On Regular Elements in an Incline

Abstract

Inclines are additively idempotent semirings in which products are less than (or) equal to either factor. Necessary and sufficient conditions for an element in an incline to be regular are obtained. It is proved that every regular incline is a distributive lattice. The existence of the Moore-Penrose inverse of an element in an incline with involution is discussed. Characterizations of the set of all generalized inverses are presented as a generalization and development of regular elements in a ^∗-regular ring.

1. Introduction

The notion of inclines and their applications are described comprehensively in Cao et al. [1]. Recently, Kim and Roush have surveyed and outlined algebraic properties of inclines and of matrices over inclines [2]. Multiplicative semigroups unlike matrices over a field are not regular; that is, it is not always possible to solve the regularity equation 𝐚𝐱𝐚=𝐚. If there exists 𝑥, 𝑥 is called a 𝑔-inverse of 𝐚 and the element 𝐚 is said to be regular. This concept of regularity of elements in a ring goes back to Neumann [3]. If every element in a ring is regular, then it is called a regular ring. Regular rings are important in many branches of mathematics, especially in matrix theory, since the regularity condition is a linear condition that solves linear equations and takes the place of canonical decomposition.

In [4], Hartwig has studied on existence and construction of various 𝑔-inverses associated with an element in a *-regular ring, that is, regular ring with an anti-automorphism and developed a technique for computing 𝑔-inverses mainly by using star cancellation law. In semirings one of the most important aspects of structure is a collection of equivalence relations called Green’s relations and the corresponding equivalence classes. In [2], it is stated that an element is regular if and only if the equivalence 𝔇 class contains an idempotent.

In this paper, we exhibit that Green’s equivalence relations on a pair of elements in an incline reduce to the equality of elements. This leads to the characterization of regular element in an incline that is, an element in an incline is regular if and only if it is idempotent and structure of set of all 𝑔-inverses of an element in an incline with involution. In Section 2, we present the basic definitions, notations, and required results on inclines. In Section 3, some characterization of regular elements in an incline are obtained as a generalization of regular elements in a *-regular ring studied by Hartwig and as a development of results available in a Fuzzy algebra. The invariance of the product 𝐛𝐚⁻𝐜for elements 𝐚,𝐛,𝐜 in a regular incline and a 𝑔-inverse 𝐚⁻ of 𝐚 is discussed. For elements in a regular incline it is proved that equality of right ideals coincides with equality of left ideals. In Section 4, equivalent conditions for the existence of the Moore-Penrose inverse of an element in an incline with involution-𝑇 are determined.

Green’s equivalence relation reduces to equality of elements. We conclude that the proofs are purely based on incline property without using star cancellation law as in the work of Hartwig [4].

2. Preliminaries

In this section, we give some definitions and notations.

Definition 2.1. An incline is a nonempty set 𝑅 with binary operations addition and multiplication denoted as +, · defined on 𝑅⋅𝑅→𝑅such that for all 𝑥, 𝑦, 𝑧∈𝑅𝑥+𝑦=𝑦+𝑥,𝑥+(𝑦+𝑧)=(𝑥+𝑦)+𝑧,𝑥(𝑦+𝑧)=𝑥𝑦+𝑥𝑧,(𝑦+𝑧)𝑥=𝑦𝑥+𝑧𝑥,𝑥(𝑦𝑧)=(𝑥𝑦)𝑧,𝑥+𝑥=𝑥,𝑥+𝑥𝑦=𝑥,𝑦+𝑥𝑦=𝑦. (2.1)

Definition 2.2. An incline 𝑅 is said to be commutative if 𝑥𝑦=𝑦𝑥 for all 𝑥,𝑦∈𝑅.

Definition 2.3. (𝑅,≤) is an incline with order relation “≤” defined on 𝑅 such that for 𝑥,𝑦∈𝑅, 𝑥≤𝑦 if and only if 𝑥+𝑦=𝑦. If 𝑥≤𝑦, then 𝑦 is said to dominate 𝑥.

Property 2.4. For 𝑥,𝑦 in an incline 𝑅, 𝑥+𝑦≥𝑥and 𝑥+𝑦≥𝑦.
For 𝑥+𝑦=(𝑥+𝑥)+𝑦=𝑥+(𝑥+𝑦), and 𝑥+𝑦=𝑥+(𝑦+𝑦)=(𝑥+𝑦)+𝑦
Thus 𝑥+𝑦≥𝑥 and 𝑥+𝑦≥𝑦.

