Additive Results for the Generalized Drazin Inverse in a Banach Algebra

Abstract

Under new conditions on Banach algebra elements a and b, we derive explicit expressions for the generalized Drazin inverse of the sum $a+b$. As an application of our results, we present new representations for the generalized Drazin inverse of a block matrix in a Banach algebra.

1 Introduction

Let ${\mathcal {A}}$ be a complex unital Banach algebra with unit 1. The sets of all invertible, nilpotent, and quasinilpotent elements of ${\mathcal {A}}$ will be denoted by ${\mathcal {A}}^{-1}$, ${\mathcal {A}}^{nil}$, and ${\mathcal {A}}^{qnil}$, respectively.

Let us recall that the generalized Drazin inverse of $a\in {\mathcal {A}}$ (or Koliha–Drazin inverse of a [9]) is the unique element $a^d\in {\mathcal {A}}$ which satisfies

$$\begin{aligned} a^daa^d=a^d,\quad aa^d=a^da,\quad a-a^2a^d\in {\mathcal {A}}^{qnil}. \end{aligned}$$

We use $a^\pi =1-aa^d$ to denote the spectral idempotent of a corresponding to the set $\{0\}$. Notice that, if $a\in {\mathcal {A}}^{qnil}$, then $a^d=0$. The set ${\mathcal {A}}^d$ consists of all generalized Drazin invertible elements of ${\mathcal {A}}$. If $a-a^2a^d\in {\mathcal {A}}^{nil}$ in the above definition, then $a^d=a^D$ is ordinary Drazin inverse. The group inverse is a special case of the Drazin inverse for which $a=aa^da$ instead of $a-a^2a^d\in {\mathcal {A}}^{nil}$. We use $a^\#$ to denote the group inverse of a, and ${\mathcal {A}}^\#$ to denote the set of all group invertible elements of ${\mathcal {A}}$.

The next auxiliary result, which is proved in [7, Lemma 2.1] for bounded linear operators, has been extended to Banach algebra elements in [4].

Lemma 1.1

[4, Lemma 2.1] Let $a, b\in {\mathcal {A}}^{qnil}$. If $ab=ba$ or $ab=0$, then $a+b\in {\mathcal {A}}^{qnil}$.

Let $p=p^2\in {\mathcal {A}}$ be an idempotent. Then we can represent element $a\in {\mathcal {A}}$ as

$$\begin{aligned} a=\left[ \begin{array}{cc} a_{11}&{}\quad a_{12}\\ a_{21}&{}\quad a_{22}\end{array}\right] , \end{aligned}$$

where $a_{11}=pap$, $a_{12}=pa(1-p)$, $a_{21} = (1-p)ap$, and $a_{22}= (1-p)a(1-p)$.

If $a\in {\mathcal {A}}^d$, we can write

$$\begin{aligned} a=\left[ \begin{array}{cc} a_{1}&{}\quad 0\\ 0&{}\quad a_{2} \end{array}\right] \end{aligned}$$

relative to $p=aa^d$, where $a_{1}\in (p{\mathcal {A}}p)^{-1}$ and $a_{2}\in ((1-p){\mathcal {A}}(1-p))^{qnil}$. Then the generalized Drazin inverse of a can be expressed as

$$\begin{aligned} a^d=\left[ \begin{array}{cc} a^d&{}\quad 0\\ 0&{}\quad 0\end{array}\right] =\left[ \begin{array}{cc} a_1^{-1}&{}\quad 0\\ 0&{}\quad 0\end{array}\right] . \end{aligned}$$

We use the following lemma.

Lemma 1.2

[2, Theorem 2.3] Let $x=\left[ \begin{array}{cc} a&{}\quad 0\\ c&{}\quad b\end{array}\right] \in {\mathcal {A}}$ relative to the idempotent $p\in {\mathcal {A}}$ and let $y=\left[ \begin{array}{cc} b&{}\quad c\\ 0&{}\quad a\end{array}\right] \in {\mathcal {A}}$ relative to the idempotent $1-p$.

• (i) If $a\in (p{\mathcal {A}}p)^d$ and $b\in ((1-p){\mathcal {A}}(1-p))^d$, then $x,y\in {\mathcal {A}}^d$ and

  $$\begin{aligned} x^d=\left[ \begin{array}{cc} a^d&{}\quad 0\\ u&{}\quad b^d\end{array}\right] , \quad y^d=\left[ \begin{array}{cc} b^d&{}\quad u\\ 0&{}\quad a^d\end{array}\right] , \end{aligned}$$

  where

  $$\begin{aligned} u=\sum \limits _{n=0}^{\infty }(b^d)^{n+2}ca^na^\pi + \sum \limits _{n=0}^{\infty }b^\pi b^{n}c(a^d)^{n+2}-b^dca^d. \end{aligned}$$

• (ii) If $x\in {\mathcal {A}}^d$ and $a\in (p{\mathcal {A}}p)^d$, then $b\in ((1-p){\mathcal {A}}(1-p))^d$ and $x^d$ is given as in part (i).

