Let $A$ be an $n \times n$ matrix with all elements positive, and $\lambda$ its largest eigenvalue, which is real, positive, and of multiplicity one.
Define $M = (1/\lambda) \, A$. Let $x$ be a vector with all positive elements with $L^1$ norm equal to one, and let $y = M \times x$. Clearly $y$ will be a positive vector.
What characteristics should $A$ (or $M$) satisfy such that $x > y$? I assume that $x$ has all elements larger than the Perron eigenvector normalized to $L^1$.
Suppose there is a vector $x \ge 0$ such that $M x < x$, i.e. $(I-M) x > 0$. Let $u$ be the Perron eigenvector for $M$. Then there is $\varepsilon > 0$ such that $(I -M) x - \varepsilon u \ge 0$. Since the set of nonnegative vectors is mapped to itself under $M$, we have $M^k ((I -M) x - \varepsilon u) \ge 0$. But $M^k u = u$ while $\lim_{k \to \infty} M^k x = c u$ for some $c > 0$. The result is $0 - \varepsilon u \ge 0$, which is impossible.
Conclusion: no such $x$ can exist.