On G-S and C-S inverses in *-rings

Abstract

The notions of the G-S inverse and the C-S inverse are extended from the set of all complex n × n matrices to *-rings. We present new characterizations of these inverses, study their properties, and apply the C-S inverse to introduce two relations on the set ${\mathcal{R}}^{⏧d}$ of all core-EP invertible elements in a *-ring with identity. We explore the connection between these two relations and the star partial order and prove that one of them is a partial order on ${\mathcal{R}}^{⏧d}$.

KEYWORDS:

• *-ring
• core-EP decomposition
• C-S inverse
• C-S partia order
• Drazin inverse

2020 MATHEMATICS SUBJECT CLASSIFICATION:

• 06A06
• 15A09
• 15A10
• 17C27

1 Introduction

Let $\mathcal{S}$ be a semigroup and $a\in \mathcal{S}$. We say that an element $a\in \mathcal{S}$ has a Drazin inverse $x=a^D\in \mathcal{S}$ if (1) $$\mathit{\text{ax}}=\mathit{\text{xa}},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}x=a{x}^2,\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{a}^k=a^{k+1}x,$$(1) for some non-negative integer k. Note that for a semigroup without the identity we have k > 0, while for a semigroup with identity we have $k\ge 0$ and for k = 0 we define $a^0=1$. If a has a Drazin inverse a^D, then we say that a is Drazin invertible and the smallest non-negative integer k in (1) is called the Drazin index i(a) of a. It is well known that there is at most one $x=a^D$ such that (1) holds (see [7]). Drazin inverse has many applications in the theories of finite Markov chains [2], singular differential and difference equations [2], cryptography [12], iterative methods in numerical analysis [17], etc. If $i(a)\le 1$, the Drazin inverse $x=a^D$ of a is known as the group inverse $a^{♯}$ of a, i.e., $x=a^D=a^{♯}$ is the group inverse of a when ax = xa, axa = a, and xax = x.

Let $\mathcal{R}$ be a ring with the (multiplicative) identity 1. Denote by ${\mathcal{R}}^D,{\mathcal{R}}^{♯}$, and $\mathcal{N}(\mathcal{R})$ the sets of all Drazin invertible, group invertible, and nilpotent elements in $\mathcal{R}$, respectively. In [14], an equivalent definition of the Drazin inverse for rings with identity was given. Namely, for $a,b\in \mathcal{R}$, (1) is equivalent to (2) $$\mathit{\text{ab}}=\mathit{\text{ba}},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}b=a{b}^2,\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}a-a^{2}b\in \mathcal{N}(\mathcal{R}).$$(2)

Moreover, the index i(a) of a is equal to the nilpotency index of $a-a^{2}b$. Suppose that $a\in {\mathcal{R}}^D$. It is known (see [21]) that we may write (3) $$a=c+n,$$(3) where $c\in {\mathcal{R}}^{♯},n\in \mathcal{N}(\mathcal{R})$ with index of nilpotency equal to i(a), and $\mathit{\text{cn}}=\mathit{\text{nc}}=0$. Then c is called the core part of a and n is the nilpotent part of a. Since $c^{♯}c{c}^{♯}=c^{♯}$ and $c^{♯}c=c{c}^{♯}$, it follows $c^{♯}n=0=n{c}^{♯}$, and therefore by (2), $a^D=c^{♯}$. Since the Drazin inverse of every element in $\mathcal{R}$ is unique if it exists, we may conclude that c and n from (3) are unique. In fact, (4) $$c=a^2{a}^D\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}\text{and}\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}n=a-a^2{a}^D.$$(4)

We refer to c + n as the core-nilpotent decomposition of a.

The Drazin inverse can be considered as a generalization of the group inverse. Let $M_n(\mathbb{F})$ be the set of all n × n matrices over a filed $\mathbb{F}$. Another generalization of the group inverse of $A\in M_n(\mathbb{C})$, where $\mathbb{C}$ is the field of complex numbers and $i(A)=k$, was introduced in [11] as the unique solution X to the following system of matrix equations: $$\mathit{\text{AX}}=\mathit{\text{XA}},\hspace{1em}{A}^{l+1}X=A^l,\hspace{1em}A{X}^{l+1}=X^l,\hspace{1em}A-X=A^l{X}^l\left(A-X\right)$$in which $l\ge k$. This solution was denoted in [13] as $A^{(S)}$ and named in [25] as the G-S inverse of A. It was proved in [11] that $A^{(S)}=A^D+N$, where $N=A-A^2{A}^D$, i.e., N is the nilpotent part of A.

Another generalized inverse which is closely related to the group inverse is called the core inverse. The core inverse was originally discussed in [20] on $M_n(\mathbb{C})$ but was later independently reintroduced (and named) in [1], and generalized in [19] to *-rings, i.e., rings that are equipped with an involution *. For a *-ring $\mathcal{R}$, we say that $a\in \mathcal{R}$ has the core inverse $a^{⏧♯}$ if $x=a^{⏧♯}$ is the unique solution to the following equations: (5) $$\mathit{\text{axa}}=a,\hspace{1em}\mathit{\text{xax}}=x,\hspace{1em}{(\mathit{\text{ax}})}^{\ast }=\mathit{\text{ax}},\hspace{1em}x{a}^2=a,\hspace{1em}\text{and}\hspace{1em}{\text{ax}}^2\text{=x}.$$(5)

It was proved in [23] that the first two equations in (5) can be omitted, that is, $a\in \mathcal{R}$ has the core inverse $x=a^{⏧♯}$ if and only if (6) $${(\mathit{\text{ax}})}^{\ast }=\mathit{\text{ax}},\hspace{1em}x{a}^2=a\hspace{1em}\text{and}\hspace{1em}a{x}^2=x.$$(6)

The set of all core invertible elements in $\mathcal{R}$ is denoted by ${\mathcal{R}}^{⏧♯}$.

Note (see [1]) that the core inverse of $A\in M_n(\mathbb{C})$ exists if and only if the index of A is less than or equal to one. Manjunatha Prasad et al. [15] extended the notion of the core inverse to the full matrix algebra $M_n(\mathbb{F})$ by introducing a new kind of matrix generalized inverse called the core-EP inverse and showed that whenever $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$ every matrix $A\in M_n(\mathbb{F})$ has the unique core-EP inverse. Let $\text{Im} A$ denote the image (i.e. the column space) of A. We say that a matrix $X\in M_n(\mathbb{F})$ is the core-EP inverse of $A\in M_n(\mathbb{F})$ if $$\mathit{\text{XAX}}=X\hspace{1em}\text{and}\hspace{1em}\mathrm{\text{Im}} X=\mathrm{\text{Im}} X^{\ast }=\mathrm{\text{Im}} A^m,$$where $m=i(A)$ is the index of A.

As a generalization for both the core inverse in a *-ring and the core-EP inverse for complex or real matrices, Gao and Chen [9] put forward the notion of the pseudo-core inverse in a * -ring. Let $\mathcal{R}$ be a *-ring and $a\in \mathcal{R}$. If there exists $x=a^{⏧d}\in \mathcal{R}$ such that (7) $$x{a}^{m+1}=a^m\text{ for some positive integer }m,\hspace{1em}a{x}^2=x,\hspace{1em}\text{ and}\hspace{1em}{(\mathit{\text{ax}})}^{\ast }=\mathit{\text{ax}},$$(7) then we say that a is pseudo-core (or core-EP) invertible and call $a^{⏧d}$ the pseudo-core (or core-EP) inverse of a.

The set of all core-EP invertible elements in $\mathcal{R}$ is denoted by ${\mathcal{R}}^{⏧d}$. It turns out (see [9, Theorem 2.2]) that $a^{⏧d}$ is unique if it exists, and that in this case the smallest positive integer m in (7), which is called the pseudo-core index of a and denoted by I(a), either equals the Drazin index i(a) of a if i(a) > 0, or is 1 if i(a) = 0 (see [9, Theorem 2.3] and note that Gao et al. defined the Drazin index of $a$ as the smallest positive integer k that satisfies (1)). By (7), for I(a) = 1, the core-EP inverse becomes the core inverse of a. Moreover, it was proved in [9] that $a\in \mathcal{R}$ has the core-EP inverse $a^{⏧d}\in \mathcal{R}$ if and only if $x=a^{⏧d}$ satisfies the following equations: (8) $$\mathit{\text{xax}}=x\mathrm{\hspace{1em}\hspace{1em}}\text{and}\mathrm{\hspace{1em}\hspace{1em}}x\mathcal{R}=x^{\ast }\mathcal{R}=a^m\mathcal{R},$$(8) for some positive integer m.

Another decomposition on $M_n(\mathbb{C})$ was introduced in [24], and generalized in [10] to *-rings. Let $\mathcal{R}$ be a *-ring. Suppose that $a\in {\mathcal{R}}^{⏧d}$ and let I(a) = k. By [10, Theorem 3.1], we may write a $=a_1+a_2$, where (9) $$i(a_1)\le 1,\hspace{1em}{a}_2^k=0,\hspace{1em}\text{and}\hspace{1em}{a}_1^{\ast }{a}_2=a_2{a}_1=0.$$(9)

It turns out that this decomposition, which we call the core-EP decomposition in $\mathcal{R}$, is unique and that $$a_1=a{a}^{⏧d}a\hspace{1em}\text{and}\hspace{1em}{a}_2=a-a{a}^{⏧d}a,$$where $a^{⏧d}$ is the core-EP inverse of a. We call a₂ the nilpotent part of the core-EP decomposition of a.

Recall that for $A\in M_n(\mathbb{C})$ with $i(A)\le 1,A^{(S)}=A^{♯}$. Motivated by the G-S inverse $A^{(S)}$ of A as a generalization of the group inverse of A, authors of [25] introduced a new generalized inverse as a generalization of the core inverse.

Definition 1.

Let $A\in M_n(\mathbb{C})$ and i(A) = k. The C-S inverse of A is defined as the solution of the following system of matrix equations: $$X{A}^{k+1}=A^k,\hspace{1em}{\left(A^k{X}^k\right)}^{\ast }=A^k{X}^k,\hspace{1em}A-X=A^k{X}^k\left(A-X\right).$$

The C-S inverse of A is denoted by $A^{⏧s}$.

It was proved in [25] that the C-S inverse of A is unique and that $A^{⏧s}=A^{⏧d}+A_2$, where $A_2=A-A{A}^{⏧d}A$, i.e., A₂ is the nilpotent part of the core-EP decomposition of A.