In this paper, we studied additive properties of the generalized Drazin inverse in a Banach algebra. Precisely, we find explicit formulae for the generalized Drazin inverse $(a+b)^d$ in the cases that $ab=b^\pi bab^\pi$ or $ab=a^\pi b^\pi bab^\pi$. Applying these expressions, some representations for the generalized Drazin inverse of a block matrix are presented.

2 Main Results

We start with an important special case of our main theorem.

Theorem 2.1

If $a\in {\mathcal {A}}^{qnil}$, $b\in {\mathcal {A}}^d$, and $ab=b^\pi bab^\pi$, then $a+b\in {\mathcal {A}}^d$ and

$$\begin{aligned} (a+b)^d=\sum _{n=0}^{\infty }(b^d)^{n+1}a^n. \end{aligned}$$

(1)

Proof

If $b\in {\mathcal {A}}^{qnil}$, by $b^\pi =1$ and $ab=b^\pi bab^\pi$, we get $ab=ba$. Applying Lemma 1.1, $a+b\in {\mathcal {A}}^{qnil}$ and the formula (1) is satisfied.

Suppose that $b\notin {\mathcal {A}}^{qnil}$. Then, relative to $p=bb^d$, we have the following representations of b and a:

$$\begin{aligned} b=\left[ \begin{array}{cc} b_{1}&{}\quad 0\\ 0&{}\quad b_{2}\end{array}\right] \quad \mathrm{and} \quad a=\left[ \begin{array}{cc} a_{1}&{}\quad a_{2}\\ a_{3}&{}\quad a_{4}\end{array}\right] , \end{aligned}$$

where $b_{1}\in (p{\mathcal {A}}p)^{-1}$ and $b_{2}\in ((1-p){\mathcal {A}}(1-p))^{qnil}$. So

$$\begin{aligned} b^d=\left[ \begin{array}{cc} b_1^{-1}&{}\quad 0\\ 0&{}\quad 0\end{array}\right] \quad \mathrm{and} \quad b^\pi =\left[ \begin{array}{cc} 0&{}\quad 0\\ 0&{}\quad 1-p\end{array}\right] . \end{aligned}$$

The equalities

$$\begin{aligned} \left[ \begin{array}{cc} a_{1}b_{1}&{}\quad a_{2}b_{2}\\ a_{3}b_{1}&{}\quad a_{4}b_{2}\end{array}\right] =ab=b^\pi bab^\pi =\left[ \begin{array}{cc} 0&{}\quad 0\\ 0&{}\quad b_{2}a_{4}\end{array}\right] \end{aligned}$$

imply $a_{1}b_{1}=0$, $a_{2}b_{2}=0$, $a_{3}b_{1}=0$, and $a_{4}b_{2}=b_{2}a_{4}$. Because $b_1$ is invertible, $a_1=0$ and $a_3=0$. Now, since $a\in {\mathcal {A}}^{qnil}$, $a_{4}\in ((1-p){\mathcal {A}}(1-p))^{qnil}$. By Lemma 1.1, $a_{4}+b_{2}\in ((1-p){\mathcal {A}}(1-p))^{qnil}$ and $(a_{4}+b_{2})^d=0$.

Using Lemma 1.2, $a_{2}b_{2}=0$ and $a_{4}b_{2}=b_{2}a_{4}$, we conclude that $a+b\in {\mathcal {A}}^d$ and

$$\begin{aligned}(a+b)^d= & {} \left[ \begin{array}{cc} b_{1}&{}\quad a_{2}\\ 0&{}\quad a_{4}+b_{2}\end{array}\right] ^d=\left[ \begin{array}{cc} b_1^{-1}&{}\quad \sum \limits _{n=0}^\infty b_1^{-(n+2)}a_2a_4^n\\ 0&{}\quad 0\end{array}\right] \\= & {} b^d+\sum _{n=0}^{\infty }(b^d)^{n+2}a^{n+1}=\sum _{n=0}^{\infty }(b^d)^{n+1}a^n. \end{aligned}$$

$\square$

Castro-González and Koliha obtained the formula (1) for $(a+b)^d$ in [2, Corollary 3.4] when $ab=0$ instead of $ab=b^\pi bab^\pi$ in Theorem 2.1.

Notice that the conditions $ab=0$ and $ab=b^\pi bab^\pi$ are independent, but in the both cases we obtain the same expressions for $(a+b)^d$. In the first example, we have that the condition $ab=0$ holds, but the condition $ab=b^\pi bab^\pi$ is not satisfied.

Example 2.1

Let ${\mathcal {A}}$ be the algebra of all complex $3\times 3$ matrices and let $a,b\in {\mathcal {A}}$ such that

$$\begin{aligned} a=\left[ \begin{array}{ccc} 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0 \end{array}\right] , \quad b=\left[ \begin{array}{ccc} 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0 \end{array}\right] . \end{aligned}$$

Thus, $ab=0$. From $b^2=0$, we have $b^\pi =1$ and

$$\begin{aligned} b^\pi bab^\pi =ba= \left[ \begin{array}{ccc} 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0 \end{array}\right] \ne ab. \end{aligned}$$

In the second example, we consider matrices a and b in the algebra ${\mathcal {A}}$ of all complex $3\times 3$ matrices such that $ab=b^\pi bab^\pi$ is satisfied but $ab=0$ does not hold.

