A Note on Representations of the Drazin Inverse for Complex Partitioned Matrices

Abstract

In this paper, we present explicit representations for Drazin inverses of the sum $P+Q$ under the conditions $P^{2}Q=0$, $Q^{2}PQ=0$, and $Q{\left(PQ\right)}^2=0$. Then, we apply our results to derive a series of representations for the Drazin inverses of a $2\times 2$ complex partitioned matrix $\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]$, where A and D are square complex matrices. Numerical examples are provided to show that the assumptions used here are genuinely weaker than those in a number of earlier results.

1. Introduction

The Drazin inverse of a block matrix appears naturally in finite Markov chains [1,2], singular linear differential equations and difference equations [3,4], iterative methods [5] and so on [6,7,8]. Lately, the related generalized inverse problem has still been investigated by many researchers [9,10,11,12,13,14].

Let $A\in {\mathbb{C}}^{n\times n}$. The Drazin inverse of A, denoted by $A^d$, is the unique matrix satisfying the equations applicable only to square matrices as follows:

$$A{A}^d=A^{d}A,A^{d}A{A}^d=A^d,A^k=A^{k+1}{A}^d,$$

where $k=\mathrm{ind}\left(A\right)$ is the index of A, i.e., the smallest non-negative integer such that $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left(A^{k+1}\right)$. We write $A^e=A{A}^d$ and $A^{\pi }=I-A^e$ for the spectral idempotent associated with the eigenvalue $\left\{0\right\}$, and define $A^0=I$, where I is the identity matrix with proper sizes.

Assume that $P,Q\in {\mathbb{C}}^{n\times n}$. In 1958, Drazin [15] first obtained the explicit formula ${(P+Q)}^d=P^d+Q^d$ under the conditions $PQ=QP=0$. After years of research by scholars, many results have been derived about ${(P+Q)}^d$ under different assumptions. Here we list several results:

1.

$PQ=0$ (see [16]);

2.

$P^{2}Q=0$ and $P{Q}^2=0$ (see [17]);

3.

$P^{2}Q=0$ and $Q^2=0$ (see [18]);

4.

$P{Q}^2=0$ and $PQP=0$ (see [19]);

5.

$P^{2}Q=0$ and $Q^{2}P=0$ (see [20]);

6.

$P^{2}QP=0,P^2{Q}^2=0$ and $QPQ=0$ (see [21]);

7.

$(P+Q)P(P+Q)Q(P+Q)=0,P^{5}Q=0$ and $PQ{P}^{3}Q=0$ (see [22]);

8.

$P^{3}Q=0,P^{2}QP=0,Q^{2}PQ=0,Q^2{P}^2=0,P{Q}^{2}P=0$ and $Q^{3}P=0$ (see [23]).

Furthermore, formulae for ${(P+Q)}^d$ are useful in computing the representations of a $2\times 2$ complex block matrix:

$$M=\left[\begin{array}{cc}A& B\\ C& D\end{array}\right],$$

(1)

where A and D are square matrices. In 1979, Campbell and Meyer [24] proposed an open problem to find an explicit representation for the Drazin inverse of M. Here we give several results for the representations of $M^d$ under the following conditions:

1.

in [25], $ABC=0$ and $DC=0$;

2.

in [26], $BC=0,DC=0$ and $BD=0$;

3.

in [27], $B{D}^{\pi }C=0,BD{D}^d=0,D{D}^{\pi }CA=0$ and $D{D}^{\pi }CB=0$;

4.

in [28], $ABD=0,BCA=0,CBD=0,BCBC=0,DCA=0$ and $D^{\pi }CBC=0$.

Motivated by previous research about the Drazin inverse of the sum of two matrices and the Drazin inverse of the block matrices, we aim to continue investigating these topics, extending known results that are frequently used in literature. Precisely, we propose the exact expressions for the Drazin inverse of the sum of two matrices under weaker assumptions than already existing ones. Applying the results in Section 3, we derive formulae for the Drazin inverse of an arbitrary partitioned matrix and generalize many famous results.

The paper is organized as follows. In Section 2, we first introduce several useful lemmas, especially the formulae on the Drazin inverse of an anti-triangular matrix. In Section 3, we derive a new explicit result for the Drazin inverse of a sum of two matrices $P,Q\in {\mathbb{C}}^{n\times n}$ under conditions $P^{2}Q=0,Q^{2}PQ=0$ and $Q{\left(PQ\right)}^2=0$, which generalizes the result for ${(P+Q)}^d$ proved in [19]. In Section 4, we apply these formulae for ${(P+Q)}^d$ to obtain the representations for the Drazin inverse of M given by (1) under conditions weaker than those used in some recent papers. We also illustrate our results with some numerical examples.

2. Key Lemma

We need the following lemmas to derive the main results. The first one is Cline’s formula.

Lemma 1 

([29]). For $A\in {\mathbb{C}}^{m\times n}$ and $B\in {\mathbb{C}}^{n\times m}$, ${\left(BA\right)}^d=B\left[{\left(AB\right)}^{2d}\right]A.$

The following identity from [16] gives the Drazin inverse of a sum of two matrices when $SR=0$.

Lemma 2 

([16]). Let $S,R\in {\mathbb{C}}^{n\times n}$. If $SR=0$, then

$${(S+R)}^d={\displaystyle \sum _{i=0}^{i_R-1}}{R}^{\pi }{R}^i{\left(S^d\right)}^{i+1}+{\displaystyle \sum _{i=0}^{i_S-1}}{\left(R^d\right)}^{i+1}{S}^i{S}^{\pi },$$

(2)

where $\mathrm{ind}\left(S\right)=i_S$ and $\mathrm{ind}\left(R\right)=i_R.$

We now state a lemma for the Drazin inverse of block triangular matrices, which we shall use later.

