On Generalized Transitive Matrices

Abstract

Transitivity of generalized fuzzy matrices over a special type of semiring is considered. The semiring is called incline algebra which generalizes Boolean algebra, fuzzy algebra, and distributive lattice. This paper studies the transitive incline matrices in detail. The transitive closure of an incline matrix is studied, and the convergence for powers of transitive incline matrices is considered. Some properties of compositions of incline matrices are also given, and a new transitive incline matrix is constructed from given incline matrices. Finally, the issue of the canonical form of a transitive incline matrix is discussed. The results obtained here generalize the corresponding ones on fuzzy matrices and lattice matrices shown in the references.

1. Introduction

Generalized transitive matrices [1] over a special type of semiring are introduced. The semiring is called incline algebra. Boolean algebra, fuzzy algebra, and distributive lattice are inclines. And the Boolean matrices, the fuzzy matrices, and the lattice matrices are the prototypical examples of the incline matrices. Inclines are useful tools in diverse areas such as design of switching circuits, automata theory, medical diagnosis, information systems, and clustering. Besides inclines are applied to nervous system, probable reasoning, dynamical programming, and decision theory.

Transitive matrices are an important type of generalized matrices which represent transitive relation (see, e.g., [2–6]). Transitive relation plays an important role in clustering, information retrieval, preference, and so on [5, 7, 8]. The transitivity problems of matrices over some special semirings have been discussed by many authors (see, e.g., [9–17]). In 1982, Kim [18] introduced the concept of transitive binary Boolean matrices. Hashimoto [11] presented the concept of transitive fuzzy matrices and considered the convergence of powers of transitive fuzzy matrices. Kołodziejczyk [10] gave the concept of s-transitive fuzzy matrices and considered the convergence of powers of s-transitive fuzzy matrices. Tan [17, 19] discussed the convergence of powers of transitive lattice matrices. Han and Li [1] studied the convergence of powers of incline matrices. In [12, 13], the canonical form of a transitive matrix over fuzzy algebra was established, and, in [14, 15, 17], the canonical form of a transitive matrix over distributive lattice was characterized. In [9, 16, 20], some properties of compositions of generalized fuzzy matrices and lattice matrices were examined.

In this paper, we continue to study transitive incline matrices. In Section 3, the transitive closure of an incline matrix is discussed. In Section 4, the convergence of powers of transitive incline matrices is considered. In Section 5, some properties of compositions of incline matrices are given and a new transitive incline matrix is constructed from given incline matrices. In Section 6, the issue of the canonical form of an incline matrix is further discussed. Some results in this paper generalize the corresponding results in [14, 17, 20].

2. Definitions and Preliminary Lemmas

In this section, we give some definitions and lemmas.

Definition 2.1 (see [1]). A nonempty set 𝐿 with two binary operations + and ⋅ is called an incline if it satisfies the following conditions:(1)(𝐿,+) is a semilattice;(2)(𝐿,⋅) is a commutative semigroup;(3)𝑥(𝑦+𝑧)=𝑥𝑦+𝑥𝑧 for all 𝑥,𝑦,𝑧∈𝐿;(4)𝑥+𝑥𝑦=𝑥 for all 𝑥,𝑦∈𝐿.

In an incline 𝐿, define a relation ≤ by 𝑥≤𝑦⇔𝑥+𝑦=𝑦. Obviously, 𝑥𝑦≤𝑥 for all 𝑥,𝑦∈𝐿. It is easy to see that ≤ is a partial order relation over 𝐿 and satisfies the following properties.

Proposition 2.2 (see [21]). Let 𝐿 be an incline and 𝑎,𝑏,𝑐∈𝐿. Then,(1)0≤𝑎≤1;(2)if 𝑎≤𝑏, then 𝑎+𝑐≤𝑏+𝑐, 𝑎𝑐≤𝑏𝑐, 𝑐𝑎≤𝑐𝑏;(3)𝑎≤𝑎+𝑏, and 𝑎+𝑏 is the least upper bound of 𝑎 and 𝑏;(4)𝑎𝑏≤𝑎, 𝑎𝑏≤𝑏. In other words, 𝑎𝑏 is a lower bound of 𝑎 and 𝑏;(5)𝑎𝑐𝑏≤𝑎𝑏;(6)𝑎+𝑏=0 if and only if 𝑎=𝑏=0;(7)𝑎𝑏=1 if and only if 𝑎=𝑏=1.

Boolean algebra ({0,1},∨,∧), fuzzy algebra ([0,1],∨,𝑇) (𝑇 is a 𝑡-norm) and distributive lattice are inclines. Let (𝐿,≤) be a poset and 𝑎,𝑏∈𝐿. If 𝑎≤𝑏 or 𝑏≤𝑎, then 𝑎 and 𝑏 are called comparable. Otherwise, 𝑎 and 𝑏 are called incomparable, in notation, 𝑎‖𝑏. If for any 𝑎,𝑏∈𝐿, 𝑎 and 𝑏 are comparable, then 𝐿 is linear and 𝐿 is called a chain. An unordered poset is a poset in which 𝑎‖𝑏 for all 𝑎≠𝑏. A chain B in a poset 𝐿 is a nonempty subset of 𝐿, which, as a subposet, is a chain. An antichain 𝐵 in a poset 𝐿 is nonempty subset which, as a subposet, is unordered. The width of a poset 𝐿, denoted by 𝑤(𝐿), is 𝑛, where 𝑛 is a natural number, iff there is an antichain in 𝐿 of 𝑛 elements and all antichains in 𝐿 have ≤𝑛 elements. A poset (𝐿,≤) is called an incline if 𝐿 satisfies Definition 2.1. It is clear that any chain is an incline, which is called a linear incline.

An element 𝑎 of an incline 𝐿 is said to be idempotent if 𝑎²=𝑎. The set of all idempotent elements in 𝐿 is denoted by 𝐼(𝐿), that is, 𝐼(𝐿)={𝑎∈𝐿∣𝑎²=𝑎}.

A matrix is called an incline matrix if its entries belong to an incline. In this paper, the incline (𝐿,≤,+,⋅) is always supposed to be a commutative incline with the least and greatest elements 0 and 1, respectively. Let 𝑀_𝑛(𝐿) be the set of all 𝑛×𝑛 matrices over 𝐿. For any 𝐴 in 𝑀_𝑛(𝐿), we will denote by 𝑎_𝑖𝑗 or 𝐴_𝑖𝑗 the element of 𝐿 which stands in the (𝑖,𝑗)th entry of 𝐴. For convenience, we will use 𝑁 to denote the set {1,2,…,𝑛}, and 𝑍₊ denotes the set of all positive integers.For any 𝐴, 𝐵, 𝐶 in 𝑀_𝑛(𝐿) and 𝑎 in 𝐿, we define𝐴+𝐵=𝐶 iff 𝑐_𝑖𝑗=𝑎_𝑖𝑗+𝑏_𝑖𝑗 for all 𝑖, 𝑗 in 𝑁;𝐴𝐵=𝐶 iff 𝑐_𝑖𝑗=∑^𝑛_𝑘=1𝑎_𝑖𝑘𝑏_𝑘𝑗 for all 𝑖, 𝑗 in 𝑁;𝐴^𝑇= C iff 𝑐_𝑖𝑗=𝑎_𝑗𝑖 for all 𝑖, 𝑗 in 𝑁;𝑎𝐴=𝐶 iff 𝑐_𝑖𝑗=𝑎𝑎_𝑖𝑗 for all 𝑖, 𝑗 in 𝑁;𝐴≤𝐵 iff 𝑎_𝑖𝑗≤𝑏_𝑖𝑗 for all 𝑖, 𝑗 in 𝑁 and 𝐴≥𝐵 iff 𝐵≤𝐴;𝐼_𝑛=(𝑡_𝑖𝑗), where𝑡_𝑖𝑗=1,𝑖=𝑗,0,𝑖≠𝑗,for𝑖,𝑗∈𝑁. (2.1)For any 𝐴 in 𝑀_𝑛(𝐿), the powers of 𝐴 are defined as follows:𝐴⁰=𝐼_𝑛, 𝐴^𝑙=𝐴^𝑙−1𝐴, 𝑙∈𝑍₊.The (𝑖,𝑗)th entry of 𝐴^𝑙 is denoted by 𝑎^𝑙_𝑖𝑗(𝑙∈𝑍₊), and obviously 𝑎^𝑙_𝑖𝑗=_{1≤𝑖1,𝑖2,…,𝑖𝑙−1≤𝑛}𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑙−1𝑗}. (2.2) The following properties will be used in this paper.(1)𝑀_𝑛(𝐿) is a semigroup with the identity element 𝐼_𝑛 with respect to the multiplication;(2)(𝑀_𝑛(𝐿),+,⋅) is a semiring.

If 𝐴²≤𝐴, then 𝐴 is called transitive; if 𝐴²=𝐴, then 𝐴 is called idempotent; if 𝐴^𝑇=𝐴, then 𝐴 is called symmetric; if 𝐴≤𝐴², then 𝐴 is called increasing; if 𝐴≥𝐼_𝑛, then 𝐴 is called reflexive; if 𝑎_𝑖𝑖=0 for all 𝑖∈𝑁, then 𝐴 is called irreflexive; if 𝐴^𝑚=0 (𝑚∈𝑍₊), then 𝐴 is called nilpotent; if 𝑎_𝑖𝑗=0 for 𝑖,𝑗=1,2,…,𝑛, then 𝐴 is called the zero matrix and denoted by 0_𝑛; 𝐴 is called a permutation matrix if exactly one of the elements of its every row and every column is 1 and the others are 0.

