Generalized Cline's Formula for G-Drazin Inverse

TL;DR

This paper extends Cline's formula to the G-Drazin inverse in associative rings, providing new conditions under which the inverse exists and exploring spectral properties of operators on Banach spaces.

Contribution

It generalizes Cline's formula for the G-Drazin inverse in rings, offering new theoretical insights and applications to operator spectral theory.

Findings

Established conditions for the existence of G-Drazin inverse of bd

Derived spectral properties of bounded linear operators

Extended Cline's formula to pseudo Drazin and Drazin inverses

Abstract

Let $R$ be an associative ring with an identity and suppose that $a,b,c,d \in R$ satisfy $bdb = bac,dbd = acd$. If $ac$ has generalized Drazin ( respectively, pseudo Drazin, Drazin) inverse, we prove that $bd$ has generalized Drazin (respectively, pseudo Drazin, Drazin) inverse. In particular, as applications, we obtain new common spectral properties of bounded linear operators over Banach spaces.

Equations

$$bdb=bac,dbd=acd.$$

$$b=bab,b\in com{m}^2(a),a-a^{2}b\in R^{nil}.$$

$$b=bab,b\in com{m}^2(a),a-a^{2}b\in R^{qnil}.$$

$$a\in {\mathcal{A}}^{qnil}\iff \underset{n\to \infty }{\lim }\parallel a^n{\parallel }^{\frac{1}{n}}=0\iff \lambda -a\in {\mathcal{A}}^{inv}\text{mbox}foranyscarlar\lambda \ne 0.$$

$$dba=aca,dbd=acd(\text{mbox}see\text{cite}[cite][\text{@@bibref}Mi,Theorem3.2]).$$

$$bdb=bac,dbd=acd(\text{mbox}see\text{cite}[cite][\text{@@bibref}M,Theorem2.5]).$$

$$b=bab,b\in com{m}^2(a),a^k-a^{k+1}b\in R^{rad}$$

$$\begin{array}[]{lll}(dx^{3}bac)ac&=&(dx^{3}bdb)ac\\ &=&(dbdx^{3})(bac)\\ &=&(acdx^{3})(bac)\\ &=&ac(dx^{3}bac).\end{array}$$

$$\begin{array}[]{lll}(1-xbd)(1+xbd+x^{2}bdbd)&=&(1+xbd+x^{2}bdbd)(1-xbd)\\ &=&1-x^{3}bdbdbd\\ &\in&R^{inv},\end{array}$$

$$ac(dtbac)=d(bd)tbac=dt(bdb)(ac)=dt(bac)(ac)=(dtbac)ac.$$

$$\begin{array}[]{lll}et&=&bh^{4}(ac)(ac)dt\\ &=&bh^{4}d(bd)^{2}t\\ &=&bh^{4}dt(bdb)d\\ &=&bh^{3}(hdtbac)d\\ &=&b(dtbac)h^{4}d\\ &=&tb(ac)^{2}h^{4}d\\ &=&te\end{array}$$

$$(db{)}^3(ac)=dbdbac(ac)=(ac)dbdbac=(db{)}^3(ac).$$

$$\begin{array}[]{lll}e(bd)e&=&bh^{2}d(bd)bh^{2}d=b(h^{3}ac)dbdbh^{2}d=bh^{3}(db)^{3}h^{2}d\\ &=&bh^{5}dbdbdbd=bh^{5}(ac)^{3}d=bh^{2}d=e.\end{array}$$

$$\begin{array}[]{lll}bd-(bd)^{2}e&=&bd-bdbdbh^{2}d=bd-bdbdbh^{2}d\\ &=&bd-bacach^{2}d=b(1-ach)d=b(pd).\end{array}$$

$$\begin{array}[]{lll}(pd)b(pd)&=&(1-ac)db(1-ac)d\\ &=&(1-ac)(dbd-dbacd)\\ &=&(1-ac)(acd-dbdbd)\\ &=&(1-ac)(acd-acacd)\\ &=&(pa)c(pd).\end{array}$$

