On the Drazin inverse of anti-triangular block matrices

Daochang Zhang; Dijana Mosić; Liangyun Chen; Daochang Zhang; Dijana Mosić; Liangyun Chen

doi:10.3934/era.2022124

Electronic Research Archive

2022, Volume 30, Issue 7: 2428-2445. doi: 10.3934/era.2022124

Previous Article Next Article

Research article

On the Drazin inverse of anti-triangular block matrices

1.
School of Mathematics and Statistics, Northeast Normal University, Changchun, China
2.
College of Sciences, Northeast Electric Power University, Jilin, China
3.
Faculty of Sciences and Mathematics, University of Niš, P.O. Box 224, 18000 Niš, Serbia

Received: 28 August 2021 Revised: 17 January 2022 Accepted: 20 January 2022 Published: 29 April 2022

Our aim is to present new expressions for the Drazin inverse of anti-triangular block matrices under some circumstances. Applying the established new formulae for anti-triangular block matrices, we derive explicit representations for the Drazin inverse of a $2\times2$ complex block matrix under corresponding assumptions. We extend several well known results in the literature in this way.

Keywords:

Citation: Daochang Zhang, Dijana Mosić, Liangyun Chen. On the Drazin inverse of anti-triangular block matrices[J]. Electronic Research Archive, 2022, 30(7): 2428-2445. doi: 10.3934/era.2022124

Related Papers:

[1]	D. Mosić, P. S. Stanimirović, L. A. Kazakovtsev . The $ m $-weak group inverse for rectangular matrices. Electronic Research Archive, 2024, 32(3): 1822-1843. doi: 10.3934/era.2024083
[2]	Yongge Tian . Characterizations of matrix equalities involving the sums and products of multiple matrices and their generalized inverse. Electronic Research Archive, 2023, 31(9): 5866-5893. doi: 10.3934/era.2023298
[3]	Jia-Min Luo, Hou-Biao Li, Wei-Bo Wei . Block splitting preconditioner for time-space fractional diffusion equations. Electronic Research Archive, 2022, 30(3): 780-797. doi: 10.3934/era.2022041
[4]	Xiuhai Fei, Haifang Zhang . Additivity of nonlinear higher anti-derivable mappings on generalized matrix algebras. Electronic Research Archive, 2023, 31(11): 6898-6912. doi: 10.3934/era.2023349
[5]	Xiu-Jian Wang, Jia-Bao Liu . Quasi-tilted property of generalized lower triangular matrix algebras. Electronic Research Archive, 2025, 33(5): 3065-3073. doi: 10.3934/era.2025134
[6]	Asima Razzaque, Abdul Razaq, Sheikh Muhammad Farooq, Ibtisam Masmali, Muhammad Iftikhar Faraz . An efficient S-box design scheme for image encryption based on the combination of a coset graph and a matrix transformer. Electronic Research Archive, 2023, 31(5): 2708-2732. doi: 10.3934/era.2023137
[7]	Wenya Shi, Xinpeng Yan, Zhan Huan . Faster free pseudoinverse greedy block Kaczmarz method for image recovery. Electronic Research Archive, 2024, 32(6): 3973-3988. doi: 10.3934/era.2024178
[8]	Xing Zhang, Xiaoyu Jiang, Zhaolin Jiang, Heejung Byun . Algorithms for solving a class of real quasi-symmetric Toeplitz linear systems and its applications. Electronic Research Archive, 2023, 31(4): 1966-1981. doi: 10.3934/era.2023101
[9]	Xinzheng Xu, Yanyan Ding, Zhenhu Lv, Zhongnian Li, Renke Sun . Optimized pointwise convolution operation by Ghost blocks. Electronic Research Archive, 2023, 31(6): 3187-3199. doi: 10.3934/era.2023161
[10]	Ranran Li, Hao Liu . On global randomized block Kaczmarz method for image reconstruction. Electronic Research Archive, 2022, 30(4): 1442-1453. doi: 10.3934/era.2022075

Abstract

1. Introduction

The Drazin inverse is a very useful tool in various fields of applied mathematics such as Markov chains, control theory, iterative methods in numerical linear algebra, singular differential and difference equations ^[1,2,3]. There are many researches about finding expressions of the Drazin inverse of a block matrix under certain conditions ^[4,5], but it is still an open problem proposed by Campbell and Meyer ^[6] in 1979.

It is well known that the Drazin inverse of a square complex matrix $A$ is the unique matrix $A^d$ for which the following equations hold

$\begin{equation*} \label{defin1} AA^d = A^dA, \; \; \; A^dAA^d = A^d, \; \; \; A^{k} = A^{k+1}A^d, \end{equation*}$

where $k$ is the index of $A$ (i.e., the smallest non-negative integer such that $\text{rank}(A^{k}) = \text{rank}(A^{k+1})$ ) and denoted by $\text{ind}(A)$ . Recall that $A^e = AA^d$ , and $A^\pi = I-A^e$ is the spectral idempotent of $A$ corresponding to $\{0\}$ . Because $A^0 = I$ , for the identity matrix $I$ of the proper size, and $(A^d)^n = (A^n)^d$ for any non-negative integer $n$ , we adopt the conventions that $A^{dn} = A^{nd} = (A^d)^n$ . Some interesting results related to the Drazin inverse can be found in ^{[7,8,9,10,11]}.

Under adequate restrictions, various representations for the Drazin inverse of a $2\times2$ complex block matrix

$\begin{equation} M = \begin{bmatrix} A&B\\C&D \end{bmatrix} \end{equation}$

(1.1)

are proved and we list some of them:

1. in ^[12], $BC = 0, BD = 0$ and $DC = 0$ ;

2. in ^[13], $BC = 0, BDC = 0$ and $BD^2 = 0$ ;

3. in ^[14], $BC = 0, DC = 0$ (or $BD = 0$ ) and $D$ is nilpotent;

4. in ^[15], $A = 0$ and $D = 0$ ;

5. in ^[16], $ABC = 0, DC = 0$ and $BD = 0$ (or $BC$ is nilpotent, or $D$ is nilpotent);

6. in ^[17], $ABC = 0, CBC = 0$ and $BD = 0$ ;

7. in ^[18], $ABC = 0$ and $BD = 0$ (or $DC = 0$ ).

Let us recall that the solutions to singular systems of differential equations are determined by the formula for the Drazin inverse of an anti-triangular block matrix. We consider the following anti-triangular block matrices:

$\begin{equation} \bar{N} = \begin{bmatrix} A&B\\I&0 \end{bmatrix} \end{equation}$

(1.2)

and

$\begin{equation} N = \begin{bmatrix} A&B\\C&0 \end{bmatrix}. \end{equation}$

(1.3)

Several interesting investigations related to the Drazin inverse of the anti-triangular block matrix $N$ partitioned as in the form (1.3) can be seen in ^{[19,20,21,22,23,24,25]}. Under assumptions $AB = 0$ and $ABC = 0$ , the formulae for the Drazin inverse of the anti-triangular block matrix $N$ given by (1.3), were respectively proved in ^[26]. Also, note that the Drazin inverse of $\tilde{N} = \begin{bmatrix} A & I \\ B & 0 \end{bmatrix}$ was studied in ^[27,28,29].

The first goal of this paper is to present a new formula for the Drazin inverse of $\bar{N}$ in the case that $A^3B = 0$ , $BAB = 0$ and $BA^2B = 0$ . Applying this new formula and the splitting of $N$ in terms of $\bar{N}$ , we get the expression for $N^d$ which recovers some earlier results from ^[26]. Then, using the obtained formula for the Drazin inverse of $N$ , we obtain explicit expressions for the Drazin inverse of $M$ under corresponding assumptions and extend several results in the literature in this manner.

The symbol $\mathbb{C}^{m\times n}$ presents the set of all $m\times n$ complex matrices and all matrices are proper sizes over $\mathbb{C}^{m\times n}$ in this paper. If the lower limit of a sum is greater than its upper limit, we define the sum to be 0, i.e., for example, the sum $\sum\limits_{n = 0}^{-1}* = 0$ . Notice that $[x]$ stands for the truncates integer of $x$ .

2. Key lemma

Some auxiliary results concerning the Drazin inverse, which will be often used, are given in this section.

Firstly, the so-called Cline's formula is stated.