Let 𝐵∈𝑀_𝑛(𝐿). The matrix 𝐵 is called the transitive closure of 𝐴 if 𝐵 is transitive and 𝐴≤𝐵, and, for any transitive matrix 𝐶 in 𝑀_𝑛(𝐿) satisfying 𝐴≤𝐶, we have 𝐵≤𝐶. The transitive closure of 𝐴 is denoted by 𝐴⁺. It is clear that if 𝐴 has a transitive closure, then it is unique.

For any 𝐴∈𝑀_𝑛(𝐿) with index, the sequence 𝐴,𝐴²,𝐴³,…,𝐴^𝑙,… (2.3) is of the form 𝐴,𝐴²,…,𝐴^𝑘−1∣𝐴^𝑘,…,𝐴^{𝑘+𝑑−1}∣𝐴^𝑘,…,𝐴^{𝑘+𝑑−1}∣⋯, (2.4) where 𝑘=𝑘(𝐴) is the least integer such that 𝐴^𝑘=𝐴^𝑘+𝑑 for some 𝑑>0. The least integers 𝑘=𝑘(𝐴), 𝑑=𝑑(𝐴) are called the index and the period of 𝐴, respectively.

The following definition will be used in this paper.

Definition 2.3. A matrix 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿) is said to be(1)row diagonally dominant if 𝑎_𝑖𝑗=𝑎_𝑖𝑖𝑎_𝑖𝑗 for all 𝑖,𝑗∈𝑁;(2)column diagonally dominant if 𝑎_𝑖𝑗=𝑎_𝑖𝑗𝑎_𝑗𝑗 for all 𝑖,𝑗∈𝑁;(3)weakly diagonally dominant if for any 𝑖∈𝑁, either 𝑎_𝑖𝑖𝑎_𝑖𝑗=𝑎_𝑖𝑗 for all 𝑗∈𝑁 or 𝑎_𝑖𝑖𝑎_𝑗𝑖=𝑎_𝑗𝑖 for all 𝑗∈𝑁;(4)strongly diagonally dominant if 𝑎_𝑖𝑗=𝑎_𝑖𝑖𝑎_𝑖𝑗=𝑎_𝑖𝑗𝑎_𝑗𝑗 for all 𝑖,𝑗∈𝑁;(5)nearly irreflexive if 𝑎_𝑖𝑖𝑎_𝑖𝑗=𝑎_𝑖𝑖 for all 𝑖,𝑗∈𝑁.

Lemma 2.4. 𝐼(𝐿) is a distributive lattice, where 𝐼(𝐿)={𝑎∈𝐿∣𝑎²=𝑎}.

The proof can be seen in [1].

3. Transitive Closure of an Incline Matrix

In this section, some properties of the transitive closure of an incline matrix are given and an algorithm for computing the transitive closure of an incline matrix is posed.

Lemma 3.1. For any 𝐴 in 𝑀_𝑛(𝐿), we have 𝐴⁺=∑^𝑛_𝑘=1𝐴^𝑘.

The proof can be seen in [21].

Lemma 3.2. Let 𝐴∈𝑀_𝑛(𝐿). Then,(1)for any 𝑖,𝑗∈𝑁 with 𝑖≠𝑗 and any 𝑘≥𝑛, there exists 𝑠(𝑠∈{1,2,…,𝑛−1}) such that 𝑎^𝑠_𝑖𝑗≥𝑎^𝑘_𝑖𝑗;(2)for any 𝑖∈𝑁 and any 𝑘≥𝑛, there exists 𝑡∈𝑁 such that 𝑎^𝑡_𝑖𝑖≥𝑎^𝑘_𝑖𝑖.

Proof. (1) Let 𝑇=𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑘−1𝑗} be any term of 𝑎^𝑘_𝑖𝑗, where 𝑘≥𝑛 and 1≤𝑖,𝑖₁,𝑖₂,…,𝑖_𝑘−1, 𝑗≤𝑛. Since the number of indices in 𝑇 is greater than 𝑛, a repetition among them must occur. Let us call the sequence of entries between two occurrences of one index a cycle. If we drop the cycle, a new expression 𝑇₁ with 𝑚₁≤𝑘 entries is obtained. If 𝑚₁≥𝑛, there must be a cycle in 𝑇₁, then we delete the cycle and obtain a new expression 𝑇₂ with 𝑚₂≤𝑚₁ entries. The deleting method can be applied repeatedly until the new expression 𝑇_𝑠 contains 𝑚_𝑠<𝑛 entries. According to properties of the operation “⋅”, 𝑇≤𝑇₁≤⋯≤𝑇_𝑠, but 𝑇_𝑠 is a term of the (𝑖,𝑗)th entry 𝑎^𝑠_𝑖𝑗 of 𝐴^𝑠 for some 𝑠<𝑛, we have 𝑇≤𝑇₁≤⋯≤𝑇_𝑠≤𝑎^𝑠_𝑖𝑗, and so 𝑎^𝑘_𝑖𝑗≤𝑎^𝑠_𝑖𝑗. This completes the proof.
(2) Let 𝑇=𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑘−1𝑖} be any term of 𝑎^𝑘_𝑖𝑖, where 𝑘≥𝑛+1 and 1≤𝑖,𝑖₁,𝑖₂,…,𝑖_𝑘−1≤𝑛. Since the number of indices in 𝑇 is greater than 𝑛, there must be two indices 𝑖_𝑢 and 𝑖_𝑣 such that 𝑖_𝑢=𝑖_𝑣 for some 𝑢, 𝑣 (𝑢<𝑣). Then, we delete 𝑎_{𝑖𝑢𝑖𝑢+1}⋯𝑎_{𝑖𝑣−1𝑖𝑣} from 𝑇 and obtain a new expression 𝑇₁=𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑢−1𝑖𝑢}𝑎_{𝑖𝑣𝑖𝑣+1}⋯𝑎_{𝑖𝑘−1𝑖} with 𝑚₁≤𝑘 entries. If 𝑚₁≥𝑛+1, there are still two identical numbers in the subscripts 𝑖,𝑖₁,…,𝑖_𝑢−1,𝑖_𝑢,𝑖_𝑣+1,…,𝑖_𝑘−1, then we apply the deleting method used in the above. The method can be applied repeatedly until the subscripts left are pairwise different. Finally, we can get a new term 𝑇_𝑡 with 𝑚_𝑡≤𝑛 entries. According to properties of the operation “⋅”, 𝑇≤𝑇₁≤⋯≤𝑇_𝑡, but 𝑇_𝑡 is a term of the (𝑖,𝑖)th entry 𝑎^𝑡_𝑖𝑖 of 𝐴^𝑡 for some 𝑡≤𝑛, we have 𝑇≤𝑇₁≤⋯≤𝑇_𝑡≤𝑎^𝑡_𝑖𝑖, and so 𝑎^𝑘_𝑖𝑖≤𝑎^𝑡_𝑖𝑖. This completes the proof.

Lemma 3.3. Let 𝐴∈𝑀_𝑛(𝐿). Then,(1)∑^𝑛_𝑙=1𝐴^𝑙≥𝐴^𝑘 for any 𝑘≥1;(2)∑^𝑛_𝑙=1𝐴^𝑙=∑^𝑚_𝑙=1𝐴^𝑙 for any 𝑚≥𝑛.

Proof. From Lemma 3.2, the proof is obvious.

Proposition 3.4. If 𝐴⁺ is reflexive, then (𝐴⁺)⁺=𝐴⁺.

Proof. Since 𝐴⁺≥𝐼_𝑛, we have 𝐴⁺≤(𝐴⁺)². On the other hand, by Lemmas 3.1 and 3.3, we see that (𝐴⁺)²≤𝐴⁺. Hence, (𝐴⁺)⁺=𝐴⁺.

Proposition 3.5. For any 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿) with index, if 𝐴 is column (or row) diagonally dominant, then 𝐴⁺=𝐴^𝑠, where 𝐴^𝑠 is transitive.

Proof. We only consider the case 𝐴 is column diagonally dominant.
For any integer 𝑙>0, we have 𝐴^𝑙𝑠≤𝐴^𝑠 since 𝐴^𝑠 is transitive, and so 𝑎^𝑙𝑠_𝑖𝑗≤𝑎^𝑠_𝑖𝑗, for all 𝑖,𝑗∈𝑁. Since 𝐴 is column diagonally dominant, we have 𝑎^𝑙_𝑖𝑗=𝑎^𝑙_𝑖𝑗𝑎_𝑗𝑗𝑎_𝑗𝑗⋯𝑎_𝑗𝑗≤𝑎^𝑙𝑠_𝑖𝑗 (because 𝑎^𝑙_𝑖𝑗𝑎_𝑗𝑗𝑎_𝑗𝑗⋯𝑎_𝑗𝑗 is the sum of some term in 𝑎^𝑙𝑠_𝑖𝑗) ≤𝑎^𝑠_𝑖𝑗. Hence, ∑^𝑛_𝑙=1𝑎^𝑙_𝑖𝑗≤𝑎^𝑠_𝑖𝑗, then 𝐴⁺≤𝐴^𝑠. On the other hand, since 𝐴⁺=∑^𝑛_𝑙=1𝐴^𝑙=∑^𝑚_𝑙=1𝐴^𝑙 for any 𝑚≥𝑛, we have 𝐴⁺≥𝐴^𝑠. Therefore, 𝐴⁺=𝐴^𝑠.

Corollary 3.6. For any 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿) with index, if 𝐴 is strongly diagonally dominant, then 𝐴⁺=𝐴^𝑠, where 𝐴^𝑠 is transitive.

Proof. Obviously, any strongly diagonally dominant matrix is column (or row) diagonally dominant. Hence, the conclusion follows from Proposition 3.5.