Example 2.2

Let ${\mathcal {A}}$ be the algebra of all complex $3\times 3$ matrices and let $a,b\in {\mathcal {A}}$ such that

$$\begin{aligned} a=b=\left[ \begin{array}{ccc} 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 1\\ 0&{}\quad 0&{}\quad 0 \end{array}\right] . \end{aligned}$$

Then

$$\begin{aligned} a^2=\left[ \begin{array}{ccc} 0&{}\quad 0&{}\quad 1\\ 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0 \end{array}\right] \end{aligned}$$

and $a^3=0$ implying $a^d=0$ and $a^\pi =1$. Hence, $0\ne ab=a^2=a^\pi a^2a^\pi =b^\pi bab^\pi$.

Now, we prove our main theorem.

Theorem 2.2

If $a,b\in {\mathcal {A}}^d$ and $ab=a^\pi b^\pi bab^\pi$, then $a+b\in {\mathcal {A}}^d$ and

$$\begin{aligned} (a+b)^d=b^\pi a^d+b^da^\pi +\sum _{n=1}^{\infty }(b^d)^{n+1}a^{n}a^\pi +\sum _{n=0}^{\infty }b^\pi (a+b)^{n}b(a^d)^{n+2}. \end{aligned}$$

(2)

Proof

First, if we assume that $a\in {\mathcal {A}}^{qnil}$, by Theorem 2.1, we get (2). For $a\in {\mathcal {A}}^{-1}$, we obtain $ab=0$ and the formula (2) holds by [2, Example 4.5].

Further, if a is neither invertible nor quasinilpotent, we have the following matrix representations of a and b relative to $p=aa^d$:

$$\begin{aligned} a=\left[ \begin{array}{cc} a_{1}&{}\quad 0\\ 0&{}\quad a_{2} \end{array}\right] , \quad b= \left[ \begin{array}{cc} b_{1}&{}\quad b_{2}\\ b_{3}&{}\quad b_{4} \end{array}\right] , \end{aligned}$$

where $a_{1}\in (p{\mathcal {A}}p)^{-1}$ and $a_{2}\in ((1-p){\mathcal {A}}(1-p))^{qnil}$.

From $ab=a^\pi b^\pi bab^\pi$, we obtain $a_1b_1=0$ and $a_1b_2=0$. Since $a_1$ is invertible, then $b_1=0$ and $b_2=0$. Now

$$\begin{aligned} b=\left[ \begin{array}{cc} 0&{}\quad 0\\ b_{3}&{}\quad b_{4} \end{array}\right] \end{aligned}$$

and, using Lemma 1.2, we observe that $b_{4}\in ((1-p){\mathcal {A}}(1-p))^{d}$,

$$\begin{aligned} b^d=\left[ \begin{array}{cc} 0&{}\quad 0\\ (b_{4}^d)^2b_{3}&{}\quad b_{4}^d \end{array}\right] \quad \mathrm{and} \quad b^\pi = \left[ \begin{array}{cc} p&{}\quad 0\\ -b_{4}^db_{3}&{}\quad b_{4}^\pi \end{array}\right] . \end{aligned}$$

Also, the equalities

$$\begin{aligned} \left[ \begin{array}{cc} 0&{}\quad 0\\ a_{2}b_3&{}\quad a_{2}b_4 \end{array}\right] =ab=a^\pi b^\pi bab^\pi = \left[ \begin{array}{cc} 0&{}\quad 0\\ b_{4}^\pi b_{3}a_1-b_{4}^\pi b_{4}a_2b_{4}^db_{3}&{}\quad b_{4}^\pi b_{4}a_2b_{4}^\pi \end{array}\right] \end{aligned}$$

give $a_{2}b_3=b_{4}^\pi b_{3}a_1-b_{4}^\pi b_{4}a_2b_{4}^db_{3}$ and $a_{2}b_4=b_{4}^\pi b_{4}a_2b_{4}^\pi$. By Theorem 2.1, we conclude that $a_2+b_4\in ((1-p){\mathcal {A}}(1-p))^{d}$ and

$$\begin{aligned} (a_2+b_4)^d=\sum _{n=0}^{\infty }(b_4^d)^{n+1}a_2^n. \end{aligned}$$

Now, from Lemma 1.2, $a+b\in {\mathcal {A}}^d$ and

$$\begin{aligned} (a+b)^d= \left[ \begin{array}{cc} a_1&{}\quad 0\\ b_{3}&{}\quad a_2+b_{4} \end{array}\right] ^d= \left[ \begin{array}{cc} a_1^{-1}&{}\quad 0\\ u&{}\quad (a_2+b_{4})^d \end{array}\right] , \end{aligned}$$

(3)

where

$$\begin{aligned} u=\sum _{n=0}^{\infty }(a_2+b_{4})^\pi (a_2+b_{4})^nb_3a_1^{-(n+2)} -(a_2+b_{4})^db_3a_1^{-1}. \end{aligned}$$

