*Article* **Strong Tolerance and Strong Universality of Interval Eigenvectors in a Max-Łukasiewicz Algebra**

**Martin Gavalec 1,\*, Zuzana N ˇemcová <sup>1</sup> and Ján Plavka <sup>2</sup>**


Received: 9 August 2020; Accepted: 2 September 2020; Published: 4 September 2020

**Abstract:** The Łukasiewicz conjunction (sometimes also considered to be a logic of absolute comparison), which is used in multivalued logic and in fuzzy set theory, is one of the most important t-norms. In combination with the binary operation 'maximum', the Łukasiewicz t-norm forms the basis for the so-called max-Łuk algebra, with applications to the investigation of systems working in discrete steps (discrete events systems; DES, in short). Similar algebras describing the work of DES's are based on other pairs of operations, such as max-min algebra, max-plus algebra, or max-*T* algebra (with a given t-norm, *T*). The investigation of the steady states in a DES leads to the study of the eigenvectors of the transition matrix in the corresponding max-algebra. In real systems, the input values are usually taken to be in some interval. Various types of interval eigenvectors of interval matrices in max-min and max-plus algebras have been described. This paper is oriented to the investigation of strong, strongly tolerable, and strongly universal interval eigenvectors in a max-Łuk algebra. The main method used in this paper is based on max-Ł linear combinations of matrices and vectors. Necessary and sufficient conditions for the recognition of strong, strongly tolerable, and strongly universal eigenvectors have been found. The theoretical results are illustrated by numerical examples.

**Keywords:** max-Łukasiewicz algebra; interval matrix; interval eigenvector; strong interval eigenvector

**MSC:** 90C15

### **1. Introduction**

A max-Łukasiewicz algebra (max-Łuk algebra, in short), is one of the so-called max-*T* fuzzy algebras, which are defined for various triangular norms.

A max-*T* fuzzy algebra contains values in the unit interval I = 0, 1 and uses the binary operation of maximum and one of the triangular norms, *T*, instead of the conventional operations of addition and multiplication. Thus, by a max-*T* fuzzy algebra we understand a triplet (I, ⊕, ⊗*T*), where I is the interval 0, 1 and ⊕ = max, ⊗*<sup>T</sup>* = *T* are binary operations on I. The symbol I(*m*, *n*), respectively, I(*n*), denotes the set of all matrices (respectively, vectors) of the given dimensions over I. The operations ⊕, ⊗*<sup>T</sup>* are extended to matrices and vectors in the standard way. The linear ordering on I induces partial orderings on I(*m*, *n*) and I(*n*).

The triangular norms (t-norms, in short) were introduced in [1], in the context of probabilistic metric spaces. The t-norms are interpreted as the conjunction in multi-valued fuzzy logics, or as the intersection of fuzzy sets. These functions are used in many fields, such as statistics and game theory, information and data fusion, decision making support, risk management, and probability theory. The t-norms (and the corresponding t-conorms) play an important role in fuzzy set theory. Many t-norms can be found in [2]).

The Łukasiewicz norm is often considered to be a logic of absolute (or metric) comparison. The Łukasiewicz conjunction is defined by

$$\mathbf{x} \otimes \mathbf{z} \text{ } y = \max\{\mathbf{x} + y - 1, 0\}. \tag{1}$$

The simplest norm is the Gödel norm, and the conjunction is defined as the minimum of the entries: the truth degrees of the constituents. Gödel logic is considered to be a logic of relative comparison.

$$\mathbf{x} \circledast\_{G} y = \min(\mathbf{x}, y) \tag{2}$$

In the particular case when *T* = min is the Gödel t-norm, we get an important max-min algebra which has useful applications to optimization and scheduling problems. Max-min algebras belong to the so-called tropical mathematics, with a wide scope of applications and interesting contributions to mathematical theory. Several monographs [3–6] and collections of papers [7–13] have dealt with tropical mathematics and related problems.

Tropical algebras can be naturally used for the study of systems working in discrete time (DES). The state of the system at time *t* is described by a vector *x*(*t*). The transitions of the system from one state to another are described by the transition matrix *A*. The next state *x*(*t* + 1) is obtained by multiplying the transition matrix and the state vector; in matrix notation we write *A* ⊗ *x*(*t*) = *x*(*t* + 1). When a DES reaches a steady state, after some time of operation, then the state vectors of the steady states are eigenvectors of *A*. In any tropical algebra, the eigenproblem for a given matrix *A* ∈ I(*n*, *n*) consists of finding an eigenvalue *λ* ∈ I and an eigenvector *x* ∈ I(*n*) fulfilling *A* ⊗ *x* = *λ* ⊗ *x*.

The eigenproblem in tropical algebra has been described in many papers, see [14]. Interesting results describing the structure of the eigenspace and several algorithms for computing the largest eigenvector of a given matrix have been published, for example, in [15,16]. The eigenvectors in a max-*T* algebra, for various triangular norms *T*, are useful in fuzzy set theory. Such eigenvectors have been studied in [17–19]. The eigenvalues and eigenvectors are interesting characteristics of the DES in fuzzy algebras. The eigenspace structures for the drastic and t-norm have been studied in [18,19]. Finally, [17] describes the case of Łukasiewicz fuzzy algebra.

#### **2. Strong Types of Interval Eigenvectors in Max-Łuk Algebras**

The investigation in this paper will be started by a simple numerical example.

