1 Introduction

Let \(\mathcal {X}\) and \(\mathcal {Y}\) be Banach spaces. Denote by \({\mathcal B}(\mathcal {X, Y})\) the set of all bounded linear operators from \(\mathcal {X}\) to \(\mathcal {Y}\), and \({\mathcal B}(\mathcal {X, X})\) is simplified to \({\mathcal B}(\mathcal {X})\). For the operator \(T\in {\mathcal B}(\mathcal {X})\), if there exists an operator \(T^D\in {\mathcal B}(\mathcal {X})\) such that

$$\begin{aligned} TT^D=T^D T,\ \ T^D=T(T^D)^2,\ \ T^{k+1}T^D =T^k \ \ \mathrm {for \ some\ integer }\ k \ge 0, \end{aligned}$$

then T is said to be Drazin invertible, and \(T^D\) is called the Drazin inverse of T. The Drazin index of T, written by \(\mathrm {ind}(T)\), is the smallest integer k such that the previous system of equations holds. In particular, if \( \mathrm {ind}(T)=0\), it can be seen that T is invertible and \(T^D=T^{-1}\). Moreover, \(T^D\) may not exist. For example, the right shift operator \(A: l^2[1,\infty )\rightarrow l^2[1,\infty )\) given by

$$\begin{aligned} A(x_1, x_2,x_3,\ldots )=(0, x_1, x_2,\ldots ),\ \ (x_1, x_2,x_3,\ldots )\in l^2[1,\infty ) \end{aligned}$$

is not Drazin invertible, since \(R(A^{k+1})\subset R(A^{k})\) and \(R(A^{k+1})\ne R(A^{k})\) for any integer k.

In 1983, Campbell [4] showed that the solutions of some second-order singular linear differential equations can be expressed by the Drazin inverse of the anti-triangular matrix \(M= \left( {\begin{matrix} A &{} B \\ C &{} 0 \end{matrix}}\right) \). From then on, many researchers began to consider the representation of the Drazin inverse for M under various conditions on A, B or C [2, 3, 5, 7, 11, 12, 14]. However, it is difficult to obtain the representation without any restriction, and it is still an open problem up to now. For more general Drazin inverse problems, we refer the reader to [1, 6, 8,9,10, 13, 15,16,17,18,19] and their references.

In this paper, the existence of the Drazin inverse for the anti-triangular operator matrix \(M= \left( {\begin{matrix} A &{} B \\ C &{} 0 \end{matrix}}\right) \) under the conditions \(A^2=A\) and \( CAB=0\) is proved. The explicit expression of \(M^D\) is further given, and the Drazin index of M is not more than \(2\ \mathrm {ind} (CB)+2\). Finally, an example is given to illustrate the obtained result.

Throughout the paper, assume that \(A\in \mathcal {B}(\mathcal {X}), B\in {\mathcal B}(\mathcal {Y, X}), C\in {\mathcal B}(\mathcal {X, Y})\) and CB is Drazin invertible with \(\mathrm {ind}(CB)=s\). We define a sum to be zero if its lower limits are larger than its upper limits. Write \(A^\pi =I-AA^D\) and

$$\begin{aligned} \begin{aligned} F_{11}&= \sum _{k=1}^{s}{A^{\pi }B(CB)^{k-1}(CB)^{\pi }CA}+\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CA^{\pi }}\\&\quad -\,\sum _{k=1}^{s}{2kAB(CB)^{k-1}(CB)^{\pi }CA}-A^{\pi }B(CB)^{D}CA-AB(CB)^{D}CA^{\pi },\\ F_{12}&= A^{\pi }B(CB)^{D}+\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }},\\ F_{21}&= (CB)^{D}CA^{\pi }+\sum _{k=1}^{s}{(CB)^{k-1}(CB)^{\pi }CA}. \end{aligned} \end{aligned}$$

2 Main Results

We start with the following lemmas, which is necessary to prove our main results.