Property 2.5. For 𝑥, 𝑦 in an incline 𝑅, 𝑥𝑦≤𝑥 and 𝑥𝑦≤𝑦.
Throughout let 𝑅 denote an incline with order relation ≤. For an element 𝑎∈𝑅, 𝑎𝑅={𝑎𝑥/𝑥∈𝑅} is the right ideal of 𝐚 and 𝑅𝑎={𝑥𝑎/𝑥∈𝑅} is the left ideal of 𝐚.

Definition 2.6 (Green’s relation [5]). For any two elements 𝑎, 𝑏 in a semigroup 𝑆.(i)𝑎ℒ𝑏 if there exist 𝑥,𝑦∈𝑆 such that 𝑥𝑎=𝑏 and 𝑦𝑏=𝑎.(ii)𝑎ℛ𝑏 if there exist 𝑥,𝑦∈𝑆 such that 𝑎𝑥=𝑏 and 𝑏𝑦=𝑎.(iii)𝑎𝒥𝑏 if there exist 𝑤,𝑥,𝑦,𝑧 such that 𝑤𝑎𝑥=𝑏, 𝑦𝑏𝑧=𝑎.(iv)𝑎ℋ𝑏 if 𝑎ℛ𝑏 and 𝑎ℒ𝑏.(v)𝑎𝔇𝑏 if there exists 𝑐∈𝑆 such that 𝑎ℛ𝑐 and 𝑐ℒ𝑏.

3. Regular Elements in an Incline

In this section, equivalent conditions for regularity of an element in an incline are obtained and it is proved that a regular commutative incline is a distributive lattice. The equality of right (left) ideals of a pair of elements in a regular incline reduces to the equality of elements. This leads to the invariance of the product 𝐛𝐚⁻𝐜 for all choice 𝐚⁻ of 𝐚 and 𝐚,𝐛,𝐜 in a regular incline. Characterization of the set of all 𝑔-inverses of an element in terms of a particular 𝑔-inverse is determined.

Just for sake of completeness we will introduce 𝑔-inverses of an element in an incline.

Definition 3.1. 𝐚∈𝑅 is said to be regular if there exists an element 𝑥∈𝑅 such that 𝐚𝐱𝐚=𝐚. Then 𝑥 is called a generalized inverse, in short 𝑔-inverse or 1-inverse of 𝐚 and is denoted as 𝐚⁻. Let 𝑎{1} denotes the set of all 1-inverses of 𝐚.

Definition 3.2. An element 𝐚∈𝑅 is called antiregular, if there exists an element 𝑥∈𝑅 such that 𝐱𝐚𝐱=𝐱. Then 𝑥 is called the 2-inverse of 𝐚. 𝐚{2} denotes the set of all 2-inverses of 𝐚.

Definition 3.3. For 𝑎∈𝑅 if there exists 𝑥∈𝑅 such that 𝐚𝐱𝐚=𝐚, 𝐱𝐚𝐱=𝐱, and 𝐚𝐱=𝐱𝐚, then 𝑥 is called the Group inverse of 𝐚. The Group inverse of 𝐚 is a commuting 1-2 inverse of 𝐚.
An incline R is said to be regular if every element of R is regular.

Example 3.4. The Fuzzy algebra ℱ with support [0,1] under the max. min. operation is an incline [2]. Each element in ℱ is regular as well as idempotent [6, page 212]. Thus ℱ is a regular incline.

Example 3.5. Let 𝐷={𝑎,𝑏,𝑐} and 𝑅=(𝒫(𝐷),∪,∩), where (𝒫(𝐷)) the power set of 𝐷 is an incline. Here for each element 𝑥∈𝒫(𝐷), 𝑥²=𝑥∩𝑥=𝑥. Hence 𝑥 is idempotent and 𝑥 is regular (refer Proposition 3.7). Thus 𝑅 is a regular incline.

Lemma 3.6. Let 𝐚∈𝑅 be regular. Then 𝑎=𝑎𝑥=𝑥a for all 𝑥∈𝑎{1}.

Proof. If 𝐚 is regular, then by Property 2.5𝑎=𝑎𝑥𝑎≤𝑎𝑥≤𝑎. (3.1)
Therefore 𝑎𝑥=𝑎.
Similarly, from 𝑎≤𝑥𝑎≤𝑎, it follows that 𝑎=𝑥𝑎. Thus, 𝑎=𝑥𝑎=𝑎𝑥 for all 𝑥∈𝑎{1}.