**Example 1.** *(Numerical illustration: Steady state vector).*

*Assume*

$$A = \begin{pmatrix} 0.5 & 0.4 & 0.1 & 0.1 & 0.5\\ 0.8 & 0.8 & 0.2 & 0.1 & 0.7\\ 0.6 & 0.6 & 0.6 & 0.2 & 0\\ 0.4 & 0.5 & 0.5 & 0.2 & 0.4\\ 0.3 & 0.1 & 0.4 & 0 & 0.8 \end{pmatrix}, \qquad \mathbf{x} = \begin{pmatrix} 0.5\\ 0.7\\ 0.5\\ 0.4\\ 0.8 \end{pmatrix}, \qquad \boldsymbol{\lambda} = 0.8\dots$$

*Then*

$$A \otimes\_L \mathbf{x} = \begin{pmatrix} 0.3\\ 0.5\\ 0.3\\ 0.2\\ 0.6 \end{pmatrix} = 0.8 \otimes\_L \mathbf{x}.$$

*That is, x is a max-Łuk eigenvector of A with the eigenvalue λ* = 0.8*.*

In practical applications, the matrix entries usually are not exact numbers, but are contained in some intervals. Interval arithmetic is an efficient way to represent matrix operations on a computer. Similarly, matrices and vectors with interval coefficients are studied in a max-Łuk algebra (or a max-min algebra, or some other tropical algebra), see [9,17,20–23]. The classification of various types of the interval eigenvectors in a max-min algebra has been investigated in [24,25].

Let *n* be a given natural number. We define *N* = {1, 2, ... , *n*}. Similarly to [21,25–27], we define the interval matrix with bounds *A*, *A* ∈ I(*n*, *n*) and the interval vector with bounds *x*, *x* ∈ I(*n*) as

$$[\underline{\mathbf{A}}, \overline{\mathbf{A}}] = \left\{ A \in \mathcal{Z}(\mathfrak{n}, \mathfrak{n}) ; \,\underline{\mathbf{A}} \le A \le \overline{\mathbf{A}} \right\}, \quad [\underline{\mathbf{x}}, \overline{\mathbf{x}}] = \left\{ \mathbf{x} \in \mathcal{Z}(\mathfrak{n}) ; \,\underline{\mathbf{x}} \le \mathbf{x} \le \overline{\mathbf{x}} \right\}.$$

Let us assume that an interval matrix **A** = [*A*, *A*] and an interval vector **X** = [*x*, *x*] have been fixed. The interval max-Łuk eigenproblem for **A** and **X** aims at recognizing whether *A* ⊗*<sup>L</sup> x* = *λ* ⊗*<sup>L</sup> x* holds true for *A* ∈ **A**, *x* ∈ **X**, *λ* ∈ I, with suitable quantifiers (e.g., for all *A* ∈ **A**, for some *A* ∈ **A**, for all *x* ∈ **X**, for some *x* ∈ **X**) and their various combinations. Various types of interval max-Łuk eigenvectors are defined, using various choices of quantifiers and their order (see [25] for the further classification types).

**Definition 1.** *Assume that an interval matrix* **A** *and an interval vector* **X** *are given. Then,* **X** *is called:*


**Remark 1.** *In general, an interval vector* **X** *is called a* tolerable *max-Łuk eigenvector of* **A** *if there is an eigenvalue λ* ∈ I *such that every x* ∈ **X** *preserves the state vector up to a multiple by λ, for some A* ∈ **A** *(in other words: A* tolerates *x with eigenvalue λ).*

*In the case when there is* one common *tolerating matrix A* ∈ **A** *for all of the vectors x* ∈ **X***, the interval vector* **X** *is called* strongly tolerable*. Otherwise, the tolerating matrix A depends on x. If also* the eigenvalue *λ depends on x, then the interval eigenvector* **X** *is usually called* weakly tolerable*.*

**Remark 2.** *Similarly, an interval vector* **X** *is called a* universal *max-Łuk eigenvector of* **A** *if there is an eigenvalue λ* ∈ I *such that for every A* ∈ **A***, some x* ∈ **X** *preserves the state vector up to a multiple by λ, (in other words: x is* universal *for A with eigenvalue λ).*

*In the case when there is* one common *universal vector x* ∈ **X** *for all matrices A* ∈ **A***, the interval vector* **X***, as well as the common universal x, are called* strongly universal*. Otherwise, the universal vector x depends on A. If also* the eigenvalue *λ depends on A, then the interval eigenvector* **X** *is usually called* weakly universal*.*

In this paper, we study in more detail the strong max-Łuk interval eigenvectors, the strongly tolerable and strongly universal max-Łuk interval eigenvectors (the remaining types of max-Łuk interval eigenvector are not considered here). Necessary and sufficient conditions are described for recognizing whether a given interval vector is a strong (strongly tolerable, strongly universal) eigenvector of a given interval matrix in a max-Łuk algebra. The results are illustrated by numerical examples.

#### **3. Strong Interval Eigenvectors in a Max-Łuk Algebra**

In this section, we assume that an interval matrix **A** = [*A*, *A*] and an interval vector **X** = [*x*, *x*] are given. For each pair *<sup>i</sup>*, *<sup>j</sup>* <sup>∈</sup> *<sup>N</sup>*, define *<sup>A</sup>*˜(*ij*) ∈ I(*n*, *<sup>n</sup>*) and *<sup>x</sup>*˜(*i*) ∈ I(*n*) by putting for every *<sup>k</sup>*, *<sup>l</sup>* <sup>∈</sup> *<sup>N</sup>*,

$$
\mathfrak{a}\_{kl}^{(ij)} = \begin{cases}
\overline{\mathfrak{a}}\_{ij\prime} & \text{for } k=i, \, l=j \\
\underline{\mathfrak{g}}\_{kl\prime} & \text{otherwise}
\end{cases}, \qquad \mathfrak{x}\_{k}^{(i)} = \begin{cases}
\overline{\mathfrak{x}}\_{i\prime} & \text{for } k=i \\
\underline{\mathfrak{x}}\_{k\prime} & \text{otherwise}
\end{cases}.
$$

It is shown in the following lemma that every *A* ∈ **A** can be written as a max-Łuk linear combination of *generators <sup>A</sup>*˜(*ij*) with *<sup>i</sup>*, *<sup>j</sup>* <sup>∈</sup> *<sup>N</sup>*. Similarly, every *<sup>x</sup>* <sup>∈</sup> **<sup>X</sup>** is equal to a max-Łuk linear combination of generators *<sup>x</sup>*˜(*i*) with *<sup>i</sup>* <sup>∈</sup> *<sup>N</sup>*.

