Some Results for the Drazin Inverses of the Sum of Two Matrices and Some Block Matrices

Abstract

We give a formula of (𝑃+𝑄)^𝐷 under the conditions 𝑃²𝑄+𝑄𝑃𝑄=0, 𝑃³𝑄=0, and 𝑃𝑄𝑃𝑄=0. Then, we apply it to give some expressions for the Drazin inverse of block matrix 𝑀=^𝐴𝐵_𝐶𝐷 (𝐴 and 𝐷 are square matrices) under some conditions, generalizing some recent results in the literature. Finally, numerical examples are given to illustrate our results.

1. Introduction

Let ℂ^𝑚×𝑛 denote the set of 𝑚×𝑛 complex matrices. The Drazin inverse of 𝐴∈ℂ^𝑛×𝑛 is the unique matrix 𝐴^𝐷, satisfying the following equation: 𝐴^𝑘+1𝐴^𝐷=𝐴^𝑘,𝐴^𝐷𝐴𝐴^𝐷=𝑋,𝐴𝐴^𝐷=𝐴^𝐷𝐴, (1) where 𝑘=ind(𝐴) is the index of 𝐴, the smallest nonnegative integer for which rank(𝐴^𝑘+1)=rank(𝐴^𝑘) (see [1]). In particular, when ind(𝐴)=1, the Drazin inverse of 𝐴 is called the group inverse of 𝐴. If 𝐴 is nonsingular, it is clear that ind(𝐴)=0 and 𝐴^𝐷=𝐴⁻¹. Throughout this paper, we denote by 𝐴^𝜋=𝐼−𝐴𝐴^𝐷 and define 𝐴⁰=𝐼, where 𝐼 is the identity matrix with proper sizes.

The importance of the Drazin inverse and its applications to singular differential equations and difference equations, Morkov chains and iterative methods, cryptography, numerical analysis, to structured matrices, and perturbation bounds for the relative eigenvalue problems can be found in [2–5].

In 1958, Drazin [6] gave a result of (𝑃+𝑄)^𝐷, where 𝑃 and 𝑄 are square matrices, and proved that (𝑃+𝑄)^𝐷=𝑃^𝐷+𝑄^𝐷when𝑃𝑄=𝑄𝑃=0. (2)

In 2001, Hartwig et al. [7] derived a result of (𝑃+𝑄)^𝐷 when 𝑃𝑄=0. In 2005, Castro-González [8] provided the representation of (𝑃+𝑄)^𝐷 when 𝑃^𝐷𝑄=0, 𝑃𝑄^𝐷=0, and 𝑄^𝜋𝑃𝑄𝑃^𝜋=0. In 2008, Castro-González et al. [9] determined the result of (𝑃+𝑄)^𝐷 when 𝑃²𝑄=0 and 𝑃𝑄²=0. In 2009, Martínez-Serrano and Castro-González [10] derived a formula of (𝑃+𝑄)^𝐷 when 𝑃²𝑄=0 and 𝑄²=0. In 2011, Yang and Liu [11] established some expressions of (𝑃+𝑄)^𝐷 when 𝑃𝑄²=0 and 𝑃𝑄𝑃=0, and in 2012, Bu et al. [12] got the explicit representation of (𝑃+𝑄)^𝐷 when 𝑃²𝑄=0, 𝑄²𝑃=0, 𝑃³𝑄=0, and 𝑄𝑃𝑄=0, 𝑄𝑃²𝑄=0, respectively. Other related results have been studied in [4, 13–19].