Lemma 2.1. ^[30] (Cline's Formula) For $A\in\mathbb{C}^{m\times n}$ and $B\in\mathbb{C}^{n\times m}$ , $(BA)^d = B[(AB)^{2d}]A.$

We cite one important result about the Drazin inverse of anti-triangular block matrices as follows.

Lemma 2.2. ^[31,32] Let $M = \left[\begin{array}{cc} A & B \\ 0 & D \end{array}\right]$ and $N = \left[\begin{array}{cc} D & 0 \\ B & A \end{array}\right]\in \mathbb{C}^{n\times n}$ , where $A$ and $D$ are square matrices such that $r = \text{ind}(A)$ and $s = \text{ind}(D)$ . Then

$M^{d} = \left[\begin{array}{cc} A^{d} & X \\ 0 & D^{d} \end{array}\right] \; \; \; \mathit{{and}}\; \; N^{d} = \left[\begin{array}{cc} D^{d} & 0 \\ X & A^{d} \end{array}\right],$

where

$X = \sum\limits_{i = 0}^{s-1}A^{(i+2)d}BD^{i}D^{\pi}+A^{\pi}\sum\limits_{i = 0}^{r-1}A^{i}BD^{(i+2)d}-A^{d}BD^{d}.$

The following expressions for the Drazin inverse of the sum of two matrices proved in ^[33], are very useful.

Lemma 2.3. [33,Theorem 2.2] Let $QPQ = 0$ and $P^2Q = 0$ , where $P, Q\in \mathbb{C}^{n\times n}$ , $\text{ind}(P) = r$ and $\text{ind}(Q) = s$ . Then

$\begin{eqnarray} (P+Q)^d& = &Q^\pi \sum\limits_{i = 0}^{s-1}Q^iP^{(i+1)d}+\sum\limits_{i = 0}^{r-1}Q^{(i+1)d}P^iP^\pi +P \sum\limits_{i = 0}^{r-1}Q^{(i+2)d}P^iP^\pi +PQ^\pi \sum\limits_{i = 0}^{s-2}Q^{i+1}P^{(i+3)d} \\ &-&PQ^dP^d-PQQ^dP^{2d}. \end{eqnarray}$

(2.1)

Lemma 2.4. [33,Theorem 2.1] Let $PQP = 0$ and $PQ^2 = 0$ , where $P, Q\in \mathbb{C}^{n\times n}$ , $\text{ind}(P) = r$ and $\text{ind}(Q) = s$ . Then

$\begin{eqnarray} (P+Q)^d& = &Q^\pi \sum\limits_{i = 0}^{s-1}Q^iP^{(i+1)d}+\sum\limits_{i = 0}^{r-1}Q^{(i+1)d}P^iP^\pi +Q^\pi \sum\limits_{i = 0}^{s-1}Q^iP^{(i+2)d}Q+\sum\limits_{i = 0}^{r-2}Q^{(i+3)d}P^{i+1}P^\pi Q\\ &-&Q^dP^dQ-Q^{2d}PP^dQ. \end{eqnarray}$

(2.2)

3. Main results

The aim of this section is to derive the representations for the Drazin inverse of $N$ expressed by (1.3) under new conditions in the literature. To establish the representations of $N^d$ , we obtain our first main result considering the Drazin inverse of $\bar{N}$ , which is given as in (1.2), in the case that $A^3B = 0$ , $BAB = 0$ and $BA^2B = 0$ .

Theorem 3.1. Let $\bar{N}$ be a matrix of the form $(1.2)$ , where $A$ and $B$ are square matrices of the same size such that $\text{ind}(A) = r$ and $\text{ind}(B) = s$ . If

$A^3B = 0, \quad BAB = 0\quad and \quad BA^2B = 0,$

then

$\begin{equation*} \bar{N}^d = \begin{bmatrix} E_{1} & A^2B^d+BB^d\\ E_{3}& AB^d \end{bmatrix}, \end{equation*}$

where

$\begin{eqnarray*} E_{1}& = &AB^\pi \sum\limits_{i = 0}^{s-1}B^iA^{(2i+2)d}+A^2B^\pi \sum\limits_{i = 0}^{s-1} B^iA^{(2i+3)d}+B^\pi \sum\limits_{i = 0}^{s-1} B^iA^{(2i+1)d}\\ &+&A^2\sum\limits_{i = 1}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i-1}A^\pi +A\sum\limits_{i = 0}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i}A^\pi+\sum\limits_{i = 1}^{[\frac{r}{2}]}B^{id}A^{2i-1}A^\pi\\ &-&A^2B^dA^d-2A^d, \\ E_{3}& = &B^\pi \sum\limits_{i = 0}^{s-1}B^iA^{(2i+2)d}+AB^\pi \sum\limits_{i = 0}^{s-1} B^iA^{(2i+3)d}+A^2B^\pi \sum\limits_{i = 0}^{s-1} B^iA^{(2i+4)d}\\ &+&\sum\limits_{i = 0}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i}A^\pi +A\sum\limits_{i = 1}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i-1}A^\pi +A^2\sum\limits_{i = 0}^{[\frac{r}{2}]}B^{(i+2)d}A^{2i}A^\pi\\ &-&AB^dA^d-A^2B^dA^{2d}-2A^{2d}. \end{eqnarray*}$

Proof. The next splitting of $\bar{N}^2$ will be used:

$\begin{eqnarray*} \bar{N}^2 = \begin{bmatrix} A^2+B&AB\\A&B \end{bmatrix} = \begin{bmatrix} A^2&0\\A&0 \end{bmatrix}+\begin{bmatrix} B&AB\\0&B \end{bmatrix}: = P+Q. \end{eqnarray*}$

By Lemma 2.2, we get

$\begin{eqnarray*} \ P^d = \begin{bmatrix} A^{2d}&0\\A^{3d}&0 \end{bmatrix}\quad{\rm and}\quad Q^d = \begin{bmatrix} B^d&AB^d\\0&B^d \end{bmatrix}. \end{eqnarray*}$

Then we verify that

$\begin{eqnarray*} \ P^\pi = \begin{bmatrix} A^\pi &0\\-A^d&I \end{bmatrix}\quad{\rm and}\quad Q^\pi = \begin{bmatrix} B^\pi &-AB^e\\0&B^\pi \end{bmatrix}. \end{eqnarray*}$

For any $n\geq 1$ , we observe that

$\begin{equation*} P^n = \begin{bmatrix} A^{2n} &0\\ A^{2n-1}& 0 \end{bmatrix}, \quad Q^n = \begin{bmatrix} B^n & AB^n\\ 0&B^n \end{bmatrix}, \end{equation*}$

and

$\begin{equation*} P^{nd} = \begin{bmatrix} A^{(2n)d} & 0\\ A^{(2n+1)d}& 0 \end{bmatrix}, \quad Q^{nd} = \begin{bmatrix} B^{nd} & AB^{nd}\\ 0& B^{nd} \end{bmatrix}. \end{equation*}$

Since $\text{ind}(A) = r$ and $\text{ind}(B) = s$ , it is clearly that $r$ and $s$ are respectively the least nonnegative integers as follows

$A^rA^\pi = 0, \; \; B^sB^\pi = 0.$

One can observe that

$r-2\leq2[\frac{r}{2}]-1\leq r-1, \; \; r-1\leq2[\frac{r}{2}]\leq r\; \; and\; \; r\leq2[\frac{r}{2}]+1\leq r+1$

for any nonnegative integer $r$ . Notice that, for $i\geq1$ ,

$\begin{eqnarray*} P^iP^\pi = \begin{bmatrix} A^{2i}&0\\A^{2i-1}&0 \end{bmatrix}\begin{bmatrix} A^\pi &0\\-A^d&I \end{bmatrix} = \begin{bmatrix} A^{2i}A^\pi &0\\A^{2i-1}A^\pi &0 \end{bmatrix}, \end{eqnarray*}$

and

$\begin{eqnarray*} Q^iQ^\pi = \begin{bmatrix} B^i&A B^i\\0&B^i \end{bmatrix}\begin{bmatrix} B^\pi &-AB^e\\0&B^\pi \end{bmatrix} = \begin{bmatrix} B^iB^\pi &AB^iB^\pi \\0&B^{i}B^\pi \end{bmatrix}. \end{eqnarray*}$

