The Weighted ω-Core Inverse and Its Associated Decomposition

1. Introduction

A Banach algebra $\mathcal{A}$ is called a Banach *-algebra if there exists an involution $*:x\to x^{*}$ satisfying ${(x+y)}^{*}=x^{*}+y^{*},{\left(\lambda x\right)}^{*}=\overline{\lambda }{x}^{*},{\left(xy\right)}^{*}=y^{*}{x}^{*},{\left(x^{*}\right)}^{*}=x$. An element a in a Banach *-algebra $\mathcal{A}$ has group inverse if and only if there exist $x\in \mathcal{A}$ such that

$$x{a}^2=a,a{x}^2=x,ax=xa.$$

If such x exists, it is unique, and denote it by $a^{\#}$.

An element a in a Banach *-algebra $\mathcal{A}$ has core inverse if and only if there exist $x\in \mathcal{A}$ such that

$$x{a}^2=a,a{x}^2=x,{\left(ax\right)}^{*}=ax.$$

If such x exists, it is unique, and denote it by $a^{\#}$. Group and core inverses are extensively studied by many authors from different views, e.g. [1,7,10,16,19].

An element $a\in \mathcal{A}$ has w-group inverse if there exist $x\in \mathcal{A}$ such that

$$x{\left(wa\right)}^2=a,a{\left(wx\right)}^2=x,awx=xwa.$$

We denote the preceding x by $a_w^{\#}$. The set of all w-group invertible elements in $\mathcal{A}$ is denoted by ${\mathcal{A}}_w^{\#}$.

In [24], Zhu et al. introduced and studied a weighted generalized inverse as a generalization of core inverse. Let $a,w\in \mathcal{A}$. An element $a\in \mathcal{A}$ is w-core invertible if there exists some $x\in \mathcal{A}$ such that

$$aw{x}^2=x,xawa=a,{\left(awx\right)}^{*}=awx.$$

Such an x is called a w-core inverse of a.

Many properties of w-core inverse are presented by many authors, e.g. [8,20,22,24]. However, a w-core invertible element in a Banach algebra is not necessarily w-group invertible. This observation motivates the introduction and study of a new weighted generalized inverse that combines core and w-group inverses.

Definition 1. 

An element $a\in \mathcal{A}$ is weighted w-core invertible if there exist $x\in \mathcal{A}$ such that

$$x{\left(wa\right)}^2=a,a{\left(wx\right)}^2=x,{\left(awxw\right)}^{*}=awxw.$$

Such x is called the weighted w-core inverse of a and denoted by $a^{w,\#}$. Let ${\mathcal{A}}^{w,\#}$ denote the set of all weighted w-core invertible elements in $\mathcal{A}$.

In Section 2, we characterize the weighted core inverse by the solution of equation systems. We establish the polar-like property of this new generalized inverse. We prove that $a\in {\mathcal{A}}^{w,\#}$ if and only if there exists a projection $p\in \mathcal{A}$ such that

$$pa=0,1-p\in \mathcal{A}wandaw+p\in {\mathcal{A}}^{-1}.$$

In Section 3, we examine the connections between the weighted w-core inverse and related generalized inverses. If $a,x$ and w satisfy the equations $a=awxwa$ and ${\left(awxw\right)}^{*}=awxw$, then x is called weighted w-$(1,3)$-inverse of a and is denoted by $a_w^{(1,3)}$. We use ${\mathcal{A}}_w^{(1,3)}$ to stand for sets of all weighted w-$(1,3)$-invertible elements in $\mathcal{A}$. We prove that $a\in {\mathcal{A}}^{w,\#}$ if and only if $a\in {\mathcal{A}}_w^{\#}\bigcap {\mathcal{A}}_w^{(1,3)}$. Consequently, many equivalent properties of the weighted w-core inverse are presented.

Finally, in Section 4, we focus on the decomposition associated with the weighted w-core inverse.

Definition 2. 

An element $a\in \mathcal{A}$ has generalized weighted w-core decomposition if there exist $z,y\in \mathcal{A}$ such that

$$a=z+y,z^{*}y=ywz=0,z\in {\mathcal{A}}^{w,\#},y\in {\mathcal{A}}^{qnil}.$$

Let ${\mathcal{A}}^{w,d}$ be the set of all elements having generalized weighted w-core decomposition in $\mathcal{A}$. We prove that $a\in {\mathcal{A}}^{w,d}$ if and only if there exists $z\in \mathcal{A}$ such that

$$\begin{array}{c}a{\left(wx\right)}^2=x,xw{\left(aw\right)}^{2}x=awx,{\left(awxw\right)}^{*}=awxw,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

The preceding x is called the generalized weighted w-core inverse of a, and denoted by $a^{w,d}$. We then characterize this generalized inverse of the preceding decomposition from multiple perspectives. An element $a\in \mathcal{A}$ has generalized w-Drazin inverse x if there exists unique $x\in \mathcal{A}$ such that

$$a{\left(wx\right)}^2=x,awx=xwa,andaw-\left(aw\right)\left(xw\right)\left(aw\right)\in {\mathcal{A}}^{qnil}.$$

We denote x by $a^{d,w}$ (see [11]). We finally establish the identity $a^{w,d}={\left[a^{d,w}w\right]}^2{\left(a^{d,w}\right)}^{w,\#}.$

2. Weighted w-Core Inverse

The aim of this section is to first introduce the weighted w-core inverse and present several of its fundamental properties.

Theorem 1. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

There exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x{\left(wa\right)}^2=a,a{\left(wx\right)}^2=x,awxwa=a,xwawx=x,{\left(awxw\right)}^{*}=awxw.\end{array}$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, there exists $x\in \mathcal{A}$ such that $x{\left(wa\right)}^2=a,a{\left(wx\right)}^2=x,{\left(awxw\right)}^{*}=awxw.$ One directly verify that

$$\begin{array}{ccc}\hfill awxwa& =\hfill & awxwx{\left(wa\right)}^2=a{\left(wx\right)}^2{\left(wa\right)}^2=x{\left(wa\right)}^2=a,\hfill \\ \hfill xwawx& =\hfill & xwawa{\left(wx\right)}^2=\left[x{\left(wa\right)}^2\right]{\left(wx\right)}^2=a{\left(wx\right)}^2=x,\hfill \end{array}$$

as required.

$\left(2\right)\Rightarrow \left(1\right)$ This is trivial. □

Corollary 1. 

Let $a\in {\mathcal{A}}^{w,\#}$. Then $a\in {\mathcal{A}}_w^{\#}$ and $a_w^{\#}={\left[a^{w,\#}w\right]}^{2}a$.

Proof. 

Let $x=a^{w,\#}$. By virtue of Theorem 2.1,

$$x{\left(wa\right)}^2=a,a{\left(wx\right)}^2=x,{\left(awxw\right)}^{*}=awxw,awxwa=a,xwawx=x.$$

Let $y={\left(xw\right)}^{2}a$. Then we verify that

$$\begin{array}{ccc}\hfill awywa& =\hfill & aw{\left(xw\right)}^{2}awa=a{\left(wx\right)}^{2}wawa=x{\left(wa\right)}^2=a,\hfill \\ \hfill ywawy& =\hfill & {\left(xw\right)}^{2}awaw{\left(xw\right)}^{2}a={\left(xw\right)}^{2}awa{\left(wx\right)}^{2}wa\hfill \\ & =\hfill & xwxw\left(awxwa\right)={\left(xw\right)}^{2}a=y,\hfill \\ \hfill awy& =\hfill & aw{\left(xw\right)}^{2}a=a{\left(wx\right)}^{2}wa=xwa\hfill \\ & =\hfill & xwx{\left(wa\right)}^2={\left(xw\right)}^{2}awa=ywa.\hfill \end{array}$$

This implies that $a_w^{\#}=y$, as desired. □

If $a,x$ and w satisfy the equations $a=awxwa$ and ${\left(awxw\right)}^{*}=awxw$, then x is called weighted w-$(1,3)$-inverse of a and is denoted by $a_w^{(1,3)}$. We use ${\mathcal{A}}_w^{(1,3)}$ to stand for sets of all weighted w-$(1,3)$-invertible elements in $\mathcal{A}$.

Theorem 2. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

$a\in {\mathcal{A}}_w^{\#}\bigcap {\mathcal{A}}_w^{(1,3)}$.

In this case, $a^{w,\#}=a_w^{\#}waw{a}_w^{(1,3)}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By virtue of Corollary 2.2, $a\in {\mathcal{A}}_w^{\#}$. Let $x=a^{w,\#}$. In view of Theorem 2.1, there exists $x\in \mathcal{A}$ such that $x{\left(wa\right)}^2=a,a{\left(wx\right)}^2=x,awxwa=a,xwawx=x,{\left(awxw\right)}^{*}=awxw.$ Hence, $a\in {\mathcal{A}}_w^{(1,3)}$, as desired.

