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IRREDUCIBLE FAMILIES OF COMPLEX MATRICES CONTAINING A RANK-ONE MATRIX
Bulletin of the Australian Mathematical Society ( IF 0.6 ) Pub Date : 2020-01-16 , DOI: 10.1017/s0004972719001448 W. E. LONGSTAFF
Bulletin of the Australian Mathematical Society ( IF 0.6 ) Pub Date : 2020-01-16 , DOI: 10.1017/s0004972719001448 W. E. LONGSTAFF
We show that an irreducible family ${\mathcal{S}}$ of complex $n\times n$ matrices satisfies Paz’s conjecture if it contains a rank-one matrix. We next investigate properties of families of rank-one matrices. If ${\mathcal{R}}$ is a linearly independent, irreducible family of rank-one matrices then (i) ${\mathcal{R}}$ has length at most $n$ , (ii) if all pairwise products are nonzero, ${\mathcal{R}}$ has length 1 or 2, (iii) if ${\mathcal{R}}$ consists of elementary matrices, its minimum spanning length $M$ is the smallest integer $M$ such that every elementary matrix belongs to the set of words in ${\mathcal{R}}$ of length at most $M$ . Finally, for any integer $k$ dividing $n-1$ , there is an irreducible family of elementary matrices with length $k+1$ .
中文翻译:
包含秩一矩阵的复矩阵的不可约族
我们证明了一个不可约的家庭${\mathcal{S}}$ 复杂的$n\次 n$ 如果矩阵包含秩一矩阵,则矩阵满足 Paz 猜想。我们接下来研究一级矩阵族的性质。如果${\mathcal{R}}$ 是一个线性独立的、不可约的一阶矩阵族,则 (i)${\mathcal{R}}$ 最多有长度$n$ , (ii) 如果所有成对乘积均非零,${\mathcal{R}}$ 长度为 1 或 2,(iii) 如果${\mathcal{R}}$ 由基本矩阵组成,其最小跨越长度$M$ 是最小的整数$M$ 使得每个基本矩阵都属于${\mathcal{R}}$ 最多长度$M$ . 最后,对于任何整数$k$ 划分$n-1$ ,有一个不可约的基本矩阵族,其长度为$k+1$ .
更新日期:2020-01-16
中文翻译:
包含秩一矩阵的复矩阵的不可约族
我们证明了一个不可约的家庭