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Let $X$ and $Y$ be square non-symmetric matrices of the same size. Assume that their eigenvalues are close in the sense that there exists a small $\varepsilon>0$ such that, for any eigenvalue $\lambda$ of $X$, there exists an eigenvalue $\mu$ of $Y$ such that $|\lambda-\mu|\le \varepsilon$.

Now, let $D$ be a diagonal matrix of bounded components. Can we say that eigenvalues of $DX$ and $DY$ are also close in the same sense (with $\varepsilon$ replaced by $C\varepsilon$ for some $C>0$)? References are greatly appreciated.

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    $\begingroup$ Is the order of quantifiers $\forall D\,\exists C\,\forall X, Y, \epsilon\,\ldots$? $\endgroup$
    – LSpice
    Commented Dec 17, 2021 at 0:18
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    $\begingroup$ In any case, $X$ and $Y$ can have the same eigenvalues, while $DX$ and $DY$ are not. $\endgroup$
    – Narutaka OZAWA
    Commented Dec 17, 2021 at 0:46
  • $\begingroup$ Wile @legon asked about $D$ being a bounded diagonal matrix, I think the question is more interesting if $D$ is a diagonal matrix of which all diagonal elements are in the interval $[a,b]$ such that $0<a<b$ $\endgroup$
    – John_Krampf
    Commented Dec 20, 2021 at 3:59

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Consider the example $X=\left(\begin{array}{cc}a\\ &b\end{array}\right)$, $Y=\left(\begin{array}{cc}b\\ &a\end{array}\right)$ and $D=\left(\begin{array}{cc}1\\ &0\end{array}\right)$.

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  • $\begingroup$ How about if $D$ is positive definite? $\endgroup$
    – John_Krampf
    Commented Dec 20, 2021 at 3:57

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