In this paper, we extend in Section 3 the notions of the G-S inverse and the C-S inverse to *-rings. We present characterizations of these inverses and study their properties. In Section 4 we apply the C-S inverse to introduce two relations on ${\mathcal{R}}^{⏧d}$, where $\mathcal{R}$ is a *-ring with identity. We explore how are these relations connected to the star partial order and prove that one of these relations is a partial order on ${\mathcal{R}}^{⏧d}$.

2 Preliminaries

Let us now present some tools which will be useful throughout the paper. Let $\mathcal{R}$ be a *-ring with identity 1. If for $p\in \mathcal{R}$, $p^2=p$, then p is said to be an idempotent. A projection $p\in \mathcal{R}$ is a self-adjoint idempotent, i.e., $p=p^2=p^{\ast }$. The equality $1=e_1+e_2+\cdots +e_n$, where $e_1,e_2,\dots ,e_n$ are idempotent elements in $\mathcal{R}$ and $e_i{e}_j=0$ for $i\ne j$, is called a decomposition of the identity of $\mathcal{R}$. Let $1=e_1+\dots +e_n$ and $1=f_1+\dots +f_n$ be two decompositions of the identity of $\mathcal{R}$. We have $$x=1·x·1=(e_1+e_2+\dots +e_n)x(f_1+f_2+\dots +f_n)=\sum _{i,j=1}^n{e}_{i}x{f}_j.$$

Then any $x\in \mathcal{R}$ can be uniquely represented in the following matrix form: (10) $$x={\left[\begin{array}{ccc}{x}_{11}& \cdots & x_{1n}\\ ⋮& \ddots & ⋮\\ x_{n1}& \cdots & x_{\mathit{\text{nn}}}\end{array}\right]}_{e\times f},$$(10) where $x_{\mathit{\text{ij}}}=e_{i}x{f}_j\in e_i\mathcal{R}{f}_j$. If $x={(x_{\mathit{\text{ij}}})}_{e\times f}$ and $y={(y_{\mathit{\text{ij}}})}_{e\times f}$, then $x+y={(x_{\mathit{\text{ij}}}+y_{\mathit{\text{ij}}})}_{e\times f}$. Moreover, if $1=g_1+\cdots +g_n$ is a decomposition of the identity of $\mathcal{R}$ and $z={(z_{\mathit{\text{ij}}})}_{f\times g}$, then, by the orthogonality of the idempotents involved, $\mathit{\text{xz}}={({\sum }_{k=1}^n{x}_{\mathit{\text{ik}}}{z}_{\mathit{\text{kj}}})}_{e\times g}$. Thus, if we have decompositions of the identity of $\mathcal{R}$, then the usual algebraic operations in $\mathcal{R}$ can be interpreted as simple operations between appropriate n × n matrices over $\mathcal{R}$. When n = 2 and $p,q\in \mathcal{A}$ are idempotent elements, we may write $$x=\mathit{\text{pxq}}+\mathit{\text{px}}(1-q)+(1-p)\mathit{\text{xq}}+(1-p)x(1-q)={\left[\begin{array}{cc}{x}_{1,1}& x_{1,2}\\ x_{2,1}& x_{2,2}\end{array}\right]}_{p\times q}.$$

Here $x_{1,1}=\mathit{\text{pxq}},$ $x_{1,2}=\mathit{\text{px}}(1-q),$ $x_{2,1}=(1-p)\mathit{\text{xq}},$ $x_{2,2}=(1-p)x(1-q)$.

By (10) may write (11) $$x^{\ast }={\left[\begin{array}{ccc}{x}_{11}^{\ast }& \cdots & x_{n1}^{\ast }\\ ⋮& \ddots & ⋮\\ x_{1n}^{\ast }& \cdots & x_{\mathit{\text{nn}}}^{\ast }\end{array}\right]}_{f^{\ast }\times e^{\ast }},$$(11) where this matrix representation of $x^{\ast }$ is given relative to the decompositions of the identity $1=f_1^{\ast }$ $+\dots +f_n^{\ast }$ and $1=e_1^{\ast }+\dots +e_n^{\ast }$.

Let $a\in \mathcal{R}$ and let $a^{°}$ denote the right annihilator of $a$, i.e., the set $a^{°}=\{x\in \mathcal{R}:\mathit{\text{ax}}=0\}$. Similarly we denote the left annihilator $°a$ of a, i.e., the set $°a=\{x\in \mathcal{R}:\mathit{\text{xa}}=0\}$. Suppose that $p,q\in \mathcal{R}$ are such idempotents that $°a=°p$ and $a^{°}=q^{°}$. Observe (or see [3, Lemma 2.2]) that $°p=\mathcal{R}(1-p)$ and $q^{°}=(1-q)\mathcal{R}$. It follows that then a = paq, i.e., (12) $$a={\left[\begin{array}{cc}a& 0\\ 0& 0\end{array}\right]}_{p\times q}.$$(12)

Let $a\in {\mathcal{R}}^{⏧d}$. Note that by (8), $a^{⏧d}a{a}^{⏧d}=a^{⏧d}$. So, $p=a{a}^{⏧d}$ (and $q=a^{⏧d}a$) is an idempotent. Let $a=a_1+a_2$ be the core-EP decomposition of a, where a₂ is the nilpotent part. It was proved in [4] that for $p=a{a}^{⏧d}$, $$a_1={\left[\begin{array}{cc}t& s\\ 0& 0\end{array}\right]}_{p\times p}\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}\text{and}\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{a}_2={\left[\begin{array}{cc}0& 0\\ 0& a_2\end{array}\right]}_{p\times p},$$where t is invertible in the ring $p\mathcal{R}p$, i.e., there exists $t^{-1}\in p\mathcal{R}p$ such that $t{t}^{-1}=t^{-1}t=p$. So, (13) $$a={\left[\begin{array}{cc}t& s\\ 0& a_2\end{array}\right]}_{p\times p}.$$(13)

Note that $t=a^2{a}^{⏧d}$ and $t^{-1}=a^{⏧d}$ (see [5]).

The following lemma will be used in the continuation.

Lemma 2.1.

Let $1=e_1+e_2+e_3$ be a decomposition of the identity of $\mathcal{R}$ for some projections $e_1,e_2,e_3\in \mathcal{R}$, and let for $b\in {\mathcal{R}}^{⏧d}$, $$b={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& n\end{array}\right]}_{e\times e},$$where t is invertible in the ring $e_1\mathcal{R}{e}_1$, t₁ is invertible in the ring $e_2\mathcal{R}{e}_2$ and n is nilpotent. Then $$b^{⏧d}={\left[\begin{array}{ccc}{t}^{-1}& -t^{-1}{s}_1{t}_1^{-1}& 0\\ 0& t_1^{-1}& 0\\ 0& 0& 0\end{array}\right]}_{e\times e}.$$

Proof.

Let $n^m=0$. Then $$b^m={\left[\begin{array}{ccc}{t}^m& U& V\\ 0& t_1^m& Z\\ 0& 0& 0\end{array}\right]}_{e\times e}$$for certain $U,V,Z\in \mathcal{R}$ and thus $$b^{m+1}={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& n\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{t}^m& U& V\\ 0& t_1^m& Z\\ 0& 0& 0\end{array}\right]}_{e\times e}={\left[\begin{array}{ccc}{t}^{m+1}& \mathit{\text{tU}}+s_1{t}_1^m& \mathit{\text{tV}}+s_{1}Z\\ 0& t_1^{m+1}& t_{1}Z\\ 0& 0& 0\end{array}\right]}_{e\times e}.$$

Let $$x={\left[\begin{array}{ccc}{t}^{-1}& -t^{-1}{s}_1{t}_1^{-1}& 0\\ 0& t_1^{-1}& 0\\ 0& 0& 0\end{array}\right]}_{e\times e}.$$

Then $$x{b}^{m+1}={\left[\begin{array}{ccc}{t}^m& U& V\\ 0& t_1^m& Z\\ 0& 0& 0\end{array}\right]}_{e\times e}$$and hence $x{b}^{m+1}=b^m$. Also, $$b{x}^2={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& n\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{t}^{-2}& -t^{-2}{s}_1{t}_1^{-1}-t^{-1}{s}_1{t}_1^{-2}& 0\\ 0& t_1^{-2}& 0\\ 0& 0& 0\end{array}\right]}_{e\times e}={\left[\begin{array}{ccc}{t}^{-1}& -t^{-1}{s}_1{t}_1^{-1}& 0\\ 0& t_1^{-1}& 0\\ 0& 0& 0\end{array}\right]}_{e\times e}=x$$ and $$\mathit{\text{bx}}={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& n\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{t}^{-1}& -t^{-1}{s}_1{t}_1^{-1}& 0\\ 0& t_1^{-1}& 0\\ 0& 0& 0\end{array}\right]}_{e\times e}={\left[\begin{array}{ccc}{e}_1& 0& 0\\ 0& e_2& 0\\ 0& 0& 0\end{array}\right]}_{e\times e}.$$

So, ${(\mathit{\text{bx}})}^{\ast }=\mathit{\text{bx}}$. It follows by (7) that $x=b^{⏧d}$. □

3 The G-S and C-S inverses in rings

In this section we extend the notions of the G-S inverse and the C-S inverse to rings.

Theorem 3.1.

Let $\mathcal{R}$ be a ring with identity and $a\in {\mathcal{R}}^D$ with $i(a)=k$, and let $a=c+n$ be the core-nilpotent decomposition of a, where n is the nilpotent part. Then $x=a^D+n$ is the unique solution to the following system of equations (14) $$\mathit{\text{ax}}=\mathit{\text{xa}},\hspace{1em}{a}^{l+1}x=a^l,\hspace{1em}a{x}^{l+1}=x^l,\hspace{1em}\text{and}\hspace{1em}a-x=a^l{x}^l\left(a-x\right),$$(14) where $l\ge k$.

Proof.

Let $x=a^D+n$. First recall that $a^D=c^{♯}$ and that $\mathit{\text{cn}}=\mathit{\text{nc}}=c^{♯}n=n{c}^{♯}=0$. Then $$\mathit{\text{ax}}=\left(c+n\right)\left(c^{♯}+n\right)=c{c}^{♯}+n^2=c^{♯}c+n^2=\left(c^{♯}+n\right)\left(c+n\right)=\mathit{\text{xa}}.$$

Since $n^l=0$, we get $$a^{l+1}x={\left(c+n\right)}^{l+1}\left(c^{♯}+n\right)=\left(c^{l+1}+n^{l+1}\right)\left(c^{♯}+n\right)=c^{l+1}\left(c^{♯}+n\right)=c^{l+1}{c}^{♯}=c^l$$ $$=c^l+n^l={\left(c+n\right)}^l=a^l.$$

Similarly we prove that $a{x}^{l+1}=x^l$ and $a^l{x}^l(a-x)=a-x$.