Because $a_{2}b_4=b_{4}^\pi b_{4}a_2b_{4}^\pi$, we get $a_2b_4^d=0$ and $b_4^da_{2}b_4=0$. By $a_{2}b_3=b_{4}^\pi b_{3}a_1-b_{4}^\pi b_{4}a_2b_{4}^db_{3}$, we obtain $a_{2}b_3=b_{4}^\pi b_{3}a_1$ and $b_4^da_{2}b_3=0$. Since

$$\begin{aligned} b_4^da_{2}a_{2}b_3=b_4^da_{2}b_{4}^\pi b_{3}a_1=b_4^da_{2}b_{3}a_1=0 \end{aligned}$$

and

$$\begin{aligned} b_4^da_{2}a_{2}b_4=b_4^da_{2}b_{4}^\pi b_{4}a_2b_{4}^\pi =b_4^da_{2}b_{4}a_2b_{4}^\pi =0, \end{aligned}$$

then $b_4^da_{2}^nb_3=0$ and $b_4^da_{2}^nb_4=0$, for all $n\ge 1$, which imply $(b_4^d)^{k+1}a_2^{k+1}(a_2+b_{4})^nb_3=0$, for all $k,n\ge 0$. Thus,

$$\begin{aligned} (a_2+b_4)^\pi= & {} (1-p)-(a_2+b_4)(a_2+b_4)^d=(1-p)-b_4\sum _{n=0}^{\infty }(b_4^d)^{n+1} a_2^n\\= & {} b_4^\pi -\sum _{n=0}^{\infty }(b_4^d)^{n+1}a_2^{n+1} \end{aligned}$$

which yields

$$\begin{aligned} u= & {} \sum _{n=0}^{\infty }b_{4}^\pi (a_2+b_{4})^nb_3a_1^{-(n+2)}- \sum _{n=0}^{\infty }\sum _{k=0}^{\infty }(b_4^d)^{k+1}a_2^{k+1} (a_2+b_{4})^nb_3a_1^{-(n+2)}\\&-b_4^db_3a_1^{-1}-\sum _{n=1}^{\infty }(b_4^d)^{n+1}a_2^{n}b_3a_1^{-1}\\= & {} \sum _{n=0}^{\infty }b_{4}^\pi (a_2+b_{4})^nb_3a_1^{-(n+2)}- b_4^db_3a_1^{-1}. \end{aligned}$$

Using the equalities

$$\begin{aligned} X_1= & {} b^da^\pi +\sum _{n=1}^{\infty }(b^d)^{n+1}a^{n}a^\pi = \left[ \begin{array}{cc} 0&{}\quad 0\\ 0&{}\quad b_{4}^d \end{array}\right] +\sum _{n=1}^{\infty } \left[ \begin{array}{cc} 0&{}\quad 0\\ 0&{}\quad (b_{4}^d)^{n+1}a_2^{n} \end{array}\right] \\= & {} \left[ \begin{array}{cc} 0&{}\quad 0\\ 0&{}\quad \sum \limits _{n=0}^{\infty }(b_{4}^d)^{n+1}a_2^{n} \end{array}\right] = \left[ \begin{array}{cc} 0&{}\quad 0\\ 0&{}\quad (a_2+b_4)^d \end{array}\right] ,\\ X_2= & {} \sum _{n=0}^{\infty }b^\pi (a+b)^{n}b(a^d)^{n+2} =\left[ \begin{array}{ll} 0&{}\quad 0\\ \sum \limits _{n=0}^{\infty }b_4^\pi (a_2+b_4)^{n}b_3a_1^{-(n+2)}&{}\quad 0 \end{array}\right] ,\\ X_3= & {} b^\pi a^d= \left[ \begin{array}{cc} a_1^{-1}&{}\quad 0\\ -b_4^db_3a_1^{-1}&{}\quad 0 \end{array}\right] \end{aligned}$$

and (3), we deduce that

$$\begin{aligned} (a+b)^d=X_1+X_2+X_3 \end{aligned}$$

and the formula (2) is satisfied. $\square$

Note that under the conditions of Theorem 2.2, it can be verified that $b^dab=0$. Precisely, $b_4^da_{2}^nb_3=0$ and $b_4^da_{2}^nb_4=0$ imply $b^da^nb=0$ for all $n\ge 1$.

In [2, Theorem 3.5], Castro-González and Koliha presented the expression for the generalized Drazin inverse of $a+b$ in the case that $a^\pi b=b$, $ab^\pi =a$, and $b^\pi aba^\pi =0$. Observe that matrices a and b introduced in Example 2.2 satisfy the conditions of Theorem 2.2, but the assumption $b^\pi aba^\pi =0$ of [2, Theorem 3.5] does not hold.

Liu and Qin [10] derived a formula for $(a+b)^d$ under the conditions $ab=a^\pi bab^\pi$. The following example describes two matrices a and b in the algebra of all complex $2\times 2$ matrices which do not satisfy the conditions of [10, Theorem 2.2], whereas the conditions of Theorem 2.2 are met, which allows us to compute $(a+b)^d$.

