The problem of the Drazin inverse was discussed widely in [1-12]. The Drazin inverse was used to be applied in sigular differential difference equations, Markov chains and numerical analysis in [1-3]. The Drazin inverse of the modified matrices was studied by many people [4-6], as a modified matrix can be seen as the sum of two matrices or a matrix added a perturbed element. In [4], Wei Yiming gave the expression for the Drazin inverse of $A-CB$; Liu Xifu weakened the condition of [4] and gave another expression in [5]; the Drazin inverse of $A-CD^{D}B$ was given in [6]. In this paper, we weaken the conditions of [4-5] and give different results.

Definition 2.1  Let $\mathbb{C}^{n\times n}$ denote the set of $n\times n$ complex matrices. The Drazin inverse of $A\in\mathbb{C}^{n\times n}$ is the unique matrix $A^{D}$ satisfying the relations:

$\begin{equation} A^{D}AA^{D}=A^{D},\ \ A^{D}A=AA^{D},\ \ A^{k+1}A^{D}=A^{k}, \end{equation}$

(2.1)

where $k$ is the smallest non-negative integer such that rank$(A^{k})=$rank$(A^{k+1}),$ i.e., $k=$ind$(A)$, the index of $A$. The case when ind$(A)=1$, the Drazin inverse is called the group inverse of $A$ and it is denoted by $A^{\#}$. We denote by $A^{\pi }$ corresponding to the eigenvalue $0$ that is given by $A^{\pi }=I-AA^{D}$.

Lemma 2.2  [4] Suppose $P=0$, $Q=0$ and $C(I-ZZ^{D})B=0$. Then

$\begin{equation} (A-CB)^{D}=A^{D}+KZ^{D}H, \end{equation}$

(2.2)

where denote $K=A^{D}C$, $H=BA^{D}$, $Z=I-BA^{D}C$, $P=(I-AA^{D})C$ and $Q=B(I-A^{D}A)$.

Lemma 2.3  [5] Let $A$ be an idempotent matrix, suppose $P=0$, ind$(Z)=k$, then

$\begin{equation} (A-CB)^{D}=A+CZ^{D}H-C(Z^{D})^{2}Q-CZ^{\pi}\sum\limits^{k-1}_{i=0}Z^{i}H, \end{equation}$

(2.3)

where denote $K=AC$, $H=BA$, $Z=I-BAC$, $P=(I-A)C$ and $Q=B(I-A)$, especially, $Z=I-BC$ at here.

Lemma 2.4  [6] Let $A,B,C$ and $D$ be complex matrices, where ind$(A)=k$. If $A^{\pi}C=0$, $CZ^{\pi}=0$, $Z^{\pi}B=0$, $CD^{\pi}=0$ and $D^{\pi}B=0$, then

$\begin{equation} (A-CD^{D}B)^{D}=A^{D}+A^{D}CZ^{D}BA^{D}-\sum\limits^{k-1}_{i=0}(A^{D}+A^{D}CZ^{D}BA^{D})^{i+1}A^{D}CZ^{D}BA^{i}A^{\pi}, \end{equation}$

(2.4)

where denote the schur complement $Z=D-BA^{D}C$, furthermore, ind$(A-CD^{D}B)\leq$ ind$(A)$.

First, we definite some notation similar to the reference [4]. Let

$\begin{equation} K=A^{D}C,\quad H=BA^{D},\quad \Gamma=HK,\quad Z=I-BA^{D}C \end{equation}$

(3.1)

and

$\begin{equation} P=(I-AA^{D})C,\quad Q=B(I-A^{D}A). \end{equation}$

(3.2)

Theorem 3.1  Let $A,B$ and $C$ be complex matrices, where ind$(A)=k$. If $A^{\pi}C=0$ and $CZ^{\pi}B=0$, then

$\begin{equation} (A-CB)^{D}=A^{D}+KZ^{D}H-\sum\limits^{k-1}_{i=0}(A^{D}+KZ^{D}H)^{i+1}KZ^{D}BA^{i}A^{\pi}. \end{equation}$

(3.3)

furthermore, ind$(A-CB)\leq {\rm ind}(A)$.