On the other hand, a related topic is to discuss a representation of the Drazin inverse of block matrix 𝑀=^𝐴𝐵_𝐶𝐷∈ℂ^𝑛×𝑛, where 𝐴 and 𝐷 are square matrices. Campbell and Meyer Jr. [2] first proposed an open problem to find an explicit formula of the Drazin inverse of block matrix 𝑀=^𝐴𝐵_𝐶𝐷∈ℂ^𝑛×𝑛, (where 𝐴 and 𝐷 are square matrices), in terms of 𝐴, 𝐵, 𝐶, and 𝐷. To find the Drazin inverse of (𝑃+𝑄) and 𝑀 in terms of 𝑃, 𝑄, 𝑃^𝐷, and 𝑄^𝐷 and 𝐴, 𝐵, 𝐶, and 𝐷, respectively, without side conditions is quite complicated and it has not been solved till now. However, many papers have been studied some special cases of this open problem and gave the representations for the Drazin inverse of (𝑃+𝑄) and 𝑀 under some conditions. Here, we list some cases of Drazin inverse of block matrix 𝑀:(i)𝐵𝐶=0, 𝐷𝐶=0 (or 𝐵𝐷=0) and 𝐷 is nilpotent (see [5]);(ii)𝐵𝐶𝐴=0, 𝐵𝐷=0, and 𝐷𝐶=0 (or 𝐵𝐶 is nilpotent) (see [9]);(iii)𝐵𝐶𝐴=0, 𝐵𝐶𝐵=0, 𝐷𝐶𝐴=0, and 𝐷𝐶𝐵=0 (see [11]);(iv)𝐵𝐶=0, 𝐵𝐷=0, and 𝐷𝐶=0 (see [20]);(v)𝐵𝐶=0 and 𝐷𝐶=0 (see [21]);(vi)𝐵𝐶𝐴=0, 𝐵𝐶𝐵=0, 𝐴𝐵𝐷=0, and 𝐶𝐵𝐷=0 (see [22]).

The generalized Schur complement of 𝐴 in 𝑀, which is stated as 𝑆=𝐷−𝐶𝐴^𝐷𝐵, is very important to find the Drazin inverse of 𝑀, when the generalized Schur complement is either zero or nonsingular. Martínez-Serrano and Castro-González [10] gave results under the conditions 𝐴²𝐴^𝜋𝐵=0, 𝐶𝐴𝐴^𝜋𝐵=0, and 𝐵𝐶𝐴^𝜋𝐵=0 and generalized that Schur complement is equal to zero. Also, they derived the expressions of 𝑀^𝐷 under the assumptions 𝐴²𝐴^𝜋𝐵=0, 𝐶𝐴𝐴^𝜋𝐵=0, and 𝐶𝐴^𝜋𝐵=0 and generalized Schur complement is nonsingular. The Drazin inverse of 𝑀 has been studied in [10, 23], when generalized Schur complement is equal to zero and also has been studied in [5, 10, 24, 25], when the generalized Schur complement is nonsingular. Some representations for the Drazin inverse of 𝑀 when the generalized Schur complement is nonsingular, including generalizations of the above mentioned results, will be derived in Section 4 under some conditions.

This paper is organized as follows. In Section 2, some helpful lemmas will be given. In Section 3, we give the explicit formula of (𝑃+𝑄)^𝐷 under the conditions 𝑃³𝑄=0, 𝑃²𝑄+𝑄𝑃𝑄=0, and 𝑃𝑄𝑃𝑄=0 and also give a numerical example to demonstrate our result. In Section 4, we use our result to find the Drazin inverse of block matrix 𝑀=^𝐴𝐵_𝐶𝐷 and also to find the expression for 𝑀^𝐷 when the generalized Schur complement is nonsingular, which can be regarded as the generalizations of some results given in [5, 20]. Finally, in Section 5, we give two numerical examples to illustrate our results of block matrices.

2. Some Lemmas

In order to prove the main results, first we need the following lemmas.

Lemma 1 (see [1]). Let 𝐴∈ℂ^𝑚×𝑛, and let 𝐵∈ℂ^𝑛×𝑚. Then, (𝐴𝐵)^𝐷=𝐴((𝐵𝐴)²)^𝐷𝐵.

Lemma 2 (see [7]). Let 𝑃,𝑄∈ℂ^𝑛×𝑛. If 𝑃𝑄=0, then (𝑃+𝑄)^𝐷=𝑄^𝜋𝑡−1_𝑖=0𝑄^𝑖𝑃^𝐷𝑖+1+^𝑡−1_𝑖=0𝑄^𝐷𝑖+1𝑃^𝑖𝑃^𝜋, (3) where 𝑡=max{ind(𝑃),ind(𝑄)}.