**Lemma 1.** *Let x* ∈ I(*n*) *and A* ∈ I(*n*, *n*)*. Then,*

(i) *x* ∈ **X** *if and only if x* = ( *<sup>β</sup><sup>i</sup>* <sup>⊗</sup>*<sup>L</sup> <sup>x</sup>*˜(*i*) *for some <sup>β</sup><sup>i</sup>* ∈ I *with xi* <sup>−</sup> *xi* <sup>+</sup> <sup>1</sup> <sup>≤</sup> *<sup>β</sup><sup>i</sup>* <sup>≤</sup> <sup>1</sup>*,*

$$\text{(iii)}\qquad A \in \mathbf{A} \text{ if and only if } A = \bigoplus\_{i,j \in N} \mathfrak{a}\_{i\bar{j}} \otimes\_L \bar{A}^{(i\bar{j})} \text{ for some } \mathfrak{a}\_{i\bar{j}} \in \mathbb{Z} \text{ with } \underline{a}\_{i\bar{j}} - \overline{a}\_{i\bar{j}} + 1 \le \mathfrak{a}\_{i\bar{j}} \le 1.$$

**Proof.** For the proof of statement (i), assume that *x* ∈ **X**: that is, *xi* ≤ *xi* ≤ *xi* for every *i* ∈ *N*. Put *β<sup>i</sup>* = *xi* − *xi* + 1 for each *i* ∈ *N*. It is easy to see that the *βi*'s satisfy the inequalities in assertion (i). Moreover, for every *j* ∈ *N*

$$\begin{split} \left( \bigoplus\_{i \in N} \bigwedge\_{j} \mathbb{R}\_{i} \otimes\_{L} \mathbb{R}^{(i)} \right)\_{j} &= \bigoplus\_{i \in N} \left( \left( \mathbf{x}\_{i} - \overline{\mathbf{x}}\_{i} + 1 \right) \otimes\_{L} \mathbb{R}^{(i)} \right)\_{j} \\ &= \left( \left( \mathbf{x}\_{j} - \overline{\mathbf{x}}\_{j} + 1 \right) \otimes\_{L} \mathbb{R}^{(j)}\_{j} \right) \oplus \bigoplus\_{i \in N \cup \{j\}} \left( \left( \mathbf{x}\_{i} - \overline{\mathbf{x}}\_{i} + 1 \right) \otimes\_{L} \mathbb{R}^{(i)}\_{j} \right) \\ &= \left( \left( \mathbf{x}\_{j} - \overline{\mathbf{x}}\_{j} + 1 \right) \otimes\_{L} \overline{\mathbf{x}}\_{j} \right) \oplus \bigoplus\_{i \in N \cup \{j\}} \left( \left( \mathbf{x}\_{i} - \overline{\mathbf{x}}\_{i} + 1 \right) \otimes\_{L} \underline{\mathbf{x}}\_{j} \right). \end{split} \tag{3}$$

In particular, (*xj* − *xj* + 1) ⊗*<sup>L</sup> xj* = (*xj* − *xj* + 1) + *xj* − 1 = *xj*, since *xj* ≥ 0. On the other hand, for *i* = *j* we have (*xi* − *xi* + 1) ⊗*<sup>L</sup> xj* = (*xi* − *xi* + 1) + *xj* − 1 ≤ *xj*, because *xi* − *xi* ≤ 0.

For the converse implication, assume that *xi* − *xi* + 1 ≤ *β<sup>i</sup>* ≤ 1 for every *i* ∈ *N* and *x* = ( *i*∈*N <sup>β</sup><sup>i</sup>* <sup>⊗</sup>*<sup>L</sup> <sup>x</sup>*˜(*i*). For every *j* ∈ *N*

$$\begin{split} \mathbf{x}\_{j} &= \left(\bigoplus\_{i \in N} \beta\_{i} \upharpoonright\_{\mathbb{O}} \mathbf{z}^{(i)}\right)\_{j} \leq \left(\bigoplus\_{i \in N} \mathbf{1} \ \mathbf{1} \ \mathbf{z}^{(i)}\right)\_{j} \\ &= \bigoplus\_{i \in N} \mathbf{x}\_{j}^{(i)} = \overline{\mathbf{x}\_{j}} \oplus \bigoplus\_{i \in N} \mathbf{(}\{j\} \ \underline{\mathbf{x}}\_{j} = \overline{\mathbf{x}}\_{j} \end{split} \tag{4}$$
 
$$\begin{split} \mathbf{x}\_{j} &= \left(\bigoplus\_{i \in N} \beta\_{i} \otimes\_{L} \widetilde{\mathbf{x}}^{(i)}\right)\_{j} \geq \bigoplus\_{i \in N} \left(\left(\underline{\mathbf{x}}\_{i} - \overline{\mathbf{x}}\_{i} + 1\right) \otimes\_{L} \widetilde{\mathbf{x}}^{(i)}\right)\_{j} \\ &= \left(\left(\underline{\mathbf{x}}\_{j} - \overline{\mathbf{x}}\_{j} + 1\right) + \overline{\mathbf{x}}\_{j} - 1\right) \oplus \bigoplus\_{i \in N} \left(\left(\underline{\mathbf{x}}\_{i} - \overline{\mathbf{x}}\_{i} + 1\right) + \underline{\mathbf{x}}\_{j} - 1\right) \\ &\geq \underline{\mathbf{x}}\_{j} \oplus \bigoplus\_{i \in N} \mathbf{(}\mathbf{(}\mathbf{j}] \ \left(\underline{\mathbf{x}}\_{i} - \overline{\mathbf{x}}\_{i} + \underline{\mathbf{x}}\_{j}\right) = \underline{\mathbf{x}}\_{j} . \end{split} \tag{5}$$

We have shown that *x* ≤ *x* ≤ *x*. That is, *x* ∈ [*x*, *x*]. The proof of (ii) is analogous.