Lemma 3 

([30,31]). 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 $\mathrm{ind}\left(A\right)=i_A$ and $\mathrm{ind}\left(D\right)=i_D$. Then

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

where

$$X={\displaystyle \sum _{i=0}^{i_D-1}}{\left(A^d\right)}^{i+2}B{D}^i{D}^{\pi }+A^{\pi }{\displaystyle \sum _{i=0}^{i_A-1}}{A}^{i}B{\left(D^d\right)}^{i+2}-A^{d}B{D}^d.$$

Some results concerning the Drazin inverse of anti-triangular matrices $\left[\begin{array}{cc}A& B\\ I& 0\end{array}\right]$ and $\left[\begin{array}{cc}A& B\\ C& 0\end{array}\right]$ are now provided, which are particularly useful in Section 4.

Lemma 4 

([32], Theorem 3.1). If the blocks A and B of a matrix $\overline{N}=\left[\begin{array}{cc}A& B\\ I& 0\end{array}\right]$ are square matrices of the same size and satisfy $A^{2}B=0,ABAB=0andA{B}^2=0$, then

$${\overline{N}}^d=\left[\begin{array}{cc}{E}_1& E_2\\ E_3& E_4\end{array}\right],$$

(3)

where $\mathrm{ind}\left(A\right)=i_A$, $\mathrm{ind}\left(B\right)=i_B$,

$$\begin{array}{ccc}\hfill E_1& =& {\displaystyle \sum _{k=0}^{i_B-1}}{B}^{\pi }{B}^k{A}^{(2k+1)d}+{\displaystyle \sum _{k=1}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{B}^{kd}{A}^{2k-1}{A}^{\pi }\hfill \\ & +& {\displaystyle \sum _{k=1}^{i_B}}{B}^{\pi }{B}^{k-1}AB{A}^{(2k+2)d}+{\displaystyle \sum _{k=2}^{\left[{\displaystyle \frac{i_A}{2}}\right]+2}}{B}^{kd}AB{A}^{2k-4}{A}^{\pi }-B^{d}AB{A}^{2d},\hfill \\ \hfill E_2& =& B^e,\hfill \\ \hfill E_3& =& {\displaystyle \sum _{k=0}^{i_B-1}}{B}^{\pi }{B}^k{A}^{(2k+2)d}+{\displaystyle \sum _{k=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{B}^{(k+1)d}{A}^{2k}{A}^{\pi }+{\displaystyle \sum _{k=1}^{i_B}}{B}^{\pi }{B}^{k-1}AB{A}^{(2k+3)d}\hfill \\ & +& {\displaystyle \sum _{k=1}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{B}^{(k+2)d}AB{A}^{2k-1}{A}^{\pi }-B^{d}AB{A}^{3d}-B^{2d}AB{A}^d,\hfill \\ \hfill E_4& =& B^{2d}AB.\hfill \end{array}$$

Lemma 5 

([32], Theorem 3.2). If the blocks A and $BC$ of a matrix $N=\left[\begin{array}{cc}A& B\\ C& 0\end{array}\right]$ are square matrices of the same size and satisfy $A^{2}BC=0,ABCABC=0andA{\left(BC\right)}^2=0,$ then

$$N^d=\left[\begin{array}{cc}{G}_1& G_2\\ G_3& G_4\end{array}\right],$$

where $\mathrm{ind}\left(A\right)=i_A$, $\mathrm{ind}\left(BC\right)=i_{BC}$,

$$\begin{array}{ccc}\hfill G_1& =& {\displaystyle \sum _{k=0}^{i_{BC}-1}}{\left(BC\right)}^{\pi }{\left(BC\right)}^k{A}^{(2k+1)d}+{\displaystyle \sum _{k=1}^{i_{BC}}}{\left(BC\right)}^{\pi }{\left(BC\right)}^{k-1}ABC{A}^{(2k+2)d}+{\displaystyle \sum _{k=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]-1}}{\left(BC\right)}^{(k+1)d}{A}^{2k+1}{A}^{\pi }\hfill \\ & +& {\displaystyle \sum _{k=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(k+2)d}ABC{A}^{2k}{A}^{\pi }-{\left(BC\right)}^{d}ABC{A}^{2d},\hfill \\ \hfill G_2& =& {\displaystyle \sum _{k=0}^{i_{BC}-1}}{\left(BC\right)}^{\pi }{\left(BC\right)}^k{A}^{(2k+2)d}B+{\displaystyle \sum _{k=1}^{i_{BC}}}{\left(BC\right)}^{\pi }{\left(BC\right)}^{k-1}ABC{A}^{(2k+3)d}B+{\displaystyle \sum _{k=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(k+1)d}{A}^{2k}{A}^{\pi }B\hfill \\ & +& {\displaystyle \sum _{k=1}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(k+2)d}ABC{A}^{2k-1}{A}^{\pi }B-{\left(BC\right)}^{2d}ABC{A}^{d}B-{\left(BC\right)}^{d}ABC{A}^{3d}B,\hfill \\ \hfill G_3& =& {\displaystyle \sum _{k=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}C{\left(BC\right)}^{(k+1)d}{A}^{2k}{A}^{\pi }+{\displaystyle \sum _{k=0}^{i_{BC}-1}}C{\left(BC\right)}^{\pi }{\left(BC\right)}^k{A}^{(2k+2)d}+{\displaystyle \sum _{k=2}^{\left[{\displaystyle \frac{i_A}{2}}\right]+1}}C{\left(BC\right)}^{(k+1)d}ABC{A}^{2k-3}{A}^{\pi }\hfill \\ & +& {\displaystyle \sum _{k=1}^{i_{BC}}}C{\left(BC\right)}^{\pi }{\left(BC\right)}^{k-1}ABC{A}^{(2k+3)d}-C{\left(BC\right)}^{2d}ABC{A}^d-C{\left(BC\right)}^{d}ABC{A}^{3d},\hfill \\ \hfill G_4& =& {\displaystyle \sum _{k=1}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}C{\left(BC\right)}^{(k+1)d}{A}^{2k-1}{A}^{\pi }B+{\displaystyle \sum _{k=0}^{i_{BC}-1}}C{\left(BC\right)}^{\pi }{\left(BC\right)}^k{A}^{(2k+3)d}B+{\displaystyle \sum _{k=2}^{\left[{\displaystyle \frac{i_A}{2}}\right]+2}}C{\left(BC\right)}^{(k+1)d}ABC{A}^{2k-4}{A}^{\pi }B\hfill \\ & +& {\displaystyle \sum _{k=1}^{i_{BC}}}C{\left(BC\right)}^{\pi }{\left(BC\right)}^{k-1}ABC{A}^{(2k+4)d}B-C{\left(BC\right)}^{2d}ABC{A}^{2d}B-C{\left(BC\right)}^{d}ABC{A}^{4d}B-C{\left(BC\right)}^d{A}^{d}B.\hfill \end{array}$$