$\left(2\right)\Rightarrow \left(1\right)$ Let $x=a_w^{\#}waw{a}_w^{(1,3)}$. Then we check that

$$\begin{array}{ccc}\hfill awxw& =\hfill & aw{a}_w^{\#}waw{a}_w^{(1,3)}w=aw{a}_w^{(1,3)}w,\hfill \\ \hfill wawx& =\hfill & waw{a}_w^{\#}waw{a}_w^{(1,3)},\hfill \\ \hfill {\left(awxw\right)}^{*}& =\hfill & awxw,\hfill \\ \hfill a{\left(wx\right)}^2& =\hfill & \left(awxw\right)x=aw{a}_w^{(1,3)}w\left(a_w^{\#}wa\right)w{a}_w^{(1,3)}=aw{a}_w^{(1,3)}w\left(aw{a}_w^{\#}\right)w{a}_w^{(1,3)}\hfill \\ & =\hfill & \left(aw{a}_w^{\#}\right)w{a}_w^{(1,3)}=\left(a_w^{\#}wa\right)w{a}_w^{(1,3)}=x,\hfill \\ \hfill x{\left(wa\right)}^2& =\hfill & a_w^{\#}waw{a}_w^{(1,3)}wawa=aw{a}_w^{\#}waw{a}_w^{(1,3)}wa=aw{a}_w^{(1,3)}wa=a.\hfill \end{array}$$

Therefore $a^{w,\#}=x$, as asserted. □

Corollary 2. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

$a\in {\mathcal{A}}_w^{\#}$ and there exists $x\in \mathcal{A}$ such that $x{\left(wa\right)}^2=a,{\left(awxw\right)}^{*}=awxw.$

(3)

$a\in {\mathcal{A}}_w^{\#}$ and there exists $x\in \mathcal{A}$ such that $x\left(wa\right)=a_w^{\#}wa,{\left(awxw\right)}^{*}=awxw.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ This is obvious by Theorem 2.1.

$\left(2\right)\Rightarrow \left(3\right)$ By hypothesis, there exists $x\in \mathcal{A}$ such that $x{\left(wa\right)}^2=a,{\left(awxw\right)}^{*}=awxw.$ Hence, $x\left(wa\right)=xw\left(aw{a}_w^{\#}wa\right)=xw{\left(aw\right)}^2{a}_w^{\#}=x{\left(wa\right)}^{2}w{a}_w^{\#}=aw{a}_w^{\#}=a_w^{\#}wa,$ as required.

$\left(3\right)\Rightarrow \left(1\right)$ By assumption, $a\in {\mathcal{A}}_w^{\#}$ and there exists $x\in \mathcal{A}$ such that $x\left(wa\right)=a_w^{\#}wa,{\left(awxw\right)}^{*}=awxw.$ Then $awxwa=aw{a}_w^{\#}wa=a$, and so $a\in {\mathcal{A}}_w^{(1,3)}$. Therefore we complete the proof by Theorem 2.3. □

Lemma 1. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{\#}$.

(2)

$a\in a^2\mathcal{A}\bigcap \mathcal{A}{a}^2$.

(3)

$a\in {\mathcal{A}}^d$ and $\mathcal{A}a=\mathcal{A}{a}^2$.

(4)

$a\in {\mathcal{A}}^d$ and $a\mathcal{A}=a^2\mathcal{A}$.

Proof.$\left(1\right)\Rightarrow \left(3\right)$ This direction is obvious.

$\left(3\right)\Rightarrow \left(2\right)$ Write $a=x{a}^2$ for some $x\in \mathcal{A}$. Then we have

$$\begin{array}{ccc}\hfill a-a^2{a}^d& =\hfill & x^n{a}^{n+1}-a^2{a}^d=x^{n}a(a^n-a^{n+1}{a}^d)+x^n{a}^{n+2}{a}^d-a^2{a}^d\hfill \\ & =\hfill & x^{n}a(a^n-a^{n+1}{a}^d)+\left(x^n{a}^{n+1}\right)a{a}^d-a^2{a}^d=x^{n}a(a^n-a^{n+1}{a}^d).\hfill \end{array}$$

Hence,

$$\Vert a-a^2{a}^d{\Vert }^{{\displaystyle \frac{1}{n}}}\le {\Vert x\Vert \Vert a\Vert }^{{\displaystyle \frac{1}{n}}}{\Vert a^n-a^{n+1}{a}^d\Vert }^{{\displaystyle \frac{1}{n}}}.$$

This implies that $\underset{n\to \infty }{lim}{\Vert a-a^2{a}^d\Vert }^{{\displaystyle \frac{1}{n}}}=0,$ and so $a=a^2{a}^d$. Thus $a\in a^2\mathcal{A}\bigcap \mathcal{A}{a}^2$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ Write $a=a^{2}r=s{a}^2$ for some $r,s\in \mathcal{A}$. We directly verify that $a^{\#}=r+s-ras$, as required.

$\left(1\right)\iff \left(4\right)$ This is obvious by the symmetry. □

Lemma 2. 

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_w^{\#}$ if and only if $aw,wa\in {\mathcal{A}}^{\#}$ and ${\left(aw\right)}^{\#}awa=a$. In this case,

$$a_w^{\#}={\left(aw\right)}^{\#}a{\left(wa\right)}^{\#},{\left(aw\right)}^2{a}_w^{\#}=aanda{\left[w{a}_w^{\#}\right]}^2=a_w^{\#}.$$

Proof. 

⟹ Let $x=a_w^{\#}$. Then

$$awxwa=a,xwawx=x,awx=xwa.$$

Hence, ${\left(aw\right)}^{2}x=aw\left(awx\right)=\left(aw\right)x\left(wa\right)=a$, and so ${\left(aw\right)}^{2}xw=aw$. Likewise, $xw{\left(aw\right)}^2=aw$. This implies that $aw\in {\mathcal{A}}^{\#}$. Moreover, we have $a{\left[w{a}_w^{\#}\right]}^2=a{\left(wx\right)}^2=\left(awx\right)wx=\left(xwa\right)wx=x,$ as desired.

⟸ Set $x={\left(aw\right)}^{\#}a{\left(wa\right)}^{\#}$. Then we verify that

$$\begin{array}{ccc}\hfill awx& =\hfill & aw{\left(aw\right)}^{\#}a{\left(wa\right)}^{\#}={\left(aw\right)}^{\#}a{\left(wa\right)}^{\#}wa=xwa,\hfill \\ \hfill x{\left(wa\right)}^2& =\hfill & \left(xwa\right)wa={\left(aw\right)}^{\#}a{\left(wa\right)}^{\#}{\left(wa\right)}^2={\left(aw\right)}^{\#}awa=a,\hfill \\ \hfill a{\left(wx\right)}^2& =\hfill & {\left(aw\right)}^{\#}a{\left(wa\right)}^{\#}w\left(awx\right)={\left(aw\right)}^{\#}a\left(wa\right)w{\left[{\left(aw\right)}^{\#}\right]}^{2}a{\left(wa\right)}^{\#}\hfill \\ & =\hfill & {\left(aw\right)}^{\#}a{\left(wa\right)}^{\#}=x.\hfill \end{array}$$

Therefore $a_w^{\#}=x$, as asserted. □

Lemma 3. 

Let $a\in \mathcal{A}$. If $a\mathcal{A}={\left(aw\right)}^2\mathcal{A}$, then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

There exists $x\in \mathcal{A}$ such that

$$x{\left(wa\right)}^2=a,xwawx=x,{\left(awxw\right)}^{*}=awxw.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ This is clear by Theorem 2.1.

$\left(2\right)\Rightarrow \left(1\right)$ Since $x{\left(wa\right)}^2=a$, we have $\mathcal{A}a=\mathcal{A}{\left(wa\right)}^2=\mathcal{A}\left(wa\right)$. Obviously, ${\left(wa\right)}^2\mathcal{A}\subseteq \left(wa\right)\mathcal{A}$. Write $a={\left(aw\right)}^{2}z$ for a $z\in \mathcal{A}$. Then $wa=w{\left(aw\right)}^{2}z={\left(wa\right)}^{2}wz\in {\left(wa\right)}^2\mathcal{A}$. This implies that $\left(wa\right)\mathcal{A}={\left(wa\right)}^2\mathcal{A}$. Hence, $wa\in {\mathcal{A}}^{\#}$. By hypothesis, we have $\left(aw\right)\mathcal{A}={\left(aw\right)}^2\mathcal{A}$. Obviously, $\mathcal{A}{\left(aw\right)}^2\subseteq \mathcal{A}\left(aw\right)$. As $a=x{\left(wa\right)}^2$, we deuce that $aw=xw{\left(aw\right)}^2\in \mathcal{A}{\left(aw\right)}^2$; hence, $\mathcal{A}\left(aw\right)\subseteq \mathcal{A}{\left(aw\right)}^2$. Thus $\mathcal{A}\left(aw\right)=\mathcal{A}{\left(aw\right)}^2$; hence, $aw\in {\mathcal{A}}^{\#}$ by Lemma 2.5. Since $xw{\left(aw\right)}^2=aw$, we have $xw\left(aw\right)=aw{\left(aw\right)}^{\#}$. Then ${\left(aw\right)}^{\#}awa=xwawa=x{\left(wa\right)}^2=a$. In light of Lemma 2.6, $a\in {\mathcal{A}}_w^{\#}$ and $a_w^{\#}={\left(aw\right)}^{\#}a{\left(wa\right)}^{\#}$.