**Theorem 1.** *The interval vector* **X** = [*x*, *x*] *is a strong max-Łuk eigenvector of the interval matrix* **A** = [*A*, *A*] *if and only if there exists λ* ∈ I, *λ* > 0, *such that for every i* ∈ *N*.

$$
\underline{\mathbf{A}} \otimes\_L \mathfrak{x}^{(i)} = \lambda \otimes\_L \mathfrak{x}^{(i)},\tag{6}
$$

$$
\overline{\mathcal{A}} \otimes\_L \tilde{\mathfrak{x}}^{(i)} = \lambda \otimes\_L \tilde{\mathfrak{x}}^{(i)}.\tag{7}
$$

**Proof.** Assume that *λ* ∈ I fulfills conditions (6) and (7), and that *x* ∈ **X** is given. Then *x* is a max-Łuk linear combination *x* = ( *<sup>i</sup>*∈*<sup>N</sup> <sup>β</sup><sup>i</sup>* <sup>⊗</sup>*<sup>L</sup> <sup>x</sup>*˜(*i*) for some coefficients *<sup>β</sup><sup>i</sup>* ∈ I, *<sup>i</sup>* <sup>∈</sup> *<sup>N</sup>* with *xi* <sup>−</sup> *xi* <sup>+</sup> <sup>1</sup> <sup>≤</sup> *<sup>β</sup><sup>i</sup>* <sup>≤</sup> 1, according to Lemma 1(i). In view of (6) we get

$$\begin{split} \underline{\mathbf{A}} \otimes\_{\boldsymbol{L}} \underline{\mathbf{x}} &= \underline{\mathbf{A}} \otimes\_{\boldsymbol{L}} \left( \bigoplus\_{i \in N} \beta\_{i} \otimes\_{\boldsymbol{L}} \tilde{\mathbf{x}}^{(i)} \right) = \bigoplus\_{i \in N} \left( \underline{\mathbf{A}} \otimes\_{\boldsymbol{L}} \beta\_{i} \otimes\_{\boldsymbol{L}} \tilde{\mathbf{x}}^{(i)} \right) \\ &= \bigoplus\_{i \in N} \beta\_{i} \otimes\_{\boldsymbol{L}} \left( \underline{\mathbf{A}} \otimes\_{\boldsymbol{L}} \tilde{\mathbf{x}}^{(i)} \right) = \bigoplus\_{i \in N} \beta\_{i} \otimes\_{\boldsymbol{L}} \left( \boldsymbol{\lambda} \otimes\_{\boldsymbol{L}} \tilde{\mathbf{x}}^{(i)} \right) \\ &= \boldsymbol{\lambda} \otimes\_{\boldsymbol{L}} \bigoplus\_{i \in N} \left( \beta\_{i} \otimes\_{\boldsymbol{L}} \tilde{\mathbf{x}}^{(i)} \right) = \boldsymbol{\lambda} \otimes\_{\boldsymbol{L}} \mathbf{x}. \end{split} \tag{8}$$

Using (7) we analogously get

$$\Delta \otimes\_L \mathbf{x} = \overline{A} \otimes\_L \left( \bigoplus\_{i \in N} \beta\_i \otimes\_L \mathfrak{x}^{(i)} \right) = \lambda \otimes\_L \bigoplus\_{i \in N} \left( \beta\_i \otimes\_L \mathfrak{x}^{(i)} \right) = \lambda \otimes\_L \mathbf{x}. \tag{9}$$

From (8) and (9) it easily follows that

$$
\lambda \otimes\_L \mathbf{x} = \underline{\mathbf{A}} \otimes\_L \mathbf{x} \le A \otimes\_L \mathbf{x} \le \overline{A} \otimes\_L \mathbf{x} = \lambda \otimes\_L \mathbf{x}, \tag{10}
$$

$$
A \otimes\_L \mathbf{x} = \lambda \otimes\_L \mathbf{x},
$$

for every *A* ∈ **A**. That is, **X** is a strong max-Łuk eigenvector of **A**. The converse implication is trivial.

**Example 2.** *(Numerical illustration: Strong max-Luk eigenvector)*

*Assume lower and upper bounds for A* ∈ [*A*, *A*] *and for x* ∈ [*x*, *x*]

$$
\underline{A} = \begin{pmatrix} 0.6 & 0.2 & 0.1 & 0.1 & 0.5 \\ 0.2 & 0.2 & 0.1 & 0 & 0.4 \\ 0.1 & 0.3 & 0.2 & 0.4 & 0.2 \\ 0.3 & 0.5 & 0.4 & 0.6 & 0.2 \\ 0 & 0.1 & 0.3 & 0 & 0.3 \end{pmatrix}, \qquad \overline{A} = \begin{pmatrix} 0.6 & 0.6 & 0.3 & 0.1 & 0.6 \\ 0.2 & 0.6 & 0.3 & 0.1 & 0.6 \\ 0.5 & 0.9 & 0.6 & 0.4 & 0.9 \\ 0.7 & 1 & 0.8 & 0.6 & 0.9 \\ 0.2 & 0.6 & 0.3 & 0.1 & 0.6 \end{pmatrix},
$$

$$
\underline{\underline{x}} = \begin{pmatrix} 0 \\ 0 \\ 0.7 \\ 0.7 \\ 0.9 \\ 0 \end{pmatrix}, \qquad \overline{\underline{x}} = \begin{pmatrix} 0.8 \\ 0.4 \\ 0.7 \\ 0.9 \\ 0.4 \end{pmatrix}.
$$

*Then*

$$\begin{aligned} \mathbf{x}^{(1)} &= \begin{pmatrix} 0.8\\ 0\\ 0.7\\ 0.9\\ 0 \end{pmatrix}, \quad \mathbf{\tilde{x}}^{(2)} = \begin{pmatrix} 0\\ 0.4\\ 0.7\\ 0.9\\ 0 \end{pmatrix}, \quad \mathbf{\tilde{x}}^{(3)} = \begin{pmatrix} 0\\ 0\\ 0.7\\ 0.9\\ 0 \end{pmatrix}, \quad \mathbf{\tilde{x}}^{(4)} = \begin{pmatrix} 0\\ 0\\ 0.7\\ 0.9\\ 0.4 \end{pmatrix}, \quad \mathbf{\tilde{x}}^{(5)} = \begin{pmatrix} 0\\ 0\\ 0.7\\ 0.9\\ 0.4 \end{pmatrix}. \end{aligned}$$