Lemma 3 (see [26]). Let 𝑀₁=^𝐴0_𝐶𝐵, and let 𝑀₂=^𝐵𝐶_0𝐴, where 𝐴 and 𝐵 are square matrices with ind(𝐴)=𝑟 and ind(𝐵)=𝑠. Then, 𝑀^𝐷₁=𝐴^𝐷0𝑋𝐵^𝐷,𝑀^𝐷₂=𝐵^𝐷𝑋0𝐴^𝐷, (4) where 𝑋=∑^𝑟−1_𝑖=0(𝐵^𝐷)^𝑖+2𝐶𝐴^𝑖𝐴^𝜋+∑^𝑠−1_𝑖−0𝐵^𝜋𝐵^𝑖𝐶(𝐴^𝐷)^𝑖+2−𝐵^𝐷𝐶𝐴^𝐷.

Lemma 4 (see [24]). Let 𝑀=^𝐴𝐵_𝐶𝐷∈𝐶^𝑛×𝑛 (𝐴 and 𝐷 are square matrices). If 𝑆=𝐷−𝐶𝐴^𝐷𝐵 is nonsingular, 𝐴^𝜋𝐵=0, and 𝐶𝐴^𝜋=0, then 𝑀^𝐷=𝐴^𝐷+𝐴^𝐷𝐵𝑆⁻¹𝐶𝐴^𝐷−𝐴^𝐷𝐵𝑆⁻¹−𝑆⁻¹𝐶𝐴^𝐷𝑆⁻¹. (5)

3. The Drazin Inverse of the Sum of Two Matrices

In this section, we first give the formula for the Drazin inverse of 𝑃+𝑄 under some conditions.

Theorem 5. Let 𝑃,𝑄∈ℂ^𝑛×𝑛. If 𝑃³𝑄=0, 𝑃²𝑄+𝑄𝑃𝑄=0, and 𝑃𝑄𝑃𝑄=0, then (𝑃+𝑄)^𝐷=𝑃^𝐷+𝑄^𝐷+𝑄𝑋𝑃+𝑃𝑄^𝐷2+𝑃²𝑄^𝐷3+𝑃𝑄𝑃^𝐷3+𝑄𝑃𝑄^𝐷3+𝑃²𝑄𝑋𝑃^𝐷+𝑃²𝑄^𝐷𝑋𝑃+𝑃𝑄²𝑋𝑃^𝐷+𝑃𝑄𝑄^𝐷𝑋𝑃+𝑄𝑃𝑄𝑋𝑃^𝐷+𝑄𝑃𝑄^𝐷𝑋𝑃, (6) where 𝑋=^𝑡−1_𝑖=0𝑄^𝐷2𝑖+4(𝑃+𝑄)𝑃^2𝑖𝑃^𝜋+^𝑡−1_𝑖=0𝑄^𝜋𝑄^2𝑖(𝑃+𝑄)𝑃^𝐷2𝑖+4−𝑄^𝐷2(𝑃+𝑄)𝑃^𝐷2,𝑡=maxind𝑃²,ind𝑄². (7)

Proof. Using Lemma 1, (𝐴𝐵)^𝐷=𝐴((𝐵𝐴)²)^𝐷𝐵, we have (𝑃+𝑄)^𝐷=𝑃𝐼𝐼𝑄^𝐷=𝐼𝑄𝑃𝑃𝑄𝐼𝑄^2𝐷𝑃𝐼=𝑃𝐼𝑄²+𝑃𝑄𝑃²𝑄+𝑃𝑄²𝑃+𝑄𝑄²+𝑃𝑄^𝐷𝑃𝐼. (8) Let 𝑃𝑀=²+𝑃𝑄𝑃²𝑄+𝑃𝑄²𝑃+𝑄𝑄²+𝑃𝑄=𝐸+𝐹, (9) where 𝑃𝐸=²0𝑃+𝑄𝑄²,𝐹=𝑃𝑄𝑃²𝑄+𝑃𝑄²0𝑃𝑄. (10) From 𝑃³𝑄=0, 𝑃²𝑄+𝑄𝑃𝑄=0 and 𝑃𝑄𝑃𝑄=0, we get 𝐸𝐹=0, and 𝐹²=0. Then, applying Lemma 2, we obtain 𝑀^𝐷=𝐸^𝐷+𝐹𝐸^𝐷2. (11) By applying Lemma 3, we have 𝐸^𝐷=𝑃^𝐷20𝑋𝑄^𝐷2, (12) where 𝑋=^𝑡−1_𝑖=0𝑄^𝐷2𝑖+4(𝑃+𝑄)𝑃^2𝑖𝑃^𝜋+^𝑡−1_𝑖=0𝑄^𝜋𝑄^2𝑖(𝑃+𝑄)𝑃^𝐷2𝑖+4−𝑄^𝐷2(𝑃+𝑄)𝑃^𝐷2,𝑡=maxind𝑃²,ind𝑄². (13) Substituting (12) into (11), we get 𝑀^𝐷=𝑃^𝐷2+𝑃𝑄𝑃^𝐷4+𝑃²𝑄𝑋𝑃^𝐷2+𝑃²𝑄^𝐷𝑋+𝑃𝑄²𝑋𝑃^𝐷2+𝑃𝑄𝑄^𝐷𝑋𝑃²𝑄^𝐷3+𝑃𝑄^𝐷2𝑋+𝑃𝑄𝑋𝑃^𝐷2+𝑃𝑄^𝐷𝑋𝑄^𝐷2+𝑃𝑄^𝐷3. (14) Substituting (14) into (8), we obtain the result.