Since $x{\left(wa\right)}^2=a$, we have $wx{\left(wa\right)}^2=wa$; and so $wx\left(wa\right)=wa{\left(wa\right)}^{\#}$. This implies that $wx{\left(wa\right)}^2=wa{\left(wa\right)}^{\#}wa=wa$. Thus, we derive that $a=a\left(wa\right)wz=a\left[wx{\left(wa\right)}^2\right]wz=awxw\left[{\left(aw\right)}^{2}z\right]=awxwa$. Thus, $a\in {\mathcal{A}}_w^{(1,3)}$. Therefore $a\in {\mathcal{A}}^{w,\#}$ by Theorem 2.3. □

Lemma 4. 

Let $a\in \mathcal{A}$. If $\mathcal{A}a=\mathcal{A}{\left(wa\right)}^2$, then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

There exists $x\in \mathcal{A}$ such that

$$a{\left(wx\right)}^2=x,awxwa=a,{\left(awxw\right)}^{*}=awxw.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ This is obvious by Theorem 2.1.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, $\mathcal{A}\left(wa\right)=\mathcal{A}a=\mathcal{A}{\left(wa\right)}^2$. Clearly, ${\left(wa\right)}^2\mathcal{A}\subseteq wa\mathcal{A}$. On the other hand, $wa=w\left(awxwa\right)=waw\left[a{\left(ax\right)}^2\right]wa\in {\left(wa\right)}^2\mathcal{A}$; hence, $wa\mathcal{A}\subseteq {\left(aw\right)}^2\mathcal{A}$. This implies that $\left(wa\right)\mathcal{A}={\left(wa\right)}^2\mathcal{A}$. Hence, $wa\in {\mathcal{A}}^{\#}$ by Lemma 2.5. Set $z=a{\left(wa\right)}^{\#}wx$. Since $\mathcal{A}a=\mathcal{A}{\left(wa\right)}^2$, we have $a{\left(wa\right)}^{\#}wa=a$. Then we verify that

$$\begin{array}{ccc}\hfill awz& =\hfill & awa{\left(wa\right)}^{\#}wx=awa{\left(wa\right)}^{\#}wa{\left(wx\right)}^2=awa{\left(wx\right)}^2=awx,\hfill \\ \hfill {\left(awzw\right)}^{*}& =\hfill & {\left(awxw\right)}^{*}=awxw=awzw,\hfill \\ \hfill a{\left(wz\right)}^2& =\hfill & \left(awz\right)wz=awxwa{\left(wa\right)}^{\#}wx=a{\left(wa\right)}^{\#}wx=z,\hfill \\ \hfill z{\left(wa\right)}^2& =\hfill & a{\left(wa\right)}^{\#}wx{\left(wa\right)}^2=a{\left[{\left(wa\right)}^{\#}\right]}^{2}w\left[awxwa\right]wa\hfill \\ & =\hfill & a{\left[{\left(wa\right)}^{\#}\right]}^2{\left(wa\right)}^2=a{\left(wa\right)}^{\#}wa=a.\hfill \end{array}$$

Therefore $a\in {\mathcal{A}}^{w,\#}$, as asserted. □

We are ready to prove:

Theorem 3. 

Let $a\in {\mathcal{A}}_w^{\#}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

There exists $x\in \mathcal{A}$ such that

$$a{\left(wx\right)}^2=x,awxwa=a,{\left(awxw\right)}^{*}=awxw.$$

(3)

There exists $x\in \mathcal{A}$ such that

$$x{\left(wa\right)}^2=a,xwawx=x,{\left(awxw\right)}^{*}=awxw.$$

Proof. 

Since $a\in {\mathcal{A}}_w^{\#}$, we see that $a\in {\left(aw\right)}^2\mathcal{A}\bigcap \mathcal{A}{\left(wa\right)}^2$. Therefore we complete the proof by Lemma 2.7 and Lemma 2.8. □

We come now to present a polar-like property for weighted w-core inverse in a Banach *-algebra.

Theorem 4. 

Let $a,w\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

There exists a projection $p\in \mathcal{A}$ such that

$$pa=0,1-p\in \mathcal{A}wand(1-p)aw(1-p)\in {\left((1-p)\mathcal{A}(1-p)\right)}^{-1}.$$

(3)

There exists a projection $p\in \mathcal{A}$ such that

$$pa=0,1-p\in \mathcal{A}wandaw+p\in {\mathcal{A}}^{-1}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Since $a\in {\mathcal{A}}^{w,\#}$, there exists $x\in \mathcal{A}$ such that

$$x{\left(wa\right)}^2=a,a{\left(wx\right)}^2=x,{\left(awxw\right)}^{*}=awxw,awxwa=a,xwawx=x.$$

Let $p=1-awxw$. Then $p^2=p=p^{*}$ and $1-p\in \mathcal{A}w$. Obviously, $pa=(1-awxw)a=0$. Moreover, we check that

$$\begin{array}{ccc}\hfill (1-p)aw(1-p)& =\hfill & aw(1-p)=awawxw,\hfill \\ \hfill (1-p)aw(1-p)xw& =\hfill & awa{\left(wx\right)}^{2}w=awxw,\hfill \\ \hfill xw(1-p)aw(1-p)& =\hfill & x{\left(wa\right)}^{2}wxw=awxw.\hfill \end{array}$$

Then ${\left((1-p)aw(1-p)\right)}^{-1}=xw\in (1-p)\mathcal{A}(1-p).$

$\left(2\right)\Rightarrow \left(3\right)$ By hypothesis, there exists a projection $p\in \mathcal{A}$ such that $pa=0,1-p\in \mathcal{A}w$ and $(1-p)aw(1-p)\in {\left((1-p)\mathcal{A}(1-p)\right)}^{-1}.$ Let $x={\left((1-p)aw(1-p)\right)}^{-1}\in (1-p)\mathcal{A}(1-p)$. Then ${\left(aw(1-p)+p\right)}^{-1}={\left((1-p)aw(1-p)\right)}^{-1}+p=x+p$. Hence, $1+(aw-1)(1-p)\in {\mathcal{A}}^{-1}$. By using Jacobson’s Lemma, $1+(1-p)(aw-1)\in {\mathcal{A}}^{-1}$. Therefore $aw+p\in {\mathcal{A}}^{-1}$, as desired.

$\left(3\right)\Rightarrow \left(1\right)$ By hypothesis, there exists a projection $p\in \mathcal{A}$ such that $pa=0,1-p\in \mathcal{A}w$ and $aw+p\in {\mathcal{A}}^{-1}$. Then $1+(1-p)aw-(1-p)=1+(1-p)(aw-1)\in {\mathcal{A}}^{-1}$. By Jacobson’s Lemma, we prove that $aw(1-p)+p=1+(aw-1)(1-p)\in {\mathcal{A}}^{-1}.$ Then $(1-p)aw(1-p)\in {\left((1-p)\mathcal{A}(1-p)\right)}^{-1}.$ As ${\left((1-p)aw(1-p)\right)}^{-1}\in \mathcal{A}(1-p)\subseteq \mathcal{A}w$. Write ${\left((1-p)aw(1-p)\right)}^{-1}=xw$ for some $x\in (1-p)\mathcal{A}$. Then $(1-p)aw(1-p)xw=(1-p)aw(1-p){\left((1-p)aw(1-p)\right)}^{-1}=1-p.$ Since $pa=0$ and $(1-p)x=x$, we have $awxw=(1-p)aw(1-p)xw=1-p$. Then ${\left(awxw\right)}^{*}={(1-p)}^{*}=1-p=awxw$. Moreover, we check that

$$\begin{array}{ccc}\hfill x{\left(wa\right)}^2& =\hfill & xwawa={\left((1-p)aw(1-p)\right)}^{-1}awa\hfill \\ & =\hfill & {\left((1-p)aw(1-p)\right)}^{-1}\left((1-p)aw(1-p)\right)a=(1-p)a=a,\hfill \\ \hfill a{\left(wx\right)}^2& =\hfill & \left(aw\right)\left(xw\right)x=\left(aw\right){\left((1-p)aw(1-p)\right)}^{-1}x\hfill \\ & =\hfill & (1-p)\left(aw\right)(1-p){\left((1-p)aw(1-p)\right)}^{-1}x=(1-p)x=x.\hfill \end{array}$$

Therefore $a^{w,\#}=x$, as required. □

Corollary 3. 