Proposition 3.7. For any 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿) with index, if 𝐴 is weakly diagonally dominant, then 𝐴⁺=𝐴^𝑠, where 𝐴^𝑠 is transitive.

Proof. Since 𝐴^𝑠 is transitive, we have 𝐴^𝑠≥𝐴^𝑙𝑠, and so 𝑎^𝑠_𝑖𝑗≥𝑎^𝑙𝑠_𝑖𝑗, for any 𝑖,𝑗∈𝑁 and any integer 𝑙>0. Since 𝐴 is weakly diagonally dominant, we have for any 𝑖∈𝑁, either 𝑎_𝑖𝑖𝑎_𝑖𝑗=𝑎_𝑖𝑗 for all 𝑗∈𝑁 or 𝑎_𝑖𝑖𝑎_𝑗𝑖=𝑎_𝑗𝑖 for all 𝑗∈𝑁.
Let 𝑎^𝑙_𝑖𝑗=_{1≤𝑖1,…,𝑖𝑙−1≤𝑛}𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑙−1𝑗}∀𝑖,𝑗∈𝑁. (3.1)
Case 1. If 𝑎_𝑖1𝑖1𝑎_𝑖1𝑗=𝑎_𝑖1𝑗 for all 𝑗∈𝑁, we have 𝑎_𝑖1𝑖1𝑎_𝑖1𝑖2=𝑎_𝑖1𝑖2. Then, 𝑎^𝑙_𝑖𝑗=_{1≤𝑖1,…,𝑖𝑙−1≤𝑛}𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑙−1𝑗}=_{1≤𝑖1,…,𝑖𝑙−1≤𝑛}𝑎_𝑖𝑖1𝑎_𝑖1𝑖1⋯𝑎_𝑖1𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑙−1𝑗}≤𝑎^𝑙𝑠_𝑖𝑗since_{1≤𝑖1,…,𝑖𝑙−1≤𝑛}𝑎_𝑖𝑖1𝑎_𝑖1𝑖1⋯𝑎_𝑖1𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑙−1𝑗}isthesumofsometermin𝑎^𝑙𝑠_𝑖𝑗≤𝑎^𝑠_𝑖𝑗. (3.2)Case 2. If 𝑎_𝑗𝑖1𝑎_𝑖1𝑖1=𝑎_𝑗𝑖1 for all 𝑗∈𝑁, we have 𝑎_𝑖𝑖1𝑎_𝑖1𝑖1=𝑎_𝑖𝑖1. Then, 𝑎^𝑙_𝑖𝑗=_{1≤𝑖1,…,𝑖𝑙−1≤𝑛}𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑙−1𝑗}=_{1≤𝑖1,…,𝑖𝑙−1≤𝑛}𝑎_𝑖𝑖1𝑎_𝑖1𝑖1⋯𝑎_𝑖1𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑙−1𝑗}≤𝑎^𝑙𝑠_𝑖𝑗since_{1≤𝑖1,…,𝑖𝑙−1≤𝑛}𝑎_𝑖𝑖1𝑎_𝑖1𝑖1⋯𝑎_𝑖1𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑙−1𝑗}isthesumofsometermin𝑎^𝑙𝑠_𝑖𝑗≤𝑎^𝑠_𝑖𝑗. (3.3)
From above, we see that 𝑎^𝑙_𝑖𝑗≤𝑎^𝑙𝑠_𝑖𝑗≤𝑎^𝑠_𝑖𝑗 for any 𝑖,𝑗∈𝑁. Hence, ∑^𝑛_𝑙=1𝑎^𝑙_𝑖𝑗≤𝑎^𝑠_𝑖𝑗, then 𝐴⁺≤𝐴^𝑠. On the other hand, since 𝐴⁺=∑^𝑛_𝑙=1𝐴^𝑙=∑^+∞_𝑙=1𝐴, we have 𝐴⁺≥𝐴^𝑠. Therefore, 𝐴⁺=𝐴^𝑠.

Proposition 3.8. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿). If the entries of 𝐴 satisfy 𝑎_𝑖𝑖𝑎_𝑗𝑘=𝑎_𝑗𝑘 (1≤𝑖,𝑗,𝑘≤𝑛), then(1)𝐴²≤𝐴³;(2)𝑎_𝑖𝑖=𝑎²_𝑖𝑖=⋯=𝑎^𝑛_𝑖𝑖 for all 𝑖∈𝑁;(3)A converges to 𝐴^𝑘(𝐴) with 𝑘(𝐴)≤𝑛−1.

Proof. (1) Since any term 𝑇 of the (𝑖,𝑗)th entry 𝑎²_𝑖𝑗 of 𝐴² is of the form 𝑎_𝑖𝑖1𝑎_𝑖1𝑗, we have 𝑎²_𝑖𝑗=∑^𝑛_𝑘=1𝑎_𝑖𝑘𝑎_𝑘𝑗 (1≤𝑖,𝑗,𝑘≤𝑛). Because the hypothesis 𝑎_𝑖𝑖𝑎_𝑗𝑘=𝑎_𝑗𝑘, we can get 𝑎²_𝑖𝑗=∑^𝑛_𝑘=1𝑎_𝑖𝑘𝑎_𝑘𝑘𝑎_𝑘𝑗. Since 𝑎_𝑖𝑘𝑎_𝑘𝑘𝑎_𝑘𝑗 is a term of 𝑎³_𝑖𝑗, we have 𝑎²_𝑖𝑗=∑^𝑛_𝑘=1𝑎_𝑖𝑘𝑎_𝑘𝑘𝑎_𝑘𝑗≤𝑎³_𝑖𝑗.
Thus, 𝐴²≤𝐴³.
(2) Because the hypothesis 𝑎_𝑖𝑖𝑎_𝑗𝑘=𝑎_𝑗𝑘, we have 𝑎_𝑖𝑖𝑎_𝑖𝑖⋯𝑎_𝑖𝑖=𝑎_𝑖𝑖. Hence,𝑎_𝑖𝑖≤𝑎^𝑠_𝑖𝑖=_{1≤𝑖1,…,𝑖𝑠−1≤𝑛}𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑠−1𝑖}≤_𝑘∈𝑁𝑎_𝑖𝑘=_𝑘∈𝑁𝑎_𝑖𝑖𝑎_𝑖𝑘≤_𝑘∈𝑁𝑎_𝑖𝑖=𝑎_𝑖𝑖, (3.4) then 𝑎_𝑖𝑖=𝑎^𝑠_𝑖𝑖 (for all 𝑠∈𝑁).
(3) Since 𝐴²≤𝐴³, we have 𝐴^𝑘−1≤𝐴^𝑘 (for any integer 𝑘≥3), and so 𝐴^𝑛−1≤𝐴^𝑛. Now we prove that 𝐴^𝑛−1≥𝐴^𝑛. By (2), it is sufficient only to show that 𝑎^𝑛−1_𝑖𝑗≥𝑎^𝑛_𝑖𝑗 for 𝑖≠𝑗. Let 𝑎^𝑛_𝑖𝑗=_{1≤𝑖1,…,𝑖𝑛−1≤𝑛}𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑛−1𝑗}∀𝑖,𝑗∈𝑁. (3.5) Since the number of indices in 𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑛−1𝑗} is 𝑛+1, there must be two indices 𝑖_𝑢 and 𝑖_𝑣 such that 𝑖_𝑢=𝑖_𝑣 for some 𝑢,𝑣(𝑢<𝑣). Then, 𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑛−1𝑗}≤𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑢−1𝑖𝑢}𝑎_{𝑖𝑣𝑖𝑣+1}⋯𝑎_{𝑖𝑛−1𝑗}. Since 𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑢−1𝑖𝑢}𝑎_{𝑖𝑣𝑖𝑣+1}⋯𝑎_{𝑖𝑛−1𝑗} is a term of 𝑎^{𝑛−(𝑣−𝑢)}_𝑖𝑗, we have 𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑛−1𝑗}≤𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑢−1𝑖𝑢}𝑎_{𝑖𝑣𝑖𝑣+1}⋯𝑎_{𝑖𝑛−1𝑗}≤𝑎^{𝑛−(𝑣−𝑢)}_𝑖𝑗 and so 𝑎^𝑛_𝑖𝑗≤𝑎^{𝑛−(𝑣−𝑢)}_𝑖𝑗, that is, 𝐴^𝑛≤𝐴^𝑚(𝑚=𝑛−(𝑣−𝑢)<𝑛)≤𝐴^𝑛−1 (because 𝐴^𝑘−1≤𝐴^𝑘 for any integer 𝑘≥3). Therefore, 𝐴^𝑛−1=𝐴^𝑛. This proves the proposition.

Lemma 3.9. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿). If the entries of 𝐴 satisfy 𝑎_𝑖𝑖𝑎_𝑗𝑘=𝑎_𝑗𝑘 (1≤𝑖,𝑗,𝑘≤𝑛), then 𝑎𝑑𝑗(𝐴)=𝐴^𝑛−1.

The proof can be seen in [22].

Proposition 3.10. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿). If the entries of 𝐴 satisfy 𝑎_𝑖𝑖𝑎_𝑗𝑘=𝑎_𝑗𝑘 (1≤𝑖,𝑗,𝑘≤𝑛), then (𝑎𝑑𝑗𝐴)⁺=𝐴^𝑛−1.

Proof. By the Proposition 3.8, we have 𝐴^𝑛−1=𝐴^𝑛. By Lemma 3.9, we can get 𝑎𝑑𝑗(𝐴)=𝐴^𝑛−1. Thus, (𝑎𝑑𝑗𝐴)⁺=(𝐴^𝑛−1)⁺=𝐴^𝑛−1.