Lemma 2.1

If \(A^{2}=A\) and \(CAB=0\), then

  1. (i)

    \(F_{11}A+F_{12}C=AF_{11}+BF_{21},\)

  2. (ii)

    \(F_{11}B+AB=AF_{12}.\)

Proof

  1. (i)

    From \(A^{2}=A\), it follows that the operator A is Drazin invertible and \(A=A^{D}\), and hence, \(AA^{\pi }=0\). Thus,

    $$\begin{aligned}&\begin{aligned} F_{11}A+F_{12}C&= \sum _{k=1}^{s}{A^{\pi }B(CB)^{k-1}(CB)^{\pi }CA}-\sum _{k=1}^{s}{2kAB(CB)^{k-1}(CB)^{\pi }CA}\\&\quad -\,A^{\pi }B(CB)^{D}CA+A^{\pi }B(CB)^{D}C +\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }C}, \end{aligned} \end{aligned}$$
    (2.1)
    $$\begin{aligned}&\begin{aligned} AF_{11}+BF_{21}&= \displaystyle \sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CA^{\pi }} -\sum _{k=1}^{s}{2kAB(CB)^{k-1}(CB)^{\pi }CA}\\&\quad -\,AB(CB)^{D}CA^{\pi }+B(CB)^{D}CA^{\pi }+\sum _{k=1}^{s}B{(CB)^{k-1}(CB)^{\pi }CA}. \end{aligned} \end{aligned}$$
    (2.2)

    By \(A^{\pi }=I-A\) and (2.1), we get

    $$\begin{aligned} \begin{aligned}&F_{11}A+F_{12}C \\&\quad = \sum _{k=1}^{s}{B(CB)^{k-1}(CB)^{\pi }CA}-\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CA}\\&\qquad -\,\sum _{k=1}^{s}{2kAB(CB)^{k-1}(CB)^{\pi }CA}-B(CB)^{D}CA+AB(CB)^{D}CA\\&\qquad +\,B(CB)^{D}C-AB(CB)^{D}C+\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }C}\\&\quad = \sum _{k=1}^{s}{B(CB)^{k-1}(CB)^{\pi }CA}+\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CA^{\pi }}\\&\qquad -\,\sum _{k=1}^{s}{2kAB(CB)^{k-1}(CB)^{\pi }CA}+B(CB)^{D}CA^{\pi }-AB(CB)^{D}CA^{\pi }. \end{aligned} \end{aligned}$$
    (2.3)

    This together with (2.2) shows that \(F_{11}A+F_{12}C=AF_{11}+BF_{21}\).

  2. (ii)

    From \(CAB=0\), we have \(CA^{\pi }B=CB\). Note that \((CB)^{s}(CB)^{\pi }=0\) by \(\text{ ind }(CB)=s\). Then

    $$\begin{aligned} F_{11}B+AB= & {} \sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CB}-AB(CB)^{D}CB+AB\\= & {} \sum _{k=1}^{s}{AB(CB)^{k}(CB)^{\pi }}+AB(CB)^{\pi }\\= & {} \sum _{k=0}^{s}{AB(CB)^{k}(CB)^{\pi }}\\= & {} \sum _{k=0}^{s-1}{AB(CB)^{k}(CB)^{\pi }}. \end{aligned}$$

    Obviously, \(F_{11}B+AB=AF_{12}\) is obtained. \(\square \)

Lemma 2.2

If \(A^{2}=A\) and \(CAB=0\), then, for \(l=2,3,\ldots \), we have

$$\begin{aligned} M^{l}=\left( \begin{array}{ll} M^{l}_{11} &{}\quad M^{l}_{12}\\ M^{l}_{21} &{}\quad M^{l}_{22} \end{array} \right) , \end{aligned}$$

where if \(l=2m\), then

$$\begin{aligned} \begin{aligned}&M^{2m}_{11}=\sum _{k=0}^{m-1}{(BC)^{k}A}+\sum _{k=1}^{m-1}{A(BC)^{k}}+\sum _{k=0}^{m-2}{(2k+1)A(BC)^{m-1-k}A}+(BC)^{m},\\&M^{2m}_{12}=\sum _{k=0}^{m-1}{AB(CB)^{k}},\quad M^{2m}_{21}=\sum _{k=0}^{m-1}{(CB)^{k}CA},\quad M^{2m}_{22}=(CB)^{m}; \end{aligned} \end{aligned}$$

if \(l=2m+1\), then

$$\begin{aligned} \begin{aligned} M^{2m+1}_{11}&=\sum _{k=0}^{m}{(BC)^{k}A}+\sum _{k=1}^{m}{A(BC)^{k}}+\sum _{k=1}^{m-1}{2kA(BC)^{m-k}A},\\ M^{2m+1}_{12}&=\sum _{k=0}^{m-1}{AB(CB)^{k}}+(BC)^{m}B, \\ M^{2m+1}_{21}&=\sum _{k=0}^{m-1}{(CB)^{k}CA}+(CB)^{m}C,\quad M^{2m+1}_{22}=0. \end{aligned} \end{aligned}$$