*The following equations hold for λ* = 0.6

*A* ⊗*<sup>L</sup> x*˜ (1) = ⎛ ⎜⎜⎜⎜⎜⎝ 0.4 0 0.3 0.5 0 ⎞ ⎟⎟⎟⎟⎟⎠ = *A* ⊗*<sup>L</sup> x*˜ (1) <sup>=</sup> 0.6 <sup>⊗</sup>*<sup>L</sup> <sup>x</sup>*˜ (1) , *A* ⊗*<sup>L</sup> x*˜ (2) = ⎛ ⎜⎜⎜⎜⎜⎝ 0 0 0.3 0.5 0 ⎞ ⎟⎟⎟⎟⎟⎠ = *A* ⊗*<sup>L</sup> x*˜ (2) <sup>=</sup> 0.6 <sup>⊗</sup>*<sup>L</sup> <sup>x</sup>*˜ (2) , *A* ⊗*<sup>L</sup> x*˜ (3) = ⎛ ⎜⎜⎜⎜⎜⎝ 0 0 0.3 0.5 0 ⎞ ⎟⎟⎟⎟⎟⎠ = *A* ⊗*<sup>L</sup> x*˜ (3) <sup>=</sup> 0.6 <sup>⊗</sup>*<sup>L</sup> <sup>x</sup>*˜ (3) , *A* ⊗*<sup>L</sup> x*˜ (4) = ⎛ ⎜⎜⎜⎜⎜⎝ 0 0 0.3 0.5 0 ⎞ ⎟⎟⎟⎟⎟⎠ = *A* ⊗*<sup>L</sup> x*˜ (4) <sup>=</sup> 0.6 <sup>⊗</sup>*<sup>L</sup> <sup>x</sup>*˜ (4) , *A* ⊗*<sup>L</sup> x*˜ (5) = ⎛ ⎜⎜⎜⎜⎜⎝ 0 0 0.3 0.5 0 ⎞ ⎟⎟⎟⎟⎟⎠ = *A* ⊗*<sup>L</sup> x*˜ (5) <sup>=</sup> 0.6 <sup>⊗</sup>*<sup>L</sup> <sup>x</sup>*˜ (5) .

*Hence,* **X** = [*x*, *x*] *is a strong max-Łuk eigenvector of A with the eigenvalue λ* = 0.6*.*

Theorem 1 leads to the following recognition problem: given **A** and **X**, recognize whether there is, or is no value *λ* ∈I\{0} such that (6) and (7) hold for every *k* ∈ *N*. If the answer is positive, then find all (or at least one) such values.

If *i*, *k* ∈ *N*, then we write, for brevity,

$$z\_{ik} = \left(\varDelta \otimes\_L \mathfrak{x}^{(i)}\right)\_k \\ \quad\tag{11}$$

$$z'\_{ik} = \left(\varDelta \otimes\_L \mathfrak{x}^{(i)}\right)\_k.$$

Furthermore, we write

$$Z^0 = \{(i, k) \in N \times N \colon z\_{ik} = 0\} \tag{12}$$

$$Z^{\triangleright} = \{ (i,k) \in N \times N \colon z\_{ik} > 0 \}\tag{13}$$

$$I = \left( 0, \min\_{(i,k)\in N\times N} \left( 1 - \mathfrak{x}\_k^{(i)} \right) \right) \tag{14}$$

**Theorem 2.** *The interval vector* **X** = [*x*, *x*] *is a strong max-Łuk eigenvector of the interval matrix* **A** = [*A*, *A*] *if and only if*


$$\text{(iii)}\quad \left(\forall (i,k), (j,l)\in Z^{>}\right) \quad \tilde{\mathbf{x}}\_{k}^{(i)} - z\_{ik} = \tilde{\mathbf{x}}\_{l}^{(j)} - z\_{jl} < 1,$$

$$\text{(iv)}\quad (\forall (i,k)\in Z^{>}\text{, }(j,l)\in Z^{0})\quad \mathfrak{x}\_{l}^{(j)}\leq \mathfrak{x}\_{k}^{(i)}-z\_{ik}.$$

**Proof.** Assume that **X** is a strong interval eigenvector of **A**. That is, there exists *λ* ∈ I, *λ* > 0 fulfilling conditions (6) and (7). The statement (i) then follows immediately. For (*i*, *<sup>k</sup>*) <sup>∈</sup> *<sup>Z</sup>*<sup>0</sup> we have *zik* <sup>=</sup> 0, which gives *λ* ⊗*<sup>L</sup> x*˜ (*i*) *<sup>k</sup>* = 0, in view of (6). Then, by definition of ⊗*L*, we have *λ* + *x*˜ (*i*) *<sup>k</sup>* − 1 ≤ 0, which implies *λ* ≤ 1 − *x*˜ (*i*) *<sup>k</sup>* . Now, statement (ii) easily follows, in view of the assumption that 0 < *λ*.

For (*i*, *<sup>k</sup>*),(*j*, *<sup>l</sup>*) <sup>∈</sup> *<sup>Z</sup>*>, we have *zik* <sup>&</sup>gt; 0, which gives *<sup>λ</sup>* <sup>⊗</sup>*<sup>L</sup> <sup>x</sup>*˜ (*i*) *<sup>k</sup>* > 0, in view of (6). Consequently, *zik* = *λ* + *x*˜ (*i*) *<sup>k</sup>* − 1. That is, *λ* = −*x*˜ (*i*) *<sup>k</sup>* + *zik* + 1. Similarly, *λ* = −*x*˜ (*j*) *<sup>l</sup>* + *zjl* + 1. In view of the assumption that *<sup>λ</sup>* <sup>&</sup>gt; 0, we get (iii) by a simple computation. Finally, assume (*i*, *<sup>k</sup>*) <sup>∈</sup> *<sup>Z</sup>*0,(*j*, *<sup>l</sup>*) <sup>∈</sup> *<sup>Z</sup>*>. By the same arguments as above, we get *λ* = −*x*˜ (*i*) *<sup>k</sup>* + *zik* + 1 ≤ 1 − *x*˜ (*j*) *<sup>l</sup>* . Then (iv) follows directly.