Proposition 3.7. For 𝑎∈𝑅, 𝐚is regular if and only if 𝐚 is idempotent.

Proof. Let 𝑎∈𝑅 be regular. Then by Lemma 3.6,  𝑎=𝑎𝑥=𝑥𝑎 for all 𝑥∈𝑎{1}. 𝑎=𝑎𝑥𝑎=(𝑎𝑥)𝑎=𝑎⋅𝑎=𝑎². Thus 𝐚 is idempotent.
Converse is trivial.

Example 3.8. Let us consider the example 7₂=([0,1],sup(𝑥,𝑦),𝑥𝑦) of an incline given in [2]. Here 𝑥𝑦 is usual multiplication of real numbers. Hence for each 𝑥∈7₂, 𝑥²≤𝑥 and 𝑥, is not idempotent. Therefore by Proposition 3.7, 7₂ is not a regular incline.

Proposition 3.9. If 𝐚 is regular, then 𝐚 is the smallest 𝑔-inverse of 𝐚, that is, 𝑎≤𝑥 for all 𝑥∈𝑎{1}.

Proof. Let 𝐚 be regular, then by Proposition 3.7, 𝑎∈𝑎{1}. By Lemma 3.6  𝑎=𝑎𝑥 for all 𝑥∈𝑎{1}. Hence by Property 2.5  𝑎≤𝑥. Thus 𝐚 is the smallest 𝑔-inverse of 𝐚.

It is well known that [7] every distributive lattice is an incline, but an incline need not be a distributive lattice. Now we shall show that regular commutative incline is a distributive lattice in the following.

Proposition 3.10 (see [1]). A commutative incline is a distributive lattice as (semiring) if and only if 𝑥²=𝑥 for all 𝑥∈𝑋.

Lemma 3.11 (see [7]). DL is a distributive lattice. (DL is the set of all idempotent elements in an incline L.)

Proposition 3.12. Let  𝑅 be a commutative incline, 𝑅 is regular ⇔𝑅 is a distributive lattice.

Proof. Let 𝑅 is commutative incline.
𝑅 is regular: ⇔ every element in 𝑅 is idempotent (by Proposition 3.7),
⇔𝐷𝑅=𝑅, where 𝐷𝑅 is the set of all idempotent elements of 𝑅,
⇒𝑅=𝐷𝑅 is distributive lattice (by [7, Lemma  2.1]).
Conversely, if 𝑅 is a distributive lattice then by Proposition 111 in [1] every element of 𝑅 is idempotent, again by Proposition 3.7𝑅 is a regular incline.

Next we shall see some characterization of regular elements in an incline.

Theorem 3.13. For 𝐚∈𝑅, the following are equivalent: (i)𝐚 is regular,(ii)𝐚 is idempotent,(iii)𝑎{1,2}={𝑎},(iv)group inverse of 𝐚 exists and coincides with 𝐚,(v)𝑎=𝑣𝑎² for some 𝑣∈𝑎{1},(vi)𝑎=𝑎²𝑢 for some 𝑢∈𝑎{1}.In either case 𝑣,𝑢,𝑣𝑎𝑢 are all 𝑔-inverses of 𝐚 and 𝑣𝑎𝑢 is invariant for all choice of 𝑢,𝑣∈𝑎{1}. 𝑣𝑎𝑢 is the smallest 𝑔-inverse of 𝐚.