So, we conclude that $\text{ind}(P) = [\frac{r}{2}]+1$ and $\text{ind}(Q) = s$ . Because the assumptions $P^2Q = 0$ and $QPQ = 0$ of Lemma 2.3 are satisfied, we calculate the following terms as in (2.1):

$\begin{equation*} Q^\pi \sum\limits_{i = 0}^{s-1}Q^iP^{(i+1)d} = \begin{bmatrix} \ B^\pi \sum\limits_{i = 0}^{s-1}B^iA^{(2i+2)d}+A\sum\limits_{i = 1}^{s-1}B^iA^{(2i+3)d}- AB^e\sum\limits_{i = 0}^{s-1}B^iA^{(2i+3)d}&0\\ \ B^\pi \sum\limits_{i = 0}^{s-1}B^iA^{(2i+3)d}& 0 \end{bmatrix}, \end{equation*}$

$\begin{equation*} \sum\limits_{i = 0}^{[\frac{r}{2}]}Q^{(i+1)d}P^iP^\pi = \begin{bmatrix} \sum\limits_{i = 0}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i}A^\pi +A\sum\limits_{i = 1}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i-1}A^\pi -AB^dA^d&AB^d\\ \sum\limits_{i = 1}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i-1}A^\pi -B^dA^d& B^d \end{bmatrix}, \end{equation*}$

$\begin{equation*} P\sum\limits_{i = 0}^{[\frac{r}{2}]}Q^{(i+2)d}P^iP^\pi = \begin{bmatrix} \ A^2\sum\limits_{i = 0}^{[\frac{r}{2}]}B^{(i+2)d}A^{2i}A^\pi &0\\ \ A\sum\limits_{i = 0}^{[\frac{r}{2}]}B^{(i+2)d}A^{2i}A^\pi +A^2\sum\limits_{i = 1}^{[\frac{r}{2}]}B^{(i+2)d}A^{2i-1}A^\pi -A^2B^{2d}A^d& A^2B^{2d} \end{bmatrix}, \end{equation*}$

$\begin{equation*} P Q^\pi\sum\limits_{i = 0}^{s-2} Q^{i+1}P^{(i+3)d} = \begin{bmatrix} A^2B^\pi\sum\limits_{i = 0}^{s-2} B^{i+1}A^{(2i+6)d} &0\\ AB^\pi\sum\limits_{i = 0}^{s-2} B^{i+1}A^{(2i+6)d}+A^2B^\pi\sum\limits_{i = 0}^{s-2}B^{i+1}A^{(2i+7)d}& 0 \end{bmatrix}, \end{equation*}$

$\begin{equation*} PQ^dP^d = \begin{bmatrix} A^2B^dA^{2d}&0\\ AB^dA^{2d}+A^2B^dA^{3d}& 0 \end{bmatrix} \end{equation*}$

and

$\begin{equation*} PQ Q^dP^{2d} = \begin{bmatrix} A^2BB^dA^{4d}&0\\ ABB^dA^{4d}+A^2BB^dA^{5d}&0 \end{bmatrix}. \end{equation*}$

Substituting the above expressions into (2.1), we get

$\begin{equation*} \bar{N}^{2d} = (P+Q)^d = \begin{bmatrix} \alpha&\beta\\ \gamma&\delta \end{bmatrix}, \end{equation*}$

where

$\begin{eqnarray*} \alpha& = &B^\pi \sum\limits_{i = 0}^{s-1}B^iA^{(2i+2)d}+A\sum\limits_{i = 1}^{s-1}B^iA^{(2i+3)d}- AB^e\sum\limits_{i = 0}^{s-1}B^iA^{(2i+3)d}+A^2B^\pi \sum\limits_{i = 0}^{s-2} B^{i+1}A^{(2i+6)d}\\ &+&A\sum\limits_{i = 1}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i-1}A^\pi +A^2\sum\limits_{i = 0}^{[\frac{r}{2}]}B^{(i+2)d}A^{2i}A^\pi +\sum\limits_{i = 0}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i}A^\pi\\ &-&A^2B^dA^{2d}-A^2BB^dA^{4d}-AB^dA^d, \\ \beta& = &AB^d, \\ \gamma& = & B^\pi \sum\limits_{i = 0}^{s-1}B^iA^{(2i+3)d}+A B^\pi\sum\limits_{i = 0}^{s-2} B^{i+1}A^{(2i+6)d}+A^2B^\pi\sum\limits_{i = 0}^{s-2}B^{i+1}A^{(2i+7)d}\\ &+&\sum\limits_{i = 1}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i-1}A^\pi +A\sum\limits_{i = 0}^{[\frac{r}{2}]}B^{(i+2)d}A^{2i}A^\pi +A^2\sum\limits_{i = 1}^{[\frac{r}{2}]}B^{(i+2)d}A^{2i-1}A^\pi \\ &-&AB^dA^{2d}-A^2B^dA^{3d} -ABB^dA^{4d}-A^2BB^dA^{5d}-A^2B^{2d}A^d-B^dA^d, \\ \delta& = &B^d+A^2B^{2d}. \end{eqnarray*}$

Computing $\bar{N}^d = \bar{N}\bar{N}^{2d}$ , we obtain

$\begin{equation*} \bar{N}^d = \begin{bmatrix} E_{1} & E_{2}\\ E_{3}& E_{4} \end{bmatrix}, \end{equation*}$

where

$\begin{eqnarray*} E_{1}& = &AB^\pi \sum\limits_{i = 0}^{s-1}B^iA^{(2i+2)d}+A^2\sum\limits_{i = 1}^{s-1}B^iA^{(2i+3)d}- A^2B^e\sum\limits_{i = 0}^{s-1}B^iA^{(2i+3)d}+B^\pi \sum\limits_{i = 0}^{s-1}B^{i+1}A^{(2i+3)d}\\ &+&A^2\sum\limits_{i = 1}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i-1}A^\pi +\sum\limits_{i = 1}^{[\frac{r}{2}]}B^{id}A^{2i-1}A^\pi+A\sum\limits_{i = 0}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i}A^\pi-A^2B^dA^d-B^eA^d, \\ E_{2}& = &A^2B^d+BB^d, \\ E_{3}& = &B^\pi \sum\limits_{i = 0}^{s-1}B^iA^{(2i+2)d}+A\sum\limits_{i = 1}^{s-1}B^iA^{(2i+3)d}- AB^e\sum\limits_{i = 0}^{s-1}B^iA^{(2i+3)d}+A^2B^\pi\sum\limits_{i = 0}^{s-2} B^{i+1}A^{(2i+6)d}\\ &+&A\sum\limits_{i = 1}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i-1}A^\pi +A^2\sum\limits_{i = 0}^{[\frac{r}{2}]}B^{(i+2)d}A^{2i}A^\pi+\sum\limits_{i = 0}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i}A^\pi \\ &-&A^2B^dA^{2d}-A^2BB^dA^{4d}-AB^dA^d, \\ E_{4}& = &AB^d.\\ \end{eqnarray*}$

We finish this proof by modulating appropriately the upper and lower limits of the corresponding sums.

By direct calculations, we can obtain the following corollary using Theorem 3.1.

Corollary 3.2. Let $\bar{N}$ be a matrix of the form $(1.2)$ , where $A$ and $B$ are square matrices of the same size such that $\text{ind}(A) = r$ and $\text{ind}(BC) = s$ . If $A^2B = 0$ and $BAB = 0$ , then

$\begin{equation*} \bar{N}^d = \begin{bmatrix} E_{1}& BB^d\\ E_{3}& AB^d \end{bmatrix}, \end{equation*}$

where

$\begin{eqnarray*} E_{1}& = &AB^\pi \sum\limits_{i = 0}^{s-1}B^iA^{(2i+2)d}+B^\pi \sum\limits_{i = 0}^{s-1}B^iA^{(2i+1)d}+A\sum\limits_{i = 0}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i}A^\pi +\sum\limits_{i = 1}^{[\frac{r}{2}]}B^{id}A^{2i-1}A^\pi-A^d, \\ E_{3}& = &B^\pi \sum\limits_{i = 0}^{s-1}B^iA^{(2i+2)d}+AB^\pi\sum\limits_{i = 0}^{s-1} B^iA^{(2i+3)d}+\sum\limits_{i = 0}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i}A^\pi +A\sum\limits_{i = 1}^{[\frac{r}{2}]}B^{(i+1)d}A^{2i-1}A^\pi \\ &-&AB^dA^d-A^{2d}.\\ \end{eqnarray*}$

A natural motivation is from the Drazin inverse of $\bar{N}$ to give a new expression for the Drazin inverse of $N$ .