Corollary 3.11. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿). If 𝐴 is reflexive, then(1)𝐴≤𝐴²;(2)𝑎_𝑖𝑖=𝑎²_𝑖𝑖=𝑎³_𝑖𝑖=⋯=𝑎^𝑛_𝑖𝑖=1 for all 𝑖∈𝑁;(3)𝐴 converges to 𝐴^𝑘(𝐴) with 𝑘(𝐴)≤𝑛−1;(4)(𝑎𝑑𝑗𝐴)⁺=𝐴^𝑛−1;(5)((𝑎𝑑𝑗𝐴)⁺)⁺=𝐴^𝑛−1.

Proof. From Propositions 3.10 and 3.8, the proof is obvious.

Proposition 3.12. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿). If 𝐿 has no nilpotent elements and 𝐴 is nilpotent, then(1)(𝑎𝑑𝑗𝐴)²=0;(2)(𝑎𝑑𝑗𝐴)⁺=𝑎𝑑𝑗𝐴.

Proof. (1) The proof can be seen in [23].
(2) Form (1), the proof is obvious.

Proposition 3.13. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿), 𝐴⁺=(𝑡_𝑖𝑗)_𝑛×𝑛, (𝐼_𝑛+𝐴)⁺=(𝑡_𝑖𝑗)_𝑛×𝑛, then(1)(𝐼_𝑛+𝐴)⁺=(𝐼_𝑛+𝐴)^𝑛=𝐼_𝑛+∑^𝑛_𝑙=1𝐴^𝑙;(2)(𝐼_𝑛+𝐴)⁺=(𝐼_𝑛+𝐴)^𝑛−1=𝐼_𝑛+∑^𝑛−1_𝑙=1𝐴^𝑙;(3)𝑡_𝑖𝑗=∑^𝑛_𝑙=1𝑎^𝑙_𝑖𝑗=𝑡_𝑖𝑗 (for all 𝑖,𝑗∈𝑁 and 𝑖≠𝑗).

Proof. (1) Since (𝐼_𝑛+𝐴) is reflexive, we have (𝐼_𝑛+𝐴) is increasing. By Lemma 3.1, we can get the conclusion.
(2) By Corollary 3.11, (𝐼_𝑛+𝐴) converges to (𝐼_𝑛+𝐴)^{𝑘(𝐼𝑛+𝐴)} with 𝑘(𝐼_𝑛+𝐴)≤𝑛−1. Thus, the conclusion is obtained.
(3) From (1) and (2), the proof is obvious.

At the end of this section, we may establish the following algorithm to find the (𝐴⁺)_𝑖𝑗 (𝑖≠𝑗).

Let 𝐴∈𝑀_𝑛(𝐿), 𝐴⁺=(𝑡_𝑖𝑗)_𝑛×𝑛,(𝐼_𝑛+𝐴)⁺=(𝑡_𝑖𝑗)_𝑛×𝑛, and 𝐵=𝐼_𝑛+𝐴.

Step 1. Compute successively 𝐵²,𝐵²²,…,𝐵^2𝑝,…,𝑝∈𝑍₊. (3.6) In (3.6), find 𝑝≤𝑘−1 such that 𝐵^2𝑝−1≠𝐵^2𝑝, but 𝐵^2𝑝=𝐵^2𝑝+1. Then, (𝐼_𝑛+𝐴)⁺=𝐵^2𝑝. Go to Step 2.

Step 2. Find (𝐴⁺)_𝑖𝑗 (𝑖≠𝑗).
Let 𝑡_𝑖𝑗=𝑡_𝑖𝑗 (𝑖≠𝑗). Stop.

4. Convergence of Powers of Transitive Incline Matrices

In this section, the convergence of powers of transitive incline matrices in 𝑀_𝑛(𝐿) will be discussed.

Definition 4.1. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿). For any 𝑖,𝑗∈𝑁, if 𝑎_𝑖𝑗𝑎_𝑖𝑘≠𝑎_𝑖𝑘⇒𝑎_𝑖𝑗𝑎_𝑘𝑗=𝑎_𝑘𝑗 holds for all 𝑘∈𝑁, then 𝐴 is called a strongly transitive matrix.

Theorem 4.2. If 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿) is a strongly transitive matrix, then 𝐴 is transitive.

Proof. Since 𝐴 is strongly transitive, for any 𝑖,𝑗∈𝑁, we have 𝑎_𝑖𝑗𝑎_𝑖𝑘=𝑎_𝑖𝑘 for all 𝑘∈𝑁 or 𝑎_𝑖𝑗𝑎_𝑖𝑘≠𝑎_𝑖𝑘 and 𝑎_𝑖𝑗𝑎_𝑘𝑗=𝑎_𝑘𝑗 for all 𝑘∈𝑁.
Case 1. For any 𝑖,𝑗∈𝑁, suppose 𝑎_𝑖𝑗𝑎_𝑖𝑘=𝑎_𝑖𝑘 for all 𝑘∈𝑁, then we can get 𝑎_𝑖𝑗≥𝑎_𝑖𝑗𝑎_𝑖𝑘=𝑎_𝑖𝑘≥𝑎_𝑖𝑘𝑎_𝑘𝑗, so 𝑎_𝑖𝑗≥∑^𝑛_𝑘=1𝑎_𝑖𝑘𝑎_𝑘𝑗=𝑎²_𝑖𝑗, which means that 𝐴 is transitive.Case 2. For any 𝑖,𝑗∈𝑁, suppose 𝑎_𝑖𝑗𝑎_𝑖𝑘≠𝑎_𝑖𝑘 and 𝑎_𝑖𝑗𝑎_𝑘𝑗=𝑎_𝑘𝑗 for all 𝑘∈𝑁, then we can get 𝑎_𝑖𝑗≥𝑎_𝑖𝑗𝑎_𝑘𝑗=𝑎_𝑘𝑗≥𝑎_𝑖𝑘𝑎_𝑘𝑗, so 𝑎_𝑖𝑗≥∑^𝑛_𝑘=1𝑎_𝑖𝑘𝑎_𝑘𝑗=𝑎²_𝑖𝑗, which means that 𝐴≥𝐴². This completes the proof.

Remark 4.3. If 𝐴 is transitive, but 𝐴 is not necessary to be a strongly transitive matrix.

Example 4.4. Consider the cline 𝐿={0,𝑎,𝑏,𝑐,1} whose diagram is as Figure 1.
Let now 𝐴=𝑐𝑎𝑏𝑐. (4.1)
Then, 𝐴²≤𝐴 which means that 𝐴 is transitive. But 𝑎₁₁𝑎₁₂=𝑐𝑎≠𝑎₁₂=𝑎 (because 𝑐𝑎<𝑎)⇏𝑎₁₁𝑎₂₁=𝑐𝑏=𝑎₂₁=𝑏 (because 𝑐𝑏<𝑏), which means that 𝐴 is not a strongly transitive matrix.

Figure 1 

Theorem 4.5. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿). If 𝐴≥𝐴² and 𝑎_𝑖𝑖∈𝐼(𝐿) (for all 𝑖∈𝑁), then(1)𝑎_𝑖𝑖=𝑎²_𝑖𝑖=⋯=𝑎^𝑛_𝑖𝑖 for all 𝑖∈𝑁;(2)𝐴 converges to 𝐴^𝑘(𝐴) with 𝑘(𝐴)≤𝑛.

Proof. (1) By the hypothesis 𝐴≥𝐴², it follows that 𝐴^𝑘≥𝐴^𝑘+1 (for all 𝑘∈𝑁). Hence, 𝐴^𝑛≥𝐴^𝑛+1 and 𝑎_𝑖𝑖≥𝑎^𝑠_𝑖𝑖 (for all 𝑠∈𝑁). Since 𝑎_𝑖𝑖∈𝐼(𝐿), we can get 𝑎_𝑖𝑖=𝑎_𝑖𝑖⋯𝑎_𝑖𝑖≤𝑎^𝑠_𝑖𝑖 (for all 𝑠∈𝑁). Thus 𝑎_𝑖𝑖=𝑎^𝑠_𝑖𝑖 and 𝑎^𝑠_𝑖𝑖∈𝐼(𝐿).
(2) By (1), it is sufficient to verify 𝐴^𝑛≤𝐴^𝑛+1.
Now, any term 𝑇 of the (𝑖,𝑗)th entry 𝑎^𝑛_𝑖𝑗 of 𝐴^𝑛 is of the form 𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑛−1𝑗}, where 1≤𝑖₁,𝑖₂,…,𝑖_𝑛−1≤𝑛. Since the number of indices in 𝑇 is greater than 𝑛, there must be two indices 𝑖_𝑢 and 𝑖_𝑣 such that 𝑖_𝑢=𝑖_𝑣 for some 𝑢,𝑣 (0≤𝑢<𝑣≤𝑛, 𝑖₀=𝑖, 𝑖_𝑛=𝑗). Then, 𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑛−1𝑗}=𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑢−1𝑖𝑢}𝑎_{𝑖𝑢𝑖𝑢+1}⋯𝑎_{𝑖𝑣−1𝑖𝑣}𝑎_{𝑖𝑣𝑖𝑣+1}⋯𝑎_{𝑖𝑛−1𝑗}≤𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑢−1𝑖𝑢}𝑎_𝑖^𝑣−𝑢_𝑢𝑖𝑢𝑎_{𝑖𝑣𝑖𝑣+1}⋯𝑎_{𝑖𝑛−1𝑗} (because 𝑎_{𝑖𝑢𝑖𝑢+1}⋯𝑎_{𝑖𝑣−1𝑖𝑣} is a term of 𝑎_𝑖^𝑣−𝑢_𝑢𝑖𝑢) =𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑢−1𝑖𝑢}𝑎_𝑖^𝑣−𝑢_𝑢𝑖𝑢𝑎_𝑖^𝑣−𝑢_𝑢𝑖𝑢𝑎_{𝑖𝑣𝑖𝑣+1}⋯𝑎_{𝑖𝑛−1𝑗} (because (1) 𝑎_𝑖𝑖=𝑎^𝑠_𝑖𝑖 (for all 𝑖,𝑠∈𝑁) and 𝑎_𝑖𝑖∈𝐼(𝐿)) ≤𝑎^{𝑛+(𝑣−𝑢)}_𝑖𝑗(𝑣−𝑢≥1) (because 𝑎_𝑖𝑖1𝑎_𝑖1𝑖2⋯𝑎_{𝑖𝑢−1𝑖𝑢}𝑎_𝑖^𝑣−𝑢_𝑢𝑖𝑢𝑎_𝑖^𝑣−𝑢_𝑢𝑖𝑢𝑎_{𝑖𝑣𝑖𝑣+1}⋯𝑎_{𝑖𝑛−1𝑗} is the sum of some term in 𝑎^{𝑛+(𝑣−𝑢)}_𝑖𝑗) ≤𝑎^𝑛+1_𝑖𝑗 (because 𝐴^𝑘≥𝐴^𝑘+1 for any 𝑘∈𝑁). Thus, 𝑇≤𝑎^𝑛+1_𝑖𝑗, and so 𝑎^𝑛_𝑖𝑗≤𝑎^𝑛+1_𝑖𝑗. Therefore, 𝐴^𝑛≤𝐴^𝑛+1. This completes the proof.