For the converse implication, assume that statements (i)–(iv) hold. We shall show that then a *λ* > 0 can be found such that (6) and (7) are satisfied. We distinguish two cases.

Case 1. *<sup>Z</sup>*<sup>&</sup>gt; <sup>=</sup> <sup>∅</sup>. Then *<sup>Z</sup>*<sup>0</sup> <sup>=</sup> *<sup>N</sup>* <sup>×</sup> *<sup>N</sup>*, and by (ii) we have 0 <sup>&</sup>lt; <sup>1</sup> <sup>−</sup> *<sup>x</sup>*˜ (*i*) *<sup>k</sup>* for every (*i*, *k*) ∈ *N* × *N*. That is, the interval *I* is non-empty. Choose an arbitrary *λ* ∈ *I*. Then, for every (*i*, *k*) ∈ *N* × *N*, we have *λ* ≤ 1 − *x*˜ (*i*) *<sup>k</sup>* which gives *λ* + *x*˜ (*i*) *<sup>k</sup>* − 1 ≤ 0. That is, *λ* ⊗*<sup>L</sup> x*˜ (*i*) *<sup>k</sup>* = 0 = *zik*. As (*i*, *k*) is arbitrary, (6) has been demonstrated. Then, (7) follows by statement (i) .

Case 2. *<sup>Z</sup>*<sup>&</sup>gt; <sup>=</sup> <sup>∅</sup>. Let (*i*, *<sup>k</sup>*) <sup>∈</sup> *<sup>Z</sup>*<sup>&</sup>gt; be fixed. By (iii) we have *<sup>x</sup>*˜ (*i*) *<sup>k</sup>* − *zik* < 1, which gives 0 < *zik* + 1 − *x*˜ (*i*) *<sup>k</sup>* . Choosing *λ* = *zik* + 1 − *x*˜ (*i*) *<sup>k</sup>* , we get *λ* > 0 and *zik* = *λ* + *x*˜ (*i*) *<sup>k</sup>* − 1. Then, the assumption that (*i*, *<sup>k</sup>*) <sup>∈</sup> *<sup>Z</sup>*<sup>&</sup>gt; implies *zik* <sup>&</sup>gt; 0 and *<sup>λ</sup>* <sup>+</sup> *<sup>x</sup>*˜ (*i*) *<sup>k</sup>* − 1 > 0. That is, *λ* + *x*˜ (*i*) *<sup>k</sup>* − 1 = *λ* ⊗*<sup>L</sup> x*˜ (*i*) *<sup>k</sup>* , which implies *zik* = *λ* ⊗*<sup>L</sup> x*˜ (*i*) *k* .

Consider an arbitrary (*j*, *<sup>l</sup>*) <sup>∈</sup> *<sup>Z</sup>*>. We have *<sup>λ</sup>* <sup>=</sup> *zik* <sup>+</sup> <sup>1</sup> <sup>−</sup> *<sup>x</sup>*˜ (*i*) *<sup>k</sup>* = *zjl* + 1 − *x*˜ (*j*) *<sup>l</sup>* , in view of (iii). That is, *zjl* = *λ* ⊗*<sup>L</sup> x*˜ (*j*) *<sup>l</sup>* , similarly as above. On the other hand, for every (*j*, *<sup>l</sup>*) <sup>∈</sup> *<sup>Z</sup>*<sup>0</sup> we have *<sup>x</sup>*˜ (*j*) *<sup>l</sup>* ≤ *x*˜ (*i*) *<sup>k</sup>* − *zik*, in view of (iv). Consequently, we get *zik* + 1 − *x*˜ (*i*) *<sup>k</sup>* ≤ 1 − *x*˜ (*j*) *<sup>l</sup>* . That is, *λ* ≤ 1 − *x*˜ (*j*) *<sup>l</sup>* , which gives *λ* + *x*˜ (*j*) *<sup>l</sup>* − 1 ≤ 0. This implies *λ* ⊗*<sup>L</sup> x*˜ (*j*) *<sup>l</sup>* = 0, i.e., *zjl* = *λ* ⊗*<sup>L</sup> x*˜ (*j*) *<sup>l</sup>* . As (*j*, *<sup>l</sup>*) <sup>∈</sup> *<sup>Z</sup>*<sup>&</sup>gt; <sup>∪</sup> *<sup>Z</sup>*<sup>0</sup> <sup>=</sup> *<sup>N</sup>* <sup>×</sup> *<sup>N</sup>* was arbitrary, we have shown that (6) is satisfied. By (i), (7) holds as well.

**Remark 3.** The proof of Theorem 2 contains a description of the set *S*(**A**, **X**) = *λ* > 0; (∀ *A* ∈ **A**)(∀ *x* ∈ **X**) *A* ⊗*<sup>L</sup> x* = *λ* ⊗*<sup>L</sup> x* . Namely


**Theorem 3.** *The recognition problem of whether a given interval vector* **X** *is a strong max-Łuk eigenvector of the interval matrix* **A** *is solvable in O*(*n*3) *time.*

**Proof.** According to Theorem 2, the problem can be solved by verifying conditions (i)–(iv). Each of them can be verified in *O* - *n*3 time. Therefore, the computational complexity is *O* - *n*3 .