Let $a,w\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

There exists a unique projection $p\in \mathcal{A}$ such that

$$pa=0,1-p\in \mathcal{A}wandaw+p\in {\mathcal{A}}^{-1}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ As in the proof of Theorem 2.10, $p=1-aw{a}^{w,\#}w$. Obviously, $a^{w,\#}w={\left(aw\right)}^{\#}$. Obviously, ${\left(aw\right)}^{\#}$ is unique (see [16]). Thus p is unique, as desired.

$\left(2\right)\Rightarrow \left(1\right)$ This is proved by Theorem 2.8. □

3. Connections to Related Generalized Inverses

In [9], Mary presented the existence criterion for the inverse along an element by the intersection of ideals, i.e, a is invertible along w if and only if $w\in waw\mathcal{A}\bigcap \mathcal{A}waw$.

Lemma 5. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

$w\in {\mathcal{A}}^{\left|\right|a}$ and $a\in {\mathcal{A}}_w^{(1,3)}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Let $x=a^{w,\#}$. In view of Theorem 2.1, we have

$$\begin{array}{c}x{\left(wa\right)}^2=a,a{\left(wx\right)}^2=x,awxwa=a,xwawx=x,{\left(awxw\right)}^{*}=awxw.\end{array}$$

Then $a=awxwa=awa{\left(wx\right)}^{2}wa=\left(awa\right)wxwxwa\in awa\mathcal{A}.$ On the other hand, we have $a=awxwa=awxwx{\left(wa\right)}^2=aw{\left(xw\right)}^2\left(awa\right)\in \mathcal{A}awa.$ Thus, $a\in awa\mathcal{A}\bigcap \mathcal{A}awa$. By virtue of [9], $w\in {\mathcal{A}}^{\left|\right|a}$. In light of Theorem 2.3, $a\in {\mathcal{A}}_w^{(1,3)}$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ Since $a\in {\mathcal{A}}_w^{(1,3)}$, we can find $x\in \mathcal{A}$ such that $a=awxwa$ and ${\left(awxw\right)}^{*}=awxw$.

As $w\in {\mathcal{A}}^{\left|\right|a}$, it follows by [9] that $w\in waw\mathcal{A}\bigcap \mathcal{A}waw$. Then $aw\in {\left(aw\right)}^2\mathcal{A}$ and $wa\in \mathcal{A}{\left(wa\right)}^2$.

Since $aw=\left(awxwa\right)w=awx\left(wa\right)w\in \mathcal{A}\left(wawa\right)w\subseteq \mathcal{A}{\left(aw\right)}^2$. Hence, $aw\in {\left(aw\right)}^2\mathcal{A}\bigcap \mathcal{A}{\left(aw\right)}^2$. In view of Lemma 2.5, $aw\in {\mathcal{A}}^{\#}$. Then $aw\in {\mathcal{A}}^d$. By using Cline’s formula, $wa\in {\mathcal{A}}^d$. By using Lemma 2.5 again, $wa\in {\mathcal{A}}^{\#}$.

Set $z={\left(aw\right)}^{\#}a{\left(wa\right)}^{\#}.$ By using Cline’s formula, we check that

$$\begin{array}{ccc}\hfill awz& =\hfill & aw{\left(aw\right)}^{\#}a{\left(wa\right)}^{\#}=a{\left(wa\right)}^{\#}=aw{\left[{\left(aw\right)}^d\right]}^{2}a={\left(aw\right)}^{d}a\hfill \\ & =\hfill & {\left(aw\right)}^{\#}a={\left(aw\right)}^{\#}a{\left(wa\right)}^{\#}wa=zwa,\hfill \\ \hfill awzwa& =\hfill & a{\left(wa\right)}^{\#}wa=a,\hfill \\ \hfill zwawz& =\hfill & {\left(aw\right)}^{\#}awz=z.\hfill \end{array}$$

Therefore $a\in {\mathcal{A}}_w^{\#}$. According to Theorem 2.3, $a\in {\mathcal{A}}^{w,\#}$, as asserted. □

Theorem 5. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

$a\in {\mathcal{A}}_w^{\#}\bigcap {\mathcal{A}}_w^{(1,3)}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ In view of Lemma 3.1, $w\in {\mathcal{A}}^{\left|\right|a}$ and $a\in {\mathcal{A}}_w^{(1,3)}$. Then we can find $x\in \mathcal{A}$ such that $a=awxwa$ and ${\left(awxw\right)}^{*}=awxw$. This implies that $a\in {\mathcal{A}}^{(1,3)}$. By virtue of [24], $a\in {\mathcal{A}}_w^{\#}$.

$\left(2\right)\Rightarrow \left(1\right)$ In view of [24], $w\in {\mathcal{A}}^{\left|\right|a}$. This completes the proof by Theorem 2.3. □

Corollary 4. 

Let $a\in \mathcal{A}$ and $w\in {\mathcal{A}}^{-1}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

$a\in {\mathcal{A}}_w^{\#}$.

Proof. 

Since $w\in {\mathcal{A}}^{-1}$, we prove that $a\in {\mathcal{A}}_w^{(1,3)}$ if and only if $aw\in {\mathcal{A}}^{(1,3)}.$ Therefore we obtain the result by Theorem 3.2. □

Corollary 5. 

Let $a\in \mathcal{A}$ and $n\ge 2$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

$a\in {\mathcal{A}}_w^{(1,3)}$ and $a\in {\left(aw\right)}^{n}a\mathcal{A}\bigcap \mathcal{A}a{\left(wa\right)}^n$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 3.2, $a\in {\mathcal{A}}_w^{\#}\bigcap {\mathcal{A}}_w^{(1,3)}$. By virtue of [8], $a\in {\left(aw\right)}^{n}a\mathcal{A}\bigcap \mathcal{A}a{\left(wa\right)}^n$.

$\left(2\right)\Rightarrow \left(1\right)$ In light of [8], $a\in {\mathcal{A}}_w^{\#}$. This completes the proof by Theorem 3.2. □

Corollary 6. 

Let $a\in \mathcal{A}$ and $n\ge 2$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

$a\in {\mathcal{A}}^{\#,w{\left(aw\right)}^{n-1}}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By virtue of Theorem 3.2, $a\in {\mathcal{A}}_w^{\#}\bigcap {\mathcal{A}}_w^{(1,3)}$. It follows by [8] that $a\in {\mathcal{A}}_{w{\left(aw\right)}^{n-1}}^{\#}$. Set $x=a^{w,\#}$. Then ${\left(aw\right)}^n{\left(xw\right)}^n=\left(aw\right)\left(xw\right)$ and ${\left(wx\right)}^n{\left(wa\right)}^n=\left(wx\right)\left(wa\right)$. Hence,

$$\begin{array}{ccc}\hfill a& =\hfill & awxwa={\left(aw\right)}^n{\left(xw\right)}^{n}a={\left(aw\right)}^n{\left(xw\right)}^{n-2}x\left(wx\right)\left(wa\right)\hfill \\ & =\hfill & {\left(aw\right)}^n{\left(xw\right)}^{n-2}x{\left(wx\right)}^n{\left(wa\right)}^n=aw{\left(aw\right)}^{n-1}{\left(xw\right)}^{n-2}{\left(xw\right)}^{n}xw{\left(aw\right)}^{n-1}a\hfill \\ & =\hfill & aw{\left(aw\right)}^{n-1}\left[{\left(xw\right)}^{n-2}{\left(xw\right)}^{n}x\right]w{\left(aw\right)}^{n-1}a\hfill \\ & =\hfill & aw{\left(aw\right)}^{n-1}zw{\left(aw\right)}^{n-1}a,\hfill \end{array}$$

where $z={\left(xw\right)}^{n-2}{\left(xw\right)}^{n}x.$ Moreover, we have

$$\begin{array}{ccc}\hfill aw{\left(aw\right)}^{n-1}zw{\left(aw\right)}^{n-1}& =\hfill & {\left(aw\right)}^n\left[{\left(xw\right)}^{n-2}{\left(xw\right)}^{n}x\right]w{\left(aw\right)}^{n-1}\hfill \\ & =\hfill & {\left(aw\right)}^n{\left(xw\right)}^n{\left(xw\right)}^{n-1}{\left(aw\right)}^{n-1}=\left(aw\right)\left(xw\right)\left(xw\right)\left(aw\right)=awxw,\hfill \end{array}$$

and so ${\left(aw{\left(aw\right)}^{n-1}zw{\left(aw\right)}^{n-1}\right)}^{*}=aw{\left(aw\right)}^{n-1}zw{\left(aw\right)}^{n-1}$. This implies that $a\in {\mathcal{A}}_{w{\left(aw\right)}^{n-1}}^{(1,3)}$. According to Theorem 3.2, $a\in {\mathcal{A}}^{\#,w{\left(aw\right)}^{n-1}}$.

