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Suppose:

  • $B_i \in \mathbb{C}^{n \times n}$, $0<w_i\in \mathbb{R}$ $(i = 0,1,2,\ldots,m)$

  • ${\rm P}(x) ={\rm{B}_m} x ^m + \cdots + B_1 x + B_0$ is a matrix polynomial, and $x $ is a complex variable.

  • ${\rm q}(x) =\rm{w}_m x^m + \cdots +w_1 x + w_0$

  • $U= \left\{ {x \in \mathbb{C}:\left\| P(x)^{-1} \right\| \ge (\varepsilon q(|x|))}^{-1} \right\}$, where $\left\| \cdot \right\|$ is any subordinate matrix norm and the boundary of $U$ denote by $\partial U$.

  • $S = \left\{ x \in \mathbb{C}:\left\| P(x )^{-1} \right\| = (\varepsilon q(|x|))^{-1} \right\}$

Can we say that $S \subseteq \partial U$?

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    $\begingroup$ If $P(x)$ and $q(x)$ are constant, and if $\epsilon$ be such that $\|P(x)^{-1}\|=(\epsilon q(|x|))^{-1}$ then the result is not correct, for, $S=U$ and $U=\mathbb C$ which its boundary is empty. $\endgroup$
    – MSMalekan
    Commented Feb 10, 2016 at 4:47
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    $\begingroup$ @MeisamSoleimaniMalekan $q(x)$ can't be constant because $\forall i\, w_i>0$. $\endgroup$
    – TZakrevskiy
    Commented Feb 10, 2016 at 18:13
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    $\begingroup$ @TZakrevskiy: I considered $q(x)$ to be "constant", not to be "zero". I assumed that $m=0$, $P(x)=B_0$, an invertible matrix, $q(x)=w_0>0$. $\endgroup$
    – MSMalekan
    Commented Feb 11, 2016 at 17:36

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