So, $x=a^D+n$ is a solution to the system of equations (14)(14) $$\mathit{\text{ax}}=\mathit{\text{xa}},\hspace{1em}{a}^{l+1}x=a^l,\hspace{1em}a{x}^{l+1}=x^l,\hspace{1em}\text{and}\hspace{1em}a-x=a^l{x}^l\left(a-x\right),$$(14) . To prove the uniqueness, let x and y be two solutions of (14). We have ay = ya and thus $a^{l}y=y{a}^l$ which yields $c^{l}y=y{c}^l$. Also, $a^{l+1}x=a^l$ and $a^{l+1}y=a^l$ imply $c^{l+1}x=c^{l+1}y$. Multiplying this equation from the left by $c^{♯}$, we get $c^{l}y=c^{l}x$. Now, $$a-x=a^l{x}^l\left(a-x\right)=a^{l+1}{x}^l-a^l{x}^{l+1}=c^{l+1}{x}^l-c^l{x}^{l+1}=c{c}^{l}x{x}^{l-1}-c^{l}x{x}^l$$ $$=c{c}^{l}y{x}^{l-1}-c^{l}y{x}^l=\mathit{\text{cy}}{c}^{l}x{x}^{l-2}-y{c}^{l}x{x}^{l-1}=\mathit{\text{cy}}{c}^{l}y{x}^{l-2}-y{c}^{l}y{x}^{l-1}$$ $$=c{y}^2{c}^l{x}^{l-2}-y^2{c}^l{x}^{l-1}=\dots =c{y}^l{c}^l-y^{l+1}{c}^l=c^{l+1}{y}^l-c^l{y}^{l+1}$$ $$=a^{l+1}{y}^l-a^l{y}^{l+1}=a^l{y}^l\left(a-y\right)=a-y$$and therefore x = y. □

Definition 2.

Let $\mathcal{R}$ be a ring with identity and $a\in {\mathcal{R}}^D$ with $i(a)=k$. The solution of (14) is called the G-S inverse of a and is denoted by $a^{(s)}$.

Unless stated otherwise, let from now on $\mathcal{R}$ be a *-ring with identity 1. Let us present an auxiliary result which extends [8, Theorem 3.7] form $M_n(\mathbb{C})$ to $\mathcal{R}$. Note (see [9, Lemma 2.1]) that if $a\in {\mathcal{R}}^{⏧d}$ then $a\in {\mathcal{R}}^D$.

Proposition 3.2.

Let $a\in {\mathcal{R}}^{⏧d}$ with I(a) = k, be written as in (13). Then $$a^D={\left[\begin{array}{cc}{t}^{-1}& t^{-(k+1)}\tilde{t}\\ 0& 0\end{array}\right]}_{p\times p},$$where $\tilde{t}={\sum }_{i=0}^{k-1}{t}^{i}s{a}_2^{k-1-i}$.

We may prove Proposition 3.2 in the same way as [8, Theorem 3.7] and thus we omit its proof. We now give a characterization of the G-S inverse (for the matrix case, see [25, p.90]). Note that (15) $$a^{(s)}=a^D+n=a^D+a-a{a}^{D}a.$$(15)

Theorem 3.3.

Let $a\in {\mathcal{R}}^{⏧d}$ with I(a) = k, be written as in (13). Then $$a^{(s)}={\left[\begin{array}{cc}{t}^{-1}& \left(t^{-(k+1)}-t^{-(k-1)}\right)\tilde{t}+s\\ 0& a_2\end{array}\right]}_{p\times p},$$where $\tilde{t}={\sum }_{i=0}^{k-1}{t}^{i}s{a}_2^{k-1-i}$.

Proof.

By Proposition 3.2, we get $$\begin{array}{c}a{a}^{D}a={\left[\begin{array}{cc}t& s\\ 0& a_2\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}{t}^{-1}& t^{-(k+1)}\tilde{t}\\ 0& 0\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}t& s\\ 0& a_2\end{array}\right]}_{p\times p}\\ ={\left[\begin{array}{cc}p& t^{-k}\tilde{t}\\ 0& 0\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}t& s\\ 0& a_2\end{array}\right]}_{p\times p}={\left[\begin{array}{cc}t& s+t^{-k}\tilde{t}{a}_2\\ 0& 0\end{array}\right]}_{p\times p}.\end{array}$$

By (15) we then obtain $$\begin{array}{c}{a}^{(s)}={\left[\begin{array}{cc}{t}^{-1}& t^{-(k+1)}\tilde{t}\\ 0& 0\end{array}\right]}_{p\times p}+{\left[\begin{array}{cc}t& s\\ 0& a_2\end{array}\right]}_{p\times p}-{\left[\begin{array}{cc}t& s+t^{-k}\tilde{t}{a}_2\\ 0& 0\end{array}\right]}_{p\times p}\\ ={\left[\begin{array}{cc}{t}^{-1}& t^{-(k+1)}\tilde{t}-t^{-k}\tilde{t}{a}_2\\ 0& a_2\end{array}\right]}_{p\times p}.\end{array}$$

We have $$t^{-k}\tilde{t}{a}_2=t^{-(k-1)}{t}^{-1}\left(s{a}_2^{k-1}+\mathit{\text{ts}}{a}_2^{k-2}+t^{2}s{a}_2^{k-3}+\dots +t^{k-2}s{a}_2+t^{k-1}s\right)a_2.$$

Since $a_2^k=0$, it follows $$t^{-k}\tilde{t}{a}_2=t^{-(k-1)}\left(s{a}_2^{k-1}+\mathit{\text{ts}}{a}_2^{k-2}+\dots +t^{k-2}s{a}_2\right)$$and thus $$t^{-k}\tilde{t}{a}_2=t^{-(k-1)}\left(\tilde{t}-t^{k-1}s\right)=t^{-(k-1)}\tilde{t}-s.$$

Therefore, $$a^{(s)}={\left[\begin{array}{cc}{t}^{-1}& \left(t^{-(k+1)}-t^{-(k-1)}\right)\tilde{t}+s\\ 0& a_2\end{array}\right]}_{p\times p}.$$

□

Let us now extend the notion of the C-S inverse to *-rings.

Theorem 3.4.

Let $a\in {\mathcal{R}}^{⏧d}$ with I(a) = k. Then the solution x of the system of equations (16) $$x{a}^{k+1}=a^k,\hspace{1em}{\left(a^k{x}^k\right)}^{\ast }=a^k{x}^k,\hspace{1em}\text{and}\hspace{1em}a-x=a^k{x}^k\left(a-x\right)$$(16) is unique. Furthermore, if a is written as in (13), the solution of (16) is $$x={\left[\begin{array}{cc}{t}^{-1}& 0\\ 0& a_2\end{array}\right]}_{p\times p}=a^{⏧d}+a_2.$$

Proof.

Let $$a={\left[\begin{array}{cc}t& s\\ 0& a_2\end{array}\right]}_{p\times p}$$be the matrix form of the core-EP decomposition of a. Since $a_2^k=0$, we obtain $$a^k={\left[\begin{array}{cc}{t}^k& \tilde{t}\\ 0& 0\end{array}\right]}_{p\times p},$$ where $\tilde{t}={\sum }_{i=0}^{k-1}{t}^{i}s{a}_2^{k-1-i}$. Let $x=a^{⏧d}+a_2$. Since $a^{⏧d}=t^{-1}$, $$x={\left[\begin{array}{cc}{t}^{-1}& 0\\ 0& a_2\end{array}\right]}_{p\times p}.$$

Thus, $$\begin{array}{c}x{a}^{k+1}={\left[\begin{array}{cc}{t}^{-1}& 0\\ 0& a_2\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}{t}^k& \tilde{t}\\ 0& 0\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}t& s\\ 0& a_2\end{array}\right]}_{p\times p}\\ ={\left[\begin{array}{cc}{t}^k& t^{k-1}s+t^{-1}\tilde{t}{a}_2\\ 0& 0\end{array}\right]}_{p\times p}.\end{array}$$

As in the proof of Theorem 3.3, we observe $t^{-k}\tilde{t}{a}_2=t^{-(k-1)}\tilde{t}-s$. By multiplying this equation from the left by $t^{k-1}$, we obtain $t^{-1}\tilde{t}{a}_2=\tilde{t}-t^{k-1}s$ and so $\tilde{t}=t^{k-1}s+t^{-1}\tilde{t}{a}_2$. Hence, $$x{a}^{k+1}={\left[\begin{array}{cc}{t}^k& \tilde{t}\\ 0& 0\end{array}\right]}_{p\times p}=a^k.$$

Also, $$a^k{x}^k={\left[\begin{array}{cc}{t}^k& \tilde{t}\\ 0& 0\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}{t}^{-k}& 0\\ 0& 0\end{array}\right]}_{p\times p}={\left[\begin{array}{cc}p& 0\\ 0& 0\end{array}\right]}_{p\times p}.$$

Since $p=a{a}^{⏧d}$ and therefore $p=p^{\ast }$, we get ${(a^k{x}^k)}^{\ast }=a^k{x}^k$. Finally, from $$a^k{x}^k\left(a-x\right)={\left[\begin{array}{cc}p& 0\\ 0& 0\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}t-t^{-1}& s\\ 0& 0\end{array}\right]}_{p\times p}={\left[\begin{array}{cc}t-t^{-1}& s\\ 0& 0\end{array}\right]}_{p\times p}=a-x.$$

For the proof of uniqueness, let $$x={\left[\begin{array}{cc}{x}_1& x_2\\ x_3& x_4\end{array}\right]}_{p\times p}$$be any solution of the system of equations (16)(16) $$x{a}^{k+1}=a^k,\hspace{1em}{\left(a^k{x}^k\right)}^{\ast }=a^k{x}^k,\hspace{1em}\text{and}\hspace{1em}a-x=a^k{x}^k\left(a-x\right)$$(16) . From $x{a}^{k+1}=a^k$, we obtain $x_1{t}^{k+1}=$ t^k and $x_3{t}^{k+1}=0$. By multiplying both equations from the right by $t^{-(k+1)}$, we get $x_1=t^{-1}$ and $x_3=0$. From $$a^k={\left[\begin{array}{cc}{t}^k& \tilde{t}\\ 0& 0\end{array}\right]}_{p\times p},$$ we have $$a^k{x}^k\left(a-x\right)={\left[\begin{array}{cc}\ast & \ast \\ 0& 0\end{array}\right]}_{p\times p}$$ and thus $(1-p)(a-x)(1-p)=(1-p)a^k{x}^k(a-x)(1-p)=0$. It follows that $$x_4=(1-p)x(1-p)=(1-p)a(1-p)=a_2.$$