Proof. (i)⇔(ii) This is precisely Proposition 3.7.
To prove the theorem it is enough to prove the following implications:
(ii)⇒(iii)⇒(iv)⇒(i); (i)⇒(v)⇒(ii) and (i)⇒(vi)⇒(ii).
(ii)⇒(iii) If 𝐚 is idempotent, then 𝑎∈𝑎{1}. For any 𝑥∈𝑎{1,2} we have 𝑥=𝑥𝑎𝑥 and by
Lemma 3.6 we get 𝑥=(𝑥𝑎)𝑥=𝑎𝑥=𝑎. Therefore 𝑎{1,2}={𝑎}.
Thus (iii) holds.
(iii)⇒(iv) If 𝑎{1,2}={𝑎} then 𝐚 is the only commuting 1-2 inverse of 𝐚.
Therefore by Definition 3.3 the Group inverse of 𝐚 exists and coincides with 𝐚.
(iv)⇒(i) This is trivial.
(i)⇒(v) Let 𝐚 be regular, then by Lemma 3.6, for some 𝑣∈𝑎{1},𝑎=(𝑎𝑣)𝑎=(𝑣𝑎)𝑎=𝑣𝑎². (3.2)
Thus (v) holds.
(v)⇒(ii) Let 𝑎=𝑣𝑎² for some 𝑣∈𝑎{1}. By Property 2.5,𝑎=𝑣𝑎²≤𝑣𝑎≤𝑎⇒𝑎=𝑣𝑎=𝑣𝑎²𝑎=𝑣𝑎²=(𝑣𝑎)𝑎=𝑎². (3.3)
Therefore 𝐚 is idempotent
Thus (ii) holds.
(i)⇒(vi)⇒(ii) can be proved along the same lines and hence omitted.
Now if 𝑎=𝑣𝑎² holds then we can show that 𝑣∈𝑎{1}𝑎=𝑣𝑎²≤𝑣𝑎≤𝑎. (3.4) Therefore 𝑎=𝑣𝑎²=𝑣𝑎𝑎²=𝑎⋅𝑎=𝑣𝑎²=𝑎, (3.5) and 𝑎𝑣𝑎=𝑎²=𝑎
Thus 𝑣∈𝑎{1}.
In a similar manner we can show 𝑢∈𝑎{1}.
Now consider, 𝑥=𝑣𝑎𝑢, where 𝑢,𝑣∈𝑎{1}. It can be verified that𝑥=𝑣𝑎𝑢∈𝑎{1,2}={𝑎}. (3.6)
Hence, 𝑎=𝑣𝑎𝑢 for all 𝑣,𝑢∈𝑎{1}.
Thus 𝑣𝑎𝑢 is invariant for all choice of 𝑔-inverse of 𝐚. By Proposition 3.9, 𝑎=𝑣𝑎𝑢 is the smallest 𝑔-inverse of 𝐚.

Remark 3.14. If 𝐚 is regular, then (i) 𝑎𝑅=𝑎²R and (ii) 𝑅𝑎=𝑅𝑎² automatically holds. The converse holds for an incline with unit.

Remark 3.15. Let us illustrate the relation between various inverses associated with an element in an incline in the following.
Let 𝑅={0,𝑎,𝑏,𝑐,𝑑,1} be a lattice ordered by the following Hasse graph. Define ·: 𝑅×𝑅→𝑅 by 𝑥⋅𝑦=𝑑 for all 𝑥,𝑦∈{1,𝑏,𝑐,𝑑} and 0 otherwise. Then (𝑅,∨, ·) is an incline which is not a distributive lattice.

In this incline R, the only two elements 0, 𝑑 are regular which satisfies the Theorem 3.13.(1)𝑑⋅𝑥⋅𝑑=𝑑 for each 𝑥∈{𝑏,𝑐,𝑑,1}.

Hence 𝑑{1}={𝑏,𝑐,𝑑,1} and 0{1}=𝑅.(2)Since 𝑑∈𝑅, 𝑥⋅𝑑⋅𝑥=𝑥 for 𝑥=0, and 𝑥=𝑑.

Hence 𝑑 is antiregular 𝑑{2}={0,𝑑}. (3.7)(3)𝑑{1}≠𝑑{2} and 𝑑{1,2}=𝑑{1}∩𝑑{2}=𝑑.

Theorem 3.16. Let 𝑅 be a regular incline. For 𝑎,𝑏,𝑐∈𝑅 the following hold:(i)𝑏=𝑦𝑎⇔𝑏=𝑏𝑎⇔𝑅𝑏⊆𝑅𝑎,𝑐=𝑎𝑥⇔𝑐=𝑎𝑐⇔𝑐𝑅⊆𝑎𝑅. (3.8)(ii)𝑥 is a 1-inverse of 𝐚⇔(𝑎𝑥)²=𝑎𝑥,𝑎𝑥𝑅=𝑎𝑅,⇔(𝑥𝑎)²=𝑥𝑎,𝑅𝑥𝑎=𝑅𝑎. (3.9)(iii)𝑥 is a 2-inverse of 𝐚⇔(𝑥𝑎)²=𝑥𝑎,𝑥𝑎𝑅=𝑥𝑅,⇔(𝑎𝑥)²=𝑎𝑥,𝑅𝑎𝑥=𝑅𝑥. (3.10)(iv)𝑞𝑎𝑅⊆𝑞𝑎𝑝𝑅 and 𝑅𝑎⊆𝑅𝑞𝑎 implies 𝑝𝑞=𝑎⁻.(v)If 𝑐=𝑎𝑥 and 𝑏=𝑦𝑎 then 𝑏𝑎⁻𝑐 is invariant under all choice of 1-inverse of 𝐚.(vi)𝑝𝑤𝑞=𝑎−⇔qapw𝑅=qa𝑅andw=(qap)⁻,𝑅qa=𝑅a. (3.11)