$$\begin{array}[]{ll}&b(pd)b\\ =&b(1-ach)db=bdb-bachdb=bac-bh^{3}(db)^{3}ac\\ =&bac-b(db)^{3}h^{3}ac=bac-(bdb)dbdbh^{3}ac=bac-b(ac)^{3}h^{3}ac\\ =&bac-bhacac=b(1-ach)ac=b(pa)c.\end{array}$$

$$\begin{array}[]{c}a=\left(\begin{array}[]{cc}0&1\\ 0&0\end{array}\right),b=\left(\begin{array}[]{cc}1&0\\ 0&0\end{array}\right),\\ c=\left(\begin{array}[]{cc}1&0\\ 1&1\end{array}\right),d=\left(\begin{array}[]{cc}1&1\\ 0&0\end{array}\right)\in R.\end{array}$$

$$bdb=\left(\begin{array}[]{cc}1&0\\ 0&0\end{array}\right)\neq\left(\begin{array}[]{cc}1&1\\ 0&0\end{array}\right)=bac,dbd=\left(\begin{array}[]{cc}1&1\\ 0&0\end{array}\right)=acd,$$

$$dba=\left(\begin{array}[]{cc}0&1\\ 0&0\end{array}\right)=aca,dbd=acd.$$

$$\begin{array}[]{c}a=\left(\begin{array}[]{cc}0&1\\ 0&0\end{array}\right),b=\left(\begin{array}[]{cc}0&0\\ 0&1\end{array}\right),\\ c=\left(\begin{array}[]{cc}1&0\\ 1&1\end{array}\right),d=\left(\begin{array}[]{cc}1&0\\ -1&0\end{array}\right)\in R.\end{array}$$

$$bdb=0=bac,dbd=0=acd,$$

$$dba=0\neq\left(\begin{array}[]{cc}0&1\\ 0&0\end{array}\right)=aca,dbd=acd.$$

$$(bd)^{k+2}(bd)^{{\dagger}}-(bd)^{k+1}=b\big{(}(ac)^{k+1}(ac)^{d}-(ac)^{k}\big{)})$$

$$a=\left(\begin{array}[]{cc}0&1\\ 0&1\end{array}\right),b=\left(\begin{array}[]{cc}1&1\\ 0&0\end{array}\right),c=\left(\begin{array}[]{cc}1&-1\\ 0&0\end{array}\right),d=a.$$

$$\begin{array}[]{c}\sigma_{D}(a)=\{\lambda\in{\mathbb{C}}~{}|~{}\lambda-a\not\in A^{D}\};\\ \sigma_{d}(a)=\{\lambda\in{\mathbb{C}}~{}|~{}\lambda-a\not\in A^{d}\}.\end{array}$$

$$(1-bd{)}^{-1}=1+b(1-ac{)}^{-1}d.$$

$${\sigma }_d(AC)={\sigma }_d(BD).$$

$$\lambda =\underset{n\to \infty }{\lim }\{{\lambda }_n|{\lambda }_{n}I-AC\notin \mathcal{L}(X{)}^{-1}\}.$$

$$\lambda =\underset{n\to \infty }{\lim }\{{\lambda }_n|{\lambda }_{n}I-BD\notin \mathcal{L}(X{)}^{-1}\}\in acc\sigma (BD)={\sigma }_d(BD).$$

$${\sigma }_d(AB)={\sigma }_d(BA).$$

Full text

[TABLE]

If $ac$ has generalized Drazin ( respectively, pseudo Drazin, Drazin) inverse, we prove that $bd$ has generalized Drazin (respectively, pseudo Drazin, Drazin) inverse. In particular, as applications, we obtain new common spectral properties of bounded linear operators over Banach spaces.

1. Introduction

Let $R$ be an associative ring with an identity. The commutant of $a\in R$ is defined by $comm(a)=\{x\in R~{}|~{}xa=ax\}$. The double commutant of $a\in R$ is defined by $comm^{2}(a)=\{x\in R~{}|~{}xy=yx~{}\mbox{for all}~{}y\in comm(a)\}$.

An element $a\in R$ has Drazin inverse in case there exists $b\in R$ such that

[TABLE]

The preceding $b$ is unique if exists, we denote it by $a^{D}$. Let $a,b\in R$. Then $ab$ has Drazin inverse if and only if $ba$ has Drazin inverse and $(ba)^{D}=b((ab)^{D})^{2}a$. This was known as Cline’s formula for Drazin inverse. Cline’s formula plays an elementary role in matrix and operator theory (see [1, 2, 5, 7, 8, 9, 12, 13, 15, 16, 17]).

