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On the $$\lambda$$-robustness of matrices over fuzzy algebra. (English) Zbl 1225.15027
Let $$(B,\leq)$$ be a non-empty, bounded, linearly ordered set. Define the operations $$a\oplus b=\max\{a,b\}$$ and $$a\otimes b=\min\{a,b\}$$ for $$a,b\in B$$. Let $$A=[a_{ij}]_{n\times n}$$ be a square matrix with coefficients in $$B$$. A column vector $$x\in B^n$$ is said to be a $$\lambda$$-eigenvector of $$A$$ for some $$\lambda\in B$$ if $$A\otimes x=\lambda\otimes x$$.
The matrix $$A$$ is called $$\lambda$$-robust if for every $$x\in B^n$$ the vector $$A^k\otimes x$$ is a $$\lambda$$-eigenvector of $$A$$ for some $$k\in{\mathbb Z}^+$$. Let $$V(A,\lambda)$$ denote the set of all $$\lambda$$-eigenvectors of $$A$$. The authors show that: When $$\lambda\geq\max\{ a_{ij}: 1\leq i,j\leq n\}$$, $$A$$ is $$\lambda$$-robust if and only if $$V(A,\lambda)=V(A^{\ell},\lambda)$$ for each $$\ell\in{\mathbb Z}^+$$. Note that $$V(A^{\ell},\lambda)=V(A^{\ell},I)$$ for $$\ell\in{\mathbb Z}^+$$ whenever $$\lambda\geq\max\{ a_{ij}: 1\leq i,j\leq n\}$$. An $$O(n^3)$$ time algorithm exists to decide whether $$A$$ is $$\lambda$$-robust.
Let $$M(A)$$ denote the set of all vectors $$x=[x_i]_{n\times 1}\in B^n$$ with each $$x_i<c(A)$$ for $$c(A)=\bigotimes_{i=1}^n\left(\bigoplus_{j=1}^n a_{ij}\right)$$. The matrix $$A$$ is called strongly $$\lambda$$-robust if for every $$x\in B^n\backslash M(A)$$ the vector $$A^k\otimes x$$ is the greatest $$\lambda$$-eigenvector $$\bigoplus_{y\in V(A,\lambda)}y$$ of $$A$$ for some $$k\in{\mathbb Z}^+$$. A main result of the paper gives equivalent conditions for $$A$$ being strongly $$\lambda$$-robust when $$\lambda > c(A)$$. Details are too involved to describe here. Basing on this, an $$O(n^3)$$ algorithm is introduced to decide whether $$A$$ is strongly $$\lambda$$-robust.

##### MSC:
 15B15 Fuzzy matrices 15A18 Eigenvalues, singular values, and eigenvectors 65F30 Other matrix algorithms (MSC2010)
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