Let <InlineEquation ID="IEq1"> <EquationSource Format="TEX">$\mathbb {F}$</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="double-struck">F</mi> </math></EquationSource> </InlineEquation> be a field. We investigate the greatest possible dimension <InlineEquation ID="IEq2"> <EquationSource Format="TEX">$t_n(\mathbb {F})$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>t</mi> <mi>n</mi> </msub> <mrow> <mo stretchy="false">(</mo> <mi mathvariant="double-struck">F</mi> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation> for a vector space of n-by-n matrices with entries in <InlineEquation ID="IEq3"> <EquationSource Format="TEX">$\mathbb {F}$</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="double-struck">F</mi> </math></EquationSource> </InlineEquation> and in which every element is triangularizable over the ground field <InlineEquation ID="IEq4"> <EquationSource Format="TEX">$\mathbb {F}$</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="double-struck">F</mi> </math></EquationSource> </InlineEquation>. It is obvious that <InlineEquation ID="IEq5"> <EquationSource Format="TEX">$t_n(\mathbb {F}) \ge \frac{n(n+1)}{2}$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>t</mi> <mi>n</mi> </msub> <mrow> <mo stretchy="false">(</mo> <mi mathvariant="double-struck">F</mi> <mo stretchy="false">)</mo> </mrow> <mo>≥</mo> <mfrac> <mrow> <mi>n</mi> <mo stretchy="false">(</mo> <mi>n</mi> <mo>+</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> <mn>2</mn> </mfrac> </mrow> </math></EquationSource> </InlineEquation>, and we prove that equality holds if and only if <InlineEquation ID="IEq6"> <EquationSource Format="TEX">$\mathbb {F}$</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="double-struck">F</mi> </math></EquationSource> </InlineEquation> is not quadratically closed or <InlineEquation ID="IEq7"> <EquationSource Format="TEX">$n=1$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>=</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>, excluding finite fields with characteristic 2. If <InlineEquation ID="IEq8"> <EquationSource Format="TEX">$\mathbb {F}$</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="double-struck">F</mi> </math></EquationSource> </InlineEquation> is infinite and not quadratically closed, we give an explicit description of the solutions with the critical dimension <InlineEquation ID="IEq9"> <EquationSource Format="TEX">$t_n(\mathbb {F})$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>t</mi> <mi>n</mi> </msub> <mrow> <mo stretchy="false">(</mo> <mi mathvariant="double-struck">F</mi> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation>, reducing the problem to the one of deciding for which integers <InlineEquation ID="IEq10"> <EquationSource Format="TEX">$k \in \mathopen {[\![}2,n\mathclose {]\!]}$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>k</mi> <mo>∈</mo> <mrow> <mo stretchy="false">[</mo> <mspace width="-0.166667em" /> <mo stretchy="false">[</mo> </mrow> <mn>2</mn> <mo>,</mo> <mi>n</mi> <mrow> <mo stretchy="false">]</mo> <mspace width="-0.166667em" /> <mo stretchy="false">]</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation> all k-by-k symmetric matrices over <InlineEquation ID="IEq11"> <EquationSource Format="TEX">$\mathbb {F}$</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="double-struck">F</mi> </math></EquationSource> </InlineEquation> are triangularizable.

Spaces of triangularizable matrices

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2025年7月24日

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Clément de Seguins Pazzis

摘要

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