So, $$x={\left[\begin{array}{cc}{t}^{-1}& x_2\\ 0& a_2\end{array}\right]}_{p\times p}$$and therefore $$x^k={\left[\begin{array}{cc}{t}^{-k}& \tilde{d}\\ 0& 0\end{array}\right]}_{p\times p},$$ where $\tilde{d}={\sum }_{i=0}^{k-1}{t}^{-k+1+i}{x}_2{a}_2^i$. Hence, $$a^k{x}^k={\left[\begin{array}{cc}{t}^k& \tilde{t}\\ 0& 0\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}{t}^{-k}& \tilde{d}\\ 0& 0\end{array}\right]}_{p\times p}={\left[\begin{array}{cc}p& t^k\tilde{d}\\ 0& 0\end{array}\right]}_{p\times p}$$ and since ${(a^k{x}^k)}^{\ast }=a^k{x}^k$ we get $t^k\tilde{d}=0$ and thus $\tilde{d}=0$. So, (17) $$0=t^{-k+1}{x}_2+t^{-k+2}{x}_2{a}_2+t^{-k+3}{x}_2{a}_2^2+\cdots +t^{-1}{x}_2{a}^{k-2}+x_2{a}_2^{k-1}.$$(17)

By multiplying this equation from the right by a₂ and from the left by $t$ and by recalling that $a_2^k=0$, we obtain (18) $$0=t^{-k+2}{x}_2{a}_2+t^{-k+3}{x}_2{a}_2^2+t^{-k+4}{x}_2{a}_2^3+\cdots +x_2{a}^{k-1}.$$(18)

Comparing equations (17)(17) $$0=t^{-k+1}{x}_2+t^{-k+2}{x}_2{a}_2+t^{-k+3}{x}_2{a}_2^2+\cdots +t^{-1}{x}_2{a}^{k-2}+x_2{a}_2^{k-1}.$$(17) and (18)(18) $$0=t^{-k+2}{x}_2{a}_2+t^{-k+3}{x}_2{a}_2^2+t^{-k+4}{x}_2{a}_2^3+\cdots +x_2{a}^{k-1}.$$(18) we get that $t^{-k+1}{x}_2=0$, and thus $x_2=0$, i.e., $$x={\left[\begin{array}{cc}{t}^{-1}& 0\\ 0& a_2\end{array}\right]}_{p\times p}.$$

□

We now generalize Definition 1 to *-rings with identity.

Definition 3.

Let $a\in {\mathcal{R}}^{⏧d}$ with I(a) = k. The solution of (16) is called the C-S inverse of a and is denoted by $a^{⏧s}$.

Remark 3.5.

For $a\in {\mathcal{R}}^{⏧d}$ with I(a) = 1, (9) implies that the nilpotent part a₂ in the core-EP decomposition of a equals zero. By Theorem 3.4 it then follows $a^{⏧s}=a^{⏧d}+a_2=a^{⏧d}$. Comparing (6) and (7) we note that when $I(a)=1,a^{⏧d}=a^{⏧♯}$ and thus $a^{⏧s}=a^{⏧♯}$, i.e., the C-S inverse of a equals the core inverse of a when I(a) = 1.

For a nilpotent element $n\in \mathcal{R}$ we denote its index of nilpotency by N(n).

Remark 3.6.

Observe that for $a\in {\mathcal{R}}^{⏧d}$ with I(a) = k, we have that the index of nilpotency $N(a_2)$ of the nilpotent part a₂ in the core-EP decomposition of a equals k. Indeed, suppose I(a) = k and $a_2^{k-1}=0$. Then, by (13), $$a^{k-1}={\left[\begin{array}{cc}{t}^{k-1}& \tilde{u}\\ 0& 0\end{array}\right]}_{p\times p},\hspace{1em}{a}^k={\left[\begin{array}{cc}{t}^k& \tilde{t}\\ 0& 0\end{array}\right]}_{p\times p},$$where $\tilde{u}={\sum }_{i=0}^{k-2}{t}^{i}s{a}_2^{k-2-i}$ and $\tilde{t}={\sum }_{i=0}^{k-1}{t}^{i}s{a}_2^{k-1-i}$, and since $$a^{⏧d}={\left[\begin{array}{cc}{t}^{-1}& 0\\ 0& 0\end{array}\right]}_{p\times p}$$ we obtain that $a^{⏧d}{a}^k=a^{k-1}$ which is a contradiction since I(a) = k.

With the next result we generalize [25, Theorem 9].

Theorem 3.7.

Let $a\in {\mathcal{R}}^{⏧d}$ with I(a) = k and let a be written as in (13). Then the following statements hold.

• $a^{⏧s}=0$ if and only if a = 0.

• $a^{⏧s}=a$ if and only if $t=t^{-1}$ and s = 0.

• $a^{⏧s}=a{a}^{⏧d}=p$ if and only if t = p and $a_2=0$.

• $a^{⏧s}=a^{\ast }$ if and only if $t^{-1}=t^{\ast }$, s = 0, and $a_2=a_2^{\ast }$.

Proof.

(i) Since $x=a^{⏧s}$ solves the equation $a-x=a^k{x}^k(a-x)$, we have that $a^{⏧s}=0$ if and only if a = 0.

(ii) By (13) and Theorem 3.4, $$a={\left[\begin{array}{cc}t& s\\ 0& a_2\end{array}\right]}_{p\times p}\hspace{1em}\text{and}\hspace{1em}{a}^{⏧s}={\left[\begin{array}{cc}{t}^{-1}& 0\\ 0& a_2\end{array}\right]}_{p\times p},$$and thus $a^{⏧s}=a$ if and only if $t=t^{-1}$ and s = 0.

(iii) Since $$a{a}^{⏧d}={\left[\begin{array}{cc}p& 0\\ 0& 0\end{array}\right]}_{p\times p},$$

$a^{⏧s}=a{a}^{⏧d}$ if and only if $t^{-1}=p$ and $a_2=0$, which is equivalent to p = pt = t and $a_2=0$.

(iv) From $$a^{\ast }={\left[\begin{array}{cc}{t}^{\ast }& 0\\ s^{\ast }& a_2^{\ast }\end{array}\right]}_{p\times p}$$we obtain that $a^{⏧s}=a^{\ast }$ if and only if $t^{-1}=t^{\ast }$, s = 0, and $a_2=a_2^{\ast }$. □

We call an element $a\in \mathcal{R}$ Moore-Penrose invertible or *-regular with respect to * if there exists $x\in \mathcal{R}$ that satisfies the following four equations: $$\mathit{\text{axa}}=a,\hspace{1em}\mathit{\text{xax}}=x,\hspace{1em}{(\mathit{\text{ax}})}^{\ast }=\mathit{\text{ax}},\hspace{1em}{(\mathit{\text{xa}})}^{\ast }=\mathit{\text{xa}}.$$

If such x exists, we write $x=a^{†}$ and call it the Moore-Penrose inverse of a. It is known that $a^{†}$ is unique if it exists. We denote set of all Moore-Penrose invertible elements in $\mathcal{R}$ by ${\mathcal{R}}^{†}$.

Lemma 3.8.

Let $\mathcal{R}$ be a *-ring and $a\in {\mathcal{R}}^{†}$. Then $°a^{\ast }=°a^{†}$.

Proof.

Let $z\in \mathcal{R}$ be such that $z{a}^{†}=0$. Then $0=z{a}^{†}a=z{(a^{†}a)}^{\ast }=z{a}^{\ast }{(a^{†})}^{\ast }$ and hence $0=$ $z{a}^{\ast }{(a^{†})}^{\ast }{a}^{\ast }=z{(a{a}^{†}a)}^{\ast }=z{a}^{\ast }$. Similarly, we prove that $z{a}^{\ast }=0$ implies $z{a}^{†}=0$. □

Remark 3.9.

If $a_2\in {\mathcal{R}}^{†}$, then by Theorem 3.7 and Lemma 3.8, $a^{⏧s}=a^{\ast }$ if and only if $t^{-1}=t^{\ast }$, s = 0, and $a_2=0$. Indeed, if $a_2=0$, then $a_2=a_2^{\ast }$. Conversely, let $a_2=a_2^{\ast }$ and suppose $I(a)=k\ge 2$. Since $a_2^k=0$, we have $a_2^{k-1}{a}_2^{\ast }=0$ and hence by Lemma 3.8, $a_2^{k-1}{a}_2^{†}=0$ which implies $a_2^{k-1}{a}_2^{†}{a}_2=0$. So, $a_2^{k-1}=0$ which is a contradiction by Remark 3.6. It follows that $I(a)=1$ and therefore $a_2=0$.