Lemma 6 

([33], Theorem 3.3). If the blocks A and $BC$ of a matrix $N=\left[\begin{array}{cc}A& B\\ C& 0\end{array}\right]$ are square matrices of the same size and satisfy $A^{3}BC=0,BCABC=0andBC{A}^{2}BC=0,$ then

$$N^d=\left[\begin{array}{cc}{F}_1& F_2\\ F_3& F_4\end{array}\right],$$

where $\mathrm{ind}\left(A\right)=i_A$, $\mathrm{ind}\left(BC\right)=i_{BC}$,

$$\begin{array}{ccc}\hfill F_1& =& A{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+2)d}+A^2{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+3)d}+{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+1)d}\hfill \\ & +& A{\displaystyle \sum _{i=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+1)d}{A}^{2i}{A}^{\pi }+A^2{\displaystyle \sum _{i=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+2)d}{A}^{2i+1}{A}^{\pi }+{\displaystyle \sum _{i=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+1)d}{A}^{2i+1}{A}^{\pi }\hfill \\ & -& A^2{\left(BC\right)}^d{A}^d-2A^d,\hfill \\ \hfill F_2& =& A{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+3)d}B+{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+2)d}B+A^2{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+4)d}B\hfill \\ & +& A{\displaystyle \sum _{i=1}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+1)d}{A}^{2i-1}{A}^{\pi }B+A^2{\displaystyle \sum _{i=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+2)d}{A}^{2i}{A}^{\pi }B+{\displaystyle \sum _{i=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+1)d}{A}^{2i}{A}^{\pi }B\hfill \\ & -& 2A^{2d}B-A^2{\left(BC\right)}^d{A}^{2d}B-A{\left(BC\right)}^d{A}^{d}B,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill F_3& =& CA{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+3)d}+C{A}^2{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{\left(BC\right)}^{\pi }{A}^{(2i+4)d}+C{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+2)d}\hfill \\ & +& C{\displaystyle \sum _{i=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+1)d}{A}^{2i}{A}^{\pi }+C{A}^2{\displaystyle \sum _{i=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+2)d}{A}^{2i}{A}^{\pi }+CA{\displaystyle \sum _{i=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+2)d}{A}^{2i+1}{A}^{\pi }\hfill \\ & -& 2C{A}^{2d}-CA{\left(BC\right)}^d{A}^d-C{A}^2{\left(BC\right)}^d{A}^{2d},\hfill \\ \hfill F_4& =& CA{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+4)d}B+C{A}^2{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{\left(BC\right)}^{\pi }{A}^{(2i+5)d}B+C{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+3)d}B\hfill \\ & +& C{\displaystyle \sum _{i=1}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+1)d}{A}^{2i-1}{A}^{\pi }B+C{A}^2{\displaystyle \sum _{i=1}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+2)d}{A}^{2i-1}{A}^{\pi }B+CA{\displaystyle \sum _{i=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+2)d}{A}^{2i}{A}^{\pi }B\hfill \\ & -& 2C{A}^{3d}B-C{\left(BC\right)}^d{A}^{d}B-CA{\left(BC\right)}^d{A}^{2d}B-C{A}^2{\left(BC\right)}^{2d}{A}^{d}B-C{A}^2{\left(BC\right)}^d{A}^{3d}B.\hfill \end{array}$$

The following lemma follows by strengthening the conditions of Lemma 6.

Lemma 7 

([33], Corollary 3.4). If the blocks A and $BC$ of a matrix $N=\left[\begin{array}{cc}A& B\\ C& 0\end{array}\right]$ are square matrices of the same size and satisfy $A^{2}BC=0$ and $CABC=0$, then

$$N^d=\left[\begin{array}{cc}{K}_1& K_2\\ K_3& K_4\end{array}\right],$$

where $\mathrm{ind}\left(A\right)=i_A$, $\mathrm{ind}\left(BC\right)=i_{BC}$,

$$\begin{array}{ccc}\hfill K_1& =& A{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+2)d}+{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+1)d}\hfill \\ & +& A{\displaystyle \sum _{i=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+1)d}{A}^{2i}{A}^{\pi }+{\displaystyle \sum _{i=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+1)d}{A}^{2i+1}{A}^{\pi }-A^d,\hfill \\ \hfill K_2& =& A{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+3)d}B+{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+2)d}B\hfill \\ & +& A{\displaystyle \sum _{i=1}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+1)d}{A}^{2i-1}{A}^{\pi }B+{\displaystyle \sum _{i=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+1)d}{A}^{2i}{A}^{\pi }B-A{\left(BC\right)}^d{A}^{d}B-A^{2d}B,\hfill \\ \hfill K_3& =& C{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+2)d}+C{\displaystyle \sum _{i=0}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+1)d}{A}^{2i}{A}^{\pi },\hfill \\ \hfill K_4& =& C{\left(BC\right)}^{\pi }{\displaystyle \sum _{i=0}^{i_{BC}-1}}{\left(BC\right)}^i{A}^{(2i+3)d}B+C{\displaystyle \sum _{i=1}^{\left[{\displaystyle \frac{i_A}{2}}\right]}}{\left(BC\right)}^{(i+1)d}{A}^{2i-1}{A}^{\pi }B-C{\left(BC\right)}^d{A}^{d}B.\hfill \end{array}$$

We remark that the above conditions were also researched in [34], Theorem 4.3.