#### **4. Strongly Tolerable Interval Eigenvectors in a Max-Łuk Algebra**

**Theorem 4.** *The interval vector* **X** = [**x**, **x**] *is a strongly tolerable max-Łuk eigenvector of the interval matrix* **A** = [*A*, *A*] *if and only if there exist an A* ∈ **A** *and λ* ∈ I *such that*

$$A \otimes\_L \tilde{\mathbf{x}}^{(k)} = \lambda \otimes\_L \tilde{\mathbf{x}}^{(k)} \quad \text{for every } k \in N. \tag{15}$$

**Proof.** Let us assume that *A* ∈ **A** and *λ* ∈ I fulfill condition (15). If *x* ∈ I(*n*) is an arbitrary vector in **X**, then *x* is a max-Łuk linear combination *x* = ( *<sup>k</sup>*∈*<sup>N</sup> <sup>β</sup><sup>k</sup>* <sup>⊗</sup>*<sup>L</sup> <sup>x</sup>*˜(*k*) for some coefficients *<sup>β</sup><sup>k</sup>* ∈ I, *<sup>k</sup>* <sup>∈</sup> *<sup>N</sup>* with *xk* − *xk* + 1 ≤ *β<sup>i</sup>* ≤ 1. According to Lemma 1 (i),

$$\begin{split} A \otimes\_{L} \mathbf{x} &= A \otimes\_{L} \left( \bigoplus\_{k \in N} \mathcal{J}\_{k} \otimes\_{L} \tilde{\mathbf{x}}^{(k)} \right) = \bigoplus\_{k \in N} \left( A \otimes\_{L} \mathcal{J}\_{k} \otimes\_{L} \tilde{\mathbf{x}}^{(k)} \right) \\ &= \bigoplus\_{k \in N} \mathcal{J}\_{k} \otimes\_{L} \left( A \otimes\_{L} \tilde{\mathbf{x}}^{(k)} \right) = \bigoplus\_{k \in N} \mathcal{J}\_{k} \otimes\_{L} \left( \lambda \otimes\_{L} \tilde{\mathbf{x}}^{(k)} \right) \\ &= \lambda \otimes\_{L} \bigoplus\_{k \in N} \left( \mathcal{J}\_{k} \otimes\_{L} \tilde{\mathbf{x}}^{(k)} \right) = \lambda \otimes\_{L} \mathbf{x}. \end{split} \tag{16}$$

By (3), **X** is a strongly tolerable eigenvector of **A**. The converse implication follows immediately.

**Remark 4.** *The property (15) can be briefly expressed in words: A is a λ*-certificate *for the strong tolerance max-Łuk problem* (**A**, **X**)*.*

**Example 3.** *(Numerical illustration: Strongly tolerable max-Luk eigenvector) Assume the lower and upper bounds for A* ∈ [*A*, *A*] *and for x* ∈ [*x*, *x*] *are*

$$
\underline{A} = \begin{pmatrix} 0.2 & 0.2 & 0.2 & 0.3 & 0 \\ 0.1 & 0.3 & 0.2 & 0.3 & 0 \\ 0 & 0.1 & 0.2 & 0.3 & 0.1 \\ 0.1 & 0.2 & 0.1 & 0.2 & 0.2 \\ 0.1 & 0.1 & 0.2 & 0.1 & 0.4 \end{pmatrix}, \qquad \overline{A} = \begin{pmatrix} 0.8 & 0.8 & 0.9 & 0.7 & 0.5 \\ 0.5 & 0.8 & 0.9 & 0.8 & 0.5 \\ 0.5 & 0.7 & 0.5 & 0.5 & 0.8 \\ 0.5 & 0.5 & 0.5 & 0.8 & 0.8 \\ 0.5 & 0.2 & 0.9 & 0.9 & 0.8 \end{pmatrix},
$$

$$
\underline{\mathbf{x}} = \begin{pmatrix} 0.6 \\ 0 \\ 0.6 \\ 0 \\ 0.7 \end{pmatrix}, \qquad \overline{\mathbf{x}} = \begin{pmatrix} 0.8 \\ 0.7 \\ 0.6 \\ 0.4 \\ 0.7 \end{pmatrix}.
$$

*Then*

$$\begin{aligned} \mathbf{\ddot{x}}^{(1)} &= \begin{pmatrix} 0.8\\ 0\\ 0.6\\ 0\\ 0.7 \end{pmatrix}, \quad \mathbf{\ddot{x}}^{(2)} = \begin{pmatrix} 0.6\\ 0.7\\ 0.6\\ 0\\ 0.7 \end{pmatrix}, \quad \mathbf{\ddot{x}}^{(3)} = \begin{pmatrix} 0.6\\ 0\\ 0.6\\ 0\\ 0.7 \end{pmatrix}, \quad \mathbf{\ddot{x}}^{(4)} = \begin{pmatrix} 0.6\\ 0\\ 0.6\\ 0.4\\ 0.7 \end{pmatrix}, \quad \mathbf{\ddot{x}}^{(5)} = \begin{pmatrix} 0.6\\ 0\\ 0.6\\ 0.7 \end{pmatrix}. \end{aligned}$$

*For λ* = 0.7 *and for given A*<sup>1</sup> ∈ **A**

$$A\_1 = \begin{pmatrix} 0.7 & 0.6 & 0.7 & 0.6 & 0.1 \\ 0.2 & 0.7 & 0.4 & 0.6 & 0.1 \\ 0.2 & 0.2 & 0.3 & 0.4 & 0.6 \\ 0.2 & 0.3 & 0.4 & 0.7 & 0.3 \\ 0.2 & 0.1 & 0.8 & 0.8 & 0.4 \end{pmatrix}$$

*the following equations hold*

$$A\_{1} \otimes\_{L} \tilde{\mathbf{x}}^{(1)} = \begin{pmatrix} 0.7 & 0.6 & 0.7 & 0.6 & 0.1 \\ 0.2 & 0.7 & 0.4 & 0.6 & 0.1 \\ 0.2 & 0.2 & 0.3 & 0.4 & 0.6 \\ 0.2 & 0.3 & 0.4 & 0.7 & 0.3 \\ 0.2 & 0.1 & 0.8 & 0.8 & 0.4 \end{pmatrix} \otimes \begin{pmatrix} 0.8 \\ 0 \\ 0.6 \\ 0 \\ 0.7 \end{pmatrix} = \begin{pmatrix} 0.5 \\ 0 \\ 0.3 \\ 0 \\ 0.4 \end{pmatrix} = 0.7 \otimes\_{L} \tilde{\mathbf{x}}^{(1)},$$