Theorem 3.3. Let $N$ be a matrix of the form $(1.3)$ , where $A$ and $BC$ are square matrices of the same size such that $\text{ind}(A) = r$ and $\text{ind}(BC) = s$ . If

$A^3BC = 0, \quad BCABC = 0\quad and\quad BCA^2BC = 0,$

then

$\begin{equation} N^d = \begin{bmatrix} F_{1} & F_{2}\\ F_{3}& F_{4} \end{bmatrix}, \end{equation}$

(3.1)

where

$\begin{eqnarray*} F_{1}& = &A(BC)^\pi\sum\limits_{i = 0}^{s-1} (BC)^{i}A^{(2i+2)d} +A^2(BC)^\pi\sum\limits_{i = 0}^{s-1} (BC)^iA^{(2i+3)d}+(BC)^\pi\sum\limits_{i = 0}^{s-1} (BC)^iA^{(2i+1)d}\\ &+&A\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i}A^\pi+A^2\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+2)d}A^{2i+1}A^\pi+\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i+1}A^\pi\\ &-&A^2(BC)^dA^d-2A^d, \\ F_{2}& = &A(BC)^\pi\sum\limits_{i = 0}^{s-1}(BC)^{i}A^{(2i+3)d}B+(BC)^\pi\sum\limits_{i = 0}^{s-1} (BC)^iA^{(2i+2)d} B +A^2(BC)^\pi\sum\limits_{i = 0}^{s-1} (BC)^iA^{(2i+4)d}B\\ &+&A\sum\limits_{i = 1}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i-1}A^\pi B+A^2\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+2)d}A^{2i}A^\pi B +\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i}A^\pi B\\ &-&2A^{2d}B-A^2(BC)^dA^{2d}B-A(BC)^dA^dB, \\ F_{3}& = &CA(BC)^\pi \sum\limits_{i = 0}^{s-1}(BC)^i A^{(2i+3)d}+CA^2\sum\limits_{i = 0}^{s-1}(BC)^i(BC)^\pi A^{(2i+4)d}+C(BC)^\pi \sum\limits_{i = 0}^{s-1}(BC)^iA^{(2i+2)d}\\ &+&C\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i}A^\pi +CA^2\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+2)d}A^{2i}A^\pi +CA\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+2)d}A^{2i+1}A^\pi\\ &-&2CA^{2d}-CA(BC)^dA^d-CA^2(BC)^dA^{2d}, \\ F_{4}& = &CA(BC)^\pi \sum\limits_{i = 0}^{s-1}(BC)^i A^{(2i+4)d}B+CA^2\sum\limits_{i = 0}^{s-1}(BC)^i(BC)^\pi A^{(2i+5)d}B+C(BC)^\pi \sum\limits_{i = 0}^{s-1}(BC)^iA^{(2i+3)d}B\\ &+&C\sum\limits_{i = 1}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i-1}A^\pi B +CA^2\sum\limits_{i = 1}^{[\frac{r}{2}]}(BC)^{(i+2)d}A^{2i-1}A^\pi B +CA\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+2)d}A^{2i}A^\pi B \\&-&2CA^{3d}B-C(BC)^dA^dB-CA(BC)^dA^{2d}B-CA^2(BC)^{2d}A^dB-CA^2(BC)^dA^{3d}B. \end{eqnarray*}$

Proof. We denote by $P$ and $Q$ , respectively, the left matrix and the right matrix of the right-hand side of the next splitting of $N$ :

$\begin{eqnarray*} N = \begin{bmatrix} I&0\\0&C \end{bmatrix}\begin{bmatrix} A&B\\I&0 \end{bmatrix}. \end{eqnarray*}$

Thus,

$\begin{eqnarray*} QP = \begin{bmatrix} A&BC\\I&0 \end{bmatrix}, \end{eqnarray*}$

and, applying Theorem 3.1, we have

$\begin{eqnarray*} (QP)^d = \begin{bmatrix} \lambda&\mu\\\nu&\xi \end{bmatrix}, \end{eqnarray*}$

where $\text{ind}(A) = r$ , $\text{ind}(BC) = s$ ,

$\begin{eqnarray*} \lambda& = &A(BC)^\pi \sum\limits_{i = 0}^{s-1}(BC)^iA^{(2i+2)d}+A^2(BC)^\pi\sum\limits_{i = 0}^{s-1} (BC)^iA^{(2i+3)d}+(BC)^\pi \sum\limits_{i = 0}^{s-1} (BC)^iA^{(2i+1)d}\\ &+&A^2\sum\limits_{i = 1}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i-1}A^\pi+A\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i}A^\pi +\sum\limits_{i = 1}^{[\frac{r}{2}]}(BC)^{id}A^{2i-1}A^\pi\\ &-&A^2(BC)^dA^d-2A^d, \\ \mu& = &A^2(BC)^d+BC(BC)^d, \\ \nu& = &(BC)^\pi \sum\limits_{i = 0}^{s-1}(BC)^iA^{(2i+2)d}+A(BC)^\pi\sum\limits_{i = 0}^{s-1} (BC)^iA^{(2i+3)d}+A^2(BC)^\pi\sum\limits_{i = 0}^{s-1} (BC)^iA^{(2i+4)d}\\ &+&\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i}A^\pi +A\sum\limits_{i = 1}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i-1}A^\pi +A^2\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+2)d}A^{2i}A^\pi\\ &-&A(BC)^dA^d-A^2(BC)^dA^{2d}-2A^{2d}, \\ \xi& = &A(BC)^d. \end{eqnarray*}$

According to Lemma 2.1, notice that

$\begin{eqnarray} N^d = P(QP)^{2d}Q = \begin{bmatrix} \lambda^2A+\mu\nu A+\lambda\mu+\mu\xi&\lambda^2B+\mu\nu B\\C\nu\lambda A+C\xi\nu A+C\nu\mu+C\xi^2&C\nu\lambda B+C\xi\nu B \end{bmatrix}. \end{eqnarray}$

(3.2)

By direct computations, we obtain $\xi^2 = 0$ , $\mu\xi = 0$ ,

$\begin{eqnarray*} \lambda^2& = &A(BC)^\pi \sum\limits_{i = 0}^{s-1}(BC)^iA^{(2i+3)d}+A^2\sum\limits_{i = 1}^{s-1}(BC)^{i}A^{(2i+4)d} - A^2(BC)^e\sum\limits_{i = 0}^{s-1}(BC)^iA^{(2i+4)d} \\&+&\sum\limits_{i = 0}^{s-1}(BC)^{i+1}(BC)^\pi A^{(2i+4)d}+A\sum\limits_{i = 1}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i-1}A^\pi\\ &-&A(BC)^dA^d-A^2(BC)^dA^{2d}-(BC)^eA^{2d}, \\ \mu\nu& = &A^2\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+2)d}A^{2i}A^\pi +\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i}A^\pi, \\ \lambda\mu & = &A(BC)^d, \\ \nu\lambda& = & A(BC)^\pi\sum\limits_{i = 0}^{s-1} (BC)^i A^{(2i+4)d}+A^2\sum\limits_{i = 0}^{s-1}(BC)^i(BC)^\pi A^{(2i+5)d}+(BC)^\pi \sum\limits_{i = 0}^{s-1}(BC)^iA^{(2i+3)d}\\ &+&\sum\limits_{i = 1}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i-1}A^\pi+A^2\sum\limits_{i = 1}^{[\frac{r}{2}]}(BC)^{(i+2)d}A^{2i-1}A^\pi \\&-&2A^{3d}-(BC)^dA^d-A(BC)^dA^{2d}-A^2(BC)^{2d}A^d-A^2(BC)^dA^{3d}, \\ \nu\mu & = &(BC)^d+A^2(BC)^{2d}, \\ \xi\nu & = &A\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+2)d}A^{2i}A^\pi.\\ \end{eqnarray*}$

The proof is finished by substituting the above expressions into (3.2).

Several particular consequences of our main result are investigated now. We combine Theorem 3.3 and routine computations to obtain the following expressions for the Drazin inverse of $N$ as in (1.3).

