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?
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 $\lambda x \le Px \le \mu x$.