Proof. Whenever two symmetric results are involved we shall prove the first leaving the second.(i)Let 𝑏=𝑦𝑎, since a is regular. 𝑏𝑎=𝑏𝑎⁻𝑎=𝑦𝑎𝑎⁻𝑎=𝑦𝑎=𝑏. (3.12) Thus 𝑏=𝑦𝑎⇒𝑏𝑎=𝑏.
Let 𝑏𝑎=𝑏 then for 𝑧∈𝑅𝑏𝑧=𝑥𝑏forsome𝑥∈𝑅=(𝑥𝑏)𝑎∈𝑅𝑎. (3.13) Thus 𝑏=𝑏𝑎⇒𝑅𝑏⊆𝑅𝑎.
Since 𝑏 is regular, by Lemma 3.6,  𝑏=𝑥𝑏∈𝑅𝑏, since 𝑅𝑏⊆𝑅𝑎, 𝑏=𝑥𝑏=𝑦𝑎. Thus 𝑅𝑏⊆𝑅𝑎⇒𝑏=𝑦𝑎.
Hence (i) holds.
(ii)Let 𝑥∈𝑎{1} then by Lemma 3.6 and Proposition 3.7 we have (𝑎𝑥)²=𝑎𝑥=(𝑥𝑎)=𝑎. (3.14) Hence, 𝑎𝑥𝑅=𝑎𝑅 and 𝑅𝑥𝑎=𝑅𝑎.
Conversely, let 𝑎𝑥𝑅=𝑎𝑅 and (𝑎𝑥)²=𝑎𝑥.
Then, 𝑎𝑅⊆𝑎𝑥𝑅⇒𝑎𝑥𝑎=𝑎 (by (i))⇒𝑥∈𝑎{1}. (3.15)(iii)Interchange 𝑥 and a in (ii) then (iii) holds.(iv)Let 𝑞𝑎𝑅⊆𝑞𝑎𝑝𝑅 and 𝑅𝑎⊆𝑅𝑞𝑎𝑞𝑎𝑅⊆𝑞𝑎𝑝𝑅⇒𝑞𝑎𝑝𝑞𝑎=𝑞𝑎(by(i)) (3.16)
𝑅𝑎⊆𝑅𝑞𝑎⇒𝑎𝑞𝑎=𝑎, that is, a is regular with 𝑞∈𝑎{1}.
Now,𝑞𝑎𝑝𝑞𝑎=𝑞𝑎,(𝑎𝑞𝑎)𝑝𝑞𝑎=𝑎𝑞𝑎,𝑎𝑝𝑞𝑎=𝑎. (3.17) Therefore 𝑝𝑞∈𝑎{1}.
Thus (iv) holds.
(v)Let 𝑐=𝑎𝑥 and 𝑏=𝑦𝑎 for some 𝑥,𝑦∈𝑅𝑏𝑎⁻𝑐=𝑦(𝑎𝑎⁻𝑎)𝑥=𝑦𝑎𝑥. (3.18) Which is independent of 𝑎⁻ and 𝑏𝑎⁻𝑐 is invariant for all choice of 𝑎⁻ of a.(vi)Let 𝑝𝑤𝑞=𝑎⁻⇒𝑎=𝑎(𝑝𝑤𝑞)𝑎 (By Definition 3.1).
From the statement (ii), we have
𝑅𝑎=𝑅𝑝𝑤𝑞𝑎⊆𝑅𝑞𝑎⊆𝑅𝑎⇒𝑅𝑎=𝑅𝑞𝑎. (3.19)
Therefore 𝑞∈𝑎{1} (by (ii)).
Now,𝑎𝑝𝑤𝑞𝑎=𝑎⇒𝑞𝑎𝑝𝑤𝑎𝑞𝑎=𝑞𝑎,𝑞𝑎𝑅⊆𝑞𝑎𝑝𝑤𝑅⊆𝑞𝑎𝑅,𝑞𝑎𝑅=𝑞𝑎𝑝𝑤𝑅,𝑞𝑎𝑝𝑤𝑞𝑎=𝑞𝑎,(𝑎𝑞𝑎)𝑝𝑤𝑞𝑎=𝑎𝑞𝑎,𝑎(𝑝𝑤𝑞)𝑎=𝑎,⇒𝑝𝑤𝑞∈𝑎{1}. (3.20)
Thus (vi) holds.