Corollary 3.4. Let $N$ be a matrix of the form $(1.3)$ , where $A$ and $BC$ are square matrices of the same size such that $\text{ind}(A) = r$ and $\text{ind}(BC) = s$ . If $A^2BC = 0$ and $CABC = 0$ , then

$\begin{equation*} N^d = \begin{bmatrix} F_{1} & F_{2}\\ F_{3}& F_{4} \end{bmatrix}, \end{equation*}$

where

$\begin{eqnarray*} F_{1}& = &A(BC)^\pi \sum\limits_{i = 0}^{s-1}(BC)^{i}A^{(2i+2)d} +(BC)^\pi \sum\limits_{i = 0}^{s-1} (BC)^iA^{(2i+1)d}\\ &+&A\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i}A^\pi+\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i+1}A^\pi -A^d, \\ F_{2}& = &A(BC)^\pi\sum\limits_{i = 0}^{s-1}(BC)^{i}A^{(2i+3)d}B +(BC)^\pi\sum\limits_{i = 0}^{s-1} (BC)^iA^{(2i+2)d} B\\ &+&A\sum\limits_{i = 1}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i-1}A^\pi B+\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i}A^\pi B\\ &-&A(BC)^dA^dB-A^{2d}B, \\ F_{3}& = & C(BC)^\pi\sum\limits_{i = 0}^{s-1} (BC)^iA^{(2i+2)d}+C\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i}A^\pi, \\ F_{4}& = &C(BC)^\pi\sum\limits_{i = 0}^{s-1} (BC)^iA^{(2i+3)d}B+C\sum\limits_{i = 1}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i-1}A^\pi B-C(BC)^dA^dB. \end{eqnarray*}$

The following corollary presents a special case of Corollary 3.4.

Corollary 3.5. Let $N$ be a matrix of the form $(1.3)$ , where $A$ and $BC$ are square matrices of the same size such that $\text{ind}(A) = r$ and $\text{ind}(BC) = s$ . If $A^2B = 0$ and $CAB = 0$ , then

$\begin{equation*} N^d = \begin{bmatrix} F_{1} & (BC)^dB\\ F_{3}& 0 \end{bmatrix}, \end{equation*}$

where

We can easily check that Corollary 3.4 extend both [26,Theorem 3.1] and [26,Theorem 3.3] as follows.

Corollary 3.6. Let $N$ be a matrix of the form $(1.3)$ , where $A$ and $BC$ are square matrices of the same size such that $\text{ind}(A) = r$ and $\text{ind}(BC) = s$ .

$\rm(i)$ [26,Theorem 3.3] If $ABC = 0$ , then

$\begin{equation*} N^d = \begin{bmatrix} XA & XB\\ CX&C[XA^d+(BC)^d(XA-A^d)]B \end{bmatrix}; \end{equation*}$

$\rm(ii)$ [26,Theorem 3.1] If $AB = 0$ , then

$\begin{equation*} N^d = \begin{bmatrix} XA& (BC)^dB\\ CX& 0 \end{bmatrix}, \end{equation*}$

where

$\begin{eqnarray} X& = &(BC)^\pi\sum\limits_{i = 0}^{s-1} (BC)^iA^{(2i+2)d}+\sum\limits_{i = 0}^{[\frac{r}{2}]}(BC)^{(i+1)d}A^{2i}A^\pi. \end{eqnarray}$

(3.3)

4. Applications to express $M^d$

Under new conditions, applying the formulae for the Drazin inverse of anti-triangular block matrices proved in Section 3, we present several representations for the Drazin inverse of a $2\times 2$ block matrix $M$ and generalize a series of results, most of whom are from the references.

Theorem 4.1. Let $M$ be a matrix of the form $(1.1)$ and $N$ be a matrix of the form $(1.3)$ , where $A$ , $D$ and $BC$ are square matrices such that $A$ and $BC$ are of the same size.If

$A^3BC = 0, \quad BCABC = 0, \quad BCA^2BC = 0, \quad BDC = 0\quad and\quad BD^2 = 0,$

then

$\begin{eqnarray*} M^d& = &\begin{bmatrix} I&0\\ 0&D^\pi \end{bmatrix}\sum\limits_{i = 0}^{s-1}\begin{bmatrix} 0&0\\ 0&D \end{bmatrix}^iN^{(i+1)d}\\&+&\sum\limits_{i = 0}^{r-1}\begin{bmatrix} 0&0\\ 0&D^{(i+1)d} \end{bmatrix}N^i\begin{bmatrix}(BC)^\pi -F_1A-A^2(BC)^d&-F_1B\\-F_3A-CA(BC)^d&I-F_3B\end{bmatrix}\\&+&\begin{bmatrix} I&0\\ 0&D^\pi \end{bmatrix} \sum\limits_{i = 0}^{s-1}\begin{bmatrix} 0&0\\ 0&D \end{bmatrix}^iN^{(i+2)d}\begin{bmatrix} 0&0\\ 0&D \end{bmatrix}\\&+&\sum\limits_{i = 0}^{r-2}\begin{bmatrix} 0&0\\ 0&D^{(i+3)d} \end{bmatrix}N^{i+1}\begin{bmatrix}0&-F_1BD\\0&(I-F_3B)D\end{bmatrix} -\begin{bmatrix}0&0\\0&D^dF_4D+D^{2d}CF_2D\end{bmatrix}, \end{eqnarray*}$

where $N^d$ and $F_{n}$ , $n = \overline{1, 4}$ , are represented as in $(3.1)$ , $\text{ind}(N) = r$ and $\text{ind}(D) = s$ .

Proof. We can write $M = N+Q$ , where

$N = \begin{bmatrix} A&B\\ C&0 \end{bmatrix}\quad{\rm and}\quad Q = \begin{bmatrix} 0&0\\ 0&D \end{bmatrix}.$

Hence,

$Q^d = \begin{bmatrix} 0&0\\ 0&D^d \end{bmatrix}\quad{\rm and}\quad Q^\pi = \begin{bmatrix} I&0\\ 0&D^\pi \end{bmatrix}.$

Using Theorem 3.3, $N^d$ is represented as in (3.1) and so

$N^\pi = \begin{bmatrix}I-F_1A-F_2C&-F_1B\\-F_3A-F_4C&I-F_3B\end{bmatrix}.$

From the equalities

$\begin{eqnarray*} F_2C& = &A^2(BC)^d+(BC)^dBC, \\ F_4C& = &CA(BC)^d, \end{eqnarray*}$

we have

$N^\pi = \begin{bmatrix}I-F_1A-A^2(BC)^d-(BC)^dBC&-F_1B\\-F_3A-CA(BC)^d&I-F_3B\end{bmatrix}.$

Because $NQN = 0$ and $NQ^2 = 0$ , the rest is clear by Lemma 2.4.

In order to illustrate the width of Theorem 4.1, we only need to list the results generalized through whose following corollary.

Corollary 4.2. Let $M$ be a matrix of the form $(1.1)$ and $N$ be a matrix of the form $(1.3)$ , where $A$ , $D$ and $BC$ are square matrices such that $A$ and $BC$ are of the same size.If

$A^2BC = 0, \quad CABC = 0, \quad BDC = 0\quad and\quad BD^2 = 0,$

then

$\begin{eqnarray*} M^d& = &\begin{bmatrix} I&0\\ 0&D^\pi \end{bmatrix}\sum\limits_{i = 0}^{s-1}\begin{bmatrix} 0&0\\ 0&D \end{bmatrix}^iN^{(i+1)d}\\&+&\sum\limits_{i = 0}^{r-1}\begin{bmatrix} 0&0\\ 0&D^{(i+1)d} \end{bmatrix}N^i\begin{bmatrix}(BC)^\pi -F_1A&-F_1B\\-F_3A&I-F_3B\end{bmatrix}\\&+&\begin{bmatrix} I&0\\ 0&D^\pi \end{bmatrix} \sum\limits_{i = 0}^{s-1}\begin{bmatrix} 0&0\\ 0&D \end{bmatrix}^iN^{(i+2)d}\begin{bmatrix} 0&0\\ 0&D \end{bmatrix}\\&+&\sum\limits_{i = 0}^{r-2}\begin{bmatrix} 0&0\\ 0&D^{(i+3)d} \end{bmatrix}N^{i+1}\begin{bmatrix}0&-F_1BD\\0&(I-F_3B)D\end{bmatrix} -\begin{bmatrix}0&0\\0&D^dF_4D+D^{2d}CF_2D\end{bmatrix}, \end{eqnarray*}$

where $N^d$ and $F_{n}$ , $n = \overline{1, 4}$ , are represented as in Corollary 3.4, $\text{ind}(N) = r$ and $\text{ind}(D) = s$ .

