矩阵张量积的数值半径
Numerical Radii for Tensor Products of Matrices
摘要：
对于$n$ × $n$和$m$ × $m$复矩阵$A$和$B$，不等式$w(A\otimes B)\le\|A\|w(B)$成立，其中$w(\cdot)$和$\|\cdot\|$分别表示矩阵的数值半径和算子范数.在本文中，我们考虑当这成为一个平等。我们证明了：（1）若$\|A\|=1$和$w(A\otimes B)=w(B)$，则要么$A$有幺正部分，要么$A$是完全非幺正的，并且$B$的数值域$W(B)$是以原点为中心的圆盘;（2）若$\|A\|=\|A^k\|=1$对某些$k$，$1\le k<\infty$，则$w(A)\ge\cos(π/(k+2))$，而且，等式成立当且仅当$A$酉类似于$(k+1)$ × $(k+1)$ Jordan块$J_{k+1}$和具有$w(B)\le\cos(π/(k+2))$的矩阵$B$的直和，（3）若$B$是实部（置换）不可约的非负矩阵，则$w(A\otimes B)=\|A\|w(B)$当且仅当$p_A=\infty$或$n_B\le p_A<\infty$和$B$置换相似于具有$k=n_B$的块移位矩阵\[[ {array}{cccc} 0 & B_1 & & & 0 & \ddots & & & \ddots & B_k & & & 0 {array} ]\]，其中$p_A=\sup\{\ell\ge 1: \|A^{\ell}\|=\|A\|^{\ell}\}$和$n_B=\sup\{\ell\ge 1 : B^{\ell}\neq 0\}$ .
英文摘要：
For $n$-by-$n$ and $m$-by-$m$ complex matrices $A$ and $B$, it is known that the inequality $w(A\otimes B)\le\|A\|w(B)$ holds, where $w(\cdot)$ and $\|\cdot\|$ denote, respectively, the numerical radius and the operator norm of a matrix. In this paper, we consider when this becomes an equality. We show that (1) if $\|A\|=1$ and $w(A\otimes B)=w(B)$, then either $A$ has a unitary part or $A$ is completely nonunitary and the numerical range $W(B)$ of $B$ is a circular disc centered at the origin, (2) if $\|A\|=\|A^k\|=1$ for some $k$, $1\le k<\infty$, then $w(A)\ge\cos(π/(k+2))$, and, moreover, the equality holds if and only if $A$ is unitarily similar to the direct sum of the $(k+1)$-by-$(k+1)$ Jordan block $J_{k+1}$ and a matrix $B$ with $w(B)\le\cos(π/(k+2))$, and (3) if $B$ is a nonnegative matrix with its real part (permutationally) irreducible, then $w(A\otimes B)=\|A\|w(B)$ if and only if either $p_A=\infty$ or $n_B\le p_A<\infty$ and $B$ is permutationally similar to a block-shift matrix \[[ {array}{cccc} 0 & B_1 & & & 0 & \ddots & & & \ddots & B_k & & & 0 {array} ]\] with $k=n_B$, where $p_A=\sup\{\ell\ge 1: \|A^{\ell}\|=\|A\|^{\ell}\}$ and $n_B=\sup\{\ell\ge 1 : B^{\ell}\neq 0\}$.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