Corollary 3.17. For a, b in a regular incline one has the following: 𝑅𝑎=𝑅𝑏⇔𝑎𝑅=𝑏𝑅⇔𝑎=𝑏. (3.21)

Proof. Since 𝑅𝑎=𝑅𝑏,𝑅𝑎⊆𝑅𝑏,𝑎𝑛𝑑𝑅𝑏⊆𝑅𝑎.
By Theorem 3.16(i) we have 𝑅𝑎⊆𝑅𝑏⇒𝑎=𝑎𝑏⇒𝑎≤𝑏(byProperty2.5)𝑅𝑏⊆𝑅𝑎⇒𝑏=𝑏𝑎⇒𝑏≤𝑎(byProperty2.5). (3.22)
Therefore 𝑎=𝑏. In a similar manner we can show 𝑎𝑅=𝑏𝑅⇒𝑎=𝑏.
On the other hand 𝑎=𝑏 automatically implies 𝑅𝑎=𝑅𝑏 and 𝑎𝑅=𝑏𝑅.

Remark 3.18. We note that Corollary 3.17 fails for regular matrices over an incline.
Let us consider 𝐵=(¹¹₁₀) and 𝐴=(¹⁰₁₁)=𝑃𝐵, where 𝑃=(⁰¹₁₀).
Since 𝑃²=𝐼₂,𝑃𝐴=𝐵. Here 𝐴 and 𝐵 are regular.
By Proposition 2.4 in [8, page 297], 𝑅(𝐴)⊆𝑅(𝐵) and 𝑅(𝐵)⊆𝑅(𝐴).
Hence 𝑅(𝐴)=𝑅(𝐵) but 𝐴≠𝐵.

It is well known that [9, page 26] if 𝑎⁻ is a particular 𝑔-inverse of a in a ring with unit, then the general solution of the equation 𝑎𝑥𝑎=𝑎 is given by 𝑎⁻+ℎ−𝑎⁻𝑎ℎ𝑎𝑎⁻, where ℎ is arbitrary. Here we shall generalize this for incline.

Theorem 3.19. Let 𝐚∈𝑅 and 𝑎⁻ be any particular 1-inverse of a then 𝑎_𝑔⁻{1}={𝑎⁻+ℎ/ℎisarbitraryelementin𝑅} is the set of all 𝑔-inverses of a dominating 𝑎⁻. Furthermore, 𝑎{1}=∪𝑎_𝑔⁻{1}, union over all 𝑔-inverses of a.

Proof. Let 𝒜 denote the set {𝑎⁻+ℎ/ℎisarbitraryelementin𝑅}. Suppose that 𝑥 is arbitrary element of 𝑎_𝑔⁻{1} then 𝑥≥𝑎⁻ which implies 𝑥+𝑘≥𝑎⁻+𝑘 for 𝑘∈𝑅 and by Property 2.4 we have 𝑎⁻+𝑘≥𝑎⁻.
Therefore 𝑥+𝑘≥𝑎⁻+𝑘≥𝑎⁻
Pre- and postmultiplication by a we get𝑎𝑥𝑎+𝑎𝑘𝑎≥𝑎(𝑎⁻+𝑘)𝑎≥𝑎𝑎⁻𝑎,𝑎+𝑎𝑘𝑎≥𝑎(𝑎⁻+𝑘)𝑎≥𝑎 (3.23) (By Definition 3.1).
By Property 2.5  𝑎𝑘𝑎≤𝑎, hence 𝑎+𝑎𝑘𝑎=𝑎.𝑎≥𝑎(𝑎⁻+𝑘)𝑎≥𝑎,𝑎(𝑎⁻+𝑘)𝑎=𝑎. (3.24) Therefore (𝑎⁻+𝑘)∈𝑎{1}.
Thus for each 𝑥∈𝑎_𝑔⁻{1} there exists an element in 𝒜. Hence 𝑎_𝑔⁻{1}⊆𝒜.
On the other hand for any 𝑦∈𝒜,𝑦=𝑎⁻+ℎ≥𝑎⁻ by Property 2.4.
From Definition 3.1 and Property 2.5, we get𝑎𝑦𝑎=𝑎(𝑎⁻+ℎ)𝑎=𝑎+𝑎ℎ𝑎=𝑎. (3.25) Hence 𝑦∈𝑎_𝑔⁻{1}, which implies 𝒜⊆𝑎_𝑔⁻{1}. Therefore 𝒜=𝑎_𝑔⁻{1}=𝑎𝑎{1}=setofallg-inversesof𝐚_𝑔−{1},unionoverallg-inversesof𝐚. (3.26)