4 Relations induced by the C-S inverse

An involution $\ast$ on a *-ring $\mathcal{R}$ is called proper if $a{a}^{\ast }=0$ implies a = 0 for every $a\in \mathcal{R}$. If a $\ast$-ring $\mathcal{R}$ is equipped with a proper involution, then it is called a proper *-ring. An example of a proper $\ast$-ring is $M_n(\mathbb{C})$. Let $\mathcal{R}$ be a proper *-ring. For $a,b\in \mathcal{R}$, we write (19) $$a{\le }^{\ast }b\hspace{1em}\text{if}\hspace{1em}{a}^{\ast }a=a^{\ast }b\hspace{1em}\text{and}\hspace{1em}a{a}^{\ast }=b{a}^{\ast }.$$(19)

If $a{\le }^{\ast }b$, we say that a is below b with respect to the star partial order. The relation ${\le }^{\ast }$ is indeed a partial order (see [6]) and on ${\mathcal{R}}^{†}$ $$a{\le }^{\ast }b\hspace{1em}\text{if and only if}\hspace{1em}{a}^{†}a=a^{†}b\hspace{1em}\text{and}\hspace{1em}a{a}^{†}=b{a}^{†}.$$

Many other partial orders are defined with various generalized inverses (see, e.g., [18]). In [25], Wang and Liu introduced the following relation on $M_n(\mathbb{C})$. For $A,B\in M_n(\mathbb{C})$ we write (20) $$A{\le }^{⏧s}B\hspace{1em}\text{if}\hspace{1em}A{\left(A^{⏧s}\right)}^{\ast }=B{\left(A^{⏧s}\right)}^{\ast }\hspace{1em}\text{and}\hspace{1em}{\left(A^{⏧s}\right)}^{\ast }A={\left(A^{⏧s}\right)}^{\ast }B.$$(20)

The relation ${\le }^{⏧s}$ is clearly reflexive. It is also antisymmetric however it is not transitive (see [25, Theorem 12 and Example 2]). Based on (20), another relation was presented in [25]. Recall that $A^{⏧d}$ denotes the core-EP inverse of $A\in M_n(\mathbb{C})$. For $A,B\in M_n(\mathbb{C})$ we write (21) $$A{\le }^{⏧\mathit{\text{cs}}}B\hspace{1em}\text{if}\hspace{1em}A{\le }^{⏧s}B\hspace{1em}\text{and}\hspace{1em}B{A}^{\ast }A{A}^{⏧d}=A{A}^{\ast }A{A}^{⏧d}.$$(21)

With [25, Theorem 14] it was shown that ${\le }^{⏧\mathit{\text{cs}}}$ is also transitive, i.e., it is a partial order on $M_n(\mathbb{C})$. The relation ${\le }^{⏧\mathit{\text{cs}}}$ is named the C-S partial order. Note that the expression $A^{\ast }A{A}^{⏧d}$, which appears in (21), is the star-core-EP matrix proposed in [22]. In what follows, we will generalize the relations ${\le }^{⏧s}$ and ${\le }^{⏧\mathit{\text{cs}}}$ to *-rings.

Definition 4.

Let $a\in {\mathcal{R}}^{⏧d}$ and $b\in \mathcal{R}$. Then we write $$a{\le }^{⏧s}b\hspace{1em}\text{if}\hspace{1em}a{\left(a^{⏧s}\right)}^{\ast }=b{\left(a^{⏧s}\right)}^{\ast }\hspace{1em}\text{and}\hspace{1em}{\left(a^{⏧s}\right)}^{\ast }a={\left(a^{⏧s}\right)}^{\ast }b$$and say that a is below b with respect to the S relation.

With the following result we generalize [25, Theorem 11].

Theorem 4.1.

Let $a\in {\mathcal{R}}^{⏧d}$ and $b\in \mathcal{R}$, and let a be written as in (13). Then $a{\le }^{⏧s}b$ if and only if $$\hspace{1em}b={\left[\begin{array}{cc}t& s\\ 0& b_4\end{array}\right]}_{p\times p},$$where $a_2{\le }^{\ast }{b}_4$.

Proof.

Let $a\in {\mathcal{R}}^{⏧d}$. We may represent a with the matrix form of the core-EP decomposition (13): $$a={\left[\begin{array}{cc}t& s\\ 0& a_2\end{array}\right]}_{p\times p}.$$

For $b\in \mathcal{R}$, let $a{\le }^{⏧s}b$. By Theorem 3.4, $$a^{⏧s}={\left[\begin{array}{cc}{t}^{-1}& 0\\ 0& a_2\end{array}\right]}_{p\times p}.$$

Let $$b={\left[\begin{array}{cc}{b}_1& b_2\\ b_3& b_4\end{array}\right]}_{p\times p}.$$

From $$a{\left(a^{⏧s}\right)}^{\ast }={\left[\begin{array}{cc}t{\left(t^{-1}\right)}^{\ast }& s{a}_2^{\ast }\\ 0& a_2{a}_2^{\ast }\end{array}\right]}_{p\times p}\hspace{1em}\text{and}\hspace{1em}\left\{b{\left(a^{⏧s}\right)}^{\ast }={\left[\begin{array}{cc}{b}_1{\left(t^{-1}\right)}^{\ast }& b_2{a}_2^{\ast }\\ b_3{\left(t^{-1}\right)}^{\ast }& b_4{a}_2^{\ast }\end{array}\right]}_{p\times p}\right\}$$and since $a{(a^{⏧s})}^{\ast }=b{(a^{⏧s})}^{\ast }$ we have $t{(t^{-1})}^{\ast }=b_1{(t^{-1})}^{\ast }$ and $b_3{(t^{-1})}^{\ast }=0$. Multiplying both equations on the right by $t^{\ast }$ we obtain $\mathit{\text{tp}}=b_{1}p$ and $b_{3}p=0$. Since $b_1,t,b_3\in \mathcal{R}p,b_1=t$ and $b_3=0$. Also, (22) $$a_2{a}_2^{\ast }=b_4{a}_2^{\ast }.$$(22)

By ${(a^{⏧s})}^{\ast }a={(a^{⏧s})}^{\ast }b$, we get $${\left[\begin{array}{cc}{\left(t^{-1}\right)}^{\ast }t& {\left(t^{-1}\right)}^{\ast }s\\ 0& a_2^{\ast }{a}_2\end{array}\right]}_{p\times p}={\left[\begin{array}{cc}{\left(t^{-1}\right)}^{\ast }{b}_1& {\left(t^{-1}\right)}^{\ast }{b}_2\\ a_2^{\ast }{b}_3& a_2^{\ast }{b}_4\end{array}\right]}_{p\times p}.$$

Hence, ${(t^{-1})}^{\ast }s={(t^{-1})}^{\ast }{b}_2$. It follows that $\mathit{\text{ps}}=p{b}_2$ and since $b_2,s\in p\mathcal{R}$, we obtain $b_2=s$. So, $$b={\left[\begin{array}{cc}t& s\\ 0& b_4\end{array}\right]}_{p\times p}.$$

Also, $a_2^{\ast }{a}_2=a_2^{\ast }{b}_4$ which together with (22) yields $a_2{\le }^{\ast }{b}_4$.

Conversely, let $$b={\left[\begin{array}{cc}t& s\\ 0& b_4\end{array}\right]}_{p\times p},$$where $a_2{\le }^{\ast }{b}_4$. By (19) we have (23) $$a_2^{\ast }{a}_2=a_2^{\ast }{b}_4\hspace{1em}\text{and}\hspace{1em}{a}_2{a}_2^{\ast }=b_4{a}_2^{\ast }.$$(23)

Since a is represented with (13), it follows by Theorem 3.4 that $$a^{⏧s}={\left[\begin{array}{cc}{t}^{-1}& 0\\ 0& a_2\end{array}\right]}_{p\times p}.$$

Then $$a{\left(a^{⏧s}\right)}^{\ast }={\left[\begin{array}{cc}t{\left(t^{-1}\right)}^{\ast }& s{a}_2^{\ast }\\ 0& a_2{a}_2^{\ast }\end{array}\right]}_{p\times p}$$and $$b{\left(a^{⏧s}\right)}^{\ast }={\left[\begin{array}{cc}t{\left(t^{-1}\right)}^{\ast }& s{a}_2^{\ast }\\ 0& b_4{a}_2^{\ast }\end{array}\right]}_{p\times p}$$

and therefore by (23), $a{(a^{⏧s})}^{\ast }=b{(a^{⏧s})}^{\ast }$. We similarly prove that ${(a^{⏧s})}^{\ast }a={(a^{⏧s})}^{\ast }b$. Thus, $a{\le }^{⏧s}b$. □

Remark 4.2.

As was shown in [25, Examples 3 and 4], the S relation ${\le }^{⏧s}$ does not in general imply the star partial order (i.e., if $a{\le }^{⏧s}b$, then it is not necessary that $a{\le }^{\ast }b$), neither does the star partial order in general imply the relation ${\le }^{⏧s}$.

Corollary 4.3.

Let $a\in {\mathcal{R}}^{⏧d}$ and $b\in \mathcal{R}$. If $a{\le }^{⏧s}b$, then $a{a}^{⏧d}a=a{a}^{⏧d}b$.

Proof.

Let $a\in {\mathcal{R}}^{⏧d}$ be written as in (13). Suppose $a{\le }^{⏧s}b$ for $b\in \mathcal{R}$. Then $$a^{⏧d}={\left[\begin{array}{cc}{t}^{-1}& 0\\ 0& 0\end{array}\right]}_{p\times p}$$and by Theorem 4.1, $$b={\left[\begin{array}{cc}t& s\\ 0& b_4\end{array}\right]}_{p\times p}.$$

Recall that $p=a{a}^{⏧d}$. Thus, $$a{a}^{⏧d}a=\mathit{\text{pa}}={\left[\begin{array}{cc}p& 0\\ 0& 0\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}t& s\\ 0& a_2\end{array}\right]}_{p\times p}={\left[\begin{array}{cc}t& s\\ 0& 0\end{array}\right]}_{p\times p}$$and $$a{a}^{⏧d}b=\mathit{\text{pb}}={\left[\begin{array}{cc}p& 0\\ 0& 0\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}t& s\\ 0& b_4\end{array}\right]}_{p\times p}={\left[\begin{array}{cc}t& s\\ 0& 0\end{array}\right]}_{p\times p}$$ which imply $a{a}^{⏧d}a=a{a}^{⏧d}b$. □

Let us now generalize the relation ${\le }^{⏧\mathit{\text{cs}}}$ form $M_n(\mathbb{C})$ to *-rings.

Definition 5.

Let $a\in {\mathcal{R}}^{⏧d}$ and $b\in \mathcal{R}$. Then we write $$a{\le }^{⏧\mathit{\text{cs}}}b\hspace{1em}\text{if}\hspace{1em}a{\le }^{⏧s}b\hspace{1em}\text{and}\hspace{1em}b{a}^{\ast }a{a}^{⏧d}=a{a}^{\ast }a{a}^{⏧d}$$and say that a is below b with respect to the C-S relation.

We will prove that the C-S relation is a partial order on ${\mathcal{R}}^{⏧d}$ but first let us show that unlike the relation ${\le }^{⏧s}$, the C-S relation implies the star partial order.

Theorem 4.4.

Let $a\in {\mathcal{R}}^{⏧d}$ and $b\in \mathcal{R}$. If $a{\le }^{⏧\mathit{\text{cs}}}b$, then $a{\le }^{\ast }b$.

Proof.