We can verify that Corollary 4.2 generalizes and unifies the following conditions about the expression for $M^d$ :

1. $BC = 0$ , $BD = 0$ and $DC = 0$ (see [12,Theorem 5.3]);

2. $BC = 0$ , $BD = 0$ and $D$ is nilpotent (see [14,Corollary 2.3]);

3. $ABC = 0, CBC = 0$ and $BD = 0$ (see [17,Corollary 3.3]);

4. $ABC = 0$ and $BD = 0$ (see [18,Theorem 2.3] ).

In addition, we utilize Corollary 4.2 to obtain the following expression for $M^d$ as in [13,Corollary 2.3].

Corollary 4.3. [13,Corollary 2.3] Let $M$ be a matrix of the form $(1.1)$ , where $A$ , $D$ and $BC$ are square matrices such that $A$ and $BC$ are of the same size.If

$BC = 0, \quad BDC = 0\quad and\quad BD^2 = 0,$

then

$\begin{eqnarray*} M^d& = &\begin{bmatrix} I&0\\ 0&D^\pi \end{bmatrix}\sum\limits_{i = 0}^{s-1}\begin{bmatrix} 0&0\\ 0&D \end{bmatrix}^i\begin{bmatrix} A^d& A^{2d}B\\ CA^{2d}& CA^{3d}B \end{bmatrix}^{i+1}+\sum\limits_{i = 0}^{r-1}\begin{bmatrix} 0&0\\ 0&D^{(i+1)d} \end{bmatrix}N^i\begin{bmatrix}A^\pi &-A^dB\\-CA^d&I-CA^{2d}B\end{bmatrix}\\&+&\begin{bmatrix} I&0\\ 0&D^\pi \end{bmatrix} \sum\limits_{i = 0}^{s-1}\begin{bmatrix} 0&0\\ 0&D \end{bmatrix}^i\begin{bmatrix} A^d& A^{2d}B\\ CA^{2d}& CA^{3d}B \end{bmatrix}^{i+2}\begin{bmatrix} 0&0\\ 0&D \end{bmatrix}\\&+&\sum\limits_{i = 0}^{r-2}\begin{bmatrix} 0&0\\ 0&D^{(i+3)d} \end{bmatrix}N^{i+1}\begin{bmatrix}0&-A^dBD\\0&(I-CA^{2d}B)D\end{bmatrix}-\begin{bmatrix}0&0\\0&D^dCA^{3d}BD+D^{2d}CA^{2d}BD\end{bmatrix}, \end{eqnarray*}$

where $\text{ind}(N) = r$ and $\text{ind}(D) = s$ .

Utilizing Corollary 4.2, we also obtain the following expression for $M^d$ .

Corollary 4.4. Let $M$ be a matrix of the form $(1.1)$ , where $A$ , $D$ and $BC$ are square matrices such that $A$ and $BC$ are of the same size.If

$A^2B = 0, \quad CAB = 0, \quad BDC = 0\quad and\quad BD^2 = 0,$

then

$\begin{eqnarray*} M^d& = &\begin{bmatrix} I&0\\ 0&D^\pi \end{bmatrix}\sum\limits_{i = 0}^{s-1}\begin{bmatrix} 0&0\\ 0&D \end{bmatrix}^iN^{(i+1)d}\\&+&\sum\limits_{i = 0}^{r-1}\begin{bmatrix} 0&0\\ 0&D^{(i+1)d} \end{bmatrix}N^i\begin{bmatrix}(BC)^\pi -F_1A&-F_1B\\-F_3A&I-F_3B\end{bmatrix}\\&+&\begin{bmatrix} I&0\\ 0&D^\pi \end{bmatrix} \sum\limits_{i = 0}^{s-1}\begin{bmatrix} 0&0\\ 0&D \end{bmatrix}^iN^{(i+2)d}\begin{bmatrix} 0&0\\ 0&D \end{bmatrix}\\&+&\sum\limits_{i = 0}^{r-2}\begin{bmatrix} 0&0\\ 0&D^{(i+3)d} \end{bmatrix}N^{i+1}\begin{bmatrix}0&-F_1BD\\0&(I-F_3B)D\end{bmatrix} -\begin{bmatrix}0&0\\0&D^{2d}C(BC)^dBD\end{bmatrix}, \end{eqnarray*}$

where $F_{1}, F_{3}$ and $N^d$ are given as in Corollary 3.5, $\text{ind}(N) = r$ and $\text{ind}(D) = s$ .

Utilizing Corollary 4.2, we obtain the expression for $M^d$ as in [16,Theorem 1].

Corollary 4.5. [16,Theorem 1] Let $M$ be a matrix of the form $(1.1)$ , where $A$ , $D$ and $BC$ are square matrices such that $A$ and $BC$ are of the same size.If

$ABC = 0, \quad BD = 0\quad and \quad DC = 0,$

then

$\begin{equation*} M^d = \begin{bmatrix} XA & XB\\ CX&D^d+C[XA^d+(BC)^d(XA-A^d)]B \end{bmatrix}, \end{equation*}$

where $X$ is represented by $(3.3)$ .

We note that Corollary 4.5 generalizes the next formula proved in [15,Theorem 2.1].

Corollary 4.6. [15,Theorem 2.1] Let $M$ be a matrix of the form $(1.1)$ , where $A$ , $D$ and $BC$ are square matrices such that $A$ and $BC$ are of the same size.If

$A = 0\quad and\quad D = 0,$

then

$\begin{eqnarray*} M^d& = &\begin{bmatrix} 0&(BC)^dB\\ C(BC)^d&0 \end{bmatrix}. \end{eqnarray*}$

The following formula, which is a dual version of Theorem 4.1, can be proved similarly.

Theorem 4.7. Let $M$ be a matrix of the form $(1.1)$ and $N$ be a matrix of the form $(1.3)$ , where $A$ , $D$ and $BC$ are square matrices such that $A$ and $BC$ are of the same size.If

$A^3BC = 0, \quad BCABC = 0, \quad BCA^2BC = 0, \quad DCA = 0\quad and\quad DCB = 0,$

then

$\begin{eqnarray*} M^d& = &\begin{bmatrix}(BC)^\pi -F_1A-A^2(BC)^d&-F_1B\\-F_3A-CA(BC)^d&I-F_3B\end{bmatrix}\sum\limits_{i = 0}^{r-1}N^i\begin{bmatrix} 0&0\\ D^{(i+2)d}C&D^{(i+1)d} \end{bmatrix}\\ &+&\sum\limits_{i = 0}^{s-1}N^{(i+1)d}\begin{bmatrix} 0&0\\ 0&D \end{bmatrix}^i\begin{bmatrix} I&0\\ 0&D^\pi \end{bmatrix}+\sum\limits_{i = 0}^{s-2}N^{(i+3)d}\begin{bmatrix} 0&0\\ D^{i+1}D^\pi C&0 \end{bmatrix}\\ &-&\begin{bmatrix} F_2D^dC&0\\ F_4D^dC&0 \end{bmatrix}-N^{2d}\begin{bmatrix} 0&0\\ DD^dC&0 \end{bmatrix}, \end{eqnarray*}$

where $N^d$ and $F_{n}$ , $n = \overline{1, 4}$ , are given by $(3.1)$ , $\text{ind}(N) = r$ and $\text{ind}(D) = s$ .

Proof. Notice that $QNQ = 0$ and $QN^2 = 0$ , where $Q$ and $N$ are given as in the proof of Theorem 4.1. As in the proof of Theorem 4.1, this proof can be finished by using Theorem 3.3 and Lemma 2.4.

Applying Theorem 4.7, we prove the next formula for $M^d$ .