4. Projection on an Incline with Involution-𝑇

In this section, the existence of the Moore-Penrose inverse of an element in an incline with involution-T is discussed as a generalization of that for elements is a *-regular ring and for elements in a Fuzzy algebra studied by Hartwig [4], Kim and Roush [8] and Meenakshi [6], respectively. Characterization of the set of all {1,3}, {1,4} inverses and a formula for Moore-Penrose are obtained analogous to those of the result established for fuzzy matrices in [6, 8].

An involution-T of an incline R is an involutary anti-automorphism, that is, (𝑎^𝑇)^𝑇=𝑎,(𝑎+𝑏)^𝑇=𝑎^𝑇+𝑏^𝑇,(𝑎𝑏)^𝑇=𝑏^𝑇𝑎^𝑇,𝑎^𝑇=0 if and only if 𝑎=0 for all 𝑎,𝑏∈𝑅.

Definition 4.1. An element 𝑎∈𝑅 is said to be a projection if 𝑎^𝑇=𝑎=𝑎², that is a is symmetric and idempotent.

Definition 4.2. For 𝑎 in an incline R with involution-T, we say that 𝑥∈𝑅 is a 3-inverse of a if (𝑎𝑥)^𝑇=𝑎𝑥, and we say that 𝑦∈𝑅 is a 4-inverse of a if (𝑦𝑎)^𝑇=𝑦𝑎.

Definition 4.3. An element 𝑥∈𝑅 is said to be Moore-Penrose inverse of 𝑎, if 𝑥 satisfies the following: (i) 𝑎𝑥𝑎=𝑎, (ii) 𝑥𝑎𝑥=𝑥, (iii) (𝑎𝑥)^𝑇=𝑎𝑥, and (iv) (𝑥𝑎)^𝑇=𝑥𝑎, denoted as 𝑎^†.
In [2] it is stated that for an element a in an incline with involution-𝑇, 𝑎^† exists if and only if aℋaa^Ta. Here we derive equivalent condition for the existence of 𝑎^† in terms of the weaker relation aℒaa^Ta.
First we shall show that Green’s equivalence relation on an incline R reduces to equality of elements in R.

Lemma 4.4. For 𝑎,𝑏∈𝑅 the following hold:
(i)𝑎ℒ𝑏⇒𝑎=𝑏, (ii)𝑎ℛ𝑏⇒𝑎=𝑏.
Converse holds for elements in a regular incline or incline with unit.

Proof. (i) If 𝑎ℒ𝑏 then by Definition 2.6 there exist 𝑥,𝑦∈𝑅 such that 𝑥𝑎=𝑏 and 𝑦𝑏=𝑎. By Property 2.5 we have 𝑥𝑎=𝑏⇒𝑏≤𝑎 and 𝑦𝑏=𝑎⇒𝑎≤𝑏.
Therefore, 𝑎ℒ𝑏⇒𝑎=𝑏.
(ii) This can be proved in a similar manner and hence omitted.
The converse holds for 𝑎 regular incline. For, if 𝑎,𝑏 are regular, then by Lemma 3.6  𝑎=𝑏=𝑦𝑏=𝑏𝑦 and 𝑏=𝑎=𝑥𝑎=𝑎𝑥 for some 𝑥,𝑦∈𝑎{1}. Hence 𝑎=𝑏⇒𝑎ℒ𝑏 and 𝑎ℛ𝑏. 𝑎ℒ𝑏 and 𝑎ℛ𝑏 trivially hold for incline with unit.

Theorem 4.5. Let R be an incline with involution-T. For 𝐚∈𝑅 the following are equivalent: (i)a is a projection,(ii)a has 1-3 inverse,(iii)a has 1-4 inverse,(iv) a^† exists and equals a,(v)a^Ta𝑥 = a^T has a solution in R,(vi)𝑥aa^T = a^T has a solution in R,(vii)a is regular and a^T∈𝑎{1},(viii)aℒaa^Ta,(ix)𝑎ℛ𝑎𝑎^𝑇𝑎,