Proof

We prove it by induction on integer \(l \ge 2\). For \(l =2\), the conclusion can be easily checked. Assume that the conclusion is satisfied for l. Next we will check it for \(l + 1\).

If \(l=2m\), then

$$\begin{aligned} M^{l+1}=M^{2m}M=\left( \begin{array}{ll} M^{2m}_{11} &{} \quad M^{2m}_{12}\\ M^{2m}_{21} &{} \quad M^{2m}_{22} \\ \end{array}\right) \left( \begin{array}{ll} A &{} \quad B \\ C &{} \quad 0 \\ \end{array} \right) =\left( \begin{array}{ll} M^{2m}_{11}A+ M^{2m}_{12}C &{} \quad M^{2m}_{11}B \\ M^{2m}_{21}A+M^{2m}_{22}C &{} \quad M^{2m}_{21}B \end{array} \right) . \end{aligned}$$

By the given conditions, we have

$$\begin{aligned} \begin{aligned} M^{2m}_{11}A+ M^{2m}_{12}C&= \sum _{k=0}^{m-1}{(BC)^{k}A}+\sum _{k=1}^{m-1}{A(BC)^{k}A}+\sum _{k=0}^{m-2}{(2k+1)A(BC)^{m-1-k}A}\\&\quad +\,(BC)^{m}A+\sum _{k=0}^{m-1}{AB(CB)^{k}C}. \end{aligned} \end{aligned}$$

Note that

$$\begin{aligned} \sum _{k=1}^{m-1}{A(BC)^{k}A}+\sum _{k=0}^{m-2}{(2k+1)A(BC)^{m-1-k}A}= & {} \displaystyle \sum _{k=0}^{m-2}{(2k+2)A(BC)^{m-1-k}}A\\= & {} \displaystyle \sum _{k=1}^{m-1}{2kA(BC)^{m-k}A} \end{aligned}$$

and

$$\begin{aligned} \sum _{k=0}^{m-1}{AB(CB)^{k}C}=\sum _{k=0}^{m-1}{A(BC)^{k+1}}=\sum _{k=1}^{m}{A(BC)^{k}}. \end{aligned}$$

So, it follows that

$$\begin{aligned} M^{2m}_{11}A+ M^{2m}_{12}C= & {} \sum _{k=0}^{m}{(BC)^{k}A}+\sum _{k=1}^{m}{A(BC)^{k}}\\&+\sum _{k=1}^{m-1}{2kA(BC)^{m-k}A}=M^{2m+1}_{11}. \end{aligned}$$

Similarly, by \(CAB=0\), we get

$$\begin{aligned} \begin{aligned} M^{2m}_{11}B&= AB+\sum _{k=1}^{m-1}{A(BC)^{k}B}+(BC)^{m}B = M^{2m+1}_{12}. \end{aligned} \end{aligned}$$

On the other hand, the equalities \(M^{2m}_{21}A+M^{2m}_{22}C=M^{2m+1}_{21}\) and \(M^{2m}_{21}B=M^{2m+1}_{22}\) are obvious. Therefore, the conclusion holds for \(l+1\).