An element $a\in R$ has g-Drazin inverse (i.e., generalized Drazin inverse) in case there exists $b\in R$ such that

[TABLE]

The preceding $b$ is unique if exists, we denote it by $a^{d}$. Here, $R^{qnil}=\{a\in R~{}|~{}1+ax\in R^{inv}~{}\mbox{for every}~{}x\in comm(a)\}$. For a Banach algebra $\mathcal{A}$ it is well known that

[TABLE]

Many papers discussed Cline’s formula for g-Drazin inverse in the setting of matrices, operators, elements of Banach algebras or rings. For any $a,b\in R$, Liao et al. proved that $ab$ has g-Drazin inverse if and only if $ba$ has g-Drazin inverse and $(ba)^{d}=b((ab)^{d})^{2}a$. This was known as Cline’s formula for g-Drazin inverse (see [8, Theorem 2.1]). In  [7, Theorem 2.3], Lian and Zeng extended Cline’s formula for a generalized Drazin inverse to the case when $aba=aca$. As a further extension, Miller and Zguitti generalized Cline’s formula for g-Drazin inverse under the condition

[TABLE]

This was independently proved by Zeng et al. (see [17, Theorem 2.7]).

For any $a,b\in R$, Jacobson’s Lemma for g-Drazin inverse states that $1+ab\in R^{d}$ if and only if $1+ba\in R^{d}$ (see [18, Theorem 2.3]). Corach et al. [4] generalized Jacobson’s Lemma to the case that $aba=aca$. Recently, Yan et al. extended Jacobson’s Lemma to the case $dba=aca,dbd=acd~{}(\mbox{see~{}\cite[cite]{[\@@bibref{}{YZ}{}{}, Theorem 3.1]}}).$ In  [10], Mosic studied the generalization of Jacobson’s Lemma for g-Drazin inverses under a new condition

[TABLE]

The motivation of this paper is to investigate whether Cline’s formula holds under the preceding Mosic’s condition. In Section 2, we proved that if $ac$ has g-Drazin inverse then $bd$ has g-Drazin inverse under the condition $bdb=bac,dbd=acd$.

Recall that an element $a$ in a ring $R$ has pseudo Drazin (i.e., p-Drazin inverse) provided that there exists $b\in R$ such that

[TABLE]

for some $k\in{\mathbb{N}}$. The preceding $b$ is unique if exists, we denote it by $a^{{\dagger}}$. The smallest $k$ for which the preceding holds is called the p-Drazin index of $a$ and denote it by $i(a)$. The p-Drazin inverse is an intermediary between the Drazin and generalized Drazin inverses. In Section 3, as consequences of our main result, we investigate the corresponding Cline’s formula for p-Drazin and Drazin inverses.

In Section 4, as applications of our main result, we determine the common spectral properties of bounded linear operators over Banach spaces. Let $A,B,C,D\in\mathcal{L}(X)$ such that $BDB=BAC,$ $DBD=ACD$. We prove that $\sigma_{d}(AC)=\sigma_{d}(BD),$ where $\sigma_{d}$ is the g-Drazin spectrum.

Throughout the paper, all rings are associative with an identity. We use $R^{inv},R^{nil}$ and $R^{rad}$ to denote the set of all units, the set of all nilpotents and the Jacobson radical of the ring $R$, respectively. $R^{D}$ and $R^{d}$ denote the sets of all Drazin and g-Drazin invertible elements in $R$. ${\mathbb{C}}$ stands for the field of all complex numbers.

2. generalized Cline’s Formula

For any elements $a,b$ in a ring $R$, it is well known that $ab\in R^{qnil}$ if and only if $ba\in R^{qnil}$ (see [7, Lemma 2.2]). We now generalized this fact as follows.

Lemma 2.1**.**

Let $R$ be a ring, and let $a,b,c,d\in R$ satisfying $bdb=bac,dbd=acd$. If $ac\in R^{qnil}$, then $bd\in R^{qnil}$.

Proof.

