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The uniform normal form of a linear mapping
Symplectic Geometry · Submitted 27 May 2014
abstract compiled · 2026-09-23 · no language-model review
The authors' main claim This paper gives a normal form for $A$, which gives a better description of the structure of $A$ than the companion matrix.
The compiled record
The paper's own sentences, arranged. Each label is drawn from a closed vocabulary; the number beside it is the probability the classifier gave that label. Numbers in the tables are the authors', checked by code. No sentence in this record was written by a model.
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The abstract, sentence by sentence
Let $V$ be a finite dimensional vector space over a field $\mathrm{k}$ of characteristic $0$. Let $A$ be a linear mapping of $V$ into itself. This paper gives a normal form for $A$, which gives a better description of the structure of $A$ than the companion matrix. The computation of this normal form uses only operations from $\mathrm{k}$ and does not require finding roots of any polynomial.
background contribution method result interpretation main claim
| # | sentence | role | the authors' own work | states the main conclusion | strength |
|---|---|---|---|---|---|
| s0 | Let $V$ be a finite dimensional vector space over a field $\mathrm{k}$ of characteristic $0$. | background 1.00 | 0.06 | 0.02 | asserted 0.82 |
| s1 | Let $A$ be a linear mapping of $V$ into itself. | background 1.00 | 0.11 | 0.02 | asserted 0.73 |
| s2 | This paper gives a normal form for $A$, which gives a better description of the structure of $A$ than the companion matrix. | contribution 0.92 | 0.93 | 0.86 | asserted 1.00 |
| s3 | The computation of this normal form uses only operations from $\mathrm{k}$ and does not require finding roots of any polynomial. | method 0.95 | 0.85 | 0.12 | asserted 1.00 |
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The paper
Original abstract
Let $V$ be a finite dimensional vector space over a field $\mathrm{k}$ of characteristic $0$. Let $A$ be a linear mapping of $V$ into itself. This paper gives a normal form for $A$, which gives a better description of the structure of $A$ than the companion matrix. The computation of this normal form uses only operations from $\mathrm{k}$ and does not require finding roots of any polynomial.
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