Let $a{\le }^{⏧\mathit{\text{cs}}}b$. Then $a{\le }^{⏧s}b$ and thus by Theorem 4.1, $$a={\left[\begin{array}{cc}t& s\\ 0& a_2\end{array}\right]}_{p\times p}\hspace{1em}\text{and}\hspace{1em}b={\left[\begin{array}{cc}t& s\\ 0& b_4\end{array}\right]}_{p\times p}.$$

Recall that $p=a{a}^{⏧d}$. Hence, $$b{a}^{\ast }a{a}^{⏧d}={\left[\begin{array}{cc}t& s\\ 0& b_4\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}{t}^{\ast }& 0\\ s^{\ast }& a_2^{\ast }\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}p& 0\\ 0& 0\end{array}\right]}_{p\times p}={\left[\begin{array}{cc}t{t}^{\ast }+s{s}^{\ast }& 0\\ b_4{s}^{\ast }& 0\end{array}\right]}_{p\times p}$$and $$a{a}^{\ast }a{a}^{⏧d}={\left[\begin{array}{cc}t& s\\ 0& a_2\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}{t}^{\ast }& 0\\ s^{\ast }& a_2^{\ast }\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}p& 0\\ 0& 0\end{array}\right]}_{p\times p}={\left[\begin{array}{cc}t{t}^{\ast }+s{s}^{\ast }& 0\\ a_2{s}^{\ast }& 0\end{array}\right]}_{p\times p}.$$

So, $a{a}^{\ast }a{a}^{⏧d}=b{a}^{\ast }a{a}^{⏧d}$ gives $$a_2{s}^{\ast }=b_4{s}^{\ast }.$$

Again, by Theorem 4.1, $a_2{\le }^{\ast }{b}_4$ and thus $$a_2^{\ast }{a}_2=a_2^{\ast }{b}_4\hspace{1em}\text{and}\hspace{1em}{a}_2{a}_2^{\ast }=b_4{a}_2^{\ast }.$$

We have $$a{a}^{\ast }={\left[\begin{array}{cc}t& s\\ 0& a_2\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}{t}^{\ast }& 0\\ s^{\ast }& a_2^{\ast }\end{array}\right]}_{p\times p}={\left[\begin{array}{cc}t{t}^{\ast }+s{s}^{\ast }& s{a}_2^{\ast }\\ a_2{s}^{\ast }& a_2{a}_2^{\ast }\end{array}\right]}_{p\times p}$$and $$b{a}^{\ast }={\left[\begin{array}{cc}t& s\\ 0& b_4\end{array}\right]}_{p\times p}{\left[\begin{array}{cc}{t}^{\ast }& 0\\ s^{\ast }& a_2^{\ast }\end{array}\right]}_{p\times p}={\left[\begin{array}{cc}t{t}^{\ast }+s{s}^{\ast }& s{a}_2^{\ast }\\ b_4{s}^{\ast }& b_4{a}_2^{\ast }\end{array}\right]}_{p\times p}$$ and therefore $a{a}^{\ast }=a{b}^{\ast }$. We similarly prove that $a^{\ast }a=a^{\ast }b$. Thus, $a{\le }^{\ast }b$. □

Note that the converse implication does not hold in general, i.e., the star partial order does not (necessarily) imply the C-S relation (see [25, Example 5]).

Corollary 4.5.

The C-S relation ${\le }^{⏧\mathit{\text{cs}}}$ is reflexive and antisymmetric on ${\mathcal{R}}^{⏧d}$.

Proof.

The relation ${\le }^{⏧\mathit{\text{cs}}}$ is clearly reflexive by Definitions 4 and 5. Let $a,b\in {\mathcal{R}}^{⏧d}$ with $a{\le }^{⏧\mathit{\text{cs}}}b$ and $b{\le }^{⏧\mathit{\text{cs}}}a$. By Theorem 4.4, $a{\le }^{\ast }b$ and $b{\le }^{\ast }a$ which yields a = b, i.e., ${\le }^{⏧\mathit{\text{cs}}}$ is antisymmetric. □

To prove that the C-S relation is also transitive, we need the following characterization of the S relation.

Theorem 4.6.

Let $a,b\in {\mathcal{R}}^{⏧d}$ with $I(a)=k$a and $p=a{a}^{⏧d}$. Then $a{\le }^{⏧s}b$ if and only if there exists a decomposition of the identity $1=e_1+e_2+e_3$ of $\mathcal{R}$ with $p=e_1$ and $e_2=e_2^{\ast }$ such that (24) $$a={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& n_1& n_2\\ 0& n_3& n_4\end{array}\right]}_{e\times e}\hspace{1em}\text{and}\hspace{1em}b={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& n_5\end{array}\right]}_{e\times e},$$(24) where $n_1+n_2+n_3+n_4{\le }^{\ast }{t}_1+z_1+n_5$, t is invertible in the ring $e_1\mathcal{R}{e}_1$, t₁ is invertible in the ring $e_2\mathcal{R}{e}_2$, and $n_1+n_2+n_3+n_4$ and n₅ are nilpotent with $N(n_1+n_2+n_3+n_4)=k$.

Proof.

Let first a and b have the form (24). Then $s_1+s_2=e_{1}a{e}_2+e_{1}a(1-e_1-e_2)=e_{1}a(1-e_1)=\mathit{\text{pa}}(1-p)$ and similarly $n_1+n_2+n_3+n_4=(1-e_1)a(1-e_1)=(1-p)a(1-p)$. Let $s=s_1+s_2$ and $a_2=n_1+n_2+n_3+n_4$. Thus, $$a={\left[\begin{array}{cc}t& s\\ 0& a_2\end{array}\right]}_{p\times p}$$and observe that this is the matrix form of the core-EP decomposition of a. Let $b_4=$ $t_1+z_1+n_5$. Note that by (24), $(1-e_1-e_2)b{e}_2=0$. Then $$b_4=t_1+z_1+0+n_5$$ $$=e_{2}b{e}_2+e_{2}b(1-e_1-e_2)+(1-e_1-e_2)b{e}_2+(1-e_1-e_2)b(1-e_1-e_2)$$ $$=e_{2}b(1-e_1)+(1-e_1-e_2)b(1-e_1)=(1-e_1)b(1-e_1)=(1-p)b(1-p)$$ and hence $$b={\left[\begin{array}{cc}t& s\\ 0& b_4\end{array}\right]}_{p\times p}.$$

By assumption, $a_2{\le }^{\ast }{b}_4$ and therefore by Theorem 4.1, $a{\le }^{⏧s}b$.

Conversely, let $a{\le }^{⏧s}b$. By Theorem 4.1 we may write a as in (13), $$b={\left[\begin{array}{cc}t& s\\ 0& b_4\end{array}\right]}_{p\times p},$$and $a_2{\le }^{\ast }{b}_4$. Since $b\in {\mathcal{R}}^{⏧d}$, it follows by [16, Lemma 2.3] that $b_4\in {\mathcal{R}}^{⏧d}$. Let $I(b_4)=l$. We may thus present b₄ in terms of the core-EP decomposition and write $$b_4={\left[\begin{array}{cc}{t}_1& z_1\\ 0& n_5\end{array}\right]}_{q\times q},$$ where $q=b_4{b}_4^{⏧d}$, n₅ is nilpotent with $N(n_5)=l$, and t₁ is invertible in the ring $q\mathcal{R}q$. Note that $q=b_4{b}_4^{⏧d}\in (1-p)\mathcal{R}$ and thus pq = 0, and since $p=p^{\ast }$ and $q=q^{\ast }$, we have $0={(\mathit{\text{pq}})}^{\ast }=q^{\ast }{p}^{\ast }=\mathit{\text{qp}}$. Observe that $$b=t+s+b_4=t+s+t_1+z_1+n_5$$ $$=\mathit{\text{pbp}}+\mathit{\text{pb}}(1-p)+q{b}_{4}q+q{b}_4(1-q)+(1-q)b_4(1-q)=\mathit{\text{pbp}}+\mathit{\text{pb}}(1-p)+$$ $$+q(1-p)b(1-p)q+q(1-p)b(1-p)(1-q)+(1-q)(1-p)b(1-p)(1-q)$$ and since $\mathit{\text{pq}}=\mathit{\text{qp}}=0$, we obtain $$b=\mathit{\text{pbp}}+\mathit{\text{pb}}(1-p)+\mathit{\text{qbq}}+\mathit{\text{qb}}(1-p-q)+(1-p-q)b(1-p-q).$$

Let $e_1=p,e_2=q$, and $e_3=1-p-q$. Note that $e_{1}b{e}_1=\mathit{\text{pbp}}=t,e_{2}b{e}_2=\mathit{\text{qbq}}=q{b}_{4}q=t_1$, and $(1-e_1-e_2)b(1-e_1-e_2)=(1-p-q)b(1-p-q)=n_5$. Also, $$s=\mathit{\text{pb}}(1-p)=\mathit{\text{pbq}}+\mathit{\text{pb}}(1-p-q)=e_{1}b{e}_2+e_{1}b(1-e_1-e_2).$$

Let us denote $s_1=e_{1}b{e}_2$ and $s_2=e_{1}b(1-e_1-e_2)$. Relative to the decomposition of the identity $1=e_1+e_2+e_3$ we may thus write $$b={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& n_5\end{array}\right]}_{e\times e}.$$

Note that $$a_2=(1-p)a(1-p)=\mathit{\text{qaq}}+\mathit{\text{qa}}(1-p-q)+(1-p-q)\mathit{\text{aq}}+(1-p-q)a(1-p-q)$$and let us denote $n_1=\mathit{\text{qaq}}=e_{2}a{e}_2,n_2=\mathit{\text{qa}}(1-p-q)=e_{2}a(1-e_1-e_2),n_3=(1-p-q)\mathit{\text{aq}}=(1-e_1-e_2)a{e}_2$, and $n_4=(1-p-q)a(1-p-q)=(1-e_1-e_2)a(1-e_1-e_2)$ to obtain $$a={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& n_1& n_2\\ 0& n_3& n_4\end{array}\right]}_{e\times e}.$$

Recall that $a_2=n_1+n_2+n_3+n_4$. So, $n_1+n_2+n_3+n_4$ is nilpotent with $N(n_1+n_2+n_3+n_4)=k$ and since $a_2{\le }^{\ast }{b}_4$, we have $n_1+n_2+n_3+n_4{\le }^{\ast }{t}_1+z_1+n_5$. □

Next we present and prove two auxiliary results.

Lemma 4.7.