$\left(2\right)\Rightarrow \left(1\right)$ In view of Theorem 3.2, $a\in {\mathcal{A}}_{w{\left(aw\right)}^{n-1}}^{\#}\bigcap {\mathcal{A}}_{w{\left(aw\right)}^{n-1}}^{(1,3)}$. By using [8], $a\in {\mathcal{A}}_w^{\#}.$ Set $x=a_{w{\left(aw\right)}^{n-1}}^{(1,3)}$. Then $aw{\left(aw\right)}^{n-1}xw{\left(aw\right)}^{n-1}a=a,{\left(aw{\left(aw\right)}^{n-1}xw{\left(aw\right)}^{n-1}\right)}^{*}=aw{\left(aw\right)}^{n-1}xw{\left(aw\right)}^{n-1}$. Let $y={\left(aw\right)}^{n-1}x{\left(wa\right)}^{n-1}$. then $awywa=a$ and ${\left(awyw\right)}^{*}=awyw$. This implies that $a\in {\mathcal{A}}_w^{(1,3)}$. By using Theorem 3.2 again, $a\in {\mathcal{A}}^{w,\#}$, as asserted. □

Lemma 6. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_w^{(1,3)}$.

(2)

$a\in {\left(w\mathcal{A})w\right)}^{*}{a}^{*}a$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Since $a\in {\mathcal{A}}_w^{(1,3)}$, we can find some $x\in \mathcal{A}$ such that $a=awxwa$ and ${\left(awxw\right)}^{*}=awxw$. Hence $a=\left(awxw\right)a={\left(awxw\right)}^{*}a={\left(wxw\right)}^{*}{a}^{*}a$$\in {\left(w\mathcal{A})w\right)}^{*}{a}^{*}a$, as desired.

$\left(2\right)\Rightarrow \left(1\right)$ Write $a={\left(wxw\right)}^{*}{a}^{*}a$ for some $x\in \mathcal{A}$. Then $a={\left(awxw\right)}^{*}a$, and so $awxw={\left(awxw\right)}^{*}awxw$. This implies that ${\left(awxw\right)}^{*}=awxw$. Therefore $a={\left(awxw\right)}^{*}a=awxwa$, as required. □

Lemma 7. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

$a\in w^{*}\mathcal{A}{\left(awa\right)}^{*}a\bigcap \mathcal{A}\left(awa\right)$.

Proof.$\left(1\right)\iff \left(2\right)$ In view of Lemma 3.1, $w\in {\mathcal{A}}^{\left|\right|a}$ and $a\in {\mathcal{A}}_w^{(1,3)}$. By virtue of [9], $a\in awa\mathcal{A}\bigcap \mathcal{A}awa$. According to Lemma 3.6, we have $a\in {\left(w\mathcal{A}w\right)}^{*}{a}^{*}a$. Write $a=awas={\left(wtw\right)}^{*}{a}^{*}a$ for some $s,t\in \mathcal{A}$. Then $a={\left(wtw\right)}^{*}{\left(awas\right)}^{*}a=w^{*}{\left(wt\right)}^{*}{s}^{*}{\left(awa\right)}^{*}a\in w^{*}\mathcal{A}{\left(awa\right)}^{*}a$, as desired.

$\left(2\right)\Rightarrow \left(1\right)$ Since $a\in w^{*}\mathcal{A}{\left(awa\right)}^{*}a\bigcap \mathcal{A}\left(awa\right)$, we have $a\in w^{*}\mathcal{A}{a}^{*}{w}^{*}{a}^{*}a;$ hence, $a\in {\left(w\mathcal{A}w\right)}^{*}{a}^{*}a$. In view of Lemma 3.6, $a\in {\mathcal{A}}_w^{(1,3)}.$

Write $a=w^{*}r{\left(awa\right)}^{*}a$ for some $r\in \mathcal{A}$. Then $a^{*}=a^{*}\left(awa\right)r^{*}$; hence, $a^{*}a=a^{*}a\left(wa{r}^{*}a\right)$. Accordingly, $a=\left[w^{*}r{\left(wa\right)}^{*}\right]\left(a^{*}a\right)=\left[w^{*}r{\left(wa\right)}^{*}\right]\left(a^{*}a\right)\left(wa{r}^{*}a\right)=\left(awa\right)r^{*}a\in awa\mathcal{A}.$ In light of [9], $w\in {\mathcal{A}}^{\left|\right|a}$. According to Lemma 3.1, $a\in {\mathcal{A}}^{w,\#}$. □

We are ready to prove:

Theorem 6. 

Let $a\in \mathcal{A}$ and $n\ge 2$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

$a\in w^{*}\mathcal{A}{\left({\left(aw\right)}^n\right)}^{*}a\bigcap \mathcal{A}{\left(aw\right)}^{n-1}a$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 3.2, $a\in {\mathcal{A}}_w^{\#}$. By using [8], $a\in \mathcal{A}{\left({\left(aw\right)}^n\right)}^{*}a\bigcap \mathcal{A}{\left(aw\right)}^{n-1}a$. It follows by Lemma 3.7 that $a\in w^{*}\mathcal{A}{\left(awa\right)}^{*}a\bigcap \mathcal{A}\left(awa\right)$. Hence, $a\in w^{*}\mathcal{A}a\subseteq w^{*}\mathcal{A}{\left({\left(aw\right)}^n\right)}^{*}a$. Therefore $a\in w^{*}\mathcal{A}{\left({\left(aw\right)}^n\right)}^{*}a\bigcap \mathcal{A}{\left(aw\right)}^{n-1}a$.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, we have $a\in \mathcal{A}{\left({\left(aw\right)}^n\right)}^{*}a\bigcap \mathcal{A}{\left(aw\right)}^{n-1}a$. In light of [8], $a\in {\mathcal{A}}_w^{\#}$. Moreover, we have $a\in w^{*}\mathcal{A}{\left({\left(aw\right)}^n\right)}^{*}a\subseteq {\left(w\mathcal{A}w\right)}^{*}\left(a^{*}a\right)$. According to Lemma 3.6, $a\in {\mathcal{A}}_w^{(1,3)}$. Therefore $a\in {\mathcal{A}}^{w,\#}$ by Theorem 3.2. □

Corollary 7. 

Let $a,w\in \mathcal{A}$ and $n\in N$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,\#}$.

(2)

There exists a projection $p\in \mathcal{A}$ such that $pa=0,1-p\in \mathcal{A}wand{\left(aw\right)}^n+p\in {\mathcal{A}}^{-1}.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Let $p=1-aw{a}^{w,\#}w$. Then $p^2=p=p^{*}\in \mathcal{A},pa=0$ and $1-p\in \mathcal{A}w$. Set $x=a^{w,\#}$. Then we verify that

$$\begin{array}{cc}& ({\left(aw\right)}^n-awxw)({\left(xw\right)}^n-xwaw)\hfill \\ \hfill =& {\left(aw\right)}^n{\left(xw\right)}^n+{\left(aw\right)}^{n-1}[aw-awxwaw]\hfill \\ \hfill -& [xw-aw{\left(xw\right)}^2]{\left(xw\right)}^{n-1}+(1-awxw)(1-xwaw)\hfill \\ \hfill =& awxw+(1-awxw)-(1-awxw)xwaw=1,\hfill \\ & ({\left(xw\right)}^n-xwaw)({\left(aw\right)}^n-awxw)\hfill \\ \hfill =& {\left(xw\right)}^n{\left(aw\right)}^n+{\left(xw\right)}^{n-1}[xw-xwawxw]\hfill \\ \hfill +& [aw-xw{\left(aw\right)}^2]{\left(aw\right)}^{n-1}+(1-xwaw)(1-awxw)\hfill \\ \hfill =& xwaw+(1-xwaw)-[aw-xw{\left(aw\right)}^2]xw=1.\hfill \end{array}$$

Therefore ${\left(aw\right)}^n+p={\left(aw\right)}^n+1-awxw\in {\mathcal{A}}^{-1}$, as desired.

$\left(2\right)\Rightarrow \left(1\right)$ Case 1. $n=1$. This is proved by Theorem 2.10.

Case 2. $n\ge 2$. By hypothesis, there exists a projection $p\in \mathcal{A}$ such that

$$pa=0,1-p\in \mathcal{A}wandu:={\left(aw\right)}^n+p\in {\mathcal{A}}^{-1}.$$

Then ${\left(aw\right)}^{n}a=ua$, and then $a=u^{-1}{\left(aw\right)}^{n}a\in mathcalA{\left(aw\right)}^{n-1}a$. Moreover, we have ${\left({\left(aw\right)}^n\right)}^{*}+p=u^{*}$; hence, ${\left({\left(aw\right)}^n\right)}^{*}a=u^{*}a$. This implies that $a={\left(u^{*}\right)}^{-1}{\left({\left(aw\right)}^n\right)}^{*}a$. Since $pa=0$, we deduce that $a=(1-p)a\in {(1-p)}^{*}{\left(u^{*}\right)}^{-1}{\left({\left(aw\right)}^n\right)}^{*}a\in w^{*}\mathcal{A}{\left({\left(aw\right)}^n\right)}^{*}a$. Thus, $a\in w^{*}\mathcal{A}{\left({\left(aw\right)}^n\right)}^{*}a\bigcap \mathcal{A}{\left(aw\right)}^{n-1}a$. Therefore we complete the proof by Theorem 3.8. □

4. The Associated Decomposition

In this section, we focus on the decomposition associated with the weighted w-core inverse. We now characterize generalized weighted w-core invertibility through a system of equations.