Corollary 4.8. Let $M$ be a matrix of the form $(1.1)$ and $N$ be a matrix of the form $(1.3)$ , where $A$ , $D$ and $BC$ are square matrices such that $A$ and $BC$ are of the same size.If

$A^2BC = 0, \quad CABC = 0, \quad DCA = 0\quad and\quad DCB = 0,$

then

$\begin{eqnarray*} M^d& = &\begin{bmatrix}-F_1A+(BC)^\pi & -F_1B\\-F_3A&I-F_3B\end{bmatrix}\sum\limits_{i = 0}^{r-1}N^i\begin{bmatrix} 0&0\\ D^{(i+2)d}C&D^{(i+1)d} \end{bmatrix}\\ &+&\sum\limits_{i = 0}^{s-1}N^{(i+1)d}\begin{bmatrix} 0&0\\ 0&D \end{bmatrix}^i\begin{bmatrix} I&0\\ 0&D^\pi \end{bmatrix}+\sum\limits_{i = 0}^{s-2}N^{(i+3)d}\begin{bmatrix} 0&0\\ D^{i+1}D^\pi C&0 \end{bmatrix}\\ &-&\begin{bmatrix} F_2D^dC&0\\ F_4D^dC&0 \end{bmatrix}-N^{2d}\begin{bmatrix} 0&0\\ DD^dC&0 \end{bmatrix}, \end{eqnarray*}$

where $N^d$ and $F_{n}$ , $n = \overline{1, 4}$ , are given as in Corollary 3.4, $\text{ind}(N) = r$ and $\text{ind}(D) = s$ .

It is worth mentioning that Corollary 4.8 recovers the formulae for the Drazin inverse of $M$ under the following assumptions:

1. $ABC = 0$ and $DC = 0$ (see [18,Theorem 2.2]);

2. $ABC = 0$ , $DC = 0$ and $BC$ is nilpotent (or $D$ is nilpotent) (see [16,Theorem 2,Theorem 3]);

3. $BC = 0$ , $DC = 0$ and $D$ is nilpotent (see [14,Lemma 2.2] ).

5. An example

In this section, we give an example to illustrate our results. Precisely, we present matrices $N$ and $M$ whose blocks are $4\times 4$ complex matrices $A$ , $B$ , $C$ and $D$ which do not satisfy the conditions of [26,Theorem 3.1 and Theorem 3.3], [12,Theorem 5.3], [,Corollary 2.3] and [,Theorem 2.3], but the assumptions of Theorem 3.3 (or Corollary 3.4) and Theorem 4.1 (or Corollary 4.4) are met, which allows us to find $N^d$ and $M^d$ .

Example 5.1. Let $M$ be a matrix of the form (1.1) and $N$ be a matrix of the form (1.3), where $A$ , $B$ , $C$ and $D$ are $4\times 4$ complex matrices given by

$A = \left[ {\begin{array}{*{20}{c}} 0&0&0&0\\a&0&0&0\\0&b&0&0\\0&0&c&0 \end{array}} \right], \quad B = \left[ {\begin{array}{*{20}{c}} 0&0&0&0\\e&0&0&0\\0&f&0&0\\0&0&g&0 \end{array}} \right],$

$C = \left[ {\begin{array}{*{20}{c}} 0&0&0&0\\u&0&0&0\\0&v&0&0\\0&0&t&0 \end{array}} \right]\quad and \quad D = \left[ {\begin{array}{*{20}{c}} 0&0&0&0\\0&0&0&0\\0&0&0&0\\d&d&d&1 \end{array}} \right],$

where $0\notin\{a, b, c, d, e, f, g, u, v, t\}$ . Notice that

$BC = \left[ {\begin{array}{*{20}{c}} 0&0&0&0\\0&0&0&0\\fu&0&0&0\\0&gv&0&0 \end{array}} \right]\neq0,$

$AB = \left[ {\begin{array}{*{20}{c}} 0&0&0&0\\0&0&0&0\\be&0&0&0\\0&cf&0&0 \end{array}} \right]\neq0\quad and \quad ABC = \left[ {\begin{array}{*{20}{c}} 0&0&0&0\\0&0&0&0\\0&0&0&0\\cfu&0&0&0 \end{array}} \right]\neq 0.$

The conclusions above imply that [26,Theorem 3.1 and Theorem 3.3], [12,Theorem 5.3], [,Corollary 2.3] and [,Theorem 2.3] can not be applied. We note that the equalities $A^2BC = 0$ , $CABC = 0$ and $BD = 0$ are satisfied, and utilize Theorem 3.3 (or Corollary 3.4) to obtain

$N^d = \left[ {\begin{array}{*{20}{c}} 0&0&0&0&0&0&0&0\\a&0&0&0&e&0&0&0\\0&b&0&0&0&f&0&0\\0&0&c&0&0&0&g&0\\ 0&0&0&0&0&0&0&0\\u&0&0&0&0&0&0&0\\0&v&0&0&0&0&0&0\\0&0&t&0&0&0&0&0 \end{array}} \right]^d = 0.$

Since $D = D^2 = D^\#$ , we apply Theorem 4.1 (or Corollary 4.2) to get

$\begin{eqnarray*} M^d& = &\left[ {\begin{array}{*{20}{c}} 0&0&0&0&0&0&0&0\\a&0&0&0&e&0&0&0\\0&b&0&0&0&f&0&0\\0&0&c&0&0&0&g&0\\ 0&0&0&0&0&0&0&0\\u&0&0&0&0&0&0&0\\0&v&0&0&0&0&0&0\\0&0&t&0&d&d&d&1 \end{array}} \right]^d\\& = &\left[ {\begin{array}{*{20}{c}} 0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\d(u+va)+t(ba+fu)&dv+tb&t&0&d+(dv+tb)e&d+tf&d&1 \end{array}} \right]. \end{eqnarray*}$

Acknowledgments

The first author (D. Zhang) is supported by the National Natural Science Foundation of China (NSFC) (No. 11901079), China Postdoctoral Science Foundation (No. 2021M700751) and the Scientific and Technological Research Program Foundation of Jilin Province (No. JJKH20190690KJ; No. 20200401085GX; No. JJKH20220091KJ). The second author (D. Mosić) is supported by the Ministry of Education, Science and Technological Development, Republic of Serbia (No. 174007/451-03-9/2021-14/200124) and the bilateral project between Serbia and Slovenia (Generalized inverses, operator equations and applications, No. 337-00-21/2020-09/32).

Conflict of interest

The authors declare there is no conflicts of interest.