Let $1=e_1+e_2+e_3$ be a decomposition of the identity of $\mathcal{R}$ for some projections $e_1,e_2,e_3\in \mathcal{R}$, and let for $b\in {\mathcal{R}}^{⏧d}$, $$b={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& n\end{array}\right]}_{e\times e},$$where t is invertible in the ring $e_1\mathcal{R}{e}_1$, t₁ is invertible in the ring $e_2\mathcal{R}{e}_2$, and n is nilpotent. Then $$b^{⏧s}={\left[\begin{array}{ccc}{t}^{-1}& -t^{-1}{s}_1{t}_1^{-1}& 0\\ 0& t_1^{-1}& 0\\ 0& 0& n\end{array}\right]}_{e\times e}.$$

Proof.

By Lemma 2.1, $$b^{⏧d}={\left[\begin{array}{ccc}{t}^{-1}& -t^{-1}{s}_1{t}_1^{-1}& 0\\ 0& t_1^{-1}& 0\\ 0& 0& 0\end{array}\right]}_{e\times e}.$$

Then $$\begin{array}{c}b(b^{⏧d}b)={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& n\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{e}_1& 0& t^{-1}{s}_2-t^{-1}{s}_1{t}_1^{-1}{z}_1\\ 0& e_2& t_1^{-1}{z}_1\\ 0& 0& 0\end{array}\right]}_{e\times e}\\ ={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& 0\end{array}\right]}_{e\times e}\end{array}$$and thus $$b-b{b}^{⏧d}b={\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& n\end{array}\right]}_{e\times e}=n.$$

So, n is the nilpotent part of the core-EP decomposition of b. By Theorem 3.4, $b^{⏧s}=b^{⏧d}+n$ and so $$b^{⏧s}={\left[\begin{array}{ccc}{t}^{-1}& -t^{-1}{s}_1{t}_1^{-1}& 0\\ 0& t_1^{-1}& 0\\ 0& 0& 0\end{array}\right]}_{e\times e}+{\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& n\end{array}\right]}_{e\times e}={\left[\begin{array}{ccc}{t}^{-1}& -t^{-1}{s}_1{t}_1^{-1}& 0\\ 0& t_1^{-1}& 0\\ 0& 0& n\end{array}\right]}_{e\times e}.$$

□

Lemma 4.8.

Let $1=e_1+e_2+e_3$ be a decomposition of the identity of $\mathcal{R}$ for some projections $e_1,e_2,e_3\in \mathcal{R}$, and let for $b\in {\mathcal{R}}^{⏧d}$, $$b={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& n\end{array}\right]}_{e\times e},$$where t is invertible in the ring $e_1\mathcal{R}{e}_1$, t₁ is invertible in the ring $e_2\mathcal{R}{e}_2$, and n is nilpotent. If $b{\le }^{⏧\mathit{\text{cs}}}c$ for some $c\in \mathcal{R}$, then $$c={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& c_9\end{array}\right]}_{e\times e},$$ where $n{\le }^{\ast }{c}_9$.

Proof.

Let $b\in {\mathcal{R}}^{⏧d}$ and $b{\le }^{⏧\mathit{\text{cs}}}c$ for some $c\in \mathcal{R}$. By Theorem 4.4, $b{\le }^{\ast }c$ and thus $b^{\ast }b=b^{\ast }c$ and $b{b}^{\ast }=c{b}^{\ast }$. Let $$c={\left[\begin{array}{ccc}{c}_1& c_2& c_3\\ c_4& c_5& c_6\\ c_7& c_8& c_9\end{array}\right]}_{e\times e}.$$

We have $$b^{\ast }b={\left[\begin{array}{ccc}{t}^{\ast }& 0& 0\\ s_1^{\ast }& t_1^{\ast }& 0\\ s_2^{\ast }& z_1^{\ast }& n^{\ast }\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& n\end{array}\right]}_{e\times e}={\left[\begin{array}{ccc}{t}^{\ast }t& t^{\ast }{s}_1& t^{\ast }{s}_2\\ s_1^{\ast }t& s_1^{\ast }{s}_1+t_1^{\ast }{t}_1& s_1^{\ast }{s}_2+t_1^{\ast }{z}_1\\ \ast & \ast & \ast \end{array}\right]}_{e\times e}$$which equals $$b^{\ast }c={\left[\begin{array}{ccc}{t}^{\ast }& 0& 0\\ s_1^{\ast }& t_1^{\ast }& 0\\ s_2^{\ast }& z_1^{\ast }& n^{\ast }\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{c}_1& c_2& c_3\\ c_4& c_5& c_6\\ c_7& c_8& c_9\end{array}\right]}_{e\times e}={\left[\begin{array}{ccc}{t}^{\ast }{c}_1& t^{\ast }{c}_2& t^{\ast }{c}_3\\ s_1^{\ast }{c}_1+t_1^{\ast }{c}_4& s_1^{\ast }{c}_2+t_1^{\ast }{c}_5& s_1^{\ast }{c}_3+t_1^{\ast }{c}_6\\ \ast & \ast & \ast \end{array}\right]}_{e\times e}.$$

So, $t^{\ast }t=t^{\ast }{c}_1,t^{\ast }{s}_1=t^{\ast }{c}_2,t^{\ast }{s}_2=t^{\ast }{c}_3$ and hence $c_1=t,c_2=s_1$, and $c_3=s_2$. Also, $s_1^{\ast }t=s_1^{\ast }{c}_1+t_1^{\ast }{c}_4=$ $s_1^{\ast }t+t_1^{\ast }{c}_4$ and therefore $t_1^{\ast }{c}_4=0$ which implies $c_4=0$. Similarly, $s_1^{\ast }{s}_1+t_1^{\ast }{t}_1=$ $s_1^{\ast }{c}_2+t_1^{\ast }{c}_5=s_1^{\ast }{s}_1+t_1^{\ast }{c}_5$ implies $c_5=t_1$, and $s_1^{\ast }{s}_2+t_1^{\ast }{z}_1=s_1^{\ast }{c}_3+t_1^{\ast }{c}_6=s_1^{\ast }{s}_2+t_1^{\ast }{c}_6$ yields $c_6=z_1$. It follows that $$c={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ c_7& c_8& c_9\end{array}\right]}_{e\times e}.$$

The assumption $b{\le }^{⏧\mathit{\text{cs}}}c$ implies $b{\le }^{⏧s}s$ and so $b{(b^{⏧s})}^{\ast }=c{(b^{⏧s})}^{\ast }$ and ${(b^{⏧s})}^{\ast }b={(b^{⏧s})}^{\ast }c$. By Lemma 4.7, $$b^{⏧s}={\left[\begin{array}{ccc}{t}^{-1}& -t^{-1}{s}_1{t}_1^{-1}& 0\\ 0& t_1^{-1}& 0\\ 0& 0& n\end{array}\right]}_{e\times e}.$$

From $$\begin{array}{c}b{\left(b^{⏧s}\right)}^{\ast }={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& n\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{\left(t^{-1}\right)}^{\ast }& 0& 0\\ {\left(-t^{-1}{s}_1{t}_1^{-1}\right)}^{\ast }& {\left(t_1^{-1}\right)}^{\ast }& 0\\ 0& 0& n^{\ast }\end{array}\right]}_{e\times e}\\ ={\left[\begin{array}{ccc}t{\left(t^{-1}\right)}^{\ast }+s_1{\left(-t^{-1}{s}_1{t}_1^{-1}\right)}^{\ast }& s_1{\left(t_1^{-1}\right)}^{\ast }& s_2{n}^{\ast }\\ t_1{\left(-t^{-1}{s}_1{t}_1^{-1}\right)}^{\ast }& t_1{\left(t_1^{-1}\right)}^{\ast }& z_1{n}^{\ast }\\ 0& 0& n{n}^{\ast }\end{array}\right]}_{e\times e}\end{array}$$which equals to $$\begin{array}{c}c{\left(b^{⏧s}\right)}^{\ast }={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ c_7& c_8& c_9\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{\left(t^{-1}\right)}^{\ast }& 0& 0\\ {\left(-t^{-1}{s}_1{t}_1^{-1}\right)}^{\ast }& {\left(t_1^{-1}\right)}^{\ast }& 0\\ 0& 0& n^{\ast }\end{array}\right]}_{e\times e}\\ ={\left[\begin{array}{ccc}t{\left(t^{-1}\right)}^{\ast }+s_1{\left(-t^{-1}{s}_1{t}_1^{-1}\right)}^{\ast }& s_1{\left(t_1^{-1}\right)}^{\ast }& s_2{n}^{\ast }\\ t_1{\left(-t^{-1}{s}_1{t}_1^{-1}\right)}^{\ast }& t_1{\left(t_1^{-1}\right)}^{\ast }& z_1{n}^{\ast }\\ c_7{\left(t^{-1}\right)}^{\ast }+c_8{\left(-t^{-1}{s}_1{t}_1^{-1}\right)}^{\ast }& c_8{\left(t_1^{-1}\right)}^{\ast }& c_9{n}^{\ast }\end{array}\right]}_{e\times e}\end{array}$$ we get $c_8{(t_1^{-1})}^{\ast }=0$. Multiplying this equation from the right by $t_1^{\ast }$ we get $c_8{e}_2^{\ast }=0$ and so $c_8=c_8{e}_2=c_8{e}_2^{\ast }=0$. Also, $0=c_7{(t^{-1})}^{\ast }+c_8{(-t^{-1}{s}_1{t}_1^{-1})}^{\ast }=c_7{(t^{-1})}^{\ast }$ implies $c_7=0$. Thus, $$c={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& c_9\end{array}\right]}_{e\times e}.$$

Moreover, $n{n}^{\ast }=c_9{n}^{\ast }$ and the equation ${(b^{⏧s})}^{\ast }b={(b^{⏧s})}^{\ast }c$ similarly yields $n^{\ast }n=n^{\ast }{c}_9$. So, $n{\le }^{\ast }{c}_9$. □

Now, we are in position to prove that the C-S relation is also transitive.

Theorem 4.9.

The C-S relation ${\le }^{⏧\mathit{\text{cs}}}$ is a partial order on ${\mathcal{R}}^{⏧d}$.

Proof.