Let $x\in comm(bd)$. Then we have

[TABLE]

Hence $dx^{3}bac\in comm(ac)$, and so $1-(dx^{3}bac)ac\in R^{inv}$. By using Jacobson’s Lemma, $1-x^{3}bdbdbd=1-x^{3}bacacd\in R^{inv}$. Then

[TABLE]

therefore $1-xbd\in R^{inv}$. This shows that $bd\in R^{qnil}.$∎

We come now to the main result of this paper.

Theorem 2.2**.**

Let $R$ be a ring, and let $a,b,c,d\in R$ satisfying $bdb=bac,dbd=acd$. If $ac\in R^{d}$, then $bd\in R^{d}$ and $(bd)^{d}=b((ac)^{d})^{2}d$.

Proof.

Suppose that $ac$ has g-Drazin inverse and $(ac)^{d}=h$. Let $e=bh^{2}d$ and $t\in comm(bd)$. Then

[TABLE]

Thus $dtbac\in comm(ac)$, and so $(dtbac)h=h(dtbac).$ Therefore we have

[TABLE]

Hence $e\in comm^{2}(bd)$.

Since $bd\in comm(bd)$, by the preceding discussion, we have $dbdbac\in comm(ac)$, and then

[TABLE]

This implies that $(db)^{3}h=h(db)^{3}$, and so

[TABLE]

Let $p=1-(ac)h$. Then $(pa)c=ac-achac=ac-(ac)^{2}h\in R^{qnil}$. Moreover, we have

[TABLE]

One easily checks that

[TABLE]

Moreover, we check that

[TABLE]

Then by Lemma 2.1, $b(pd)\in R^{qnil}$. Hence $bd$ has g-Drazin inverse $e$. That is, $e=bh^{2}a=(bd)^{d},$ as desired.∎

In the case that $c=b$ and $d=a$, we recover the Cline’s formula for g-Drazin inverse.

Corollary 2.3**.**

Let $R$ be a ring, and let $a,b\in R$. If $ab\in R^{d}$, then $ba\in R^{d}$ and $(ba)^{d}=b((ab)^{d})^{2}a$.

The following examples show that the preceding theorem is independent from [9, Theorem 3.2] and  [17, Theorem 2.7].

Example 2.4**.**

Let $R=M_{2}({\mathbb{C}})$. Choose

[TABLE]

Then we check that

[TABLE]

but

[TABLE]

In this case, $ac\in R^{D}$.

Example 2.5**.**

Let $R=M_{2}({\mathbb{C}})$. Choose

[TABLE]

Then we check that

[TABLE]

but

[TABLE]

In this case, $ac\in R^{D}$.

3. p-Drazin inverses

In this section, we investigate the Cline’s formula for the p-Drazin inverse. The following lemma is crucial.

Lemma 3.1**.**

Let $R$ be a ring, and let $a\in R$. If $a\in R^{{\dagger}}$, then $a\in R^{d}$ and $a^{{\dagger}}=a^{d}$.

Proof.

By hypothesis, there exists $b\in comm^{2}(a)$ such that $b=b^{2}a,a^{k}-a^{k+1}b\in R^{rad}$ for some $k\in{\mathbb{N}}$. Hence, $(a-a^{2}b)^{k}=a^{k}(1-ab)^{k}=a^{k}(1-ab)=a^{k}-a^{k+1}b\in R^{rad}$. Thus, $a-a^{2}b\in R^{qnil}$. Therefore $a$ has g-Drazin inverse $b$. In light of [6, Lemma 2.4], $b$ is unique, and so $b=a^{{\dagger}}$, as required.∎

Theorem 3.2**.**

Let $R$ be a ring, and let $a,b,c,d\in R$ satisfying $bdb=bac,dbd=acd$. If $ac\in R^{{\dagger}}$, $bd\in R^{{\dagger}}$, then $(bd)^{{\dagger}}=b((ac)^{{\dagger}})^{2}d$ and $i(bd)\leq i(ac)+1$.

Proof.