Corollary 4.6. If 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿) is strongly transitive, then 𝐴 converges to 𝐴^𝑘(𝐴) with 𝑘(𝐴)≤𝑛.

Proof. Since 𝐴 is strongly transitive, by Theorem 4.2, we have 𝐴≥𝐴² and 𝑎_𝑖𝑖∈𝐼(𝐿) (for all 𝑖∈𝑁), the conclusion is obtained.

Theorem 4.7. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿) be transitive. If 𝐵=(𝑏_𝑖𝑗)∈𝑀_𝑛(𝐼(𝐿)) and diag(𝐴)≤𝐵≤𝐴, where diag(𝐴)=(𝑐_𝑖𝑗) with 𝑐_𝑖𝑖=𝑎_𝑖𝑖 (for all 𝑖∈𝑁) and 𝑐_𝑖𝑗=0(𝑖≠𝑗,𝑖,𝑗∈𝑁). Then,(1)𝐵 converges to 𝐵^𝑘(𝐵) with 𝑘(𝐵)≤𝑛;(2)if 𝐴 satisfies 𝑎_𝑘𝑘≥∑^𝑛_𝑖=1𝑎_𝑘𝑖 (or 𝑎_𝑘𝑘≥∑^𝑛_𝑖=1𝑎_𝑖𝑘) for some 𝑘 in 𝑁, then 𝐵 converges to 𝐵^𝑘(𝐵) with 𝑘(𝐵)≤𝑛−1;(3)if 𝐵 satisfies 𝑏_𝑘𝑘≥∑^𝑛_𝑖=1𝑏_𝑘𝑖 (or 𝑏_𝑘𝑘≥∑^𝑛_𝑖=1𝑏_𝑖𝑘) for some 𝑘 in 𝑁, then 𝐵 converges to 𝐵^𝑘(𝐵) with 𝑘(𝐵)≤𝑛−1.

Proof. (1) Firstly, 𝑏_𝑖𝑖=𝑎_𝑖𝑖 (for all 𝑖∈𝑁) since diag(𝐴)≤𝐵≤𝐴. Since 𝑏_𝑖𝑖∈𝐼(𝐿), we have 𝑎_𝑖𝑖∈𝐼(𝐿). By Theorem 4.5, we can get 𝑎_𝑖𝑖=𝑎²_𝑖𝑖=⋯=𝑎^𝑛_𝑖𝑖 for all 𝑖∈𝑁.
It follows that any term 𝑇 of the (𝑖,𝑗)th entry 𝑏^𝑛_𝑖𝑗 of 𝐵^𝑛 is of the form 𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑛−1𝑗}, where 1≤𝑖₁,𝑖₂,…,𝑖_𝑛−1≤𝑛. Since {𝑖,𝑖₁,𝑖₂,…,𝑖_𝑛−1,𝑗}⊂{1,2,…,𝑛} and 𝑛+1>𝑛, there are 𝑢, 𝑣 such that 𝑖_𝑢=𝑖_𝑣 (0≤𝑢<𝑣≤𝑛, 𝑖₀=𝑖,𝑖_𝑛=𝑗). Then, 𝑇≤𝑏_{𝑖𝑢𝑖𝑢+1}⋯𝑏_{𝑖𝑣−1𝑖𝑣} and 𝑇≤𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑢−1𝑖𝑢}𝑏_{𝑖𝑣𝑖𝑣+1}⋯𝑏_{𝑖𝑛−1𝑗}. Since 𝐴 is transitive, we have 𝐴≥𝐴^𝑘 for all 𝑘≥1, and so 𝑎_𝑖𝑗≥𝑎^𝑘_𝑖𝑗 for all 𝑖,𝑗∈𝑁. Thus, 𝑏_𝑖^{𝑣−𝑢−1}_𝑢𝑖𝑢≥𝑏_{𝑖𝑢𝑖𝑢} (because 𝐵∈𝐼(𝐿)) =𝑎_{𝑖𝑢𝑖𝑢}=𝑎_𝑖^𝑣−𝑢_𝑢𝑖𝑢≥𝑏_𝑖^𝑣−𝑢_𝑢𝑖𝑢 (because 𝐴≥𝐵) ≥𝑏_{𝑖𝑢𝑖𝑢+1}⋯𝑏_{𝑖𝑣−1𝑖𝑣}≥𝑇. Since 𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑢−1𝑖𝑢}𝑏_𝑖^{𝑣−𝑢−1}_𝑢𝑖𝑢𝑏_{𝑖𝑣𝑖𝑣+1}⋯𝑏_{𝑖𝑛−1𝑗} is the sum of some term in 𝑏^𝑛−1_𝑖𝑗, we have 𝑏^𝑛−1_𝑖𝑗≥𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑢−1𝑖𝑢}𝑏_𝑖^{𝑣−𝑢−1}_𝑢𝑖𝑢𝑏_{𝑖𝑣𝑖𝑣+1}⋯𝑏_{𝑖𝑛−1𝑗}≥𝑇 (because 𝐵∈𝐼_𝑛(𝐿)), and so 𝑏^𝑛−1_𝑖𝑗≥𝑏^𝑛_𝑖𝑗 (because 𝑇 is any term of 𝑏^𝑛_𝑖𝑗). Thus, 𝐵^𝑛−1≥𝐵^𝑛.
Certainly, 𝐵^𝑛≥𝐵^𝑛+1≥𝐵^𝑛+2≥⋯.
On the other hand, for any term 𝑇=𝑏_𝑖𝑖1⋯𝑏_{𝑖𝑛−1𝑗} of 𝑏^𝑛_𝑖𝑗, there must be two indices 𝑖_𝑢 and 𝑖_𝑣 such that 𝑖_𝑢=𝑖_𝑣 for some 𝑢,𝑣 (0≤𝑢<𝑣≤𝑛, 𝑖₀=𝑖, 𝑖_𝑛=𝑗). Then, 𝑇=𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑛−1𝑗}=𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑢−1𝑖𝑢}𝑏_{𝑖𝑢𝑖𝑢+1}⋯𝑏_{𝑖𝑣−1𝑖𝑣}𝑏_{𝑖𝑣𝑖𝑣+1}⋯𝑏_{𝑖𝑛−1𝑗}≤𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑢−1𝑖𝑢}𝑏_𝑖^𝑣−𝑢_𝑢𝑖𝑢𝑏_{𝑖𝑣𝑖𝑣+1}⋯𝑏_{𝑖𝑛−1𝑗} (because 𝑏_{𝑖𝑢𝑖𝑢+1}⋯𝑏_{𝑖𝑣−1𝑖𝑣} is a term of 𝑏_𝑖^𝑣−𝑢_𝑢𝑖𝑢) =𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑢−1𝑖𝑢}𝑏_𝑖^𝑣−𝑢_𝑢𝑖𝑢𝑏_𝑖^𝑣−𝑢_𝑢𝑖𝑢𝑏_{𝑖𝑣𝑖𝑣+1}⋯𝑏_{𝑖𝑛−1𝑗}(because 𝐵∈𝑀_𝑛(𝐼(𝐿))) ≤𝑏^{𝑛+(𝑣−𝑢)}_𝑖𝑗 (because 𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑢−1𝑖𝑢}𝑏_𝑖^𝑣−𝑢_𝑢𝑖𝑢𝑏_𝑖^𝑣−𝑢_𝑢𝑖𝑢𝑏_{𝑖𝑣𝑖𝑣+1}⋯𝑏_{𝑖𝑛−1𝑗} is the sum of some term in 𝑏^{𝑛+(𝑣−𝑢)}_𝑖𝑗;𝑣−𝑢≥1) ≤𝑏^𝑛+1_𝑖𝑗 (because 𝐵^𝑘≥𝐵^𝑘+1 for any 𝑘≥𝑛−1). Thus, 𝑇≤𝑏^𝑛+1_𝑖𝑗, and so 𝑏^𝑛_𝑖𝑗≤𝑏^𝑛+1_𝑖𝑗. Therefore, 𝐵^𝑛≤𝐵^𝑛+1.
Consequently, we have 𝐵^𝑛=𝐵^𝑛+1. This completes the proof.
(2) By the proof of (1), we have 𝐵^𝑛≤𝐵^𝑛−1. Hence, 𝐵^𝑛−1≥𝐵^𝑛≥𝐵^𝑛+1≥𝐵^𝑛+2≥⋯. In the following, we will show that 𝐵^𝑛−1≤𝐵^𝑛. It is clear that any term 𝑇 of the (𝑖,𝑗)th entry 𝑏^𝑛−1_𝑖𝑗 of 𝐵^𝑛−1 is of the form 𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑛−2𝑗}, where 1≤𝑖,𝑖_𝑖,𝑖₂,…,𝑖_𝑛−2, 𝑗≤𝑛. Let 𝑖₀=𝑖 and 𝑖_𝑛−1=𝑗.
Case 1. If 𝑖_𝑢=𝑖_𝑣 for some 𝑢 and 𝑣 (𝑢<𝑣,𝑣−𝑢≥1), then 𝑇=𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑛−2𝑗}=𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑢−1𝑖𝑢}𝑏_{𝑖𝑢𝑖𝑢+1}⋯𝑏_{𝑖𝑣−1𝑖𝑣}𝑏_{𝑖𝑣𝑖𝑣+1}⋯𝑏_{𝑖𝑛−2𝑗}≤𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑢−1𝑖𝑢}𝑏_𝑖^𝑣−𝑢_𝑢𝑖𝑢𝑏_{𝑖𝑣𝑖𝑣+1}⋯𝑏_{𝑖𝑛−2𝑗} (because 𝑏_{𝑖𝑢𝑖𝑢+1}⋯𝑏_{𝑖𝑣−1𝑖𝑣} is a term of 𝑏_𝑖^𝑣−𝑢_𝑢𝑖𝑢) =𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑢−1𝑖𝑢}𝑏_𝑖^𝑣−𝑢_𝑢𝑖𝑢𝑏_𝑖^𝑣−𝑢_𝑢𝑖𝑢𝑏_{𝑖𝑣𝑖𝑣+1}⋯𝑏_{𝑖𝑛−2𝑗} (because 𝐵∈𝑀_𝑛(𝐼(𝐿))) ≤𝑏^{𝑛−1+(𝑣−𝑢)}_𝑖𝑗 (because 𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑢−1𝑖𝑢}𝑏_𝑖^𝑣−𝑢_𝑢𝑖𝑢𝑏_𝑖^𝑣−𝑢_𝑢𝑖𝑢𝑏_{𝑖𝑣𝑖𝑣+1}⋯𝑏_{𝑖𝑛−2𝑗} is the sum of some term in 𝑏^{𝑛−1+(𝑣−𝑢)}_𝑖𝑗) ≤𝑏^𝑛_𝑖𝑗 (because 𝐵^𝑛−1≥𝐵^𝑛≥𝐵^𝑛+1≥⋯). Thus, 𝑇≤𝑏^𝑛_𝑖𝑗, and so 𝑏^𝑛−1_𝑖𝑗≤𝑏^𝑛_𝑖𝑗. Therefore, 𝐵^𝑛−1≤𝐵^𝑛.Case 2. Suppose that 𝑖_𝑢≠𝑖_𝑣 for all 𝑢≠𝑣. By the hypothesis 𝑎_𝑘𝑘≥∑^𝑛_𝑖=1𝑎_𝑘𝑖 for some 𝑘, we have 𝑏_{𝑖𝑢𝑖𝑢}=𝑎_{𝑖𝑢𝑖𝑢}≥∑^𝑛_𝑖=1𝑎_𝑖𝑢𝑖≥∑^𝑛_𝑖=1𝑏_𝑖𝑢𝑖 (because 𝐴≥𝐵) ≥𝑏_{𝑖𝑢𝑖𝑢+1}. Then, 𝑇=𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑛−2𝑗}=𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑢−1𝑖𝑢}𝑏_{𝑖𝑢𝑖𝑢+1}𝑏_{𝑖𝑢+1𝑖𝑢+2}⋯𝑏_{𝑖𝑛−2𝑗}=𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑢−1𝑖𝑢}𝑏_{𝑖𝑢𝑖𝑢+1}𝑏_{𝑖𝑢𝑖𝑢+1}𝑏_{𝑖𝑢+1𝑖𝑢+2}⋯𝑏_{𝑖𝑛−2𝑗} (because 𝐵∈𝑀_𝑛(𝐼(𝐿))) ≤𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑢−1𝑖𝑢}𝑏_{𝑖𝑢𝑖𝑢+1}𝑏_{𝑖𝑢𝑖𝑢}𝑏_{𝑖𝑢+1𝑖𝑢+2}⋯𝑏_{𝑖𝑛−2𝑗}(because 𝑏_{𝑖𝑢𝑖𝑢}≥𝑏_{𝑖𝑢𝑖𝑢+1}) ≤𝑏^𝑛_𝑖𝑗 (because 𝑏_𝑖𝑖1𝑏_𝑖1𝑖2⋯𝑏_{𝑖𝑢−1𝑖𝑢}𝑏_{𝑖𝑢𝑖𝑢+1}𝑏_{𝑖𝑢𝑖𝑢}𝑏_{𝑖𝑢+1𝑖𝑢+2}⋯𝑏_{𝑖𝑛−2𝑗} is a term of 𝑏^𝑛_𝑖𝑗). Thus, 𝑇≤𝑏^𝑛_𝑖𝑗, and so 𝑏^𝑛−1_𝑖𝑗≤𝑏^𝑛_𝑖𝑗. Therefore, 𝐵^𝑛−1≤𝐵^𝑛.
The case of 𝑎_𝑘𝑘≥∑^𝑛_𝑖=1𝑎_𝑖𝑘 is similar to that of the hypothesis 𝑎_𝑘𝑘≥∑^𝑛_𝑖=1𝑎_𝑘𝑖.
Consequently, we have 𝐵^𝑛−1=𝐵^𝑛.
(3) The proof of (3) is similar to that of (2). This completes the proof.