*A*<sup>1</sup> ⊗*<sup>L</sup> x*˜ (2) = ⎛ ⎜⎜⎜⎜⎜⎝ 0.7 0.6 0.7 0.6 0.1 0.2 0.7 0.4 0.6 0.1 0.2 0.2 0.3 0.4 0.6 0.2 0.3 0.4 0.7 0.3 0.2 0.1 0.8 0.8 0.4 ⎞ ⎟⎟⎟⎟⎟⎠ ⊗ ⎛ ⎜⎜⎜⎜⎜⎝ 0.6 0.7 0.6 0 0.7 ⎞ ⎟⎟⎟⎟⎟⎠ = ⎛ ⎜⎜⎜⎜⎜⎝ 0.3 0.4 0.3 0 0.4 ⎞ ⎟⎟⎟⎟⎟⎠ = 0.7 ⊗*<sup>L</sup> x*˜ (2) , *A*<sup>1</sup> ⊗*<sup>L</sup> x*˜ (3) = ⎛ ⎜⎜⎜⎜⎜⎝ 0.7 0.6 0.7 0.6 0.1 0.2 0.7 0.4 0.6 0.1 0.2 0.2 0.3 0.4 0.6 0.2 0.3 0.4 0.7 0.3 0.2 0.1 0.8 0.8 0.4 ⎞ ⎟⎟⎟⎟⎟⎠ ⊗ ⎛ ⎜⎜⎜⎜⎜⎝ 0.6 0 0.6 0 0.7 ⎞ ⎟⎟⎟⎟⎟⎠ = ⎛ ⎜⎜⎜⎜⎜⎝ 0.3 0 0.3 0 0.4 ⎞ ⎟⎟⎟⎟⎟⎠ = 0.7 ⊗*<sup>L</sup> x*˜ (3) , *A*<sup>1</sup> ⊗*<sup>L</sup> x*˜ (4) = ⎛ ⎜⎜⎜⎜⎜⎝ 0.7 0.6 0.7 0.6 0.1 0.2 0.7 0.4 0.6 0.1 0.2 0.2 0.3 0.4 0.6 0.2 0.3 0.4 0.7 0.3 0.2 0.1 0.8 0.8 0.4 ⎞ ⎟⎟⎟⎟⎟⎠ ⊗ ⎛ ⎜⎜⎜⎜⎜⎝ 0.6 0 0.6 0.4 0.7 ⎞ ⎟⎟⎟⎟⎟⎠ = ⎛ ⎜⎜⎜⎜⎜⎝ 0.3 0 0.3 0.1 0.4 ⎞ ⎟⎟⎟⎟⎟⎠ = 0.7 ⊗*<sup>L</sup> x*˜ (4) , *A*<sup>1</sup> ⊗*<sup>L</sup> x*˜ (5) = ⎛ ⎜⎜⎜⎜⎜⎝ 0.7 0.6 0.7 0.6 0.1 0.2 0.7 0.4 0.6 0.1 0.2 0.2 0.3 0.4 0.6 0.2 0.3 0.4 0.7 0.3 0.2 0.1 0.8 0.8 0.4 ⎞ ⎟⎟⎟⎟⎟⎠ ⊗ ⎛ ⎜⎜⎜⎜⎜⎝ 0.6 0 0.6 0 0.7 ⎞ ⎟⎟⎟⎟⎟⎠ = ⎛ ⎜⎜⎜⎜⎜⎝ 0.3 0 0.3 0 0.4 ⎞ ⎟⎟⎟⎟⎟⎠ = 0.7 ⊗*<sup>L</sup> x*˜ (5) ,

*Hence,* **X** = [*x*, *x*] *is a strongly tolerable max-Łuk eigenvector of* **A** *with the eigenvalue λ* = 0.7*, and A*<sup>1</sup> *is the λ-certificate for the strong tolerance max-Łuk problem* (**A**, **X**)*.*

**Remark 5.** *In general, not every matrix A* ∈ **A** *is a λ-certificate for* (**A**, **X**)*, for some λ. Take, e.g.,*

$$A\_2 = \begin{pmatrix} 0.8 & 0.8 & 0.9 & 0.7 & 0.5 \\ 0.5 & 0.8 & 0.9 & 0.8 & 0.5 \\ 0.5 & 0.7 & 0.5 & 0.5 & 0.8 \\ 0.5 & 0.5 & 0.5 & 0.8 & 0.8 \\ 0.5 & 0.2 & 0.9 & 0.9 & 0.8 \end{pmatrix}$$

*and x*˜(1)*, x*˜(2)*, x*˜(3)*, x*˜(4)*, x*˜(5) *from Example 3. Then*

$$A\_2 \otimes\_L \mathfrak{x}^{(1)} = A\_2 \otimes \begin{pmatrix} 0.8\\0\\0.6\\0\\0.7 \end{pmatrix} = \begin{pmatrix} 0\\0\\0\\0\\0.1 \end{pmatrix} \neq \lambda \otimes\_L \mathfrak{x}^{(1)}.$$

*It is easy to see that the equality in the last position cannot hold for any λ* ∈ I*. That is, A*<sup>2</sup> *is not a λ-certificate in Example 3 with any λ* ∈ I*.*

In Example 3, the certificate *A*<sup>1</sup> was given. Now the question arises of how to find a certificate (or to show that no certificate exists) for a given instance (**A**, **X**). In other words, how do we recognize whether or not **X** is a strongly tolerable interval eigenvector of **A**?

A method for solving the strong tolerance interval eigenproblem in a max-Łuk algebra for instances with a natural additional condition is described in the rest of this section. We start with a simple lemma.
