{"paper":{"title":"On completion of a linearly independent set to a basis with shifts of a fixed vector","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":["math.AG"],"primary_cat":"math.RA","authors_text":"Marek Rychlik","submitted_at":"2019-05-27T17:35:26Z","abstract_excerpt":"Let $\\mathbb{F}$ be an infinite field. Let $n$ be a positive integer and let $1\\leq d\\leq n$. Let $\\vec{f}_1, \\vec{f}_2, \\ldots, \\vec{f}_{d-1} \\in \\mathbb{F}^{n}$ be $d-1$ linearly independent vectors. Let $\\vec{x}=(x_1,x_2,\\ldots,x_{d},0,0,\\ldots,0)\\in\\mathbb{F}^{n}$, with $n-d$ zeros at the end. Let $\\vec{R}: \\mathbb{F}^n \\to\\mathbb{F}^n$ be the cyclic shift operator to the right, e.g. $\\vec{R}\\,\\vec{x} = (0,x_1,x_2,\\ldots,x_{d},0,0,\\ldots,0)$. Is there a vector $\\vec{x} \\in \\mathbb{F}^n$, such that the $n-d+1$ vectors $\\vec{x},\\vec{R}\\vec{x}, \\ldots ,\\vec{R}^{n-d}\\vec{x}$ complete the set $"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1905.11812","kind":"arxiv","version":1},"verdict":{"id":null,"model_set":{},"created_at":null,"strongest_claim":"","one_line_summary":"","pipeline_version":null,"weakest_assumption":"","pith_extraction_headline":""},"references":{"count":0,"sample":[],"resolved_work":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57","internal_anchors":0},"formal_canon":{"evidence_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"author_claims":{"count":0,"strong_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"builder_version":"pith-number-builder-2026-05-17-v1"}