Similarly, we give a symmetrical form of Theorem 5.

Theorem 6. Let 𝑃,𝑄∈ℂ^𝑛×𝑛. If 𝑃𝑄³=0, 𝑃𝑄²+𝑃𝑄𝑃=0, and 𝑃𝑄𝑃𝑄=0, then (𝑃+𝑄)^𝐷=𝑃^𝐷+𝑄^𝐷+𝑄𝑋𝑃+𝑃^𝐷2𝑄+𝑃^𝐷3𝑄²+𝑃^𝐷3+𝑄𝑄𝑃_𝐷3𝑃𝑄+𝑄^𝐷𝑋𝑃𝑄²+𝑄^𝐷𝑋𝑃²𝑄+𝑄𝑋𝑃^𝐷𝑄²+𝑄𝑋𝑃^𝐷𝑃𝑄+𝑄^𝐷𝑋𝑃𝑄𝑃+𝑄𝑋𝑃^𝐷𝑄𝑃, (15) where 𝑋=^𝑡−1_𝑖=0𝑄^𝐷2𝑖+4(𝑃+𝑄)𝑃^2𝑖𝑃^𝜋+^𝑡−1_𝑖=0𝑄^𝜋𝑄^2𝑖(𝑃+𝑄)𝑃^𝐷2𝑖+4−𝑄^𝐷2(𝑃+𝑄)𝑃^𝐷2,𝑡=maxind𝑃²,ind𝑄². (16)

Next, we present a numerical example to illustrate Theorem 5. This numerical example describes neither the matrices 𝑃 and 𝑄 which do not satisfy the conditions of [10, Theorem 2.2] nor the conditions of [11, Theorem 2.1], but they satisfy the conditions of Theorem 5. Therefore, we can apply the formula given in Theorem 5 to obtain the Drazin inverse of 𝑃+𝑄.

Numerical Example. Consider the matrices 𝑃,𝑄∈ℂ^4×4, where 𝑃=1000000100000000,𝑄=0000001000101000. (17) Since 𝑃²𝑄=0, 𝑄²≠0 and 𝑃𝑄²=0, 𝑃𝑄𝑃≠0, we know that the conditions of Theorem  2.2 in [10] and Theorem 2.1 in [11] do not hold, respectively. But it satisfies 𝑃³𝑄=0, 𝑃²𝑄+𝑄𝑃𝑄=0, and 𝑃𝑄𝑃𝑄=0. Also, we have ind𝑃²=1,ind𝑄²𝑃=1,_𝐷=1000000000000000,𝑄^𝐷=,.0000001000100000𝑋=1000001000101000 (18) So, applying Theorem 5, we get (𝑃+𝑄)^𝐷=𝑃^𝐷+𝑄^𝐷+𝑃𝑄𝑃^𝐷3=.+𝑄𝑋𝑃1000101000101000 (19)

Remark 7. The above example shows that the conditions given in Theorem 5 are satisfied, but the conditions given in [10, 11] are not satisfied.