3. Main Results

This section begins with the formula for the Drazin inverse of $P+Q$ under conditions $P^{2}Q=0,Q^{2}PQ=0$ and $Q{\left(PQ\right)}^2=0$, which generalizes the result presented in [19] under conditions $P^{2}Q=0$ and $QPQ=0$. Likewise, we give another corollary from Lemma 8.

Lemma 8.

Let $P^{2}Q=0$, $Q^{2}PQ=0$ and $Q{\left(PQ\right)}^2=0$, where $P,Q\in {\mathbb{C}}^{n\times n}$ are such that $\mathrm{ind}\left(Q\right)=i_Q$ and $\mathrm{ind}\left(P\right)=i_P$. Then

$$\begin{array}{ccc}\hfill {(P+Q)}^d& =& {\displaystyle \sum _{i=0}^{i_P-1}}{Q}^{(i+1)d}{P}^i{P}^{\pi }+{\displaystyle \sum _{i=0}^{i_P-1}}QP{Q}^{(i+3)d}{P}^i{P}^{\pi }+{\displaystyle \sum _{i=0}^{i_Q-1}}{Q}^{\pi }{Q}^i{P}^{(i+1)d}+{\displaystyle \sum _{i=0}^{i_Q-1}}PQP{Q}^i{Q}^{\pi }{P}^{(i+4)d}\hfill \\ & +& {\displaystyle \sum _{i=0}^{i_Q-1}}QP{Q}^i{Q}^{\pi }{P}^{(i+3)d}+{\displaystyle \sum _{i=0}^{i_Q-1}}P{Q}^i{Q}^{\pi }{P}^{(i+2)d}+{\displaystyle \sum _{i=0}^{i_P-1}}PQP{Q}^{(i+4)d}{P}^i{P}^{\pi }+{\displaystyle \sum _{i=0}^{i_P-1}}P{Q}^{(i+2)d}{P}^i{P}^{\pi }\hfill \\ & -& QP{Q}^d(P^d+Q^d)P^d-PQP{Q}^d(Q^{2d}+Q^d{P}^d+P^{2d})P^d-P^d-Q{P}^{2d}-PQ{P}^{3d}-P{Q}^d{P}^d.\hfill \end{array}$$

Proof. 

We note that $P+Q=\left[\begin{array}{cc}Q& I\end{array}\right]\left[\begin{array}{c}I\\ P\end{array}\right]$. Due to Lemma 1, we have

$${(P+Q)}^d=\left[\begin{array}{cc}Q& I\end{array}\right]{\left[\begin{array}{cc}Q& I\\ PQ& P\end{array}\right]}^{2d}\left[\begin{array}{c}I\\ P\end{array}\right].$$

The next splitting of $\left[\begin{array}{cc}Q& I\\ PQ& P\end{array}\right]$ will be used:

$$\left[\begin{array}{cc}Q& I\\ PQ& P\end{array}\right]=\left[\begin{array}{cc}Q& I\\ PQ& 0\end{array}\right]+\left[\begin{array}{cc}0& 0\\ 0& P\end{array}\right]:=R+S.$$

Since $P^{2}Q=0$, we obtain $SR=0$. Hence, Lemma 2 can be utilized. Now, we calculate $R^d$ as follows:

$$\begin{array}{ccc}\hfill {\left[\begin{array}{cc}Q& I\\ PQ& 0\end{array}\right]}^d& =& {\left(H\left[\begin{array}{cc}Q& PQ\\ I& 0\end{array}\right]H^{-1}\right)}^d=H{\left[\begin{array}{cc}Q& PQ\\ I& 0\end{array}\right]}^d{H}^{-1}\hfill \\ & =& \left[\begin{array}{cc}{E}_{3}Q+E_4& E_3\\ E_{1}Q-Q{E}_{3}Q+E_2-Q{E}_4& E_1-Q{E}_3\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}{Q}^d+P{Q}^{2d}+PQP{Q}^{4d}+QP{Q}^{3d}& Q^{2d}+P{Q}^{3d}+PQP{Q}^{5d}+QP{Q}^{4d}\\ P{Q}^d+PQP{Q}^{3d}& P{Q}^{2d}+PQP{Q}^{4d}\end{array}\right],\hfill \end{array}$$

where $H=\left[\begin{array}{cc}0& I\\ I& -Q\end{array}\right]$ and $H^{-1}=\left[\begin{array}{cc}Q& I\\ I& 0\end{array}\right].$ Furthermore, ${\left[\begin{array}{cc}Q& PQ\\ I& 0\end{array}\right]}^d$ and $E_n$, $n=\overline{1,4}$ are represented as in (3). In addition, we get the expression for $R^{\pi }$ as

$$R^{\pi }=\left[\begin{array}{cc}{Q}^{\pi }-QP{Q}^{2d}-P{Q}^d-PQP{Q}^{3d}& -Q^d-QP{Q}^{3d}-P{Q}^{2d}-PQP{Q}^{4d}\\ -PQ{Q}^d-PQP{Q}^{2d}& I-P{Q}^d-PQP{Q}^{3d}\end{array}\right].$$

Then, we prove, for $n\ge 5$,

$$R^n=\left[\begin{array}{cc}{Q}^n+QP{Q}^{n-2}+P{Q}^{n-1}+PQP{Q}^{n-3}& Q^{n-1}+QP{Q}^{n-3}+P{Q}^{n-2}+PQP{Q}^{n-4}\\ P{Q}^n+PQP{Q}^{n-2}& P{Q}^{n-1}+PQP{Q}^{n-3}\end{array}\right],$$

and for $n\ge 1$,

$$R^{nd}=\left[\begin{array}{cc}{Q}^{nd}+QP{Q}^{(n+2)d}+P{Q}^{(n+1)d}+PQP{Q}^{(n+3)d}& Q^{(n+1)d}+QP{Q}^{(n+3)d}+P{Q}^{(n+2)d}+PQP{Q}^{(n+4)d}\\ P{Q}^{nd}+PQP{Q}^{(n+2)d}& P{Q}^{(n+1)d}+PQP{Q}^{(n+3)d}\end{array}\right].$$

Thus, with the above expressions substituted into (2), the proof is complete. □

The following corollary is obtained by strengthening the conditions of Lemma 8.