Theorem 4.7 generalizes and develops Theorem  4.1 of Tan [17].

5. Compositions of Transitive Matrices

In this section, we construct a transitive matrix from given incline matrices and construct a new incline matrix with some special properties from transitive matrices. We know that any idempotent matrix is also a transitive matrix.

Theorem 5.1. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿). If 𝐴 is transitive and row diagonally dominant, then 𝐵=(𝑏_𝑖𝑗) is idempotent, where 𝑏_𝑖𝑗=𝑎_𝑖𝑗𝑎_𝑗𝑖.

Proof. For any 𝑖, 𝑗 in 𝑁, 𝑏²_𝑖𝑗=∑^𝑛_𝑘=1𝑏_𝑖𝑘𝑏_𝑘𝑗=∑^𝑛_𝑘=1𝑎_𝑖𝑘𝑎_𝑘𝑖𝑎_𝑘𝑗𝑎_𝑗𝑘=∑^𝑛_𝑘=1(𝑎_𝑖𝑘𝑎_𝑘𝑗)(𝑎_𝑗𝑘𝑎_𝑘𝑖)≤∑^𝑛_𝑘=1(𝑎_𝑖𝑗𝑎_𝑗𝑖) (because 𝐴≥𝐴²) =∑^𝑛_𝑘=1𝑏_𝑖𝑗=𝑏_𝑖𝑗. Thus, 𝐵²≤𝐵. On the other hand, since 𝑏²_𝑖𝑗=∑^𝑛_𝑘=1𝑏_𝑖𝑘𝑏_𝑘𝑗≥𝑏_𝑖𝑖𝑏_𝑖𝑗=𝑎_𝑖𝑖𝑎_𝑖𝑖𝑎_𝑖𝑗𝑎_𝑗𝑖=𝑎_𝑖𝑗𝑎_𝑗𝑖 (because 𝐴 is row diagonally dominant) =𝑏_𝑖𝑗. Thus, 𝐵²≥𝐵. Therefore, 𝐵²=𝐵. This completes the proof.

Definition 5.2. 𝐴∘𝐷=𝐶 iff 𝑐_𝑖𝑗=∏^𝑛_𝑘=1(𝑎_𝑖𝑘+𝑑_𝑘𝑗) for any 𝑖,𝑗∈𝑁.

Theorem 5.3. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿) be a symmetric and nearly irreflexive matrix. Then, the matrix 𝑀=𝐶∘𝐴 is idempotent, where 𝐶=(𝑐_𝑖𝑗) with 𝑐_𝑖𝑖=𝑐_𝑖∈𝐿, 𝑐_𝑖𝑗=0 (𝑖≠𝑗).