Theorem 7. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,d}.$

(2)

There exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}a{\left(wx\right)}^2=x,xw{\left(aw\right)}^{2}x=awx,{\left(awxw\right)}^{*}=awxw,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, there exist $z,y\in \mathcal{A}$ such that

$$a=z+y,z^{*}y=ywz=0,z\in {\mathcal{A}}^{w,\#},y\in {\mathcal{A}}_w^{qnil}.$$

Set $x=z^{w,\#}$. Then

$$\begin{array}{ccc}\hfill awxw& =\hfill & (z+y)w{z}^{w,\#}w=zw{z}^{w,\#}w,\hfill \\ \hfill {\left(awxw\right)}^{*}& =\hfill & awxw,\hfill \\ \hfill a{\left(wx\right)}^2& =\hfill & (z+y){\left(w{z}^{w,\#}\right)}^2=z{\left(w{z}^{w,\#}\right)}^2=x.\hfill \end{array}$$

Moreover, we see that

$$\begin{array}{cc}& xw{\left(aw\right)}^{2}x=z^{w,\#}w(z+y)w(z+y)w{z}^{w,\#}\hfill \\ \hfill =& z^{w,\#}w(z+y)wzw{z}^{w,\#}=z^{w,\#}{\left(wz\right)}^{2}w{z}^{w,\#}\hfill \\ \hfill =& zw{z}^{w,\#}=(z+y)w{z}^{w,\#}=awx.\hfill \end{array}$$

It is easy to verify that

$$\begin{array}{ccc}\hfill (1-awxw)aw& =\hfill & (1-zw{z}^{w,\#}w)(z+y)w\hfill \\ & =\hfill & (1-zw{z}^{w,\#}w)zw+(1-zw{z}^{w,\#}w)yw\hfill \\ & =\hfill & yw-{\left(zw{z}^{w,\#}w\right)}^{*}yw\hfill \\ & =\hfill & yw-{\left(w{z}^{w,\#}w\right)}^{*}\left(z^{*}y\right)=yw.\hfill \end{array}$$

Furthermore, we have

$$\begin{array}{cc}& {\left|\right|\left(aw\right)}^n-\left(aw\right)xw{\left(aw\right)}^n\left|\right|={\left|\right|(1-awxw)\left(aw\right)}^n\left|\right|={\left|\right|yw\left(aw\right)}^{n-1}\left|\right|\hfill \\ \hfill =& {\left|\right|yw(z+y)w\left(aw\right)}^{n-2}\left|\right|={\left|\right|\left(yw\right)}^2{\left(aw\right)}^{n-2}\left|\right|=\dots ={\left|\right|\left(yw\right)}^n\left|\right|.\hfill \end{array}$$

Since $y\in {\mathcal{A}}_w^{qnil}$, we see that $\underset{n\to \infty }{lim}{\left|\right|\left(yw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0$. Therefore $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)xw{\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0,$ as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypotheses, there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,xw{\left(aw\right)}^{2}x=awx,{\left(awxw\right)}^{*}=awxw,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-awxw{\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

Then we check that

$$awxw{\left(aw\right)}^{2}x={\left(aw\right)}^{2}x.$$

Set $z=awxwa$ and $y=a-awxwa$. We verify that

$$\begin{array}{ccc}\hfill ywz& =\hfill & (a-awxwa)w\left(awxwa\right)=awawxwa-awxw{\left(aw\right)}^{2}xwa\hfill \\ & =\hfill & {\left(aw\right)}^{2}xwa-{\left(aw\right)}^{2}xwa=0,\hfill \\ \hfill z^{*}y& =\hfill & {\left(awxwa\right)}^{*}y=a^{*}\left(awxw\right)y=a^{*}\left(awxw\right)(a-awxwa)\hfill \\ & =\hfill & a^{*}[awxwa-\left(awxwawx\right)wa]=a^{*}[awxwa-\left(awx\right)wa]=0.\hfill \end{array}$$

We claim that $z\in {\mathcal{A}}^{w,\#}$ and $z^{w,\#}=x$.

Claim 1. $x=z{\left(wx\right)}^2$. We verify that $z{\left(wx\right)}^2=awxwa{\left(wx\right)}^2=awxwx=x.$

Claim 2. ${\left(zwxw\right)}^{*}=zwxw$. Clearly, we have $zwxw=\left(awxwaw\right)xw=awxw$, and then ${\left(zwxw\right)}^{*}={\left(awxw\right)}^{*}=awxw=zwxw$.

Claim 3. $z=x{\left(wz\right)}^2$. One checks that $x{\left(wz\right)}^2=x\left(wawxwa\right)\left(wawxwa\right)=\left[xw{\left(aw\right)}^{2}xw\right]a=awxwa=z.$

Therefore $z\in {\mathcal{A}}^{w,\#}$. Moreover, we see that

$$\begin{array}{ccc}\hfill {\left|\right|\left(aw\right)}^n-awxw{\left(aw\right)}^n\left|\right|& =\hfill & {\left|\right|(a-awxwa)w\left(aw\right)}^{n-1}\left|\right|\hfill \\ & =\hfill & {\left|\right|yw\left(aw\right)}^{n-1}\left|\right|={\left|\right|ywaw\left(aw\right)}^{n-2}\left|\right|\hfill \\ & =\hfill & {\left|\right|yw(z+y)w\left(aw\right)}^{n-2}\left|\right|={\left|\right|\left(yw\right)}^2{\left(aw\right)}^{n-2}\left|\right|\hfill \\ & =\hfill & \dots ={\left|\right|\left(yw\right)}^n\left|\right|.\hfill \end{array}$$

Therefore

$$\underset{n\to \infty }{lim}{\left|\right|\left(yw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0,$$

and then $y\in {\mathcal{A}}_w^{qnil}$. This completes the proof. □

We denote x in Theorem 4.1 by $a^{w,d}$, and call it a generalized weighted w-core inverse of a.

Corollary 8. 

Let $a\in {\mathcal{A}}^{w,d}$. Then $aw\in {\mathcal{A}}^d$ and ${\left(aw\right)}^d=a^{w,d}w$.

Proof. 

In view of Theorem 4.1, there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}a{\left(wx\right)}^2=x,xw{\left(aw\right)}^{2}x=awx,{\left(awxw\right)}^{*}=awxw,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

Then $aw{\left(xw\right)}^2=xw$. By hypothesis, there exist $z,y\in \mathcal{A}$ such that

$$a=z+y,z^{*}y=ywz=0,z\in {\mathcal{A}}^{w,\#},y\in {\mathcal{A}}_w^{qnil}.$$

Hence $aw=zw+yw,zw\in {\mathcal{A}}^{\#},yw\in {\mathcal{A}}^{qnil}$ and $\left(yw\right)\left(zw\right)=0$. In view of [4], $aw\in {\mathcal{A}}^d$. Therefore $aw\in {\mathcal{A}}^d$ by [5]. In this case, ${\left(aw\right)}^d={\left(zw\right)}^{\#}={\left(awxwaw\right)}^{\#}=xw=a^{w,d}w$. □

Corollary 9. 

Let $a,w\in \mathcal{A}$. Then $a\in {\mathcal{A}}^{w,\#}$ if and only if $a\in {\mathcal{A}}_w^{\#}\bigcap {\mathcal{A}}^{w,d}$.

Proof. 

This is obvious by Theorem 4.1 and Corollary 2.4. □

We come now to present the polar-like property of elements associated the weighted w-core inverse.

Theorem 8. 

Let $a,w\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,d}$.

(2)

There exists a projection $p\in \mathcal{A}$ such that

$$1-p\in \mathcal{A}w,paw=pawp\in {\mathcal{A}}^{qnil},andaw+p\in {\mathcal{A}}^{-1}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Since $a\in {\mathcal{A}}^{w,\#}$, there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}a{\left(wx\right)}^2=x,xw{\left(aw\right)}^{2}x=awx,{\left(awxw\right)}^{*}=awxw,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

In view of Corollary 4.2, $aw\in {\mathcal{A}}^d$ and ${\left(aw\right)}^d=xw$. In light of [6], there exists a projection $p\in \mathcal{A}$ such that

$$paw=pawp\in {\mathcal{A}}^{qnil},andaw+p\in {\mathcal{A}}^{-1}.$$

Exactly, we have $p=1-aw{\left(aw\right)}^d$. Then $1-p=aw{\left(aw\right)}^d=awxw\in \mathcal{A}w$, as desired.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exists a projection $p\in \mathcal{A}$ such that

$$u:=aw+p\in {\mathcal{A}}^{-1},1-p\in \mathcal{A}w,paw=pawp\in {\mathcal{A}}^{qnil}.$$

Set $z=(1-p)a$ and $y=pa$. Then

$$\begin{array}{ccc}\hfill z^{*}y& =\hfill & {\left[(1-p)a\right]}^{*}\left(pa\right)=\left[a^{*}{(1-p)}^{*}\right]pa=a^{*}(1-p)pa=0,\hfill \\ \hfill ywz& =\hfill & paw(1-p)a=0,yw=paw\in {\mathcal{A}}^{qnil}.\hfill \end{array}$$

Claim 1. $z\in {\mathcal{A}}^{w,\#}$. Write $1-p=sw$ and $(1-p)s=s$. Set $x=u^{-1}s$. Then we verify that

Since $(1-p)aw=(1-p)u$, we have $(1-p)aw{u}^{-1}(1-p)=1-p$. Then $zwxw=1-p$. Then ${\left(zwxw\right)}^{*}=1-p=zwxw$.