Corollary 1.

Let $P^{2}Q=0$, $Q^{2}PQ=0$ and ${\left(PQ\right)}^2=0$, where $P,Q\in {\mathbb{C}}^{n\times n}$ are such that $\mathrm{ind}\left(Q\right)=i_Q$ and $\mathrm{ind}\left(P\right)=i_P$. Then

$$\begin{array}{ccc}\hfill {(P+Q)}^d& =& {\displaystyle \sum _{i=0}^{i_P-1}}{Q}^{(i+1)d}{P}^i{P}^{\pi }+{\displaystyle \sum _{i=0}^{i_P-1}}QP{Q}^{(i+3)d}{P}^i{P}^{\pi }+{\displaystyle \sum _{i=0}^{i_Q-1}}{Q}^{\pi }{Q}^i{P}^{(i+1)d}+{\displaystyle \sum _{i=0}^{i_Q-1}}QP{Q}^i{Q}^{\pi }{P}^{(i+3)d}\hfill \\ & +& {\displaystyle \sum _{i=0}^{i_P-1}}P{Q}^{(i+2)d}{P}^i{P}^{\pi }+{\displaystyle \sum _{i=0}^{i_Q-1}}P{Q}^i{Q}^{\pi }{P}^{(i+2)d}-P^d-Q{P}^{2d}-QP{Q}^{2d}{P}^d-QP{Q}^d{P}^{2d}-P{Q}^d{P}^d.\hfill \end{array}$$

Additionally, we generalize Theorem 2.2 in [19] given by Yang and Liu, and we obtain Corollary 2.

Corollary 2.

Let $P^{2}Q=0$ and $QPQ=0$, where $P,Q\in {\mathbb{C}}^{n\times n}$ are such that $\mathrm{ind}\left(Q\right)=i_Q$ and $\mathrm{ind}\left(P\right)=i_P$. Then

$$\begin{array}{ccc}\hfill {(P+Q)}^d& =& {\displaystyle \sum _{i=0}^{i_P-1}}{Q}^{(i+1)d}{P}^i{P}^{\pi }+{\displaystyle \sum _{i=0}^{i_Q-1}}{Q}^{\pi }{Q}^i{P}^{(i+1)d}+{\displaystyle \sum _{i=0}^{i_P-1}}P{Q}^{(i+2)d}{P}^i{P}^{\pi }+{\displaystyle \sum _{i=0}^{i_Q-2}}P{Q}^{i+1}{Q}^{\pi }{P}^{(i+3)d}\hfill \\ & -& P{Q}^d{P}^d-PQ{Q}^d{P}^{2d}.\hfill \end{array}$$

The following example illustrates that our results are extensions of Corollary 2.

Example 1.

Consider the following choice of P and Q, which satisfies the hypotheses of Lemma 8 yet does not satisfy those of Corollary 2:

$$P=\left[\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]andQ=\left[\begin{array}{cccc}0& 2& a& b\\ 0& 0& c& d\\ 0& 0& 0& 0\\ 0& 0& 2& 2\end{array}\right].$$

Then, after simple calculations, we have $PQP=0$,

$$PQ=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 2& 2\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]\ne 0,andQPQ=\left[\begin{array}{cccc}0& 0& 4& 4\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]\ne 0.$$

Further, we obtain $P^2=0$, $Q^{2d}P=0$, $Q^{2}PQ=0$ and $Q{\left(PQ\right)}^2=0$. According to Lemma 8, we simplify the formula of ${(P+Q)}^d$ as follows:

$$\begin{array}{c}\hfill {(P+Q)}^d=Q^d+QP{Q}^{3d}+P{Q}^{2d}.\end{array}$$

(4)

By applying Lemma 3, we obtain

$$Q^d=\left[\begin{array}{cccc}0& 0& {\displaystyle \frac{1}{4}}(b+d)& {\displaystyle \frac{1}{4}}(b+d)\\ 0& 0& {\displaystyle \frac{1}{4}}d& {\displaystyle \frac{1}{4}}d\\ 0& 0& 0& 0\\ 0& 0& {\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{2}}\end{array}\right].$$

After substituting the above matrices into (4), we get

$${(P+Q)}^d=\left[\begin{array}{cccc}0& 0& {\displaystyle \frac{1}{4}}(b+d)+{\displaystyle \frac{1}{4}}& {\displaystyle \frac{1}{4}}(b+d)+{\displaystyle \frac{1}{4}}\\ 0& 0& {\displaystyle \frac{1}{4}}d+{\displaystyle \frac{1}{4}}& {\displaystyle \frac{1}{4}}d+{\displaystyle \frac{1}{4}}\\ 0& 0& 0& 0\\ 0& 0& {\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{2}}\end{array}\right].$$

4. Applications to Block Matrices

The Drazin inverses of anti-triangular matrices from Section 2 play a key role in deriving the explicit expression of $M^d$ here. We now apply Lemma 8 to present a sequence of main formulae for the Drazin inverse of a block matrix M given by (1).

Theorem 1.

