发表机构
The University of Jordan; Korea University; Catholic University of Rio de Janeiro; Ferdowsi University of Mashhad（约旦大学; 高丽大学; 里约热内卢天主教大学; 马什哈德费尔多西大学）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本文研究可逆正算子换位子的刻画，利用谱隙证明有界序列蕴含块对角结构，并推广至正规算子及投影加权和，同时给出正定矩阵为恒等正倍数的充要条件的简单证明。
AI 中文摘要
Radjabalipour 的一个重要结果断言：若 ${A\in\BB(\H)}$ 是可逆正算子且 ${T\in\BB(\H)}$ 是任意的，则算子序列 $\{A^{-n}TA^n\}_{n\in\ZZ}$ 有界当且仅当 ${TA=AT}.$ 利用谱隙的概念，我们证明若 $A$ 具有谱隙且序列 $\{A^{-n}TA^n\}_{n\in\ZZ}$ 有界，则 $T$ 必关于 $\H$ 的某个正交分解为块对角。我们还给出了 Radjabalipour 定理的另一种证明。我们将该定理推广到一些丰富的算子类，包括可逆正规算子和可逆的投影加权和。此外，我们为以下事实提供了简单证明：正定矩阵 ${A\in\MM_N}$ 是恒等算子的正倍数当且仅当对每个 ${T\in\MM_N}$，序列 $\{A^{-n}TA^n\}_{n\in\ZZ}$ 有界。
英文摘要
An important result due to Radjabalipour asserts that if ${A\in\BB(\H)}$ is an invertible positive operator and ${T\in\BB(\H)}$ is arbitrary, then the operator sequence $\{A^{-n}TA^n\}_{n\in\ZZ}$ is norm-bounded if and only if ${TA=AT}.$ Using the concept of spectral gaps, we show that if $A$ possesses a spectral gap and the sequence $\{A^{-n}TA^n\}_{n\in\ZZ}$ is norm-bounded, then $T$ must be block diagonal with respect to a certain orthogonal decomposition of $\H$. We also provide an alternative proof of Radjabalipour's theorem. We also extend the theorem to some rich classes of operators, including invertible normal operators and invertible weighted sums of projections. Moreover, we provide simple proofs for the fact that a positive definite matrix ${A\in\MM_N}$ is a positive multiple of the identity if and only if the sequence $\{A^{-n}TA^n\}_{n\in\ZZ}$ is norm-bounded for every~${T\in\MM_N}.$
Comments13 pages, to appear in Positivity