Example 2.3

Let ${\mathcal {A}}$ be the algebra of all complex $2\times 2$ matrices and let $a,b\in {\mathcal {A}}$ such that

$$\begin{aligned} a=\left[ \begin{array}{cc} 0&{}\quad 2\\ 0&{}\quad 0 \end{array}\right] , \quad b= \left[ \begin{array}{cc} 1&{}\quad 0\\ 0&{}\quad 0 \end{array}\right] . \end{aligned}$$

Since $a^2=0$ and $b^2=b$, then $a^\pi =1$ and

$$\begin{aligned} b^\pi =1-b= \left[ \begin{array}{cc} 0&{}\quad 0\\ 0&{}\quad 1 \end{array}\right] . \end{aligned}$$

Thus, $0=ab\ne a^\pi bab^\pi =a$ and $ab=0=a^\pi b^\pi bab^\pi$.

3 Applications

The problem of finding explicit representations for the Drazin inverse of a $2\times 2$ block matrix in terms of its blocks was posed by Campbell and Meyer [1]. There have been many papers on this subject, under different conditions [3, 5, 8, 11–14], but it is still a hard problem to find an explicit formula for the Drazin inverse of a block matrix.

In this section, as an application of Theorem 2.2, we obtain representations for the generalized Drazin inverse of a block matrix $x\in {\mathcal {A}}$ given by

$$\begin{aligned} x=\left[ \begin{array}{cc} a&{}\quad b\\ c&{}\quad d \end{array}\right] \end{aligned}$$

(4)

relative to the idempotent $p\in {\mathcal {A}}$, where $a\in (p{\mathcal {A}}p)^d$ and $d\in ((1-p){\mathcal {A}}(1-p))^d$. Throughout this section, if the lower limit of a sum is greater than its upper limit, we always define the sum to be 0. For example, the sum $\sum \nolimits _{k=1}^0*=0$.

Theorem 3.1

Let x be defined as in (4). If

$$\begin{aligned} a^\pi bc=0,\quad a^\pi bd=ab,\quad \mathrm{{and}} \quad \sum _{n=0}^\infty (d^d)^nca^nb=0, \end{aligned}$$

(5)

then $x\in {\mathcal {A}}^d$ and

$$\begin{aligned} x^d= \left[ \begin{array}{cc} a^d&{}\quad 0\\ u&{}\quad d^d \end{array}\right] +\sum _{n=0}^{\infty }x^{n} \left[ \begin{array}{cc} i_n-\sum \limits _{k=1}^{n}b(d^d)^{k+1}c(a^d)^{n+2-k}&{}\quad b(d^d)^{n+2}\\ 0&{}\quad 0 \end{array}\right] , \end{aligned}$$

(6)

where

$$\begin{aligned} u= & {} \sum \limits _{n=0}^{\infty }(d^d)^{n+2}ca^na^\pi + \sum \limits _{n=0}^{\infty }d^\pi d^{n}c(a^d)^{n+2}-d^dca^d,\nonumber \\ i_n= & {} \sum \limits _{k=0}^{\infty }bd^\pi d^{k}c(a^d)^{n+k+3}\!-\!bd^dc(a^d)^{n+2}\!+\!\sum \limits _{k=0}^{\infty }b(d^d)^{n+k+3} ca^ka^\pi -b(d^d)^{n+2}ca^d,\nonumber \\ \end{aligned}$$

(7)

for $n\ge 0$.

Proof

Let

$$\begin{aligned} x=\left[ \begin{array}{cc} a&{}\quad 0\\ c&{}\quad d \end{array}\right] + \left[ \begin{array}{cc} 0&{}\quad b\\ 0&{}\quad 0 \end{array}\right] :=y+z. \end{aligned}$$

(8)

Applying Lemma 1.2, we have that $y\in {\mathcal {A}}^d$,

$$\begin{aligned} y^d=\left[ \begin{array}{cc} a^d&{}\quad 0\\ u&{}\quad d^d \end{array}\right] \quad \mathrm{and}\quad y^\pi = \left[ \begin{array}{cc} a^\pi &{}\quad 0\\ -ca^d-du&{}\quad d^\pi \end{array}\right] , \end{aligned}$$

where u is defined as in (7). Observe that $z^2=0$ implies $z^d=0$ and $z^\pi =1$.

Then

$$\begin{aligned} yz=\left[ \begin{array}{cc} 0&{}\quad ab\\ 0&{}\quad cb \end{array}\right] \quad \mathrm{and}\quad y^\pi z^\pi zyz^\pi = \left[ \begin{array}{cc} a^\pi bc&{}\quad a^\pi bd\\ (-ca^d-du)bc&{}\quad (-ca^d-du)bd \end{array}\right] . \end{aligned}$$

The hypothesis $a^\pi bd=ab$ gives $a^db=0$ and $bd=ab$. Now, from $a^\pi bc=0$, we get $bc=0$. By the third equality in (5), we obtain

$$\begin{aligned} cb+dubd= & {} cb+\sum \limits _{n=0}^{\infty }(d^d)^{n+1}ca^nbd =cb+\sum \limits _{n=0}^{\infty }(d^d)^{n+1}ca^{n+1}b\\= & {} \sum \limits _{n=0}^\infty (d^d)^nca^nb=0. \end{aligned}$$