Proof. By the definitions of the matrices 𝐴, 𝐶, and 𝑀, we have 𝑚_𝑖𝑗=_𝑙≠𝑖𝑎_𝑙𝑗𝑐_𝑖+𝑎_𝑖𝑗=_𝑙≠𝑖𝑎_𝑗𝑙𝑐_𝑖+𝑎_𝑖𝑗=𝑎_𝑖𝑖+_𝑙≠𝑖𝑎_𝑙𝑖𝑐_𝑖𝑎,𝑖=𝑗,_𝑗𝑗,𝑖≠𝑗. (5.1) The (𝑖,𝑗)th entry of 𝑀² is 𝑚²_𝑖𝑗=∑^𝑛_𝑘=1𝑚_𝑖𝑘𝑚_𝑘𝑗.
Case 1 (𝑖≠𝑗). In this case, 𝑚²_𝑖𝑗=∑_{𝑘≠𝑖,𝑗}𝑚_𝑖𝑘𝑚_𝑘𝑗+𝑚_𝑖𝑖𝑚_𝑖𝑗+𝑚_𝑖𝑗𝑚_𝑗𝑗=∑_{𝑘≠𝑖,𝑗}𝑎_𝑘𝑘𝑎_𝑗𝑗+(𝑎_𝑖𝑖+∏_𝑙≠𝑖𝑎_𝑙𝑖𝑐_𝑖)𝑎_𝑗𝑗+𝑎_𝑗𝑗(𝑎_𝑗𝑗+∏_𝑙≠𝑗𝑎_𝑙𝑗𝑐_𝑗)=𝑎_𝑗𝑗(∑^𝑛_𝑘=1𝑎_𝑘𝑘+∏_𝑙≠𝑖𝑎_𝑙𝑖𝑐_𝑖+∏_𝑙≠𝑗𝑎_𝑙𝑗𝑐_𝑗)=𝑎_𝑗𝑗 (because 𝐴 is nearly irreflexive) =𝑚_𝑖𝑗.Case 2 (𝑖=𝑗). In this case, 𝑚²_𝑖𝑖=∑_𝑙≠𝑖𝑚_𝑖𝑙𝑚_𝑙𝑖+𝑚_𝑖𝑖=∑_𝑙≠𝑖𝑎_𝑙𝑙𝑎_𝑖𝑖+𝑎_𝑖𝑖+∏_𝑘≠𝑖𝑎_𝑘𝑖𝑐_𝑖=𝑎_𝑖𝑖+∏_𝑘≠𝑖𝑎_𝑘𝑖𝑐_𝑖 (because 𝐴 is nearly irreflexive) =𝑚_𝑖𝑖.
Consequently, we can get 𝑀²=𝑀. Therefore, 𝑀=𝐶∘𝐴 is idempotent.

Theorem 5.3 generalizes Theorem  3.5 of Tan [20].

Lemma 5.4. Let 𝐴,𝐵∈𝑀_𝑚×𝑛(𝐿), 𝐶∈𝑀_𝑛×𝑙(𝐿) and 𝐷∈𝑀_𝑝×𝑚(𝐿). Then,(1)(𝐵∘𝐶)^𝑇=𝐶^𝑇∘𝐵^𝑇;(2)If 𝐴≤𝐵, then 𝐷∘𝐴≤𝐷∘𝐵 and 𝐴∘𝐶≤𝐵∘𝐶.

The proof is trivial.

Lemma 5.5. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑚×𝑛(𝐿) be a nearly irreflexive matrix, then 𝐴∘𝐴^𝑇 is nearly irreflexive and symmetric.

Proof. Let 𝑆=𝐴∘𝐴^𝑇. Then, 𝑠_𝑖𝑖=∏^𝑚_𝑙=1(𝑎_𝑖𝑙+𝑎_𝑖𝑙)=∏^𝑚_𝑙=1𝑎_𝑖𝑙=𝑎_𝑖𝑖 (because 𝐴 is nearly irreflexive). Thus, 𝑠_𝑖𝑖𝑠_𝑖𝑗=𝑎_𝑖𝑖(∏^𝑚_𝑙=1(𝑎_𝑖𝑙+𝑎_𝑗𝑙))=𝑎_𝑖𝑖=𝑠_𝑖𝑖, that is, 𝑆=𝐴∘𝐴^𝑇 is nearly irreflexive. By Lemma 5.4, we have (𝐴∘𝐴^𝑇)^𝑇=(𝐴^𝑇)^𝑇∘𝐴^𝑇=𝐴∘𝐴^𝑇.

Lemma 5.5 generalizes Theorem  3.3 of Tan [20].

Corollary 5.6. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿) be a nearly irreflexive matrix. Then, 𝐶∘(𝐴∘𝐴^𝑇) is idempotent, where 𝐶=(𝑐_𝑖𝑗) with 𝑐_𝑖𝑖=𝑐_𝑖∈𝐿, 𝑐_𝑖𝑗=0 (𝑖≠𝑗).

Proof. It follows from Theorem 5.3 and Lemma 5.5.

Proposition 5.7. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿) be a symmetric and nearly irreflexive matrix. Then,(1)𝐴∘𝐴≤𝐴;(2)𝐴∘𝐴 is symmetric and nearly irreflexive.

Proof. (1) Let 𝑅=𝐴∘𝐴. Then, 𝑟_𝑖𝑗=∏^𝑛_𝑙=1(𝑎_𝑖𝑙+𝑎_𝑙𝑗)≤𝑎_𝑖𝑖+𝑎_𝑖𝑗=𝑎_𝑖𝑖𝑎_𝑖𝑗+𝑎_𝑖𝑗 (because 𝐴 is nearly irreflexive) =𝑎_𝑖𝑗, so that 𝐴∘𝐴≤𝐴.
(2) Let 𝑅=𝐴∘𝐴. Since 𝑟_𝑗𝑖=∏^𝑛_𝑙=1(𝑎_𝑗𝑙+𝑎_𝑙𝑖)=∏^𝑛_𝑙=1(𝑎_𝑖𝑙+𝑎_𝑙𝑗)=𝑟_𝑖𝑗, we have 𝑅 is symmetric. Since 𝑟_𝑖𝑖=∏^𝑛_𝑙=1(𝑎_𝑖𝑙+𝑎_𝑙𝑖)=∏^𝑛_𝑙=1𝑎_𝑖𝑙=𝑎_𝑖𝑖 (because 𝐴 is nearly irreflexive), we can get 𝑟_𝑖𝑖𝑟_𝑖𝑗=𝑎_𝑖𝑖(∏^𝑛_𝑙=1(𝑎_𝑖𝑙+𝑎_𝑙𝑗))=𝑎_𝑖𝑖=𝑟_𝑖𝑖. Thus, 𝑅 is nearly irreflexive.

Proposition 5.7 generalizes Proposition  3.1 of Tan [20].

Corollary 5.8. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿) be a symmetric and nearly irreflexive matrix. Then, 𝐶∘(𝐴∘𝐴) is idempotent, where 𝐶=(𝑐_𝑖𝑗) with 𝑐_𝑖𝑖=𝑐_𝑖∈𝐿, 𝑐_𝑖𝑗=0 (𝑖≠𝑗).

Proof. It follows from Theorem 5.3 and Proposition 5.7.

Corollary 5.8 generalizes Corollary  3.7 (1) of Tan [20].

Corollary 5.9. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿) be a nearly irreflexive matrix. Then, (𝐴∘𝐴^𝑇)∘𝐶 is idempotent, where 𝐶=(𝑐_𝑖𝑗) with 𝑐_𝑖𝑖=𝑐_𝑖∈𝐿, 𝑐_𝑖𝑗=0 (𝑖≠𝑗).

The proof is trivial.

Corollary 5.9 generalizes Corollary  3.8 of Tan [20].

Proposition 5.10. Let 𝐴∈𝑀_𝑛(𝐿) be irreflexive and transitive. Then,(1)𝐴∘𝐴^𝑇=0, 0∈𝑀_𝑛(𝐿);(2)𝐴^𝑇∘𝐴=0, 0∈𝑀_𝑛(𝐿).

Proof. (1) Let 𝑅=𝐴∘𝐴^𝑇. Then, 𝑟_𝑖𝑗=∏^𝑛_𝑙=1(𝑎_𝑖𝑙+𝑎_𝑗𝑙)≤(𝑎_𝑖𝑖+𝑎_𝑗𝑖)(𝑎_𝑖𝑗+𝑎_𝑗𝑗) (because 𝐴 is irreflexive) =𝑎_𝑖𝑗𝑎_𝑗𝑖≤𝑎_𝑖𝑖 (because 𝐴 is transitive) = 0. Therefore, 𝑅=0.
The proof of (2) is similar to that of (1). This proves the Proposition.

Proposition 5.10 generalizes Proposition  3.9 of Tan [20].

Definition 5.11. An incline 𝐿 is said to be a Brouwerian incline if for any 𝑎,𝑏∈𝐿, there exists an element 𝑏→𝑎∈𝐿 such that 𝑏𝑥≤𝑎⇔𝑥≤𝑏→𝑎.

Obviously, 𝑏→𝑎 is the largest element 𝑥∈𝐿 satisfying 𝑏𝑥≤𝑎.

Definition 5.12. 𝐴←𝐷=𝐶 iff 𝑐_𝑖𝑗=∏^𝑛_𝑘=1(𝑑_𝑘𝑗→𝑎_𝑖𝑘) for any 𝑖,𝑗∈𝑁.

Lemma 5.13. Let 𝐿 be a Brouwerian incline. Then, for any 𝑎∈𝐿,(1)𝑎→𝑎=1;(2)1→𝑎=𝑎.

The proof is trivial.

Theorem 5.14. Let 𝐴∈𝑀_𝑚×𝑛(𝐿). Then, 𝐴←𝐴^𝑇 is reflexive and transitive.

Proof. Let 𝑅=𝐴←𝐴^𝑇. Then, 𝑟_𝑖𝑗=∏^𝑛_𝑘=1(𝑎_𝑗𝑘→𝑎_𝑖𝑘). Obviously, 𝑟_𝑖𝑖=∏^𝑛_𝑘=1(𝑎_𝑖𝑘→𝑎_𝑖𝑘)=1 (by Lemma 5.13 (1)). Thus, 𝑅 is reflexive. Furthermore, since 𝑎_𝑗𝑘(𝑎_𝑙𝑘→𝑎_𝑖𝑘)(𝑎_𝑗𝑘→𝑎_𝑙𝑘)=(𝑎_𝑙𝑘→𝑎_𝑖𝑘)(𝑎_𝑗𝑘(𝑎_𝑗𝑘→𝑎_𝑙𝑘)≤(𝑎_𝑙𝑘→𝑎_𝑖𝑘)𝑎_𝑙𝑘≤𝑎_𝑖𝑘, we have (𝑎_𝑙𝑘→𝑎_𝑖𝑘)(𝑎_𝑗𝑘→𝑎_𝑙𝑘)≤𝑎_𝑗𝑘→𝑎_𝑖𝑘. Therefore, 𝑟_𝑖𝑙𝑟_𝑙𝑗=∏^𝑛_𝑘=1(𝑎_𝑙𝑘→𝑎_𝑖𝑘)(𝑎_𝑗𝑘→𝑎_𝑙𝑘)≤∏^𝑛_𝑘=1(𝑎_𝑗𝑘→𝑎_𝑖𝑘)=𝑟_𝑖𝑗, and so 𝑅²≤𝑅. This proves the Theorem.