Proof Let $X=A^{D}+KZ^{D}H-\sum\limits^{k-1}\limits_{i=0}(A^{D}+KZ^{D}H)^{i+1}KZ^{D}BA^{i}A^{\pi}$. Then,

$\begin{eqnarray} && (A-CB)X = (A-CB)[A^{D}+KZ^{D}H-\sum\limits^{k-1}_{i=0}(A^{D}+KZ^{D}H)^{i+1}KZ^{D}BA^{i}A^{\pi}] \nonumber\\ & = & (A-CB)(A^{D}+KZ^{D}H) -(A-CB)(A^{D}+KZ^{D}H)\sum\limits^{k-1}_{i=0}(A^{D}+KZ^{D}H)^{i}KZ^{D}BA^{i}A^{\pi} \nonumber \\ & = & AA^{D}-AA^{D}\sum\limits^{k-1}_{i=0}(A^{D}+KZ^{D}H)^{i}KZ^{D}BA^{i}A^{\pi} \nonumber\\ & = & AA^{D}-\sum\limits^{k-1}_{i=0}(A^{D}+KZ^{D}H)^{i}KZ^{D}BA^{i}A^{\pi}. \end{eqnarray}$

(3.4)

At the same time, we get

$\begin{eqnarray} X(A-CB)& = & [A^{D}+KZ^{D}H-\sum\limits^{k-1}_{i=0}(A^{D}+KZ^{D}H)^{i+1}KZ^{D}BA^{i}A^{\pi}](A-CB) \nonumber\\ & = & (A^{D}+KZ^{D}H)(A-CB) \nonumber\\ &&-\sum\limits^{k-1}_{i=0}(A^{D}+KZ^{D}H)^{i+1}KZ^{D}BA^{i}A^{\pi}(A-CB) \nonumber\\ & = & AA^{D}-A^{D}CZ^{D}BA^{\pi}-\sum\limits^{k-1}_{i=0}(A^{D}+KZ^{D}H)^{i+1}KZ^{D}BA^{i+1}A^{\pi} \nonumber\\ & = & AA^{D}-\sum\limits^{k-1}_{i=0}(A^{D}+KZ^{D}H)^{i}KZ^{D}BA^{i}A^{\pi}. \end{eqnarray}$

(3.5)

From (3.4) and (3.5) it follows that $(A-CB)X=X(A-CB)$.

Now, using (3.5) and $A^{\pi}X=0$, we obtain

$\begin{equation} (X(A-CB)-I)X=0, \nonumber \end{equation}$

i.e., $X(A-CB)X=X$.

Finally, we will prove that $(A-CB)-(A-CB)^{2}X$ is a nilpotent matrix. Using $A^{\pi}C=0,CZ^{\pi}B=0$, and expressions (3.4) conveniently, it can be proved that

$\begin{equation} (A-CB)-(A-CB)^{2}X=AA^{\pi}+\sum\limits^{k-1}_{i=0}(A^{D}+KZ^{D}H)^{i}KZ^{D}BA^{i+1}A^{\pi}. \nonumber \end{equation}$

by induction on integer $j\geq1$, we have

$\begin{equation} [(A-CB)-(A-CB)^{2}X]^{j}=A^{j}A^{\pi}+\sum\limits^{k-1}_{i=0}(A^{D}+KZ^{D}H)^{i}KZ^{D}BA^{i+j}A^{\pi}. \nonumber \end{equation}$

Then we get

$\begin{equation} [(A-CB)-(A-CB)^{2}X]^{k}=0. \end{equation}$

(3.6)

where $k={\rm ind}(A)$. Therefore, we get $(A-CB)^{k+1}X=(A-CB)^{k}$ and ${\rm ind}(A-CB)\leq {\rm ind}(A)$. The theorem is proved completely.

Corollary 3.2 Let $A,B$ and $C$ be complex matrices, where ind$(A)=k$. If $BA^{\pi}=0$ and $CZ^{\pi}B=0$, then

$\begin{equation} (A-CB)^{D}=A^{D}+KZ^{D}H-\sum\limits^{k-1}_{i=0}KZ^{D}BA^{i}A^{\pi}(A^{D}+KZ^{D}H)^{i+1}, \end{equation}$

(3.7)

furthermore, ind$(A-CB)\leq$ ind$(A)$.

Proof  The proof is similar to Theorem 3.1.