Hence, $yz=y^\pi z^\pi zyz^\pi$ which yields, by Theorem 2.2, $x\in {\mathcal {A}}^d$ and

$$\begin{aligned}x^d= & {} y^d +\sum _{n=0}^{\infty }x^{n}z(y^d)^{n+2}\\= & {} \left[ \begin{array}{cc} a^d&{}\quad 0\\ u&{}\quad d^d \end{array}\right] +\sum _{n=0}^{\infty }x^{n} \left[ \begin{array}{cc} \sum \limits _{k=0}^{n+1}b(d^d)^{k}u(a^d)^{n+1-k}&{}\quad b(d^d)^{n+2}\\ 0&{}\quad 0 \end{array}\right] \\= & {} \left[ \begin{array}{cc} a^d&{}\quad 0\\ u&{}\quad d^d \end{array}\right] \\&+\sum _{n=0}^{\infty }x^{n} \left[ \begin{array}{cc} bu(a^d)^{n+1}-\sum \limits _{k=1}^{n}b(d^d)^{k+1}c(a^d)^{n+2-k}+b(d^d)^{n+1}u &{}\quad b(d^d)^{n+2}\\ 0&{}\quad 0 \end{array}\right] \end{aligned}$$

implying that the equality (6) is satisfied. $\square$

Observe that the third equality in (5) can be replaced by weaker assumption $cb=0$, since the equalities $cb=0$ and $bd=ab$ give $ca^nb=0$, for $n\ge 0$.

In the following theorem, we derive a formula for the generalized Drazin inverse of x under some rather complicated conditions, but the theorem itself will have useful consequences which will include much simpler conditions.

Theorem 3.2

Let x be defined as in (4) and let u be defined as in (7). If

$$\begin{aligned} a^\pi abd^\pi =bd,\quad (-ca^d-du)ab+d^\pi cb=0,\quad \mathrm{{and}} \quad \sum _{n=0}^\infty bd^{n}c(a^d)^{n}=0, \end{aligned}$$

(9)

then $x\in {\mathcal {A}}^d$ and

$$\begin{aligned} x^d=\left[ \begin{array}{cc} a^d&{}\quad (a^d)^{2}b\\ u&{}\quad d^d+i\end{array}\right] , \end{aligned}$$

(10)

where

$$\begin{aligned} i=\sum \limits _{n=0}^{\infty }d^\pi d^{n}c(a^d)^{n+3}b-d^dc(a^d)^{2}b+\sum \limits _{n=0}^{\infty }(d^d)^{n+3}ca^na^\pi b-(d^d)^{2}ca^db. \end{aligned}$$

Proof

If we suppose that x is represented as in (8) and denote by $t=(-ca^d-du)ab+d^\pi cb$, we have

$$\begin{aligned} zy=\left[ \begin{array}{cc} bc&{}\quad bd\\ 0&{}\quad 0 \end{array}\right] \quad \mathrm{and}\quad y^\pi yzy^\pi = \left[ \begin{array}{cc} a^\pi ab(-ca^d-du)&{}a^\pi abd^\pi \\ t(-ca^d-du)&{}td^\pi \end{array}\right] . \end{aligned}$$

From the condition $a^\pi abd^\pi =bd$, we obtain $bd^d=0$ and $a^\pi ab=bd$. So, using the assumptions (9), note that $t=0$,

$$\begin{aligned} a^\pi ab(-ca^d-du)= & {} -bdca^d-bd^2u=-bdca^d-\sum _{n=1}^\infty bd^{n+1}c(a^d)^{n+1}\\= & {} -\sum _{n=1}^\infty bd^{n}c(a^d)^{n}=bc \end{aligned}$$

and $zy=z^\pi y^\pi yzy^\pi$. Applying Theorem 2.1, we conclude that $x\in {\mathcal {A}}^d$ and

$$\begin{aligned} x^d=y^d+(y^d)^2z=\left[ \begin{array}{cc} a^d&{}\quad (a^d)^{2}b\\ u&{}\quad d^d+ua^db+d^dub\end{array}\right] , \end{aligned}$$

which yields that (10) holds. $\square$

The following corollary presents conditions weaker than those given in Theorem 3.2 under which we have simpler expression for $x^d$.

Corollary 3.1

Let x be defined as in (4) and let u be defined as in (7).

• (i) If $a^\pi abd^\pi =bd$, $ca^dab=d^\pi cb$, $dca=0$, and $bc=0$, then $x\in {\mathcal {A}}^d$ and

  $$\begin{aligned} x^d=\left[ \begin{array}{cc} a^d&{}\quad (a^d)^2b\\ (d^d)^2c+c(a^d)^2&{}\quad d^d+c(a^d)^3b+(d^d)^3cb\end{array}\right] . \end{aligned}$$

• (ii) If $a^\pi abd^\pi =bd$, $ca^\pi b=0$, $dc=0$, and $bc=0$, then $x\in {\mathcal {A}}^d$ and

  $$\begin{aligned} x^d=\left[ \begin{array}{cc} a^d&{}\quad (a^d)^2b\\ c(a^d)^2&{}\quad d^d+c(a^d)^3b\end{array}\right] . \end{aligned}$$