Theorem 5.14 generalizes Lemma  4.1 of Tan [20].

Theorem 5.15. Let 𝐴∈𝑀_𝑛(𝐼(𝐿)). Then, the following conditions are equivalent.(1)𝐴 is reflexive and transitive;(2)𝐴←𝐴^𝑇=𝐴.

Proof. (1) ⇒ (2). Let 𝑅=𝐴←𝐴^𝑇. Then, 𝑟_𝑖𝑗=∏^𝑛_𝑘=1(𝑎_𝑗𝑘→𝑎_𝑖𝑘)≤𝑎_𝑗𝑗→𝑎_𝑖𝑗=1→𝑎_𝑖𝑗=𝑎_𝑖𝑗 (by Lemma 5.13(2)), and so 𝑅≤𝐴. On the other hand, since 𝑎_𝑖𝑘≥𝑎_𝑖𝑗𝑎_𝑗𝑘 for all 𝑖,𝑗,𝑘∈𝑁, we have 𝑎_𝑖𝑗≤𝑎_𝑗𝑘→𝑎_𝑖𝑘, and so 𝑎_𝑖𝑗≤∏^𝑛_𝑘=1(𝑎_𝑗𝑘→𝑎_𝑖𝑘) (because 𝐴∈𝑀_𝑛(𝐼(𝐿)))=𝑟_𝑖𝑗, that is, 𝐴≤𝑅. Consequently, we have 𝑅=𝐴←𝐴^𝑇=𝐴.
(2) ⇒ (1). It follows from Theorem 5.14.

Theorem 5.15 generalizes Proposition  4.2 of Tan [20].

Theorem 5.16. Let 𝐴∈𝑀_𝑚×𝑛(𝐿). Then, (𝐴←𝐴^𝑇)𝐴=𝐴.

Proof. Let 𝑅=(𝐴←𝐴^𝑇)𝐴. Then, 𝑟_𝑖𝑗=∑^𝑚_𝑘=1∏^𝑛_𝑙=1(𝑎_𝑘𝑙→𝑎_𝑖𝑙)𝑎_𝑘𝑗≤∑^𝑚_𝑘=1𝑎_𝑘𝑗(𝑎_𝑘𝑗→𝑎_𝑖𝑗)≤∑^𝑚_𝑘=1𝑎_𝑖𝑗=𝑎_𝑖𝑗, so that 𝑅≤𝐴. On the other hand, since 𝐴←𝐴^𝑇 is reflexive (by Theorem 5.14), we have 𝑅=(𝐴←𝐴^𝑇)𝐴≥𝐴. Therefore, (𝐴←𝐴^𝑇)𝐴=𝐴.

Theorem 5.16 generalizes Lemma  4.3 of Tan [20].

6. On Canonical Form of an Incline Matrix

In this section, we will discuss the canonical form of an incline matrix. Let 𝐴 be an 𝑛×𝑛 incline matrix. If there exists an 𝑛×𝑛 permutation matrix 𝑃 such that 𝐹=𝑃𝐴𝑃^𝑇=(𝑓_𝑖𝑗) satisfies 𝑓_𝑖𝑗0𝑥0008𝑒<𝑓̸_𝑗𝑖 for 𝑖>𝑗 then, 𝐹 is called a canonical form of 𝐴. The main results obtained here generalize the previous results on canonical form of a lattice matrix (see, e.g., [17]) and a fuzzy matrix (see, e.g., [12]).

Definition 6.1. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿). For any 𝑖,𝑗,𝑘∈𝑁 with 𝑖≠𝑗, 𝑖≠𝑘, 𝑗≠𝑘, if 𝑎_𝑖𝑘>𝑎_𝑘𝑖 and 𝑎_𝑘𝑗>𝑎_𝑗𝑘, we have 𝑎_𝑖𝑗>𝑎_𝑗𝑖. Then, 𝐴 is called a especially strongly transitive matrix.

Lemma 6.2. If 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿) is especially strongly transitive matrix and 𝐴 is put in the block forms 𝑎𝐴=¹¹𝛼₁𝛼₂𝐴₁=𝐴₂𝛽₁𝛽₂𝑎_𝑛𝑛, (6.1) where 𝛼₁,𝛼^𝑇₂,𝛽₁,𝛽^𝑇₂∈𝐿^{1×(𝑛−1)} and 𝐴₁,𝐴₂∈𝑀_𝑛−1(𝐿), then 𝐴₁, 𝐴₂, and 𝑃𝐴𝑃^𝑇 are especially strongly transitive matrices for any permutation matrix 𝑃.

The proof is similar to that of Lemma  3.1 in [14].

Lemma 6.3. Let 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑚×𝑛(𝐿) and 𝐵=(𝑏_𝑖𝑗)∈𝑀_𝑛×𝑚(𝐿). If 𝐴0𝑥0008𝑒<𝐵̸^𝑇, then 𝐴𝑃^𝑇<0̸𝑥0008𝑒𝐵^𝑇𝑃^𝑇 holds for any 𝑛×𝑛 permutation matrix 𝑃.

The proof is omitted.

Theorem 6.4. If 𝐴=(𝑎_𝑖𝑗)∈𝑀_𝑛(𝐿) is especially strongly transitive, then 𝐴 has a canonical form.

The proof is similar to that of Theorem  3.1 in [14].

Theorem 6.5. Let 𝐿 be an incline and 𝑛 an integer with 𝑛≥4. Then, for any 𝑛×𝑛 transitive matrix 𝐴 over 𝐿, there exists an 𝑛×𝑛 permutation matrix 𝑃 such that 𝐹=𝑃𝐴𝑃^𝑇 satisfies 𝑓_𝑖𝑗<0̸𝑥0008𝑒𝑓_𝑗𝑖 for 𝑖>𝑗 only if 𝐿 is a linear incline.

Proof. Suppose that 𝐿 is not a linear incline. Then, 𝑤(𝐿)≥2, and so that there must be two elements 𝑎 and 𝑐 in 𝐿 such that 𝑎‖𝑐. Therefore, 𝑎𝑐<𝑎<𝑎+𝑐 and 𝑎𝑐<𝑐<𝑎+𝑐. Now, let 𝐴=𝑎𝐸₁₂+𝑐𝐸₂₃+𝑎𝐸₃₄+𝑐𝐸₄₁+𝑎𝑐𝐽_𝑛∈𝑀_𝑛(𝐿), where 𝐽_𝑛=(1)_𝑛×𝑛.
It is easy to see that 𝐴²≤𝐴. This means 𝐴 is transitive. Let 𝑃 be any 𝑛×𝑛 permutation matrix. Then, there exists a unique permutation 𝜏 of the set {1,2,…,𝑛} such that 𝑃=∑^𝑛_𝑖=1𝐸_{𝜏(𝑖)𝑖}, and so 𝑃^𝑇=∑^𝑛_𝑖=1𝐸_{𝑖𝜏(𝑖)}. Therefore, 𝐹=(𝑓_𝑖𝑗)_𝑛×𝑛=𝑃𝐴𝑃^𝑇=𝑎𝐸_{𝜏(1)𝜏(2)}+𝑐𝐸_{𝜏(2)𝜏(3)}+𝑎𝐸_{𝜏(3)𝜏(4)}+𝑐𝐸_{𝜏(4)𝜏(1)}+𝑎𝑐𝐽_𝑛. Thus, 𝑓_{𝜏(1)𝜏(2)}=𝑓_{𝜏(3)𝜏(4)}=𝑎, 𝑓_{𝜏(2)𝜏(3)}=𝑓_{𝜏(4)𝜏(1)}=𝑐 and 𝑓_{𝜏(2)𝜏(1)}=𝑓_{𝜏(4)𝜏(3)}=𝑓_{𝜏(3)𝜏(2)}=𝑓_{𝜏(1)𝜏(4)}=𝑎𝑐. Since 𝜏 is a permutation, we have 𝜏(𝑖)≠𝜏(𝑗)(𝑖≠𝑗). By the hypothesis 𝐹=𝑃𝐴𝑃^𝑇 satisfies 𝑓_𝑖𝑗<0̸𝑥0008𝑒𝑓_𝑗𝑖 for 𝑖>𝑗, we have 𝜏(1)>𝜏(2), 𝜏(2)>𝜏(3), 𝜏(3)>𝜏(4), and 𝜏(4)>𝜏(1). This implies 𝜏(1)>𝜏(1), which leads to a contradiction. This proves the Theorem.

Theorem 6.5 generalizes Theorem  5.2 of Tan [17].

Remark 6.6. It is easy to verify that 𝐴 always has a canonical form for any transitive matrix 𝐴∈𝑀₂(𝐿).

Acknowledgments

This work was supported by the Foundation of National Nature Science of China (Grant no. 11071 178) and the Fostering Plan for Young and Middle Age Leading Research of UESTC (Grant no. Y020 18023601033).