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Operators which are polynomially isometric to a normal operator

Research paper by Laurent Marcoux, Yuanhang Zhang

Indexed on: 05 Sep '20Published on: 19 Aug '19Published in: arXiv - Mathematics - Functional Analysis



Abstract

Let $\mathcal{H}$ be a complex, separable Hilbert space and $\mathcal{B}(\mathcal{H})$ denote the algebra of all bounded linear operators acting on $\mathcal{H}$. Given a unitarily-invariant norm $\| \cdot \|_u$ on $\mathcal{B}(\mathcal{H})$ and two linear operators $A$ and $B$ in $\mathcal{B}(\mathcal{H})$, we shall say that $A$ and $B$ are \emph{polynomially isometric relative to} $\| \cdot \|_u$ if $\| p(A) \|_u = \| p(B) \|_u$ for all polynomials $p$. In this paper, we examine to what extent an operator $A$ being polynomially isometric to a normal operator $N$ implies that $A$ is itself normal. More explicitly, we first show that if $\| \cdot \|_u$ is any unitarily-invariant norm on $\mathbb{M}_n(\mathbb{C})$, if $A, N \in \mathbb{M}_n(\mathbb{C})$ are polynomially isometric and $N$ is normal, then $A$ is normal. We then extend this result to the infinite-dimensional setting by showing that if $A, N \in \mathcal{B}(\mathcal{H})$ are polynomially isometric relative to the operator norm and $N$ is a normal operator whose spectrum neither disconnects the plane nor has interior, then $A$ is normal, while if the spectrum of $N$ is not of this form, then there always exists a non-normal operator $B$ such that $B$ and $N$ are polynomially isometric. Finally, we show that if $A$ and $N$ are compact operators with $N$ normal, and if $A$ and $N$ are polynomially isometric with respect to the $(c,p)$-norm studied by Chan, Li and Tu, then $A$ is again normal.