• (iii) If $ab=0$, $bd=0$, $d^\pi cb=0$, and $bc=0$, then $x\in {\mathcal {A}}^d$ and

  $$\begin{aligned} x^d=\left[ \begin{array}{cc} a^d&{}\quad 0\\ u&{}\quad d^d+(d^d)^3cb\end{array}\right] . \end{aligned}$$

• (iv) If $ab=0$, $bd=0$, $cb=0$, and $bc=0$, then $x\in {\mathcal {A}}^d$ and

  $$\begin{aligned} x^d=\left[ \begin{array}{cc} a^d&{}\quad 0\\ u&{}\quad d^d\end{array}\right] . \end{aligned}$$

The formula for the generalized Drazin inverse given in part (ii) of Corollary 3.1 was obtained for operator matrices in [6, Theorem 5.3] in the case that $bd=0$, $dc=0$, and $bc=0$.

If we suppose that a and d are group invertible in Theorems 3.1 and 3.2, we get the following representations of $x^d$.

Corollary 3.2

Let x be defined as in (4), $a\in (p{\mathcal {A}}p)^\#$ and $d\in ((1-p){\mathcal {A}}(1-p))^\#$.

• (i) If the equalities (5) hold, then $x\in {\mathcal {A}}^d$ and

  $$\begin{aligned} x^d= & {} \left[ \begin{array}{cc} a^\#&{}\quad 0\\ (d^\#)^2ca^\pi +d^\pi c(a^\#)^2-d^\#ca^\#&{}\quad d^\#\end{array}\right] \\&+\sum _{n=0}^{\infty }x^n\left[ \begin{array}{cc} i'_n-\sum \limits _{k=1}^{n}b(d^\#)^{k+1}c(a^\#)^{n+2-k}&{}\quad b(d^\#)^{n+2}\\ 0&{}\quad 0 \end{array}\right] , \end{aligned}$$

  where $i'_n=bd^\pi c(a^\#)^{n+3}-bd^\#c(a^\#)^{n+2}+b(d^\#)^{n+3}ca^\pi -b(d^\#)^{n+2}ca^\#$, for $n\ge 0$.

• (ii) If $bd=0$, $(-ca^\#-dd^\#ca^\#)ab+d^\pi cb=0$, and $bc=0$, then $x\in {\mathcal {A}}^d$ and

  $$\begin{aligned} x^d=\left[ \begin{array}{cc} a^\#&{}\quad (a^\#)^{2}b\\ (d^\#)^2ca^\pi +d^\pi c(a^\#)^2-d^\#ca^\#&{}\quad d^\#+i'\end{array}\right] , \end{aligned}$$

  where $i'=d^\pi c(a^\#)^{3}b-d^\#c(a^\#)^{2}b+(d^\#)^{3}ca^\pi b-(d^\#)^{2}ca^\#b.$

In a similar way as it was done in the previous theorems, using the another splitting, we present new expressions for the generalized Drazin inverse of a block matrix in a Banach algebra.

Theorem 3.3

Let x be defined as in (4). If

$$\begin{aligned} d^\pi cb=0,\quad d^\pi ca=dc\quad \mathrm{{and}} \quad \sum _{n=0}^\infty (a^d)^nbd^nc=0, \end{aligned}$$

(11)

then $x\in {\mathcal {A}}^d$ and

$$\begin{aligned} x^d=\left[ \begin{array}{cc} a^d&{}\quad v\\ 0&{}\quad d^d \end{array}\right] +\sum _{n=0}^{\infty }x^{n} \left[ \begin{array}{cc} 0&{}\quad 0\\ c(a^d)^{n+2}&{}\quad j_n-\sum \limits _{k=1}^{n}c(a^d)^{n+2-k}b(d^d)^{k+1} \end{array}\right] , \end{aligned}$$

(12)

where

$$\begin{aligned} v= & {} \sum \limits _{n=0}^{\infty }(a^d)^{n+2}bd^nd^\pi + \sum \limits _{n=0}^{\infty }a^\pi a^{n}b(d^d)^{n+2}-a^dbd^d,\nonumber \\ j_n= & {} \sum \limits _{k=0}^{\infty }c(a^d)^{n+k+3}bd^kd^\pi -c(a^d)^{n+2}bd^d \!+\!\sum \limits _{k=0}^{\infty }ca^\pi a^{k}b(d^d)^{n+k+3}\!-\!ca^db(d^d)^{n+2},\nonumber \\ \end{aligned}$$

(13)

for $n\ge 0$.

Proof

If we write

$$\begin{aligned} x=\left[ \begin{array}{cc} a&{}\quad b\\ 0&{}\quad d \end{array}\right] +\left[ \begin{array}{cc} 0&{}\quad 0\\ c&{}\quad 0 \end{array}\right] :=y+z, \end{aligned}$$

(14)

then $z^2=0=z^d$, $z^\pi =0$,

$$\begin{aligned} y^d=\left[ \begin{array}{cc} a^d&{}\quad v\\ 0&{}\quad d^d \end{array}\right] \quad \mathrm{and}\quad y^\pi =\left[ \begin{array}{cc} a^\pi &{}\quad -av-bd^d\\ 0&{}\quad d^\pi \end{array}\right] . \end{aligned}$$

The equalities (11) give $yz=y^\pi z^\pi zyz^\pi$. By Theorem 2.2, similarly as in the proof of Theorem 3.1, we have that $x\in {\mathcal {A}}^d$ and $x^d$ is represented as in (12). $\square$

Instead of the third condition of (11), we can assume that weaker condition $bc=0$ holds.