If \(l=2m+1\), then the proof is analogous, and we omit it. \(\square \)

Lemma 2.3

If \(A^{2}=A\) and \(CAB=0\), then the following statements are true:

  1. (i)

    \( M^{2(s+1)+1}_{11}(F_{11}+A)+M^{2(s+1)+1}_{12}F_{21}= M^{2(s+1)}_{11},\)

  2. (ii)

    \(M^{2(s+1)+1}_{21}(F_{11}+A)+M^{2(s+1)+1}_{22}F_{21}=M^{2(s+1)}_{21}.\)

Proof

  1. (i)

    By \(AA^{\pi }=0, (CB)^{s}(CB)^{\pi }=0\) and \( (CB)^{s+1}(CB)^{D}=(CB)^{s},\) we obtain

    $$\begin{aligned} {} \begin{aligned}&M^{2(s+1)+1}_{11}(F_{11}+A)+M^{2(s+1)+1}_{12}F_{21}\\&\quad =\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CA^{\pi }}-\sum _{k=1}^{s}{2kAB(CB)^{k-1}(CB)^{\pi }CA}-AB(CB)^{D}CA^{\pi }\\&\qquad +\,\sum _{k=1}^{s+1}{A(BC)^{k}}\cdot \sum _{k=1}^{s}{A^{\pi }B(CB)^{k-1}(CB)^{\pi }CA}-\sum _{k=1}^{s+1}{A(BC)^{k}}\cdot A^{\pi }B(CB)^{D}CA\\&\qquad +\,\sum _{k=0}^{s+1}{(BC)^{k}A}+\sum _{k=1}^{s+1}{A(BC)^{k}A}+\sum _{k=1}^{s}{2kA(BC)^{s+1-k}A}\\&\qquad +\sum _{k=0}^{s}{AB(CB)^{k}(CB)^{D}CA^{\pi }}\\&\qquad +\,\sum _{k=0}^{s}{AB(CB)^{k}}\cdot \sum _{k=1}^{s}{(CB)^{k-1}(CB)^{\pi }CA}+B(CB)^{s}CA^{\pi }. \end{aligned} \end{aligned}$$
    (2.4)

    From \((CB)^{\pi }=I-CB(CB)^{D}\), it follows that

    $$\begin{aligned} {}\nonumber&\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CA^{\pi }}-AB(CB)^{D}CA^{\pi }+\sum _{k=0}^{s}{AB(CB)^{k}(CB)^{D}CA^{\pi }}\\\nonumber&\quad =\sum _{k=1}^{s}{AB(CB)^{k-1}(I-CB(CB)^{D})CA^{\pi }}+\sum _{k=1}^{s}{AB(CB)^{k}(CB)^{D}CA^{\pi }}\\&\quad =\sum _{k=1}^{s}{AB(CB)^{k-1}CA^{\pi }}. \end{aligned}$$
    (2.5)

    Since \(CA^{\pi }B=CB \), we have

    $$\begin{aligned}&\sum _{k=1}^{s+1}{A(BC)^{k}}\cdot \sum _{k=1}^{s}{A^{\pi }B(CB)^{k-1}(CB)^{\pi }CA}\nonumber \\&\qquad +\sum _{k=0}^{s}{AB(CB)^{k}}\cdot \sum _{k=1}^{s}{(CB)^{k-1}(CB)^{\pi }CA}\nonumber \\&\quad =\sum _{k=1}^{s-1}{AB(CB)^{k}}\cdot \sum _{k=1}^{s}{(CB)^{k-1}(CB)^{\pi }CA}\nonumber \\&\qquad +\sum _{k=0}^{s-1}{AB(CB)^{k}}\cdot \sum _{k=1}^{s}{(CB)^{k-1}(CB)^{\pi }CA}\nonumber \\&\quad =\sum _{k=1}^{s-1}{AB(CB)^{k}}\cdot \sum _{k=1}^{s}{(CB)^{k-1}(CB)^{\pi }CA}\nonumber \\&\qquad +\sum _{k=1}^{s-1}{AB(CB)^{k}}\cdot \sum _{k=1}^{s}{(CB)^{k-1}(CB)^{\pi }CA}\nonumber \\&\qquad +\,\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CA}\nonumber \\&\quad =2\sum _{k=1}^{s-1}{AB(CB)^{k}}\cdot \sum _{k=1}^{s}{(CB)^{k-1}(CB)^{\pi }CA}+\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CA}\nonumber \\ \end{aligned}$$
    (2.6)

    and

    $$\begin{aligned} \nonumber \sum _{k=1}^{s+1}{A(BC)^{k}}\cdot A^{\pi }B(CB)^{D}CA= & {} \sum _{k=1}^{s+1}{AB(CB)^{k}(CB)^{D}CA}\nonumber \\= & {} \sum _{k=1}^{s}{AB(CB)^{k}(CB)^{D}CA}+AB(CB)^{s}CA\nonumber \\= & {} \sum _{k=1}^{s}{AB(CB)^{k}(CB)^{D}CA}+A(BC)^{s+1}A.\nonumber \\ \end{aligned}$$
    (2.7)

