Spaces of triangularizable matrices

IF 0.6 Q3 MATHEMATICS
Clément de Seguins Pazzis
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引用次数: 0

Abstract

Let \(\mathbb {F}\) be a field. We investigate the greatest possible dimension \(t_n(\mathbb {F})\) for a vector space of n-by-n matrices with entries in \(\mathbb {F}\) and in which every element is triangularizable over the ground field \(\mathbb {F}\). It is obvious that \(t_n(\mathbb {F}) \ge \frac{n(n+1)}{2}\), and we prove that equality holds if and only if \(\mathbb {F}\) is not quadratically closed or \(n=1\), excluding finite fields with characteristic 2. If \(\mathbb {F}\) is infinite and not quadratically closed, we give an explicit description of the solutions with the critical dimension \(t_n(\mathbb {F})\), reducing the problem to the one of deciding for which integers \(k \in \mathopen {[\![}2,n\mathclose {]\!]}\) all k-by-k symmetric matrices over \(\mathbb {F}\) are triangularizable.

可三角化矩阵的空间
让 \(\mathbb {F}\) 成为一个领域。我们研究最大可能的维度 \(t_n(\mathbb {F})\) 对于一个n × n矩阵的向量空间 \(\mathbb {F}\) 其中每个元素都可以在地面上三角化 \(\mathbb {F}\). 很明显, \(t_n(\mathbb {F}) \ge \frac{n(n+1)}{2}\),我们证明等式成立当且仅当 \(\mathbb {F}\) 是不是二次闭的 \(n=1\),排除特征为2的有限域。如果 \(\mathbb {F}\) 是无限且非二次闭的,我们给出了具有临界维数的解的显式描述 \(t_n(\mathbb {F})\),将问题简化为决定使用哪些整数 \(k \in \mathopen {[\![}2,n\mathclose {]\!]}\) 所有的k × k对称矩阵 \(\mathbb {F}\) 是可以三角化的。
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来源期刊
CiteScore
1.00
自引率
0.00%
发文量
39
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