Let M be defined in (1). If

$$A^{3}BC=0,BCABC=0,BC{A}^{2}BC=0,ABDC=0,CBDC=0and{D}^{2}C=0,$$

then

$$\begin{array}{ccc}\hfill M^d& =& {\displaystyle \sum _{i=0}^{i_D-1}}{N}^{(i+1)d}{\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]}^i\left[\begin{array}{cc}I& 0\\ 0& D^{\pi }\end{array}\right]+{\displaystyle \sum _{i=0}^{i_D-1}}\left[\begin{array}{cc}0& BD\\ 0& 0\end{array}\right]N^{(i+3)d}{\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]}^i\left[\begin{array}{cc}I& 0\\ 0& D^{\pi }\end{array}\right]\hfill \\ & +& {\displaystyle \sum _{i=1}^{i_N-1}}{N}^{\pi }{N}^i\left[\begin{array}{cc}0& 0\\ 0& D^{(i+1)d}\end{array}\right]+{\displaystyle \sum _{i=0}^{i_D-1}}\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]N^{(i+2)d}{\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]}^i\left[\begin{array}{cc}I& 0\\ 0& D^{\pi }\end{array}\right]\hfill \\ & +& {\displaystyle \sum _{i=0}^{i_N-1}}\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]N^i{N}^{\pi }\left[\begin{array}{cc}0& 0\\ 0& D^{(i+2)d}\end{array}\right]+{\displaystyle \sum _{i=1}^{i_N-1}}\left[\begin{array}{cc}0& BD\\ 0& 0\end{array}\right]N^i{N}^{\pi }\left[\begin{array}{cc}0& 0\\ 0& D^{(i+3)d}\end{array}\right]\hfill \\ & -& \left[\begin{array}{cc}0& A{F}_2{D}^d+B{F}_4{D}^d+BD{F}_3{F}_2{D}^d+BD{F}_4^2{D}^d+BD{F}_4{D}^{2d}+BDC{F}_2{D}^{3d}\\ 0& C{F}_2{D}^d+D{F}_4{D}^d\end{array}\right],\hfill \end{array}$$

where $N^d$ and $F_n$, $n=\overline{2,4}$ are given by Lemma 6 such that $\mathrm{ind}\left(N\right)=i_N$ and $\mathrm{ind}\left(D\right)=i_D$.

Proof. 

We consider the splitting

$$M=\left[\begin{array}{cc}A& B\\ C& 0\end{array}\right]+\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]:=N+P.$$

We have $PNP=0$,

$$P^{2}N=\left[\begin{array}{cc}0& 0\\ D^{2}C& 0\end{array}\right]=0,N^{2}PN=\left[\begin{array}{cc}ABDC& 0\\ CBDC& 0\end{array}\right]=0,$$

and $N{\left(PN\right)}^2=0$. Hence, we attain the following three conditions about D: $ABDC=0$, $CBDC=0$ and $D^{2}C=0$. We further integrate conditions $A^{3}BC=0$, $BCABC=0$ and $BC{A}^{2}BC=0$ on the Drazin inverse of anti-triangular matrix N. By applying Lemma 6, this gives $N^d$ and leads to the following representation.

$$N^{\pi }=\left[\begin{array}{cc}I-F_{1}A-F_{2}C& -F_{1}B\\ -F_{3}A-F_{4}C& I-F_{3}B\end{array}\right].$$

Meanwhile, by applying Lemma 3, we note that

$$P^d=\left[\begin{array}{cc}0& 0\\ 0& D^d\end{array}\right]\mathrm{and}{P}^{\pi }=\left[\begin{array}{cc}I& 0\\ 0& D^{\pi }\end{array}\right].$$

After that $\mathrm{ind}\left(P\right)=i_D,$ for $i\ge 1$,

$$\begin{array}{c}\hfill P^i{P}^{\pi }=\left[\begin{array}{cc}0& 0\\ 0& D^i{D}^{\pi }\end{array}\right].\end{array}$$

Applying Lemma 8 then completes the proof. □

Strengthening the conditions of Theorem 1 yields the following deduction, which facilitates application.

Corollary 3.

Let M be defined in (1). If

$$A^{3}BC=0,BCABC=0,BC{A}^{2}BC=0,BDC=0and{D}^{2}C=0,$$

then

$$\begin{array}{ccc}\hfill M^d& =& {\displaystyle \sum _{i=0}^{i_D-1}}{N}^{(i+1)d}{\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]}^i\left[\begin{array}{cc}I& 0\\ 0& D^{\pi }\end{array}\right]+{\displaystyle \sum _{i=0}^{i_D-1}}\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]N^{(i+2)d}{\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]}^i\left[\begin{array}{cc}I& 0\\ 0& D^{\pi }\end{array}\right]\hfill \\ & +& {\displaystyle \sum _{i=0}^{i_N-1}}\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]N^i{N}^{\pi }\left[\begin{array}{cc}0& 0\\ 0& D^{(i+2)d}\end{array}\right]+{\displaystyle \sum _{i=1}^{i_N-1}}{N}^{\pi }{N}^i\left[\begin{array}{cc}0& 0\\ 0& D^{(i+1)d}\end{array}\right]-\left[\begin{array}{cc}0& A{F}_2{D}^d+B{F}_4{D}^d\\ 0& C{F}_2{D}^d+D{F}_4{D}^d\end{array}\right],\hfill \end{array}$$

where $N^d$, $F_2$ and $F_4$ are given by Lemma 6 such that $\mathrm{ind}\left(N\right)=i_N$ and $\mathrm{ind}\left(D\right)=i_D$.

Corollary 4.