    Again,

    $$\begin{aligned} \nonumber&\sum _{k=1}^{s+1}{A(BC)^{k}A}+\sum _{k=1}^{s}{2kA(BC)^{s+1-k}A}\\\nonumber&\quad =A(BC)^{s+1}A+2\sum _{k=1}^{s}{A(BC)^{k}A}+\sum _{k=1}^{s}{(2k-1)A(BC)^{s+1-k}A}\\&\quad =A(BC)^{s+1}A+2\sum _{k=1}^{s}{A(BC)^{k}A}+\sum _{k=0}^{s-1}{(2k+1)A(BC)^{s-k}A}. \end{aligned}$$
    (2.8)

    Thus, instituting (2.5), (2.6), (2.7) and (2.8) into (2.4) yields

    $$\begin{aligned} \nonumber&M^{2(s+1)+1}_{11}(F_{11}+A)+M^{2(s+1)+1}_{12}F_{21}\\\nonumber&\quad =\sum _{k=1}^{s}{AB(CB)^{k-1}CA^{\pi }}-\sum _{k=1}^{s}{2kAB(CB)^{k-1}(CB)^{\pi }CA}\\\nonumber&\qquad +\,2\sum _{k=1}^{s-1}{AB(CB)^{k}}\cdot \sum _{k=1}^{s}{(CB)^{k-1}(CB)^{\pi }CA}+\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CA}\\\nonumber&\qquad -\,\sum _{k=1}^{s}{AB(CB)^{k}(CB)^{D}CA}+\sum _{k=0}^{s}{(BC)^{k}A}+(BC)^{s+1}A\\&\qquad +\,\sum _{k=0}^{s-1}{(2k+1)A(BC)^{s-k}A}+2\sum _{k=1}^{s}{A(BC)^{k}A}+B(CB)^{s}CA^{\pi }. \end{aligned}$$
    (2.9)

    Note that

    $$\begin{aligned} \nonumber&2\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CA}+2\sum _{k=1}^{s-1}{AB(CB)^{k}}\cdot \sum _{k=1}^{s}{(CB)^{k-1}(CB)^{\pi }CA}\\&\quad =\sum _{k=1}^{s}{2kAB(CB)^{k-1}(CB)^{\pi }CA} \end{aligned}$$
    (2.10)

    and

    $$\begin{aligned}&\sum _{k=1}^{s}{AB(CB)^{k-1}CA^{\pi }}+\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CA}-\sum _{k=1}^{s}{AB(CB)^{k}(CB)^{D}CA}\nonumber \\&\qquad +\,2\sum _{k=1}^{s}{A(BC)^{k}A}\nonumber \\&\quad =\sum _{k=1}^{s}{AB(CB)^{k-1}C}-2\sum _{k=1}^{s}{AB(CB)^{k}(CB)^{D}CA}+2\sum _{k=1}^{s}{A(BC)^{k}A}\nonumber \\&\quad =\sum _{k=1}^{s}{A(BC)^{k}}-2\sum _{k=1}^{s}{AB(CB)^{k}(CB)^{D}CA}+2\sum _{k=1}^{s}{AB(CB)^{k-1}CA}\nonumber \\&\quad =\sum _{k=1}^{s}{A(BC)^{k}}+2\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CA} \end{aligned}$$
    (2.11)

    by \(A^{\pi }=I-A\) and \((CB)^{\pi }=I-CB(CB)^{D}\). Combining (2.9), (2.11) and (2.10), we have

    $$\begin{aligned}&M^{2(s+1)+1}_{11}(F_{11}+A)+M^{2(s+1)+1}_{12}F_{21}\\&\quad =\sum _{k=0}^{s}{(BC)^{k}A}+(BC)^{s+1}A+\sum _{k=1}^{s}{A(BC)^{k}}+\sum _{k=0}^{s-1}{(2k+1)A(BC)^{s-k}A}\\&\qquad +(BC)^{s+1}A^{\pi }\\&\quad =\sum _{k=0}^{s-1}{(BC)^{k}A}+\sum _{k=1}^{s}{A(BC)^{k}}+\sum _{k=0}^{s-1}{(2k+1)A(BC)^{s-k}A}+(BC)^{s+1}\\&\quad =M^{2(s+1)}_{11}. \end{aligned}$$
  2. (ii)