4. Drazin Inverse of Some Block Matrices

In this section, we apply our formula to give the representations for the Drazin inverse of block matrix 𝑀=^𝐴𝐵_𝐶𝐷 (𝐴 and 𝐷 are square matrices). First, we give the expression of 𝑀^𝐷 under the conditions 𝐴𝐵𝐶=0, 𝐵𝐷𝐶=0, 𝐷²𝐶=0, and 𝐶𝐵𝐶=0, which generalizes the results in [5, 20].

Theorem 8. Let 𝑀=^𝐴𝐵_𝐶𝐷 (𝐴 and 𝐷 are square matrices), such that ind(𝐴)=𝑟 and ind(𝐷)=𝑠. If 𝐴𝐵𝐶=0, 𝐵𝐷𝐶=0, 𝐷²𝐶=0, and 𝐶𝐵𝐶=0, then 𝑀^𝐷=𝑃^𝐷+𝑃00𝐶0^𝐷2+𝑃𝐵𝐶0𝐷𝐶0^𝐷3, (20) where 𝑃^𝐷𝑖=𝐴∑^{𝐷𝑖𝑖−1}_𝑗=0𝐴^𝐷𝑗𝑋𝐷^{𝐷𝑖−𝑗−1}0𝐷^𝐷𝑖,𝑓𝑜𝑟𝑖≥1,𝑋=^𝑠−1_𝑖=0𝐴^𝐷𝑖+2𝐵𝐷^𝑖𝐷^𝜋+^𝑟−1_𝑖=0𝐴^𝜋𝐴^𝑖𝐵𝐷^𝐷𝑖+2−𝐴^𝐷𝐵𝐷^𝐷. (21)

Proof. Let 𝑀=𝐴𝐵𝐶𝐷=𝑃+𝑄, (22) where 𝑃=𝐴𝐵0𝐷,𝑄=00𝐶0. (23) From 𝐴𝐵𝐶=0, 𝐵𝐷𝐶=0, 𝐷²𝐶=0, and 𝐶𝐵𝐶=0, we have 𝑃³𝑄=0, 𝑃²𝑄+𝑄𝑃𝑄=0, and 𝑃𝑄𝑃𝑄=0. We can see that 𝑄 is 2-nilpotent, so we get 𝑄^𝐷=0 and 𝑄^𝜋=𝐼. Applying Theorem 5, we have 𝑀^𝐷=𝑃^𝐷+𝑄𝑃^𝐷2+𝑃𝑄𝑃^𝐷3. (24) Using Lemma 3, we obtain the following result: 𝑃^𝐷𝑖=𝐴∑^{𝐷𝑖𝑖−1}_𝑗=0𝐴^𝐷𝑗𝑋𝐷^{𝐷𝑖−𝑗−1}0𝐷^𝐷𝑖,for𝑖≥1, (25) where 𝑋=∑^𝑠−1_𝑖=0(𝐴^𝐷)^𝑖+2𝐵𝐷^𝑖𝐷^𝜋+∑^𝑟−1_𝑖=0𝐴^𝜋𝐴^𝑖𝐵(𝐷^𝐷)^𝑖+2−𝐴^𝐷𝐵𝐷^𝐷.
Substituting (25) into (24), we get the result.

Similarly, we consider another splitting of the block matrix 𝑀 and state another theorem.

Theorem 9. Let 𝑀=^𝐴𝐵_𝐶𝐷 (𝐴 and 𝐷 are square matrices), such that ind(𝐴)=𝑟 and ind(𝐷)=𝑠. If 𝐷𝐶𝐵=0, 𝐵𝐶𝐵=0, 𝐴²𝐵=0, and 𝐶𝐴𝐵=0, then 𝑀^𝐷=𝑃^𝐷+𝑃0𝐵00^𝐷2+𝑃0𝐴𝐵0𝐶𝐵^𝐷3, (26) where 𝑃^𝐷𝑖=𝐴^𝐷𝑖0_𝑖−1𝑗∑₌₀𝐷^𝐷𝑗𝑋𝐴^{𝐷𝑖−𝑗−1}𝐷^𝐷𝑖,𝑓𝑜𝑟𝑖≥1,𝑋=^𝑟−1_𝑖=0𝐷^𝐷𝑖+2𝐶𝐴^𝑖𝐴^𝜋+^𝑠−1_𝑖=0𝐷^𝜋𝐷^𝑖𝐶𝐴^𝐷𝑖+2−𝐷^𝐷𝐶𝐴^𝐷. (27)

Proof. Let 𝑀=𝐴𝐵𝐶𝐷=𝑃+𝑄, (28) where 𝑃=𝐴0𝐶𝐷,𝑄=0𝐵00. (29) The remaining proof follows directly from Theorem 8.