In the reference [5], it was discussed the Drazin inverse of a modified matrix $A-CB$, where $A$ is an idempotent matrix (Lemma 2.3). In this paper we will consider the consequence when $A$ is a $k$-idempotent matrix.

When $A$ is a $k$-idempotent matrix, we can easily proof $A^{D}=A^{k-2}$. Then, we can change notations (3.1) and (3.2) to be

$\begin{eqnarray} && K=A^{k-2}C,\quad H=BA^{k-2},\quad \Gamma=HK,\quad Z=I-BA^{k-2}C, \end{eqnarray}$

(E3.8)

$\begin{eqnarray} && P=(I-A^{k-1})C,\quad Q=B(I-A^{k-1}). \end{eqnarray}$

(3.9)

Theorem 3.3  Let $A$ be a $k$-idempotent matrix, suppose $AC=C$, ind$(Z)=k$, then

$\begin{equation} (A-CB)^{D}=A^{k-2}+KZ^{D}H-K(Z^{D})^{2}Q-KZ^{\pi}\sum\limits^{k-1}_{i=0}Z^{i}H. \end{equation}$

(3.10)

Proof  The proof is similar to Theorem 3.1.

Corollary 3.4  Let $A$ be a $k$-idempotent matrix, suppose $BA=B$, ind$(Z)=k$, then

$\begin{equation} (A-CB)^{D}=A^{k-2}+KZ^{D}H-P(Z^{D})^{2}H-\sum\limits^{k-1}_{i=0}KZ^{i}Z^{\pi}H. \end{equation}$

(3.11)

Proof  The proof is similar to Theorem 3.1.

Theorem 3.5  Let $A$, $B$, $C$ be diagonalizable. Suppose $A$, $B$, $C$ commute,

$ {\rm rank}(A)={\rm rank}(B)={\rm rank}(C)$

and $\sigma(A) \cap \sigma(BC) = \varnothing$, then

$\begin{equation} (A-CB)^{D}=A^{D}+A^{D}C(I-BA^{D}C)^{D}BA^{D}, \end{equation}$

(3.12)

where $\sigma(A)$ is the eigenvalues of $A$.

Proof  Because $A$, $B$ and $C$ are diagonalizable and they can commute, there is a nonsingular matrix $S$ such that $S^{-1}AS$, $S^{-1}BS$, $S^{-1}CS$ are diagonal. We denote

$\begin{equation} S^{-1}AS= \left[ {\begin{array}{*{20}{c}} {{\Lambda _1}}&o\\ o&o \end{array}} \right]\ \ S^{-1}BS= \left[ {\begin{array}{*{20}{c}} {{\Lambda _2}}&o\\ o&o \end{array}} \right]\ \ S^{-1}CS= \left[ {\begin{array}{*{20}{c}} {{\Lambda _3}}&o\\ o&o \end{array}} \right], \end{equation}$

where each of matrices $\Lambda _1$, $\Lambda _2$ and $\Lambda _3$ is full rank diagonal and its diagonal line elements are eigenvalues. Their rank is equal to $A$. Then

$ (A-CB)^{D}=S \left[ {\begin{array}{*{20}{c}} {{(\Lambda _1-\Lambda _3\Lambda _2)^{-1}}}&o\\ o&o \end{array}} \right]\ S^{-1}, $

as $\sigma(A) \cap \sigma(BC) = \varnothing$, $\Lambda _1-\Lambda _3\Lambda _2$ is nonsingular. At he same time, we have

$ A^D+A^{D}C(I-BA^{D}C)^{D}BA^{D}=S \left[ {\begin{array}{*{20}{c}} {{\Lambda _1^{-1}+\Lambda _1^{-1}\Lambda _3(I-\Lambda _2\Lambda _1^{-1}\Lambda _3)^{-1}\Lambda _2\Lambda _1^{-1}}}&o\\ o&o \end{array}} \right]\ S^{-1} $

and

$\begin{equation} (\Lambda _1-\Lambda _3\Lambda _2)^{-1}=\Lambda _1^{-1}+\Lambda _1^{-1}\Lambda _3(I-\Lambda _2\Lambda _1^{-1}\Lambda _3)^{-1}\Lambda _2\Lambda _1^{-1}, \end{equation}$

(3.13)

so we have $(A-CB)^{D}=A^{D}+A^{D}C(I-BA^{D}C)^{D}BA^{D}$.