    The facts

    $$\begin{aligned} \begin{aligned} M^{2(s+1)+1}_{21}F_{11}&=-(CB)^{s+1}CA,\\ M^{2(s+1)+1}_{21}A&=\sum _{k=0}^{s}{(CB)^{k}CA}+(CB)^{s+1}CA,\\ M^{2(s+1)+1}_{22}F_{21}&=0 \end{aligned} \end{aligned}$$

    can be obtained from \(CAB=0, AA^{\pi }=0, (CB)^{s}(CB)^{\pi }=0\) and \((CB)^{s+1}(CB)^{D}=(CB)^{s}\). Thus, claim (ii) follows immediately. The proof is completed. \(\square \)

The following is the main result of this section.

Theorem 2.1

For the anti-triangular operator matrix \(M=\left( \begin{array}{ll} A &{} B\\ C &{} 0 \\ \end{array} \right) \), if \(A^{2}=A\), \(CAB=0\), and CB is Drazin invertible, then M is Drazin invertible, and the representation of the Drazin inverse \(M^D\) is given by

\(M^{D}=\left( \begin{array}{ll} F_{11}+A &{} \quad F_{12} \\ F_{21}&{} \quad 0\\ \end{array} \right) \).

Moreover, \(\mathrm {ind} (M)\le 2s+2\).

Proof

Let \(X=\left( \begin{array}{ll} F_{11}+A &{} F_{12} \\ F_{21} &{} 0 \\ \end{array}\right) \). In the following, we prove that X satisfies the definition of Drazin inverse, namely \(X = M^D\).

Firstly, we verify \(MX=XM\). By computation,

$$\begin{aligned} \begin{aligned}&MX=\left( \begin{array}{ll} AF_{11}+A+BF_{21} &{} \quad AF_{12} \\ CF_{11}+CA &{} \quad CF_{12} \\ \end{array} \right) ,\quad \\&XM=\left( \begin{array}{ll} F_{11}A+A+F_{12}C &{} \quad F_{11}B+AB \\ F_{21}A &{} \quad F_{21}B \\ \end{array} \right) . \end{aligned} \end{aligned}$$
(2.12)

By \(CA^{\pi }B=CB\), we get

$$\begin{aligned} \begin{aligned} CF_{11}+CA&=\sum _{k=1}^{s}{(CB)^{k}(CB)^{\pi }CA+(CB)^{\pi }CA}&=\sum _{k=0}^{s-1}{(CB)^{k}(CB)^{\pi }CA}. \end{aligned} \end{aligned}$$

Moreover, \(F_{21}A=\sum _{k=0}^{s-1}{(CB)^{k}(CB)^{\pi }CA}\) from \(AA^{\pi }=0\). It is clear that \(CF_{11}+CA= F_{21}A\). So, \(MX=XM\) follows from Lemma 2.1 together with \(CF_{12}=CB(CB)^{D}=F_{21}B.\)

Secondly, we verify \(XMX = X\). By (2.12),

$$\begin{aligned} XMX=\left( \begin{array}{ll} T &{} \quad P \\ R &{} \quad Q \\ \end{array}\right) \left( \begin{array}{ll} F_{11}+A &{} \quad F_{12}\\ F_{21} &{} \quad 0 \\ \end{array}\right) =\left( \begin{array}{ll} TF_{11}+TA+PF_{21}&{}\quad TF_{12}\\ RF_{11}+RA+QF_{21} &{}\quad RF_{12}\\ \end{array} \right) , \end{aligned}$$

where \(T =F_{11}A+A+F_{12}C,\ \ P =F_{11}B+AB,\ \ R =F_{21}A \) and \(Q =F_{21}B.\)

We only prove \(TF_{11}+TA+PF_{21}=F_{11}+A\) and \(TF_{12}=F_{12}\). In the same way, \( RF_{11}+RA+QF_{21} =F_{21}\) and \( RF_{12} =0\) can be proved.

