂ which implies that 𝑟(𝐴)=𝑟(𝐵+𝐸)≥𝑟(𝐵). (30) Since ‖𝐴^⊕‖₀₂≠0, then we also have ‖𝐸‖₂<1/‖𝐴^⊕‖₀₂ which implies that 𝑟(𝐵)=𝑟(𝐴−𝐸)≥𝑟(𝐴). (31) Now, the result follows by using (30) and (31).

If 𝐴∈𝑀_𝑚,𝑛 and 𝜆₀ is the largest eigenvalue of 𝐴^∼𝐴, then ‖𝐴‖₀₂=𝜆₀ in the Minkowski space 𝜇. Here, if 𝑃 is the 𝜇-symmetric projector, then it is easy to show that 𝑃^⊕=𝑃. Since the eigenvalues of a projector are only 0 and 1, then we have ‖𝑃‖₀₂=‖𝑃^⊕‖₀₂=1, and by applying Corollary 12, we get the following result.

Corollary 13. Let 𝑃 and 𝑄 be two 𝜇-symmetric projectors such that ‖𝑃−𝑄‖₀₂<1. Then, 𝑟(𝑃)=𝑟(𝑄). (32)

Theorem 14. The matrix 𝐴∈𝑀_𝑚,𝑛 can be written by using SVD as in the form 𝐴=𝑈𝑆𝑉^∗ with 𝑈^∗𝑈=𝐼_𝑚, 𝑉^∗𝑉=𝐼_𝑛, and 𝑆 is a diagonal if and only if the following conditions hold:(i)𝜎(𝐴^∼𝐴) are nonnegative real numbers,(ii)𝐴^∼𝐴 is diagonalizable,(iii)𝑁(𝐴^∼𝐴)=𝑁(𝐴).

If only assumption (i) is violated but (ii) and (iii) hold of Theorem 14, then we can still get singular value decomposition (SVD). But in the Minkowski space, each of the assumptions can fail even if the other two hold. This is illustrated by the following three counterexamples [18].

Example 15. Let 𝐴=⁻¹¹₁₁. Then, 𝐴^∼=⁻¹⁻¹₋₁₋₁, and 𝐴𝐴^∼=⁰⁻²₂₀. Hence, 𝜎(𝐴^∼𝐴)={±2𝑖} which are not real numbers.

Example 16. Let 𝐴=^1.51_0.51. Then, 𝐴^∼=^1.5−0.5₋₁₁, and 𝐴𝐴^∼=²¹₋₁₀. Hence, 𝜎(𝐴^∼𝐴)={1} and cannot be diagonalized.

Example 17. Let 𝐴=¹⁻¹₋₁₁. Then, 𝐴^∼=¹¹₁₁, and 𝐴𝐴^∼=⁰⁰₀₀. Hence, 𝑁(𝐴^∼𝐴)≠𝑁(𝐴).

The following result gives the equivalent conditions for the continuity to be held for the Minkowski inverse of any rectangular matrix.

Theorem 18. Let (𝐴_𝑛)_𝑛∈ℕ∈𝑀_𝑚,𝑝 be a sequence of matrices such that lim_𝑛→∞𝐴_𝑛=𝐴. Then, lim_𝑛→∞𝐴^⊕_𝑛=𝐴^⊕=(lim_𝑛→∞𝐴_𝑛)^⊕ if and only if 𝑟(𝐴_𝑛)=𝑟(𝐴).