Since $paw(1-p)=0$ and $paw\in {\mathcal{A}}^d$, it follows by [21] that $aw(1-p)\in {\mathcal{A}}^d$, and so ${\left(aw(1-p)\right)}^d={\left(aw\right)}^d(1-p)$. By virtue of Cline’s formula, $(1-p)aw\in {\mathcal{A}}^d$. Then $(1-p)u\in {\mathcal{A}}^d$. By Cline’s formula again, $u(1-p)\in {\mathcal{A}}^d$. Since $pu(1-p)=paw(1-p)=0$, it follows by [21] that ${\left[u(1-p)\right]}^d=u^{-1}(1-p)$. Thus, $px=p{u}^{-1}s=\left[p{u}^{-1}(1-p)\right]s=pu(1-p){\left[u{(1-p)}^d\right]}^{2}s=0$, and then $(1-p)x=x$. Therefore we verify that

$$\begin{array}{ccc}\hfill x{\left(wz\right)}^2& =\hfill & u^{-1}\left(sw\right)zwz=u^{-1}\left[(1-p)z\right]wz=u^{-1}zwz=u^{-1}(z+y)wz\hfill \\ & =\hfill & u^{-1}aw(1-p)z=u^{-1}(aw+p)(1-p)z=(1-p)z=z,\hfill \\ \hfill z{\left(wx\right)}^2& =\hfill & zwxwx=\left[(1-p)aw{u}^{-1}\right]swx=\left[(1-p)s\right]wx=swx=(1-p)x=x.\hfill \end{array}$$

Therefore $z\in {\mathcal{A}}^{w,\#}$ and $x=z^{w,\#}$.

Claim 2. $a\in {\mathcal{A}}^{w,d}$. We verify that

$$\begin{array}{ccc}\hfill a{\left(wx\right)}^2& =\hfill & (z+y){\left(wx\right)}^2=z{\left(wx\right)}^2=x,\hfill \\ \hfill awxw& =\hfill & (z+y)wxw=zwxw,\hfill \\ \hfill {\left(awxw\right)}^{*}& =\hfill & {\left(zwxw\right)}^{*}=zwxw=awxw,\hfill \\ \hfill x{\left(aw\right)}^{2}x& =\hfill & x(z+y)w(z+y)wx=x{\left(zw\right)}^{2}x=zwx=awx,\hfill \end{array}$$

Moreover, we see that

$$\begin{array}{cc}& {\left|\right|\left(aw\right)}^n-awxw{\left(aw\right)}^n\left|\right|={\left|\right|(aw-awxwaw)\left(aw\right)}^{n-1}\left|\right|\hfill \\ \hfill =& {\left|\right|[aw-(z+y)wxwaw]\left(aw\right)}^{n-1}\left|\right|={\left|\right|[aw-\left(zwxw\right)(z+y)w]\left(aw\right)}^{n-1}\left|\right|\hfill \\ \hfill =& \left|\right|[aw-{\left(wxw\right)}^{*}{z}^{*}(z+y)w]{\left(aw\right)}^{n-1}\left|\right|=\left|\right|[aw-{\left(wxw\right)}^{*}{z}^{*}zw]{\left(aw\right)}^{n-1}\left|\right|\hfill \\ \hfill =& \left|\right|[aw-{\left(zwxw\right)}^{*}zw]{\left(aw\right)}^{n-1}\left|\right|=\left|\right|[aw-\left(zwxwz\right)w]{\left(aw\right)}^{n-1}\left|\right|\hfill \\ \hfill =& {\left|\right|(aw-zw)\left(aw\right)}^{n-1}\left|\right|={\left|\right|yw\left(aw\right)}^{n-1}\left|\right|={\left|\right|yw(z+y)w\left(aw\right)}^{n-2}\left|\right|\hfill \\ \hfill =& {\left|\right|\left(yw\right)}^2{\left(aw\right)}^{n-2}\left|\right|=\dots ={\left|\right|\left(yw\right)}^n\left|\right|.\hfill \end{array}$$

Since $yw\in {\mathcal{A}}^{qnil}$, we have $\underset{n\to \infty }{lim}{\left|\right|\left(yw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$ Then $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-awxw{\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$ Therefore $a\in {\mathcal{A}}^{w,d}$, as asserted. □

Corollary 10. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^d$.

(2)

There exists a projection $p\in \mathcal{A}$ such that $pa=pap\in {\mathcal{A}}^{qnil},anda+p\in {\mathcal{A}}^{-1}.$

Proof. 

This is obvious by choosing $w=1$ in Theorem 4.4. □

Lemma 8. 

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}^{w,d}$ if and only if

(1)

$a\in {\mathcal{A}}^{d,w}$;

(2)

There exists $x\in \mathcal{A}$ such that

$$xwawx=x,x\mathcal{A}=a^{d,w}\mathcal{A},\mathcal{A}xw=\mathcal{A}{\left(a^{d,w}\right)}^{*}.$$

In this case, $a^{w,d}=x$.

Proof. 

⟹ Set $x=a^{w,d}$. By virtue of Corollary 4.2, $aw\in {\mathcal{A}}^d$ and $xw={\left(aw\right)}^d$. In view of [5], $aw\in {\mathcal{A}}^d$ and

$$xw\left(aw\right)xw=xw,xw\mathcal{A}={\left(aw\right)}^d\mathcal{A},\mathcal{A}xw=\mathcal{A}{\left({\left(aw\right)}^d\right)}^{*}.$$

Since $a^{d,w}={\left[{\left(aw\right)}^d\right]}^{2}a=a{\left[{\left(wa\right)}^d\right]}^2={\left(aw\right)}^{d}a{\left(wa\right)}^d$, we easily check that ${\left(aw\right)}^d={\left[{\left(aw\right)}^d\right]}^{2}aw=a^{d,w}a,$ and then ${\left(aw\right)}^d\mathcal{A}=a^{d,w}\mathcal{A}.$ On the other hand, we have ${\left(a^{d,w}\right)}^{*}={\left[{\left(aw\right)}^{d}aw\right]}^{*}{\left[{\left(aw\right)}^d\right]}^{*}$ and ${\left[{\left(aw\right)}^d\right]}^{*}={\left[{\left({\left(aw\right)}^d\right)}^{2}aw\right]}^{*}=w^{*}{\left(a^{d,w}\right)}^{*}.$ Thus, $\mathcal{A}{\left[{\left(aw\right)}^d\right]}^{*}=\mathcal{A}{\left(a^{d,w}\right)}^{*}.$ Therefore $x\mathcal{A}=xw\mathcal{A}=a^{d,w}\mathcal{A},\mathcal{A}xw=\mathcal{A}{\left(a^{d,w}\right)}^{*}$.

⟸ By hypothesis, There exists $x\in \mathcal{A}$ such that

$$xwawx=x,x\mathcal{A}=a^{d,w}\mathcal{A},\mathcal{A}xw=\mathcal{A}{\left(a^{d,w}\right)}^{*}.$$

As the argument above, we have

$${\left(aw\right)}^d\mathcal{A}=a^{d,w}\mathcal{A},\mathcal{A}{\left[{\left(aw\right)}^d\right]}^{*}=\mathcal{A}{\left(a^{d,w}\right)}^{*}.$$

Therefore we get

$$xwawx=x,xw\mathcal{A}=x\mathcal{A}={\left(aw\right)}^d\mathcal{A},\mathcal{A}xw=\mathcal{A}{\left({\left(aw\right)}^d\right)}^{*}.$$

By virtue of [5], $aw\in {\mathcal{A}}^d$ and ${\left(aw\right)}^d=xw$. Therefore we have

$$xw=aw{\left(xw\right)}^2=xwawxw,{\left(awxw\right)}^{*}=awxw,\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

Since $xw=aw{\left(xw\right)}^2$ and $x\in xw\mathcal{A}$, we have $x=awxwx=a{\left(wx\right)}^2$. This implies that $x=awxwx={\left(aw\right)}^{2}x{\left(wx\right)}^2={\left(aw\right)}^{n-1}x{\left(wx\right)}^{n-1}$ for any $n\in N$. Hence $awx={\left(aw\right)}^{n}x{\left(wx\right)}^{n-1}$. This implies that

$$\begin{array}{ccc}\hfill \left|\right|awx-xw{\left(aw\right)}^{2}x{\left|\right|}^{{\displaystyle \frac{1}{n}}}& =\hfill & {\left|\right|\left(aw\right)}^{n}x{\left(wx\right)}^{n-1}-\left[xw{\left(aw\right)}^{n+1}\right]x{\left(wx\right)}^{n-1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ & \le \hfill & {\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}{\left|\right|x\left(wx\right)}^{n-1}{\left|\right|}^{{\displaystyle \frac{1}{n}}};\hfill \end{array}$$

whence $\underset{n\to \infty }{lim}\left|\right|awx-xw{\left(aw\right)}^{2}x{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$ Thus $awx=xw{\left(aw\right)}^{2}x$. Therefore $a^{w,d}=x$, as required. □

We are now prepared to prove:

Theorem 9. 

