Makoto Suwama | Searcharxiv

arXiv2021

On Trace Zero Matrices and Commutators

Given any commutative ring $R$, a commutator of two $n\times n$ matrices over $R$ has trace $0$. In this paper, we study the converse: whether every $n \times n$ trace $0$ matrix is a commutator. We show that if $R$ is a Bézout domain with algebraically closed quotient field, then every $n\times n$ trace $0$ matrix is a commutator. We also show that if $R$ is a regular ring with large enough Krull dimension relative to $n$, then there exist a $n\times n$ trace $0$ matrix that is not a commutator. This improves on a result of Lissner by increasing the size of the matrix allowed for a fixed $R$. We also give an example of a Noetherian dimension $1$ commutative domain $R$ that admits a $n\times n$ trace $0$ non-commutator for any $n\ge 2$.

math.RA↗

arXiv2021

Integer Dynamics

Let $b \geq 2$ be an integer, and write the base $b$ expansion of any non-negative integer $n$ as $n=x_0+x_1b+\dots+ x_{d}b^{d}$, with $x_d>0$ and $ 0 \leq x_i < b$ for $i=0,\dots,d$. Let $ϕ(x)$ denote an integer polynomial such that $ϕ(n) >0$ for all $n>0$. Consider the map $S_{ϕ,b}: {\mathbb Z}_{\geq 0} \to {\mathbb Z}_{\geq 0}$, with $ S_{ϕ,b}(n) := ϕ(x_0)+ \dots + ϕ(x_d)$. It is known that the orbit set $\{n,S_{ϕ,b}(n), S_{ϕ,b}(S_{ϕ,b}(n)), \dots \}$ is finite for all $n>0$. Each orbit contains a finite cycle, and for a given $b$, the union of such cycles over all orbit sets is finite. Fix now an integer $\ell\geq 1$ and let $ϕ(x)=x^2$. We show that the set of bases $b\geq 2$ which have at least one cycle of length $\ell$ always contains an arithmetic progression and thus has positive lower density. We also show that a 1978 conjecture of Hasse and Prichett on the set of bases with exactly two cycles needs to be modified, raising the possibility that this set might not be finite.

math.NT↗