Proof. Suppose that lim_𝑛→∞𝐴^⊕_𝑛=𝐴^⊕=(lim_𝑛→∞𝐴_𝑛)^⊕, and set 𝐴_𝑛=𝐴+𝐸_𝑛such that lim_𝑛→∞𝐸_𝑛=0 and lim_𝑛→∞(𝐴+𝐸_𝑛)^⊕=𝐴^⊕. Then, lim_𝑛→∞𝐴+𝐸_𝑛𝐴+𝐸_𝑛^⊕=𝐴𝐴^⊕, (33) which means that there exists 𝑛₀ such that, for any 𝑛≥𝑛₀, 𝐴+𝐸_𝑛𝐴+𝐸_𝑛^⊕−𝐴𝐴^⊕₀₂<1. (34) Since (𝐴+𝐸_𝑛)(𝐴+𝐸_𝑛)^⊕ and 𝐴𝐴^⊕ are 𝜇-symmetric projectors, then, by Corollary 13, we have 𝑟𝐴+𝐸_𝑛=𝑟𝐴+𝐸_𝑛𝐴+𝐸_𝑛^⊕=𝑟𝐴𝐴^⊕=𝑟(𝐴). (35) Conversely, suppose that 𝐴 satisfies the SVD conditions as in Theorem 14; then, 𝐵=𝑈^∗𝐴𝑉=^𝐷𝑟0₀₀, where 𝐷_𝑟 is a diagonal matrix and 𝑈 and 𝑉 are unitary matrices. Suppose also that lim_𝑛→∞𝐸_𝑛=0 and 𝑟(𝐴+𝐸_𝑛)=𝑟(𝐴)=𝑟 for any 𝑛≥𝑛₀. Now, set 𝐹_𝑛=𝑈^∗𝐸_𝑛𝐹𝑉=^(𝑛)11𝐹^(𝑛)₁₂𝐹^(𝑛)₂₁𝐹^(𝑛)₂₂. (36) Then, 𝑟𝐵+𝐹_𝑛=𝑟𝐴+𝐸_𝑛=𝑟,𝐵^⊕=𝑉^∼𝐴^⊕𝑈^∗∼,𝐵+𝐹_𝑛^⊕=𝑉^∼𝐴+𝐸_𝑛^⊕𝑈^∗∼. (37) Since lim_𝑛→∞𝐹_𝑛=0, then, for ‖𝐷_𝑟⁻¹‖₀₂≠0 and ‖𝐴^⊕‖₀₂≠0, we have 𝐹_𝑛02<1𝐷_𝑟⁻¹₀₂=1𝐴^⊕₀₂. (38) But ‖𝐹_𝑛‖₀₂≥sup_𝑖,𝑗{‖𝐹^(𝑛)_𝑖𝑗‖₀₂}, and then, by Lemma 11, we have 𝑟𝐷_𝑟+𝐹^(𝑛)₁₁≥𝑟. (39) On the other hand, 𝑟=𝑟𝐵+𝐹_𝑛≥𝑟𝐷_𝑟+𝐹^(𝑛)₁₁≥𝑟. (40) Now, by using (39) and (40), we get that 𝑟(𝐷_𝑟+𝐹^(𝑛)₁₁)=𝑟, which means that (𝐷_𝑟+𝐹^(𝑛)₁₁)⁻¹ exists. Also by using the Schur complement, we have 𝐵+𝐹_𝑛=𝐷_𝑟+𝐹^(𝑛)₁₁𝐹^(𝑛)₁₂𝐹^(𝑛)₂₁𝐹^(𝑛)₂₁𝐷_𝑟+𝐹^(𝑛)₁₁⁻¹𝐹^(𝑛)₁₂=𝐼_𝑟𝑀𝐷_𝑟+𝐹^(𝑛)₁₁𝐼_𝑟𝑁, (41) where 𝑀=𝐹^(𝑛)₂₁(𝐷_𝑟+𝐹^(𝑛)₁₁)⁻¹ and 𝑁=(𝐷_𝑟+𝐹^(𝑛)₁₁)⁻¹𝐹^(𝑛)₁₂. Now, if 𝑀^∼𝑀≠−𝐼_𝑟 and 𝑁𝑁^∼≠−𝐼_𝑟, then, by Definition 2, we can see that 𝐵+𝐹_𝑛^⊕=𝐼_𝑟𝑁^∼𝐼_𝑟+𝑀^∼𝑀𝐷_𝑟+𝐹^(𝑛)₁₁𝐼_𝑟+𝑁𝑁^∼−1×𝐼_𝑟𝑀^∼ (42) is the Minkowski inverse of 𝐵+𝐹_𝑛. Since lim_𝑛→∞𝐹_𝑛=0, then lim_𝑛→∞𝐹^(𝑛)₁₁=lim_𝑛→∞𝐹^(𝑛)₂₁=lim_𝑛→∞𝐹^(𝑛)₁₂=0. Therefore, lim_𝑛→∞(𝐷_𝑟+𝐹^(𝑛)₁₁)⁻¹=𝐷_𝑟, lim_𝑛→∞𝑀=0=lim_𝑛→∞𝑀^∼, and lim_𝑛→∞𝑁=0=lim_𝑛→∞𝑁^∼. Now, by using the fact that the map which transforms an invertible matrix to its inverse is continuous, consequently, we find that lim_𝑛→∞𝐵+𝐹_𝑛^⊕=𝐼_𝑟0𝐷_𝑟⁻¹𝐼_𝑟0=𝐷_𝑟⁻¹000=𝐵^⊕, (43) which completes the proof of Theorem 18.

5. Conclusion

Several attractive properties and conditions of the Minkowski inverse 𝐴^⊕ in the Minkowski space 𝜇 are presented. In our opinion, it is worth extending these properties and establishing some necessary and sufficient conditions for the reverse order rule of the weighted Minkowski inverse 𝐴^⊕_𝑀,𝑁 in the Minkowski space 𝜇 of two and multiple matrix products.

Acknowledgment

The authors express their sincere thanks to the referees for the careful reading of the paper and several helpful suggestions.