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}^{w,d}$.

(2)

$a\in {\mathcal{A}}^{d,w}$ and $a^{d,w}\in {\mathcal{A}}^{w,\#}$.

In this case, $a^{w,d}={\left[a^{d,w}w\right]}^2{\left(a^{d,w}\right)}^{w,\#}.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ As in the proof of Corollary 4.2, $aw\in {\mathcal{A}}^d$. By using Cline’s formula, $wa\in {\mathcal{A}}^d$. Hence, $a\in {\mathcal{A}}^{d,w}$. Let $x=a^{w,d}$. Then we have

$$\begin{array}{c}a{\left(wx\right)}^2=x,xw{\left(aw\right)}^{2}x=awx,{\left(awxw\right)}^{*}=awxw,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

Obviously, $x={\left(aw\right)}^{n}x{\left(wx\right)}^n$ for any $n\in N$. Then

$$\begin{array}{ccc}\hfill \left|\right|x-aw{\left(aw\right)}^{d}x\left|\right|& =\hfill & {\left|\right|\left(aw\right)}^{n}x{\left(wx\right)}^n-\left(aw\right){\left(aw\right)}^d{\left(aw\right)}^{n}x{\left(wx\right)}^n\left|\right|\hfill \\ & \le \hfill & {\left|\right|\left(aw\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}{\left|\right|\left|\right|x\left(wx\right)}^n\left|\right|.\hfill \end{array}$$

Since $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0$, we get $\underset{n\to \infty }{lim}\left|\right|x-aw{\left(aw\right)}^{d}x{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$ Hence $\left(aw\right){\left(aw\right)}^{d}x=x$. Furthermore, we have

$$\begin{array}{ccc}\hfill {\left|\right|aw\left(aw\right)}^d-awxw\left(aw\right){\left(aw\right)}^d\left|\right|& =\hfill & {\left|\right|\left(aw\right)}^n{\left[{\left(aw\right)}^d\right]}^n-awx{\left(aw\right)}^n{\left[{\left(aw\right)}^d\right]}^n\left|\right|\hfill \\ & \le \hfill & {\left|\right|\left(aw\right)}^n-awxw{\left(aw\right)}^n{\left|\right|\left|\right|\left(aw\right)}^d{]}^n\left|\right|.\hfill \end{array}$$

Since $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-awxw{\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0$, we deduce that

$$\underset{n\to \infty }{lim}{\left|\right|aw\left(aw\right)}^d-awxw\left(aw\right){\left(aw\right)}^d{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

Hence $awxw\left(aw\right){\left(aw\right)}^d=aw{\left(aw\right)}^d$. Let $z={\left(aw\right)}^{2}x$. Then

$$\begin{array}{ccc}\hfill a^{d,w}wzw& =\hfill & a^{d,w}w{\left(aw\right)}^{2}xw={\left[{\left(aw\right)}^d\right]}^{2}aw{\left(aw\right)}^{2}xw=awxw,\hfill \\ \hfill a^{d,w}{\left(wz\right)}^2& =\hfill & \left(awxw\right)z=\left(awxw\right){\left(aw\right)}^{2}x=aw\left[xw{\left(aw\right)}^{2}x\right]={\left(aw\right)}^{2}x=z,\hfill \\ \hfill {\left(a^{d,w}wzw\right)}^{*}& =\hfill & {\left(awxw\right)}^{*}=awxw=a^{d,w}wzw,\hfill \\ \hfill z{\left(w{a}^{d,w}\right)}^2& =\hfill & {\left(aw\right)}^{2}xw{a}^{d,w}w{a}^{d,w}=aw\left[awxw\left(aw\right){\left(aw\right)}^d\right]a^{d,w}w{a}^{d,w}\hfill \\ & =\hfill & aw\left[\left(aw\right){\left(aw\right)}^d\right]a^{d,w}w{a}^{d,w}=aw\left[\left(aw\right){\left(aw\right)}^d\right]{\left[{\left(aw\right)}^d\right]}^{2}aw{a}^{d,w}\hfill \\ & =\hfill & aw{\left(aw\right)}^d{a}^{d,w}=a^{d,w}.\hfill \end{array}$$

Accordingly, $a^{d,w}\in {\mathcal{A}}^{w,\#}$ and ${\left(a^{d,w}\right)}^{w,\#}=z={\left(aw\right)}^2{a}^{w,d},$ as required.

$\left(2\right)\Rightarrow \left(1\right)$ Set $z={\left(a^{d,w}\right)}^{w,\#}$. Then we have $a^{d,w}wzw{a}^{d,w}=a^{d,w},{\left[a^{d,w}wzw\right]}^{*}=a^{d,w}wzw.$ Let $x={\left(a^{d,w}w\right)}^{2}z$. Then we verify that

$$\begin{array}{ccc}\hfill xwawx& =\hfill & \left[a^{d,w}w{a}^{d,w}wz\right]waw\left[a^{d,w}w{a}^{d,w}wz\right]\hfill \\ & =\hfill & a^{d,w}w\left[a^{d,w}wzw{a}^{d,w}\right]waw{a}^{d,w}wz]\hfill \\ & =\hfill & a^{d,w}w{a}^{d,w}waw{a}^{d,w}wz]\hfill \\ & =\hfill & a^{d,w}w{\left(aw\right)}^{d}aw{a}^{d,w}wz]\hfill \\ & =\hfill & a^{d,w}w{a}^{d,w}wz=x.\hfill \end{array}$$

Clearly, $x\mathcal{A}\subseteq a^{d,w}\mathcal{A}$. Also we see that

$$a^{d,w}={\left(a^{d,w}w\right)}^{2}a=a^{d,w}w\left[a^{d,w}wzw{a}^{d,w}\right]wa=\left[{\left(a^{d,w}w\right)}^{2}z\right]w{a}^{d,w}wa;$$

whence, $a^{d,w}\mathcal{A}\subseteq x\mathcal{A}.$ Then $x\mathcal{A}=a^{d,w}\mathcal{A}$.

We easily verify that

$$\begin{array}{ccc}\hfill xw& =\hfill & a^{d,w}w{a}^{d,w}wzw=a^{d,w}w\left[a^{d,w}wzw\right]\hfill \\ & =\hfill & a^{d,w}w{\left[a^{d,w}wzw\right]}^{*}=a^{d,w}w{\left(wzw\right)}^{*}{\left(a^{d,w}\right)}^{*};\hfill \end{array}$$

and then, $\mathcal{A}xw\subseteq \mathcal{A}{\left(a^{d,w}\right)}^{*}$. On the other hand, we have

$$\begin{array}{ccc}\hfill {\left(a^{d,w}\right)}^{*}& =\hfill & {\left[a^{d,w}wzw{a}^{d,w}\right]}^{*}={\left[{\left(a^{d,w}wzw\right)}^{*}{a}^{d,w}\right]}^{*}\hfill \\ & =\hfill & {\left(a^{d,w}\right)}^{*}{a}^{d,w}wzw\hfill \\ & =\hfill & {\left(a^{d,w}\right)}^{*}\left[a^{d,w}waw{a}^{d,w}\right]wzw\hfill \\ & =\hfill & {\left(a^{d,w}\right)}^{*}\left[a^{d,w}wa\right]w{a}^{d,w}wzw\hfill \\ & =\hfill & {\left(a^{d,w}\right)}^{*}\left[aw{a}^{d,w}\right]w{a}^{d,w}wzw\hfill \\ & =\hfill & {\left(a^{d,w}\right)}^{*}aw\left[a^{d,w}w{a}^{d,w}wz\right]w\hfill \\ & =\hfill & {\left(a^{d,w}\right)}^{*}aw\left(xw\right),\hfill \end{array}$$

and so $\mathcal{A}{\left(a^{d,w}\right)}^{*}\subseteq \mathcal{A}xw$. Hence $\mathcal{A}xw=\mathcal{A}{\left(a^{d,w}\right)}^{*}$. Therefore $a\in {\mathcal{A}}^{w,d}$ by Lemma 4.6. In this case, $a^{w,d}=x={\left[a^{d,w}w\right]}^2{\left(a^{d,w}\right)}^{w,\#}.$ □