Theorem 3.4

Let x be defined as in (4) and let v be defined as in (13). If

$$\begin{aligned} d^\pi dca^\pi =ca,\quad (-av-bd^d)dc+a^\pi bc=0,\quad \mathrm{{and}} \quad \sum _{n=0}^\infty ca^{n}b(d^d)^{n}=0, \end{aligned}$$

then $x\in {\mathcal {A}}^d$ and

$$\begin{aligned} x^d=\left[ \begin{array}{cc} a^d+j&{}\quad v\\ (d^d)^{2}c&{}\quad d^d \end{array}\right] , \end{aligned}$$

where

$$\begin{aligned} j=\sum \limits _{n=0}^{\infty }(a^d)^{n+3}bd^nd^\pi c-(a^d)^2bd^dc +\sum \limits _{n=0}^{\infty }a^\pi a^{n}b(d^d)^{n+3}c-a^db(d^d)^2c. \end{aligned}$$

Proof

Using the decomposition (14) of x, as Theorem 3.2, we prove this result. $\square$

Remark that, if $b=0$ in Theorem 3.1 (or Theorem 3.2) and $c=0$ in Theorem 3.3 (or Theorem 3.4), we obtain Lemma 1.2 (i).

As a consequence of Theorem 3.4, we get the next result.

Corollary 3.3

Let x be defined as in (4) and let v be defined as in (13).

• (i) If $d^\pi dca^\pi =ca$, $bd^ddc=a^\pi bc$, $abd=0$, and $cb=0$, then $x\in {\mathcal {A}}^d$ and

  $$\begin{aligned} x^d=\left[ \begin{array}{cc} a^d+(a^d)^3bc+b(d^d)^3c&{}\quad (a^d)^2b+b(d^d)^2\\ (d^d)^2c&{}\quad d^d\end{array}\right] . \end{aligned}$$

• (ii) If $d^\pi dca^\pi =ca$, $bd^\pi c=0$, $ab=0$, and $bc=0$, then $x\in {\mathcal {A}}^d$ and

  $$\begin{aligned} x^d=\left[ \begin{array}{cc} a^d+b(d^d)^3c&{}\quad b(d^d)^2\\ (d^d)^2c&{}\quad d^d\end{array}\right] . \end{aligned}$$

• (iii) If $dc=0$, $ca=0$, $a^\pi bc=0$, and $cb=0$, then $x\in {\mathcal {A}}^d$ and

  $$\begin{aligned} x^d=\left[ \begin{array}{cc} a^d+(a^d)^3bc&{}\quad v\\ 0&{}\quad d^d \end{array}\right] . \end{aligned}$$

• (iv) If $dc=0$, $ca=0$, $bc=0$, and $cb=0$, then $x\in {\mathcal {A}}^d$ and

  $$\begin{aligned} x^d=\left[ \begin{array}{cc} a^d&{}\quad v\\ 0&{}\quad d^d\end{array}\right] . \end{aligned}$$

Applying Theorems 3.3 and 3.4, we verify the following corollary.

Corollary 3.4

Let x be defined as in (4), $a\in (p{\mathcal {A}}p)^\#$ and $d\in ((1-p){\mathcal {A}}(1-p))^\#$.

• (i) If the equalities (11) hold, then $x\in {\mathcal {A}}^d$ and

  $$\begin{aligned} x^d= & {} \left[ \begin{array}{cc} a^\#&{}\quad (a^\#)^2bd^\pi +a^\pi b(d^\#)^2-a^\#bd^\#\\ 0&{}\quad d^\#\end{array}\right] \\&+\sum _{n=0}^{\infty }x^{n} \left[ \begin{array}{cc} 0&{}\quad 0\\ c(a^\#)^{n+2}&{}j'_n-\sum \limits _{k=1}^{n}c(a^\#)^{n+2-k}b(d^\#)^{k+1}\end{array}\right] , \end{aligned}$$

  where $j'_n=c(a^\#)^{n+3}bd^\pi -c(a^\#)^{n+2}bd^\#+ca^\pi b(d^\#)^{n+3}-ca^\#b(d^\#)^{n+2}$, for $n\ge 0$.

• (ii) If $ca=0$, $(-aa^\#bd^\#-bd^\#)dc+a^\pi bc=0$, and $cb=0$, then $x\in {\mathcal {A}}^d$ and

  $$\begin{aligned} x^d=\left[ \begin{array}{cc} a^\#+j'&{}\quad (a^\#)^2bd^\pi +a^\pi b(d^\#)^2-a^\#bd^\#\\ (d^\#)^{2}c&{}\quad d^\#\end{array}\right] , \end{aligned}$$

  where $j'=(a^\#)^{3}bd^\pi c-(a^\#)^2bd^\#c +a^\pi b(d^\#)^{3}c-a^\#b(d^\#)^2c.$