Now, we give the representation for 𝑀^𝐷 when the generalized Schur complement is nonsingular, which generalizes the result in [5].

Theorem 10. Let 𝑀=^𝐴𝐵_𝐶𝐷 (𝐴 and 𝐷 are square matrices). If 𝑆=𝐷−𝐶𝐴^𝐷𝐵 is nonsingular, 𝐵𝐶𝐴^𝜋𝐵=0, 𝐷𝐶𝐴^𝜋𝐵=0, 𝐴²𝐴^𝜋𝐵=0, and 𝐶𝐴𝐴^𝜋𝐵=0, then 𝑀^𝐷=𝑃^𝐷+0𝐴^𝜋𝐵𝑃00^𝐷2+0𝐴𝐴^𝜋𝐵0𝐶𝐴^𝜋𝐵𝑃^𝐷3, (30) where 𝑃^𝐷𝑗=𝑃^𝐷₁^𝑗+^𝑘_𝑖=1𝑃^𝐷₁^𝑖+𝑗𝐴^𝑖𝐴^𝜋0𝐶𝐴^𝑖−1𝐴^𝜋0,𝑃^𝐷₁=𝐴^𝐷+𝐴^𝐷𝐵𝑆⁻¹𝐶𝐴^𝐷−𝐴^𝐷𝐵𝑆⁻¹−𝑆⁻¹𝐶𝐴^𝐷𝑆⁻¹,𝑘=ind(𝐴). (31)

Proof. Let 𝑀=𝐴𝐵𝐶𝐷=𝑃+𝑄, (32) where 𝑃=𝐴𝐴𝐴^𝐷𝐵𝐶𝐷,𝑄=0𝐴^𝜋𝐵00. (33) From 𝐵𝐶𝐴^𝜋𝐵=0, 𝐷𝐶𝐴^𝜋𝐵=0, 𝐴²𝐴^𝜋𝐵=0, and 𝐶𝐴𝐴^𝜋𝐵=0, it is obvious that 𝑃³𝑄=0, 𝑃²𝑄+𝑄𝑃𝑄=0, and 𝑃𝑄𝑃𝑄=0. We can see that 𝑄 is 2-nilpotent, so we get 𝑄^𝐷=0 and 𝑄^𝜋=𝐼. Applying Theorem 5, we have 𝑀^𝐷=𝑃^𝐷+𝑄𝑃^𝐷2+𝑃𝑄𝑃^𝐷3. (34) Let 𝑃=𝑃₁+𝑃₂, where 𝑃₁=𝐴²𝐴^𝐷𝐴𝐴^𝐷𝐵𝐶𝐴𝐴^𝐷𝐷,𝑃₂=𝐴𝐴^𝜋0𝐶𝐴^𝜋0. (35) Obviously, 𝑃₂𝑃₁=0 and 𝑃₂^𝑘+1=0, where 𝑘=ind(𝐴). By Lemma 2, we have 𝑃^𝐷𝑗=𝑃^𝐷₁^𝑗+^𝑘_𝑖=1𝑃^𝐷₁^𝑖+𝑗𝑃^𝑖₂,𝑃^𝑖₂=𝐴^𝑖𝐴^𝜋0𝐶𝐴^𝑖−1𝐴^𝜋0,𝑖≥1. (36) For 𝑃₁, we get that 𝑆=𝐷−𝐶𝐴𝐴^𝐷(𝐴²𝐴^𝐷)^𝐷𝐴𝐴^𝐷𝐵=𝐷−𝐶𝐴^𝐷𝐵 is nonsingular, (𝐴²𝐴^𝐷)^𝜋𝐴𝐴^𝐷𝐵=0, and 𝐶𝐴𝐴^𝐷(𝐴²𝐴^𝐷)^𝜋=0. Using Lemma 4, we obtain the following result: 𝑃^𝐷₁=𝐴^𝐷+𝐴^𝐷𝐵𝑆⁻¹𝐶𝐴^𝐷−𝐴^𝐷𝐵𝑆⁻¹−𝑆⁻¹𝐶𝐴^𝐷𝑆⁻¹. (37) From the above equation, we obtain the result in Theorem 10.