Suppose that $ac\in R^{{\dagger}}$. By virtue of Lemma 3.1, $(ac)^{d}=(ac)^{{\dagger}}$. In view of Theorem 2.2, $bd\in R^{d}$ and $(bd)^{d}=b((ac)^{{\dagger}})^{2}d$. One easily checks that

[TABLE]

for all $k\in{\mathbb{N}}$. Therefore $(bd)^{m}-(bd)^{m+1}(bd)^{{\dagger}}\in R^{rad}$ for some $m\in{\mathbb{N}}$, and so $bd\in R^{{\dagger}}$. Moreover, $(bd)^{{\dagger}}=(bd)^{d}=b((ac)^{{\dagger}})^{2}d$, as desired.∎

Corollary 3.3**.**

Let $R$ be a ring, and let $a,b\in R$. If $ab\in R^{{\dagger}}$, then $ba\in R^{{\dagger}}$ and $(ba)^{{\dagger}}=b((ab)^{{\dagger}})^{2}a$.

Proof.

This is obvious by choosing $c=b$ and $d=a$ in Theorem 3.2.∎

Theorem 3.4**.**

Let $R$ be a ring, and let $a,b,c,d\in R$ satisfying $bdb=bac,dbd=acd$. If $ac\in R^{D}$, then $(bd)^{D}=b((ac)^{D})^{2}d$ and $i(bd)\leq i(ac)+1$.

Proof.

Suppose that $ac\in R^{D}$. Then $ac\in R^{dag}$. In view of Theorem 3.2, $bd\in R^{{\dagger}}$, $(bd)^{{\dagger}}=b((ac)^{D})^{2}d$ and $i(bd)\leq i(ac)+1$. As in the proof of Theorem 3.2, (bd)^{k+2}(bd)^{{\dagger}}-(bd)^{k+1}=b\big{(}(ac)^{k+1}(ac)^{D}-(ac)^{k}\big{)}) for all $k\in{\mathbb{N}}$. Therefore $(bd)^{m}=(bd)^{m+1}(bd)^{{\dagger}}$ for some $m\in{\mathbb{N}}$, and so $bd\in R^{D}$. Furthermore, $(bd)^{D}=(bd)^{{\dagger}}=b((ac)^{D})^{2}d$, as required.∎

Recall that $a\in R$ has group inverse if $a$ has Drazin inverse with index $1$, and we denote the group inverse of $a$ by $a^{\#}$. We now derive

Corollary 3.5**.**

Let $R$ be a ring, and let $a,b,c,d\in R$ satisfying $bdb=bac,dbd=acd$. If $ac$ has group inverse, then

1. (1)

$bd\in R^{inv}$; or 2. (2)

$bd$* has group inverse and $(bd)^{\#}=b((ac)^{\#})^{2}d$; or* 3. (3)

$bd\in R^{d}$* and $(bd)^{d}=b((ac)^{\#})^{2}d$.*

Proof.

Since $ac$ has group inverse, it follows by Theorem 3.4 that $i(bd)\leq 2$. Hence, $i(bd)=0,1,2$. This completes the proof.∎

Example 3.6**.**

Let $R=M_{2}({\mathbb{Z}})$ be the ring of all $2\times 2$ integer matrices. Choose

[TABLE]

Then $bdb=bac,dbd=acd$. In this case, $ac=0$ has group inverse, but bd=\left(\begin{array}[]{cc}0&2\\ 0&0\end{array}\right) has no group inverse in $R$.

4. Applications

Let $X$ be Banach space, and let $\mathcal{L}(X)$ denote the set of all bounded linear operators from Banach space to itself, and let $a\in\mathcal{L}(X)$. The Drazin spectrum $\sigma_{D}(a)$ and g-Drazin spectrum $\sigma_{d}(a)$ are defined by

[TABLE]

The goal of this section is concern on common spectrum properties of $\mathcal{L}(X)$. We now record the generalized Jacobson’s Lemma as follows.

Lemma 4.1**.**

([10]) Let $R$ be a ring, and let $a,b,c,d\in R$ satisfying $bdb=bac,dbd=acd$. If $1-ac\in R^{inv}$, then $1-bd\in R^{inv}$ and

[TABLE]

We now ready to prove the following.

Theorem 4.2**.**

Let $A,B,C,D\in\mathcal{L}(X)$ such that $BDB=BAC,$ $DBD=ACD$, then

[TABLE]

Proof.