Let M be defined in (1). If

$$A^{2}BC=0,CABC=0,ABDC=0,CBDC=0and{D}^{2}C=0,$$

then

$$\begin{array}{ccc}\hfill M^d& =& {\displaystyle \sum _{i=0}^{i_D-1}}{N}^{(i+1)d}{\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]}^i\left[\begin{array}{cc}I& 0\\ 0& D^{\pi }\end{array}\right]+{\displaystyle \sum _{i=0}^{i_D-1}}\left[\begin{array}{cc}0& BD\\ 0& 0\end{array}\right]N^{(i+3)d}{\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]}^i\left[\begin{array}{cc}I& 0\\ 0& D^{\pi }\end{array}\right]\hfill \\ & +& {\displaystyle \sum _{i=1}^{i_N-1}}{N}^{\pi }{N}^i\left[\begin{array}{cc}0& 0\\ 0& D^{(i+1)d}\end{array}\right]+{\displaystyle \sum _{i=0}^{i_D-1}}\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]N^{(i+2)d}{\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]}^i\left[\begin{array}{cc}I& 0\\ 0& D^{\pi }\end{array}\right]\hfill \\ & +& {\displaystyle \sum _{i=0}^{i_N-1}}\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]N^i{N}^{\pi }\left[\begin{array}{cc}0& 0\\ 0& D^{(i+2)d}\end{array}\right]+{\displaystyle \sum _{i=1}^{i_N-1}}\left[\begin{array}{cc}0& BD\\ 0& 0\end{array}\right]N^i{N}^{\pi }\left[\begin{array}{cc}0& 0\\ 0& D^{(i+3)d}\end{array}\right]\hfill \\ & -& \left[\begin{array}{cc}0& A{K}_2{D}^d+B{K}_4{D}^d+BD{K}_3{K}_2{D}^d+BD{K}_4^2{D}^d+BD{K}_4{D}^{2d}+BDC{K}_2{D}^{3d}\\ 0& C{K}_2{D}^d+D{K}_4{D}^d\end{array}\right],\hfill \end{array}$$

where $N^d$ and $K_n$, $n=\overline{2,4}$ are given by Lemma 7 such that $\mathrm{ind}\left(N\right)=i_N$ and $\mathrm{ind}\left(D\right)=i_D$.

The assumptions in Theorem 1 are critical. They cover a broader class of matrices, including cases where existing results do not apply. The specific generalizations are as follows.

Remark 1.

A list of results extended by Theorem 1 is given below:

 1. 

In [35], Theorem 2.1, $A=0$ and $D=0$;

 2. 

In [26], Theorem 5.3, $BC=0,BD=0$ and $DC=0$;

 3. 

In [36], Lemma 2.2, $BC=0,DC=0$ and D is nilpotent;

 4. 

In [37], Theorem 2.2, $ABC=0$ and $DC=0$;

 5. 

In [38], Theorem 1, $ABC=0,BD=0$ and $DC=0$;

 6. 

In [38], Theorems 2 and 3, $ABC=0,DC=0$ and $BC$ is nilpotent (or D is nilpotent).

Example 2.

Let A, B, C and D be the following $3\times 3$ complex matrices:

$$A=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right],B=\left[\begin{array}{ccc}1& 0& 0\\ 1& 0& 0\\ 1& 0& 0\end{array}\right],C=\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]andD=\left[\begin{array}{ccc}0& 0& 0\\ 1& 0& 0\\ 1& 0& 0\end{array}\right].$$

We observe that the matrices A, B, C and D satisfy the conditions of Theorem 1 but not the conditions of Remark 1, since

$$BC=\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 1\\ 0& 0& 1\end{array}\right]\ne 0,AB=\left[\begin{array}{ccc}1& 0& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}\right]\ne 0andDC=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ 0& 0& 1\end{array}\right]\ne 0.$$

Since $A^3=0$, $D^2=0$, $BD=0$, $CA=0$ and ${\left(BC\right)}^2=BC$, we deduce that $A^d=0$, $D^d=0$ and ${\left(BC\right)}^d=BC$. By elementary computations, we check that $A^{3}BC=0$, $BCABC=0$, $BC{A}^{2}BC=0$, $ABDC=0$, $CBDC=0$ and $D^{2}C=0$. So, we apply Theorem 1 to get

$$M^d=\left[\begin{array}{cc}{F}_1& F_2\\ F_3+D{F}_3{F}_1& F_4+D{F}_3{F}_2\end{array}\right]=\left[\begin{array}{cccccc}0& 0& 1& 2& 0& 0\\ 0& 0& 1& 1& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 1& 0& 0\end{array}\right].$$

The following result provides another representation of $M^d$.

Theorem 2.

Let M be defined in (1). If

$$A^{3}BC=0,BCABC=0,BC{A}^{2}BC=0,ABD=0andCBD=0,$$

then

$$\begin{array}{ccc}\hfill M^d& =& {\displaystyle \sum _{i=0}^{i_N-1}}\left[\begin{array}{cc}0& 0\\ 0& D^{(i+1)d}\end{array}\right]N^i{N}^{\pi }+{\displaystyle \sum _{i=1}^{i_D-1}}\left[\begin{array}{cc}0& 0\\ 0& D^i{D}^{\pi }\end{array}\right]N^{(i+1)d}\hfill \\ & +& {\displaystyle \sum _{i=0}^{i_N-1}}\left[\begin{array}{cc}0& B{D}^{(i+2)d}\\ 0& 0\end{array}\right]N^i{N}^{\pi }+{\displaystyle \sum _{i=1}^{i_D-1}}\left[\begin{array}{cc}0& B{D}^i{D}^{\pi }\\ 0& 0\end{array}\right]N^{(i+2)d}\hfill \\ & +& \left[\begin{array}{cc}A& B{D}^{\pi }\\ C& 0\end{array}\right]N^{2d}-\left[\begin{array}{cc}B{D}^d{F}_3& B{D}^d{F}_4\\ D{D}^D{F}_3& D{D}^d{F}_4\end{array}\right],\hfill \end{array}$$

where $N^d$, $F_3$ and $F_4$ are given by Lemma 6 such that $\mathrm{ind}\left(N\right)=i_N$ and $\mathrm{ind}\left(D\right)=i_D$.

