Iranian Mathematical Society Journal of the Iranian Mathematical Society 2717-1612 4 1 2023 01 01 Commutators and hyponormal operators on a Hilbert space 67 78 172367 10.30504/jims.2023.393155.1106 EN M. Akhmadiev Kazan National Research Technological University, Kazan, Russia H. Alhasan Department of Mathematics and Mechanics, Kazan Federal University, Kazan, Russia A. Bikchentaev Department of Mathematics and Mechanics, Kazan Federal University, Kazan, Russia P. Ivanshin Department of Mathematics and Mechanics, Kazan Federal University, Kazan, Russia Journal Article 2023 04 14 Let $\mathcal{H}$ be an infinite-dimensional Hilbert space over the field $\mathbb{C}$, $\mathcal{B}(\mathcal{H})$ be the $\ast$-algebra of all linear bounded operators on $\mathcal{H}$, let $|X|=\sqrt{X^*X}$ for $X\in \mathcal{B}(\mathcal{H})$. An operator $A\in \mathcal{B}(\mathcal{H})$ is a commutator if $A=[S, T]=ST-TS$ for some $S, T\in \mathcal{B}(\mathcal{H})$. Let $X, Y \in \mathcal{B}(\mathcal{H})$ and $X\geq 0$. If the operator $XY$ is a non-commutator, then $X^pYX^{1-p}$ is a non-commutator for every $0<p<1$. Let $A \in \mathcal{B}(\mathcal{H})$ be $p$-hyponormal for some $0<p\leq 1$. If $|A^*|^r$ is a non-commutator for some $r>0$ then $|A|^q$ is a non-commutator <br />for every $q>0$. Let $\mathcal{H}$ be separable and $A \in \mathcal{B}(\mathcal{H})$ be a non-commutator. If $A$ is hyponormal (or cohyponormal) then $A$ is normal. We also present results in the case of a finite-dimensional Hilbert space. https://jims.ims.ir/article_172367_a9842b3b7e517ce205d1dd371c5e83b0.pdf