Case 1. $0\in\sigma_{d}(AC)$. Then $AC\not\in A^{d}$. In view of Theorem 2.2, $BD\in A^{d}$. Thus $0\in\sigma_{d}(BD)$.

Case 2. $0\not\in\lambda\in\sigma_{d}(AC)$. Then $\lambda\in acc\sigma(AC)$. Thus, we see that

[TABLE]

For $\lambda_{n}\neq 0$, it follows by Lemma 4.1 that $I-(\frac{1}{\lambda_{n}}A)C\in\mathcal{L}(X)^{-1}$ if and only if $I-B(\frac{1}{\lambda_{n}}D)\in\mathcal{L}(X)^{-1}$. Therefore

[TABLE]

Therefore $\sigma_{d}(AC)\subseteq\sigma_{d}(BD).$ Likewise, $\sigma_{d}(BD)\subseteq\sigma_{d}(AC)$, as required.∎

Corollary 4.3**.**

Let $A,B\in\mathcal{L}(X)$ , then

[TABLE]

Proof.

This is obvious by choosing $C=B$ and $D=A$ in Theorem 4.2.∎

For the Drazin spectrum $\sigma_{D}(a)$, we now derive

Theorem 4.4**.**

Let $A,B,C,D\in\mathcal{L}(X)$ such that $BDB=BAC,$ $DBD=ACD$, then

[TABLE]

Proof.

In view of Theorem 3.4, $AC\in\mathcal{L}(X)^{D}$ if and only if $BD\in\mathcal{L}(X)^{D}$, and therefore we complete the proof by [12, Theorem 3.1].∎

A bounded linear operator $T\in\mathcal{L}(X)$ is Fredholm operator if $dimN(T)$ and $codimR(T)$ are finite, where $N(T)$ and $R(T)$ are the null space and the range of $T$ respectively. If furthermore the Fredholm index $ind(T)=0$, then $T$ is said to be Weyl operator. For each nonnegative integer $n$ define $T_{|n|}$ to be the restriction of $T$ to $R(T^{n})$. If for some $n$, $R(T^{n})$ is closed and $T_{|n|}$ is a Fredholm operator then $T$ is called a $B$-Fredholm operator. $T$ is said to be a $B$-Weyl operator if $T_{|n|}$ is a Fredholm operator of index zero (see [1]). The $B$-Fredholm and $B$-Weyl spectrums of $T$ are defined by

[TABLE]

Corollary 4.5**.**

Let $A,B,C,D\in\mathcal{L}(X)$ such that $BDB=BAC,$ $DBD=ACD$, then

[TABLE]

Proof.

Let $\pi:\mathcal{L}(X)\to\mathcal{L}(X)/F(X)$ be the canonical map and $F(X)$ be the ideal of finite rank operators in $\mathcal{L}(X)$. As in well known, $T\in\mathcal{L}(X)$ is $B$-Fredholm if and only if $\pi(T)$ has Drazin inverse. By hypothesis, we see that

[TABLE]

According to Theorem 3.4, for every scalar $\lambda$, we have

[TABLE]

This completes the proof.∎

Corollary 4.6**.**

Let $A,B,C,D\in\mathcal{L}(X)$ such that $BDB=BAC,$ $DBD=ACD$, then

[TABLE]

Proof.

If $T$ is $B$-Fredholm then for $\lambda\neq 0$ small enough, $T-\lambda I$ is Fredholm and $ind(T)=ind(T-\lambda I)$. As in the proof of [12, Lemma 2.3, Lemma 2.4], we see that $I-AC$ is Fredholm if and only if $I-BD$ is Fredholm and in this case, $ind(I-AC)=ind(I-BD)$. Therefore we complete the proof by Theorem 4.4.∎

An element $a\in\mathcal{L}(X)$ is algebraic if there exists a non-zero complex polynomial $p$ such that $p(a)=0$. As an immediate consequence of Theorem 4.4, we have

Corollary 4.7**.**

Let $A,B,C,D\in\mathcal{L}(X)$ such that $BDB=BAC,$ $DBD=ACD$, then $AC$ is algebraic if and only if $BD$ is algebraic.

Proof.

It follows immediately from  [3, Theorem 2.1] and Theorem 4.4.∎