Since \((CB)^{\pi }(CB)^{\pi }=(CB)^{\pi }\), by (2.10), we obtain

$$\begin{aligned} \begin{aligned}&TF_{11}+TA+PF_{21}\\&=2\sum _{k=1}^{s-1}{AB(CB)^{k}}\cdot \sum _{k=1}^{s}{(CB)^{k-1}(CB)^{\pi }CA} -A^{\pi }B(CB)^{D}CA-AB(CB)^{D}CA^{\pi }\\&\quad +\,\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CA^{\pi }}+\sum _{k=1}^{s}{A^{\pi }B(CB)^{k-1}(CB)^{\pi }CA}\\&\quad +\,2\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CA}-\sum _{k=1}^{s}{4kAB(CB)^{k-1}(CB)^{\pi }CA}+A\\&=-A^{\pi }B(CB)^{D}CA-AB(CB)^{D}CA^{\pi }+\sum _{k=1}^{s}{A^{\pi }B(CB)^{k-1}(CB)^{\pi }CA}\\&\quad +\,\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }CA^{\pi }}-\sum _{k=1}^{s}{2kAB(CB)^{k-1}(CB)^{\pi }CA}+A\\&=F_{11}+A. \end{aligned} \end{aligned}$$

Again, from \(((CB)^{D})^{2}CB=(CB)^{D}\), it follows that

$$\begin{aligned} \begin{aligned} TF_{12}&=A^{\pi }B(CB)^{D}CB(CB)^{D}+\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }}\\&= A^{\pi }B(CB)^{D}+\sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }}\\&= F_{12}. \end{aligned} \end{aligned}$$

Thus, we conclude that \(XMX=X\).

Finally, we prove that there exists an integer r such that \(M^{r+1}X=M^{r}\). We set \(r=2s+2\), then

$$\begin{aligned} \begin{aligned} M^{r+1}X=\left( \begin{array}{ll} M^{2(s+1)+1}_{11}(F_{11}+A)+M^{2(s+1)+1}_{12}F_{21} &{} \quad M^{2(s+1)+1}_{11}F_{12} \\ M^{2(s+1)+1}_{21}(F_{11}+A)+M^{2(s+1)+1}_{22}F_{21}&{} \quad M^{2(s+1)+1}_{21}F_{12} \\ \end{array} \right) . \end{aligned} \end{aligned}$$

By Lemma 2.3, we only need to prove that

$$\begin{aligned} M^{2(s+1)+1}_{11}F_{12}=M^{2(s+1)}_{12} \quad \text{ and } \ \ \ M^{2(s+1)+1}_{21}F_{12}=M^{2(s+1)}_{22}. \end{aligned}$$

A simple computation shows that

$$\begin{aligned} \begin{aligned} M^{2(s+1)+1}_{11}F_{12}&= \sum _{k=1}^{s+1}{AB(CB)^{k-1}(CB)^{\pi }}+\sum _{k=1}^{s+1}{A(BC)^{k}\cdot A^{\pi }B(CB)^{D}}\\&= \sum _{k=1}^{s+1}{AB(CB)^{k-1}(CB)^{\pi }}+\sum _{k=1}^{s+1}{AB(CB)^{k}(CB)^{D}}\\&= \sum _{k=1}^{s+1}{AB(CB)^{k-1}}\\&= M^{2(s+1)}_{12} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} M^{2(s+1)+1}_{21}F_{12} = (CB)^{s+2}(CB)^{D}=(CB)^{s+1}=M^{2(s+1)}_{22}. \end{aligned} \end{aligned}$$

So, \(M^{r+1}X=M^{r}\) holds for \(r = 2s+2\)

Hence, by the definition of Drazin inverse, we have \(X=M^{D}\), which shows that the operator matrix M is Drazin invertible and \(\mathrm {ind} (M)\le 2s+2\). This completes the proof. \(\square \)

The following corollaries are obvious by Theorem 2.1.