In the same way, we consider another splitting of the block matrix 𝑀 and present next theorem.

Theorem 11. Let 𝑀=^𝐴𝐵_𝐶𝐷 (𝐴 and 𝐷 are square matrices). If 𝑆=𝐷−𝐶𝐴^𝐷𝐵 is nonsingular, 𝐴𝐵𝐶𝐴^𝜋=0, 𝐵𝐷𝐶𝐴^𝜋=0, 𝐷²𝐶𝐴^𝜋=0, and 𝐴^𝜋𝐵𝐶𝐴^𝜋=0, then 𝑀^𝐷=𝑃^𝐷+00𝐶𝐴^𝜋0𝑃^𝐷2+𝐵𝐶𝐴^𝜋0𝐷𝐶𝐴^𝜋0𝑃^𝐷3, (38) where 𝑃^𝐷𝑗=𝑃^𝐷₁^𝑗+^𝑘_𝑖=1𝐴^𝑖𝐴^𝜋𝐴^𝑖−1𝐴^𝜋𝐵𝑃00^{𝐷1𝑖+𝑗},𝑃^𝐷₁=𝐴^𝐷+𝐴^𝐷𝐵𝑆⁻¹𝐶𝐴^𝐷−𝐴^𝐷𝐵𝑆⁻¹−𝑆⁻¹𝐶𝐴^𝐷𝑆⁻¹,𝑘=ind(𝐴). (39)

Proof. Let 𝑀=𝐴𝐵𝐶𝐷=𝑃+𝑄, (40) where 𝑃=𝐴𝐵𝐶𝐴𝐴^𝐷𝐷,𝑄=00𝐶𝐴^𝜋0. (41) The remaining proof is similar to that of Theorem 10.

5. Numerical Examples

In this section, two numerical examples are given to illustrate Theorems 8 and 10.

Example 1. Consider the block matrix 𝑀=^𝐴𝐵_𝐶𝐷∈ℂ^4×4, where ,.𝐴=1000,𝐵=0001𝐶=0010,𝐷=1010 (42) Since 𝐵𝐶≠0 and 𝐵𝐶𝐴≠0, the representation for 𝑀^𝐷 fail to apply in [9, 11, 20–22], respectively. But it satisfies 𝐴𝐵𝐶=0, 𝐵𝐷𝐶=0, 𝐷²𝐶=0, and 𝐶𝐵𝐶=0. Also, we have 𝐴^𝐷=1000,𝐴^𝜋=,𝐷0001_𝐷=1010,𝐷^𝜋=.00−11 (43) Then, applying Theorem 8, we obtain 𝑀^𝐷=1000101000101010. (44)

Remark 12. The above example shows that the conditions given in Theorem 8 are satisfied, but the conditions given in [9, 11, 20–22] are not satisfied.

Example 2. Consider the block matrix 𝑀=^𝐴𝐵_𝐶𝐷∈ℂ^5×5, where ,.𝐴=1100,𝐵=121000𝐶=121001,𝐷=221131001 (45) By computing, we get that 𝑆=𝐷−𝐶𝐴^𝐷𝐵 is nonsingular and 𝐵𝐶𝐴^𝜋𝐵=0, 𝐷𝐶𝐴^𝜋𝐵=0, 𝐴²𝐴^𝜋𝐵=0, and 𝐶𝐴𝐴^𝜋𝐵=0. Also, we have ind(𝐴)=1,𝐴^𝐷=,𝐴1100_𝜋=0−101,𝑆⁻¹=,𝑃100010001_𝐷1=,𝑃44−1−2−100000−1−1100−1−101000001_𝐷=.44−1−2−100000−10100−1−201001001 (46) Then, applying Theorem 10, we get 𝑀^𝐷=44−1−2−100000−10100−1−201001001. (47)

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors thank the reviewers for the valuable comments. This work was supported by the Ph.D. Programs Foundation of Ministry of Education of China (Grant no. 20110191110033).