Proof. 

Consider the splitting

$$M=\left[\begin{array}{cc}A& B\\ C& 0\end{array}\right]+\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]:=N+Q.$$

By Lemma 8, we derive $N^{2}Q=\left[\begin{array}{cc}0& ABD\\ 0& CBD\end{array}\right]=0,$ $Q^{2}NQ=0$ and $Q{\left(NQ\right)}^2=0$.

• The proof of the new formula for the Drazin inverse of block matrix M parallels that of Theorem 1. □

To illustrate the generality of Theorem 2, we further compare it with existing ones as follows.

Remark 2.

Here we list some existing conditions:

 1. 

in [39], Corollary 2.3, $BC=0$ and $BD=0$;

 2. 

in [36], Corollary 2.3, $BC=0,BD=0$ and D is nilpotent;

 3. 

in [19], Corollary 3.4, $AB$ and $CB=0$;

 4. 

in [37], Theorem 2.3, $ABC=0$ and $BD=0$;

 5. 

in [40], Corollary 3.3, $ABC=0,CBC=0$ and $BD=0$;

 6. 

in [19], Theorem 3.2, $ABC=0,CBC=0,ABD=0$ and $CBD=0$.

Theorem 3.

Let M be defined in (1). If

$$A^{2}BC=0,ABCABC=0,A{\left(BC\right)}^2=0,ABDC=0,CBDC=0and{D}^{2}C=0,$$

then

$$\begin{array}{ccc}\hfill M^d& =& {\displaystyle \sum _{i=0}^{i_D-1}}{N}^{(i+1)d}{\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]}^i\left[\begin{array}{cc}I& 0\\ 0& D^{\pi }\end{array}\right]+{\displaystyle \sum _{i=0}^{i_D-1}}\left[\begin{array}{cc}0& BD\\ 0& 0\end{array}\right]N^{(i+3)d}{\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]}^i\left[\begin{array}{cc}I& 0\\ 0& D^{\pi }\end{array}\right]\hfill \\ & +& {\displaystyle \sum _{i=1}^{i_N-1}}{N}^{\pi }{N}^i\left[\begin{array}{cc}0& 0\\ 0& D^{(i+1)d}\end{array}\right]+{\displaystyle \sum _{i=0}^{i_D-1}}\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]N^{(i+2)d}{\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]}^i\left[\begin{array}{cc}I& 0\\ 0& D^{\pi }\end{array}\right]\hfill \\ & +& {\displaystyle \sum _{i=0}^{i_N-1}}\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]N^i{N}^{\pi }\left[\begin{array}{cc}0& 0\\ 0& D^{(i+2)d}\end{array}\right]+{\displaystyle \sum _{i=1}^{i_N-1}}\left[\begin{array}{cc}0& BD\\ 0& 0\end{array}\right]N^i{N}^{\pi }\left[\begin{array}{cc}0& 0\\ 0& D^{(i+3)d}\end{array}\right]\hfill \\ & -& \left[\begin{array}{cc}0& A{G}_2{D}^d+B{G}_4{D}^d+BD{G}_3{G}_2{D}^d+BD{G}_4^2{D}^d+BD{G}_4{D}^{2d}+BDC{G}_2{D}^{3d}\\ 0& C{G}_2{D}^d+D{G}_4{D}^d\end{array}\right],\hfill \end{array}$$

where $N^d$ and $G_n$, $n=\overline{2,4}$ are given by Lemma 5 such that $\mathrm{ind}\left(N\right)=i_N$ and $\mathrm{ind}\left(D\right)=i_D$.

Proof. 

We consider the splitting

$$M=\left[\begin{array}{cc}A& B\\ C& 0\end{array}\right]+\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]:=N+P.$$

This expression for $M^d$ is checked in the same manner as the proof of Theorem 1. □

Corollary 5.

Let M be defined in (1). If

$$A^{2}BC=0,ABCABC=0,A{\left(BC\right)}^2=0,BDC=0and{D}^{2}C=0,$$

then

$$\begin{array}{ccc}\hfill M^d& =& {\displaystyle \sum _{i=0}^{i_D-1}}{N}^{(i+1)d}{\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]}^i\left[\begin{array}{cc}I& 0\\ 0& D^{\pi }\end{array}\right]+{\displaystyle \sum _{i=0}^{i_D-1}}\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]N^{(i+2)d}{\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]}^i\left[\begin{array}{cc}I& 0\\ 0& D^{\pi }\end{array}\right]\hfill \\ & +& {\displaystyle \sum _{i=0}^{i_N-1}}\left[\begin{array}{cc}0& 0\\ 0& D\end{array}\right]N^i{N}^{\pi }\left[\begin{array}{cc}0& 0\\ 0& D^{(i+2)d}\end{array}\right]+{\displaystyle \sum _{i=1}^{i_N-1}}{N}^{\pi }{N}^i\left[\begin{array}{cc}0& 0\\ 0& D^{(i+1)d}\end{array}\right]-\left[\begin{array}{cc}0& A{G}_2{D}^d+B{G}_4{D}^d\\ 0& C{G}_2{D}^d+D{G}_4{D}^d\end{array}\right],\hfill \end{array}$$

where $N^d$, $G_2$ and $G_4$ are given by Lemma 5 such that $\mathrm{ind}\left(N\right)=i_N$ and $\mathrm{ind}\left(D\right)=i_D$.

Moreover, Theorem 3 generalizes some known results for $M^d$ under the following assumptions.

Remark 3.

 1. 

$BC=0$ and $DC=0$ (see [41], Corollary 2.1);

 2. 

$ABC=0$ and $DC=0$ (given in [37], Theorem 2.2, [25], Theorem 2.2);

 3. 

$ABC=0,DC=0$ and $BC$ is nilpotent (considered in [38], Theorem 2);

 4. 

$ABC=0,DC=0$ and D is nilpotent (see [38], Theorem 3).