Prove that operator is completely continuous

Question

Let's consider Banach space $\ell^\infty$ of bounded sequences $x = \{ \xi_n\}_{n=1}^\infty$: $$ ||x|| = \sup_{n \in \mathbb N} |\xi_n|. $$ Suppose matrix $||a_{i j}||_1^\infty$ specifies operator $A$ in $\ell^\infty$ such that vector $x = (\xi_1, \xi_2, \dots)$ transforms into vector $y = Ax = (\eta_1, \eta_2, \dots)$, where $$ \eta_i = \sum_{j = 1}^\infty a_{i j} \xi_j. $$ How to prove operator $A$ is completely continuous if it's matrix is nonnegative and has finite row-wise sums: $$ a_{i \star} = \sum_{j = 1}^\infty a_{i j} < \infty \, ? $$ If necessary we can suppose sequence $\{a_{i \star}\}_{i=1}^\infty$ of row-wise sums is bounded. We can also suppose the matrix of $A$ is irreducible, or in other words some power of $A$ is strictly positive.