Corollary 2.1

For the anti-triangular operator matrix \(M=\left( \begin{array}{ll} A &{} B\\ C &{} 0 \\ \end{array} \right) \), if \(A^{2}=A\), \(CA=0\), and CB is Drazin invertible, then M is Drazin invertible, and

$$\begin{aligned}M^{D}=\left( \begin{array}{ll} \displaystyle \sum _{k=1}^{s}AB(CB)^{k-1}C-AB(CB)^{D}C+A &{} \quad A^{\pi }B(CB)^{D}+\displaystyle \sum _{k=1}^{s}{AB(CB)^{k-1}(CB)^{\pi }} \\ (CB)^{D}C&{} \quad 0\\ \end{array} \right) .\end{aligned}$$

Corollary 2.2

For the anti-triangular operator matrix \(M=\left( \begin{array}{ll} A &{} B\\ C &{} 0 \\ \end{array} \right) \), if \(A^{2}=A\), \(AB=0\), and CB is Drazin invertible, then M is Drazin invertible, and

$$\begin{aligned}M^{D}=\left( \begin{array}{ll} \displaystyle \sum _{k=1}^{s}B(CB)^{k-1}(CB)^{\pi }CA-B(CB)^{D}CA+A &{} \quad B(CB)^{D} \\ (CB)^{D}CA^{\pi }+\sum _{k=1}^{s}{(CB)^{k-1}(CB)^{\pi }CA}&{} \quad 0\\ \end{array} \right) . \end{aligned}$$

3 Numerical Example

In order to illustrate the representation of \(M^D\) in Theorem 2.1, we give the following example.

Example

Let \(M=\left( \begin{array}{ll} A &{} B\\ C &{} 0 \\ \end{array} \right) \), where

$$\begin{aligned} A=\left( \begin{array}{ccc} 0 &{} \quad 1 &{} \quad -1\\ 1 &{} \quad 0 &{} \quad 1 \\ 1 &{} \quad -1 &{} \quad 2 \\ \end{array} \right) , \quad B=\left( \begin{array}{cc} 1 &{} \quad -1 \\ 1 &{} \quad 1 \\ -1 &{} \quad 1 \\ \end{array} \right) , \quad C=\left( \begin{array}{lll} 1 &{} \quad 1 &{} \quad 1 \\ 2 &{} \quad 0 &{} \quad 2 \\ \end{array} \right) . \end{aligned}$$

By computation, it follows that \(A^2=A, CAB=0\). So all the conditions of Theorem 2.1 are satisfied. On the other hand,

$$\begin{aligned} A=A^D,\ \ CB=\left( \begin{array}{ll} 1 &{} \quad 1 \\ 0 &{} \quad 0 \\ \end{array} \right) ,\ \ \ \ (CB)^{D}=\left( \begin{array}{ll} 1 &{} \quad 1 \\ 0 &{} \quad 0 \\ \end{array} \right) . \end{aligned}$$

Thus,

$$\begin{aligned} F_{11}=\left( \begin{array}{ccc} 14 &{} \quad -2 &{} \quad 14\\ -4 &{} \quad 0 &{} \quad -4 \\ -14 &{} \quad 2 &{} \quad -14 \\ \end{array} \right) ,\quad F_{12}=\left( \begin{array}{cc} -1 &{} \quad -3\\ 1 &{} \quad 1 \\ 1&{} \quad 3 \\ \end{array} \right) , \quad F_{21}=\left( \begin{array}{ccc} -3 &{} \quad 1 &{} \quad -3 \\ 2 &{} \quad 0 &{} \quad 2 \\ \end{array} \right) . \end{aligned}$$

Therefore, by Theorem 2.1, the Drazin inverse of M is

$$\begin{aligned} M^{D}=\left( \begin{array}{ccccc} 14 &{} \quad -1 &{} \quad 13 &{} \quad -1 &{} \quad -3 \\ -3 &{} \quad 0 &{} \quad -3 &{} \quad 1 &{} \quad 1 \\ -13 &{} \quad 1 &{} \quad -12 &{} \quad 1 &{} \quad 3 \\ -3 &{} \quad 1 &{} \quad -3 &{} \quad 0 &{} \quad 0 \\ 2 &{} \quad 0 &{} \quad 2 &{} \quad 0 &{} \quad 0 \\ \end{array} \right) . \end{aligned}$$

Remark

In the above example, \(ABC\ne 0 \) and \(BCA\ne 0 \). So the Drazin inverse of M cannot be obtained by [9, Theorem 3] and [6, Theorem 4.5].