Proof. (i)⇔(ii) Let a be a projection, by Definition 4.1  a is symmetric idempotent. a is regular follows from Proposition 3.7. Thus a has 1-inverse 𝑥 (say) and by Lemma 3.6  𝑎=𝑎𝑥=𝑥𝑎. Since a is symmetric, 𝑎=𝑎^𝑇. Therefore 𝑥 is a 1-3 inverse of a. Thus a has 1–3 inverses. Coverersly if a has 1–3 inverses, then again by Lemma 3.6 there exists 𝑥∈𝑅, such that 𝑎=𝑎𝑥=𝑥𝑎 and 𝑎𝑥=(𝑎𝑥)^𝑇. Hence a is symmetric idempotent. Thus (i) holds.(i)⇔(iii) This can be proved along the same lines as that of (i)⇔(iii), hence omitted.(i)⇔(iv) This equivalence can be proved directly by verifying that a satisfy the four equations in Definition 4.3.(ii)⇔(v) Let a has 1–3 inverses, 𝑥 (say) then 𝑎^𝑇𝑎𝑥=𝑎^𝑇(𝑎𝑥)=𝑎^𝑇(𝑎𝑥)^𝑇=𝑎^𝑇𝑥^𝑇𝑎^𝑇=(𝑎𝑥𝑎)^𝑇=𝑎^𝑇,𝑎^𝑇𝑎𝑥=𝑎^𝑇, (4.1) (by Definition 4.2).
Conversely, if 𝑎^𝑇𝑎𝑥=𝑎^𝑇, then 𝑎^𝑇≤𝑎^𝑇𝑎≤𝑎^𝑇⇒𝑎^𝑇=𝑎^𝑇𝑎 and therefore 𝑎=𝑎^𝑇𝑎 and a is symmetric. Hence the given condition 𝑎^𝑇𝑎𝑥=𝑎^𝑇 reduces to a 𝑥=𝑎^𝑇
Now, 𝑎𝑥𝑎=(𝑎𝑥)𝑎=𝑎^𝑇𝑎=𝑎 and 𝑎𝑥=𝑎^𝑇=𝑎=(𝑎𝑥)^𝑇
𝑥∈𝑎{1,3}. Thus a has 1–3 inverses.
(iii)⇔(vi) This can be proved in the same manner and hence omitted.(vii)⇔(i)a is regular and 𝑎^𝑇∈𝑎{1}⇔ a is regular and 𝑎=𝑎𝑎^𝑇𝑎⇔ a is idempotent and 𝑎=𝑎^𝑇𝑎=𝑎𝑎^𝑇 (by Proposition 3.7 and Lemma 3.6) ⇔a is symmetric and idempotent⇔a is a projection.
(vii)⇔(viii)⇔(ix) follow from Lemma 4.4.

Remark 4.6. It is well known that [4] for an element a in a *-regular ring if 𝑎^† exists then 𝑎^†=𝑎^(1,4)𝑎𝑎^(1,3). We observe that for an element 𝑎 in an incline with involution-T if 𝑎 is regular, then by Lemma 3.6 it follows that 𝑎{1,3}=𝑎{1,4}. If 𝑎^† exists it is unique and given by 𝑎^†=𝑎^(1,3)𝑎𝑎^(1,3).

Remark 4.7. Let us consider the incline R in Remark 3.15 under the identity involution-T on R. Here each element in R is symmetric and the 3-inverse of the element 𝑑 is R and 4-inverse also the same.
Hence 0,𝑑 are the only projections in 𝑅𝑑^†=𝑑=𝑑{1,2}≠𝑑{1,3}=𝑑{1,4}=𝑑{1}={𝑏,𝑐,𝑑,1},𝑏⋅𝑑⋅𝑐=𝑑=𝑑^†∀𝑏,𝑐∈𝑑{1,4}. (4.2)

Theorem 4.8. Let R be an incline with involution-T. For any element 𝑎∈𝑅 and 𝑥∈𝑎{1,3} given, then 𝑎_𝑔{1,3}={𝑥+ℎ/ℎisarbitraryelementin𝑅} is the set of all {1,3} inverses of 𝐚  dominating 𝑥.

Proof. This can be proved along the same lines as that of Theorem 3.19 and hence omitted.

5. Conclusion

The main results in the present paper are the generalization of the available results shown in the reference for elements in a *-regular ring [4] and for elements in a Fuzzy algebra [8]. We have proved the results by using Property 2.5 without using star cancellation law.

In [2] it is stated that an element is regular if and only if 𝔇 class contains an idempotent. By Lemma 4.4 the 𝔇 class {𝑏/𝑏𝔇𝑎}={𝑎} and by Proposition 3.1  a is regular if and only if a is idempotent.

Figure 1