By Corollary 4.5, the C-S relation is reflexive and antisymmetric on ${\mathcal{R}}^{⏧d}$. Let us show that it is also transitive. Let $a,b\in {\mathcal{R}}^{⏧d}$ with $a{\le }^{⏧\mathit{\text{cs}}}b$ and $b{\le }^{⏧\mathit{\text{cs}}}c$ for some $c\in \mathcal{R}$. Let $p=$ $a{a}^{⏧d}$. By Definition 5, Theorem 4.6, and Lemma 4.8 we may write $$a={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& n_1& n_2\\ 0& n_3& n_4\end{array}\right]}_{e\times e},b={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& n_5\end{array}\right]}_{e\times e},\hspace{1em}\text{and}\hspace{1em}c={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& c_9\end{array}\right]}_{e\times e}$$relative to the decomposition of the identity $1=e_1+e_2+e_3$, where $e_1=p$ and $e_2=e_2^{\ast }$. Here t is invertible in the ring $e_1\mathcal{R}{e}_1$, t₁ is invertible in the ring $e_2\mathcal{R}{e}_2$, $n_1+n_2+n_3+n_4$ and n₅ are nilpotent, $n_5{\le }^{\ast }{c}_9$, and $n_1+n_2+n_3+n_4{\le }^{\ast }{t}_1+z_1+n_5$. Note that a is represented with the matrix form of the core-EP decomposition, where $n_1+n_2+n_3+n_4$ is the nilpotent part. By Theorem 3.4, we obtain $$a^{⏧s}={\left[\begin{array}{ccc}{t}^{-1}& 0& 0\\ 0& n_1& n_2\\ 0& n_3& n_4\end{array}\right]}_{e\times e}.$$

Let $N=n_1+n_2+n_3+n_4$ and $M=t_1+z_1+c_9$. We have $$\begin{array}{c}a{\left(a^{⏧s}\right)}^{\ast }={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& n_1& n_2\\ 0& n_3& n_4\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{(t^{-1})}^{\ast }& 0& 0\\ 0& n_1^{\ast }& n_3^{\ast }\\ 0& n_2^{\ast }& n_4^{\ast }\end{array}\right]}_{e\times e}\\ ={\left[\begin{array}{ccc}t{(t^{-1})}^{\ast }& s_1{n}_1^{\ast }+s_2{n}_2^{\ast }& s_1{n}_3^{\ast }+s_2{n}_4^{\ast }\\ 0& & N{N}^{\ast }\\ 0& & \end{array}\right]}_{e\times e}\end{array}$$and $$\begin{array}{c}c{\left(a^{⏧s}\right)}^{\ast }={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& c_9\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{(t^{-1})}^{\ast }& 0& 0\\ 0& n_1^{\ast }& n_3^{\ast }\\ 0& n_2^{\ast }& n_4^{\ast }\end{array}\right]}_{e\times e}\\ ={\left[\begin{array}{ccc}t{(t^{-1})}^{\ast }& s_1{n}_1^{\ast }+s_2{n}_2^{\ast }& s_1{n}_3^{\ast }+s_2{n}_4^{\ast }\\ 0& & M{N}^{\ast }\\ 0& & \end{array}\right]}_{e\times e}.\end{array}$$

Since $n_1+n_2+n_3+n_4{\le }^{\ast }{t}_1+z_1+n_5$, we have $N{N}^{\ast }=M{N}^{\ast }$ and $N^{\ast }N=N^{\ast }M$. It follows that $a{(a^{⏧s})}^{\ast }=$ $c{(a^{⏧s})}^{\ast }$. Similarly we prove that ${(a^{⏧s})}^{\ast }a={(a^{⏧s})}^{\ast }c$. Thus, $a{\le }^{⏧s}c$.

Let us now prove that $c{a}^{\ast }a{a}^{⏧d}=a{a}^{\ast }a{a}^{⏧d}$. From $a{\le }^{⏧\mathit{\text{cs}}}b$ and $b{\le }^{⏧\mathit{\text{cs}}}c$, we get (25) $$b{a}^{\ast }a{a}^{⏧d}=a{a}^{\ast }a{a}^{⏧d}\hspace{1em}\text{and}\hspace{1em}c{b}^{\ast }b{b}^{⏧d}=b{b}^{\ast }b{b}^{⏧d}.$$(25)

Note that $$a^{⏧d}={\left[\begin{array}{ccc}{t}^{-1}& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]}_{e\times e}$$and by Lemma 2.1, $$b^{⏧d}={\left[\begin{array}{ccc}{t}^{-1}& -t^{-1}{s}_1{t}_1^{-1}& 0\\ 0& t_1^{-1}& 0\\ 0& 0& 0\end{array}\right]}_{e\times e}.$$

We have $$a^{\ast }\left(a{a}^{⏧d}\right)={\left[\begin{array}{ccc}{t}^{\ast }& 0& 0\\ s_1^{\ast }& n_1^{\ast }& n_3^{\ast }\\ s_2^{\ast }& n_2^{\ast }& n_4^{\ast }\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{e}_1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]}_{e\times e}={\left[\begin{array}{ccc}{t}^{\ast }& 0& 0\\ s_1^{\ast }& 0& 0\\ s_2^{\ast }& 0& 0\end{array}\right]}_{e\times e}$$and therefore (26) $$a{a}^{\ast }a{a}^{⏧d}={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& n_1& n_2\\ 0& n_3& n_4\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{t}^{\ast }& 0& 0\\ s_1^{\ast }& 0& 0\\ s_2^{\ast }& 0& 0\end{array}\right]}_{e\times e}={\left[\begin{array}{ccc}t{t}^{\ast }+s_1{s}_1^{\ast }+s_2{s}_2^{\ast }& 0& 0\\ n_1{s}_1^{\ast }+n_2{s}_2^{\ast }& 0& 0\\ n_3{s}_2^{\ast }+n_4{s}_2^{\ast }& 0& 0\end{array}\right]}_{e\times e}$$(26) and $$b{a}^{\ast }a{a}^{⏧d}={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& n_5\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{t}^{\ast }& 0& 0\\ s_1^{\ast }& 0& 0\\ s_2^{\ast }& 0& 0\end{array}\right]}_{e\times e}={\left[\begin{array}{ccc}t{t}^{\ast }+s_1{s}_1^{\ast }+s_2{s}_2^{\ast }& 0& 0\\ t_1{s}_1^{\ast }+z_1{s}_2^{\ast }& 0& 0\\ n_5{s}_2^{\ast }& 0& 0\end{array}\right]}_{e\times e}.$$

From (25) we get $n_1{s}_1^{\ast }+n_2{s}_2^{\ast }=t_1{s}_1^{\ast }+z_1{s}_2^{\ast }$ and $n_3{s}_2^{\ast }+n_4{s}_2^{\ast }=n_5{s}_2^{\ast }$. By $$b^{\ast }\left(b{b}^{⏧d}\right)={\left[\begin{array}{ccc}{t}^{\ast }& 0& 0\\ s_1^{\ast }& t_1^{\ast }& 0\\ s_2^{\ast }& z_1^{\ast }& n_5^{\ast }\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{e}_1& 0& 0\\ 0& e_2& 0\\ 0& 0& 0\end{array}\right]}_{e\times e}={\left[\begin{array}{ccc}{t}^{\ast }& 0& 0\\ s_1^{\ast }& t_1^{\ast }& 0\\ s_2^{\ast }& z_1^{\ast }& 0\end{array}\right]}_{e\times e}$$we have $$\begin{array}{c}b{b}^{\ast }b{b}^{⏧d}={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& n_5\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{t}^{\ast }& 0& 0\\ s_1^{\ast }& t_1^{\ast }& 0\\ s_2^{\ast }& z_1^{\ast }& 0\end{array}\right]}_{e\times e}\\ ={\left[\begin{array}{ccc}t{t}^{\ast }+s_1{s}_1^{\ast }+s_2{s}_2^{\ast }& s_1{t}_1^{\ast }+s_2{z}_1^{\ast }& 0\\ t_1{s}_1^{\ast }+z_1{s}_2^{\ast }& t_1{t}_1^{\ast }+z_1{z}_1^{\ast }& 0\\ n_5{s}_2^{\ast }& n_5{z}_1^{\ast }& 0\end{array}\right]}_{e\times e}\end{array}$$ and $$\begin{array}{c}c{b}^{\ast }b{b}^{⏧d}={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& c_9\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{t}^{\ast }& 0& 0\\ s_1^{\ast }& t_1^{\ast }& 0\\ s_2^{\ast }& z_1^{\ast }& 0\end{array}\right]}_{e\times e}\\ ={\left[\begin{array}{ccc}t{t}^{\ast }+s_1{s}_1^{\ast }+s_2{s}_2^{\ast }& s_1{t}_1^{\ast }+s_2{z}_1^{\ast }& 0\\ t_1{s}_1^{\ast }+z_1{s}_2^{\ast }& t_1{t}_1^{\ast }+z_1{z}_1^{\ast }& 0\\ c_9{s}_2^{\ast }& c_9{z}_1^{\ast }& 0\end{array}\right]}_{e\times e}.\end{array}$$

Again, by (25) it follows that $n_5{s}_2^{\ast }=c_9{s}_2^{\ast }$ and $n_5{z}_1^{\ast }=c_9{z}_1^{\ast }$. Now, (27) $$c{a}^{\ast }a{a}^{⏧d}={\left[\begin{array}{ccc}t& s_1& s_2\\ 0& t_1& z_1\\ 0& 0& c_9\end{array}\right]}_{e\times e}{\left[\begin{array}{ccc}{t}^{\ast }& 0& 0\\ s_1^{\ast }& 0& 0\\ s_2^{\ast }& 0& 0\end{array}\right]}_{e\times e}={\left[\begin{array}{ccc}t{t}^{\ast }+s_1{s}_1^{\ast }+s_2{s}_2^{\ast }& 0& 0\\ t_1{s}_1^{\ast }+z_1{s}_2^{\ast }& 0& 0\\ c_9{s}_2^{\ast }& 0& 0\end{array}\right]}_{e\times e}.$$(27)

Since $n_1{s}_1^{\ast }+n_2{s}_2^{\ast }=t_1{s}_1^{\ast }+z_1{s}_2^{\ast }$ and $n_3{s}_2^{\ast }+n_4{s}_2^{\ast }=n_5{s}_2^{\ast }=c_9{s}_2^{\ast }$, we may conclude that $c{a}^{\ast }a{a}^{⏧d}=a{a}^{\ast }a{a}^{⏧d}$. Therefore, $a{\le }^{⏧\mathit{\text{cs}}}c$. □

Additional information

Funding

The first author acknowledges the financial support from the Slovenian Research and Innovation Agency, ARIS (research program P1-0288). The second author is supported by the Ministry of Education, Science and Technological Development, Republic of Serbia, grant no. 451-03-65/2024-03/200124.