Atkinson-Lloyd conjecture for primitively intransitive spaces of linear operators
Let and be finite-dimensional vector spaces over a field . Let be a linear subspace of . We say that it is intransitive when for all , and we say that it is primitively intransitive when, in addition, there is no proper linear subspace of such that is an intransitive subspace of , where stands for the canonical projection.
Atkinson and Lloyd have proved that if and is primitively intransitive then . The conjecture states that the result holds without the cardinality assumption on .
Progress summary
The original bound is known for sufficiently large fields, while a recent paper appears to extend related results but does not clearly settle the conjecture for every field.
Atkinson and Lloyd conjectured that every primitively intransitive operator space satisfies the stated dimension bound without any restriction on the field. Their published theorem imposed the hypothesis .
Known results
- Atkinson and Lloyd: if , then .
Recent related result, reported September 2026
The paper Spaces of matrices with few eigenvalues (II) states an Atkinson-type theorem for primitively intransitive spaces and says its main results hold beyond the usual field-cardinality assumptions, apart from the cases . The retrieved text does not explicitly identify this as a proof of the exact Atkinson–Lloyd conjecture, so this is claimed progress rather than a verified resolution.
Current status (as of September 2026): The conjecture is proved under , while its unrestricted formulation remains open; a recent paper may cover related cases for but does not explicitly establish this exact statement.
Sources
- dsp.prod.free.fr
- arxiv.org
- arxiv.org
- cambridge.org
- arxiv.org
- mathoverflow.net
- quantamagazine.org
- byjus.com
- math.stackexchange.com
- quantamagazine.org
- quantamagazine.org
- quantamagazine.org
- ar5iv.labs.arxiv.org
- ar5iv.labs.arxiv.org
- ar5iv.labs.arxiv.org
- ar5iv.labs.arxiv.org
- mathstodon.xyz
- mathstodon.xyz
- mathstodon.xyz
- mathstodon.xyz
Solutions 0
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