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Let $P$ be a positive $n\times n$ matrix. Given these two definitions:

Denote by $A(P)$ the set of all nonnegative numbers $\lambda$ for which there is a nonnegative vector $x\neq 0$ such that $Px\geq \lambda x$.

Denote by $B(P)$ the set of all nonnegative numbers $\mu$ for which there is a nonnegative vector $x\neq 0$ such that $Px\leq \mu x$.

Can I do the following:

"Fix some nonnegative vector $x\neq 0$, we get $\lambda x\leq Px\leq \mu x$?"

Does "there is a nonnegative vector" mean inequalities hold for all nonnegative vectors?

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    $\begingroup$ Yes because it means "Fix some ARBITARY nonnegative vector ..." $\endgroup$ – Peter Apr 14 '18 at 16:27
  • $\begingroup$ @Peter: But what is a nonnegative vector? $\endgroup$ – Postal Model Apr 14 '18 at 16:38
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    $\begingroup$ @Niing I guess a vector with non-negative entries. $\endgroup$ – Peter Apr 14 '18 at 16:40
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No, "there is" does not mean "for all". "There is" means only that you can find one nonnegative vector for which the inequalities hold.

To add a point of clarification:

You can fix a nonnegative $x \ne 0$ such that $\lambda x \le Px \le \mu x$, but you cannot fix a nonnegative $x \ne 0$ (with no further assumptions) and get $\lambda x \le Px \le \mu x$.

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  • $\begingroup$ Then, I guess I cannot fix some nonnegative $x\neq 0$ and get $\lambda x\leq Px\leq \mu x$. $\endgroup$ – user539442 Apr 14 '18 at 19:42
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    $\begingroup$ Right. You can fix a nonnegative $x \ne 0$ such that $\lambda x \le Px \le \mu x$, but you cannot fix a nonnegative $x \ne 0$ and get $\lambda x \le Px \le \mu x$. $\endgroup$ – Ted Apr 14 '18 at 20:41

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