References

[1]	C. D. Meyer, The role of the group generalized inverse in the theory of finite Markov chains, SIAM Review., 17 (1975), 443–464. https://doi.org/10.1137/1017044 doi: 10.1137/1017044
[2]	C. D. Meyer, The condition number of a finite Markov chains and perturbation bounds for the limitimg probabilities, SIAM J. Alg. Dis. Methods, 1 (1980), 273–283. https://doi.org/10.1137/0601031 doi: 10.1137/0601031
[3]	C. D. Meyer, R. J. Plemmons, Convergent powers of a matrix with applications to iterative methods for singular systems of linear systems, SIAM J. Numer. Anal., 14 (1977), 699–705. https://doi.org/10.1137/0714047 doi: 10.1137/0714047
[4]	Q. Xu, C. Song, L. He, Representations for the Drazin inverse of an anti-triangular block operator matrix $E$ with $ind(E)\leq2$ , Linear Multilinear Algebra, 66 (2018), 1026–1045. https://doi.org/10.1080/03081087.2017.1335688 doi: 10.1080/03081087.2017.1335688
[5]	D. Zhang, D. Mosić, T. Tam, On the existence of group inverses of Peirce corner matrices, Linear Algebra Appl., 582 (2019), 482–498. https://doi.org/10.1016/j.laa.2019.07.033 doi: 10.1016/j.laa.2019.07.033
[6]	S. L. Campbell, C. D. Meyer, Generalized inverses of linear Transformations, London, Pitman, 1979, Reprint, Dover, New York, 1991.
[7]	I. Kyrchei, Explicit formulas for determinantal representations of the Drazin inverse solutions of some matrix and differential matrix equations, Appl. Math. Comput., 219 (2013), 7632–7644. https://doi.org/10.1016/j.amc.2013.01.050 doi: 10.1016/j.amc.2013.01.050
[8]	I. Kyrchei, Determinantal representations of the Drazin inverse over the quaternion skew field with applications to some matrix equations, Appl. Math. Comput., 238 (2014), 193–207. https://doi.org/10.1016/j.amc.2014.03.125 doi: 10.1016/j.amc.2014.03.125
[9]	J. Rafael Sendra, J. Sendra, Symbolic computation of Drazin inverses by specializations, J. Comput. Anal. Appl., 301 (2016), 201–212. https://doi.org/10.1016/j.cam.2016.01.059 doi: 10.1016/j.cam.2016.01.059
[10]	P. S. Stanimirović, V. N. Katsikis, S. Srivastava, D. Pappas, A class of quadratically convergent iterative methods, RACSAM, 113 (2019), 3125–3146. https://doi.org/10.1007/s13398-019-00681-w doi: 10.1007/s13398-019-00681-w
[11]	P. S. Stanimirović, M. D. Petković, D. Gerontitis, Gradient neural network with nonlinear activation for computing inner inverses and the Drazin inverse, Neural Process Lett, 48 (2018), 109–133. https://doi.org/10.1007/s11063-017-9705-4 doi: 10.1007/s11063-017-9705-4
[12]	D. S. Djordjević, P. S. Stanimirović, On the generalized Drazin inverse and generalized resolvent, Czechoslovak Math. J., 51 (2001), 617–634. https://doi.org/10.1023/A:1013792207970 doi: 10.1023/A:1013792207970
[13]	E. Dopazo, M. F. Martínez-Serrano, Further results on the representation of the Drazin inverse of a $2\times2$ block matrix, Linear Algebra Appl., 432 (2010), 1896–1904.
[14]	R. E. Hartwig, X. Li, Y. Wei, Representations for the Drazin inverse of a $2\times2$ block matrix, SIAM J. Matrix Anal. Appl., 27 (2006), 757–771. https://doi.org/10.1137/040606685 doi: 10.1137/040606685
[15]	M. Catral, D. D. Olesky, P. Van Den Driessche, Block representations of the Drazin inverse of a bipartite matrix, Electron. J. Linear Algebra, 18 (2009), 98–107. https://doi.org/10.13001/1081-3810.1297 doi: 10.13001/1081-3810.1297
[16]	A. S. Cvetković, G. V. Milovanović, On Drazin inverse of operator matrices, J. Math. Anal. Appl., 375 (2011), 331–335. https://doi.org/10.1016/j.jmaa.2010.08.080 doi: 10.1016/j.jmaa.2010.08.080
[17]	L. Guo, J. Chen, H. Zou, Representations for the Drazin inverse of the sum of two matrices and its applications, Bull. Iran. Math. Soc., 45 (2019), 683–699. https://doi.org/10.1007/s41980-018-0159-x doi: 10.1007/s41980-018-0159-x
[18]	C. Bu, K. Zhang, The explicit representations of the Drazin inverses of a class of block matrices, Electron. J. Linear Algebra, 20 (2010), 406–418. https://doi.org/10.13001/1081-3810.1384 doi: 10.13001/1081-3810.1384
[19]	C. Bu, C. Feng, S. Bai, Representations for the Drazin inverses of the sum of two matrices and some block matrices, Appl. Math. Comput., 218 (2012), 10226–10237. https://doi.org/10.1016/j.amc.2012.03.102 doi: 10.1016/j.amc.2012.03.102
[20]	C. Bu, J. Zhao, J. Tang, Representation of the Drazin inverse for special block matrix, Appl. Math. Comput., 217 (2011), 4935–4943. https://doi.org/10.1016/j.amc.2010.11.042 doi: 10.1016/j.amc.2010.11.042
[21]	N. Castro-González, E. Dopazo, Representations of the Drazin inverse for a class of block matrices, Linear Algebra Appl., 400 (2005), 253–269. https://doi.org/10.1016/j.laa.2004.12.027 doi: 10.1016/j.laa.2004.12.027
[22]	C. Deng, Generalized Drazin inverses of anti-triangular block matrices, J. Math. Anal. Appl., 368 (2010), 1–8. https://doi.org/10.1016/j.jmaa.2010.03.003 doi: 10.1016/j.jmaa.2010.03.003
[23]	E. Dopazo, M. F. Martínez-Serrano, J. Robles, Block representations for the Drazin inverse of anti-triangular matrices, Filomat, 30 (2016), 3897–3906. https://doi.org/10.2298/FIL1614897D doi: 10.2298/FIL1614897D
[24]	J. Huang, Y. Shi, A. Chen, The representation of the Drazin inverse of anti-triangular operator matrices based on resolvent expansions, Appl. Math. Comput., 242 (2014), 196–201. https://doi.org/10.1016/j.amc.2014.05.053 doi: 10.1016/j.amc.2014.05.053
[25]	X. Liu, H. Yang, Further results on the group inverses and Drazin inverses of anti-triangular block matrices, Appl. Math. Comput., 218 (2012), 8978–8986. https://doi.org/10.1016/j.amc.2012.02.058 doi: 10.1016/j.amc.2012.02.058
[26]	C. Deng, Y. Wei, A note on the Drazin inverse of an anti-triangular matrix, Linear Algebra Appl., 431 (2009), 1910–1922. https://doi.org/10.1016/j.laa.2009.06.030 doi: 10.1016/j.laa.2009.06.030
[27]	C. Bu, K. Zhang, J. Zhao, Representation of the Drazin inverse on solution of a class singular differential equations, Linear Multilinear Algebra, 59 (2011), 863–877. https://doi.org/10.1080/03081087.2010.512291 doi: 10.1080/03081087.2010.512291
[28]	P. Patrício, R. E. Hartwig, The (2, 2, 0) Drazin inverse problem, Linear Algebra Appl., 437 (2012), 2755–2772. https://doi.org/10.1016/j.laa.2012.07.008
[29]	D. Zhang, D. Mosić, Explicit formulae for the generalized Drazin inverse of block matrices over a Banach algebra, Filomat, 32 (2018), 5907–5917. https://doi.org/10.2298/FIL1817907Z doi: 10.2298/FIL1817907Z
[30]	R. E. Cline, An application of representation for the generalized inverse of a matrix, MRC Technical Report 592, 1965.
[31]	R. E. Hartwig, J. M. Shoaf, Group inverses and Drazin inverses of bidiagonal and triangular Toeplitz matrices, J. Aust. Math. Soc., 24 (1977), 10–34. https://doi.org/10.1017/S1446788700020036 doi: 10.1017/S1446788700020036
[32]	C. D. Meyer, N. J. Rose, The index and the Drazin inverse of block triangular matrices, SIAM J. Appl. Math., 33 (1977), 1–7. https://doi.org/10.1137/0133001 doi: 10.1137/0133001
[33]	H. Yang, X. Liu, The Drazin inverse of the sum of two matrices and its applications, J. Comput. Appl. Math., 235 (2011), 1412–1417. https://doi.org/10.1016/j.cam.2010.08.027 doi: 10.1016/j.cam.2010.08.027

This article has been cited by:

1.	Huanyin Chen, Marjan Sheibani Abdolyousefi, The generalized Drazin inverse of an operator matrix with commuting entries, 2024, 31, 1072-947X, 195, 10.1515/gmj-2023-2074
2.	Huanyin Chen, Marjan Sheibani, The Drazin inverse for perturbed block-operator matrices, 2024, 38, 0354-5180, 2311, 10.2298/FIL2407311C
3.	Huanyin Chen, Marjan Sheibani, g-Drazin inverse and group inverse for the anti-triangular block-operator matrices, 2024, 38, 0354-5180, 4973, 10.2298/FIL2414973C

Reader Comments

Your name:*

Email:*
© 2022 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0)

通讯作者: 陈斌, bchen63@163.com

1.
沈阳化工大学材料科学与工程学院沈阳 110142

Electronic Research Archive

1 1.7

Metrics

Article views(1825) PDF downloads(119) Cited by(3)

Preview PDF

Download XML

Export Citation

Article outline

Show full outline

Electronic Research Archive

On the Drazin inverse of anti-triangular block matrices

Related Papers:

Abstract

1. Introduction

2. Key lemma

3. Main results

4. Applications to express $M^d$

5. An example

Acknowledgments

Conflict of interest

References

This article has been cited by:

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Other Articles By Authors

Catalog

Electronic Research Archive

On the Drazin inverse of anti-triangular block matrices

Related Papers:

Abstract

1. Introduction

2. Key lemma

3. Main results

4. Applications to express Md M^d

5. An example

Acknowledgments

Conflict of interest

References

This article has been cited by:

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Other Articles By Authors

Related pages

Tools

Export File

Citation

Format

Content

Catalog

4. Applications to express $M^d$