This is mentioned in these slides.
A non-negative square matrix $T$ is called primitive if there is a $k$ such that all the entries of $T^ k$ are positive.
It is called irreducible if for any$ i, j$ there is a $k = k(i, j)$ such that$ (T^ k )_{ij} > 0.$
We have to show that if T is irreducible then I + T is primitive.
In the slides the author just does the binomial expansion since $I$ and $T$ commute and then says for large enough $k$ the entries are positive. I do not understand why? Can somebody explain this. Thanks
When you expand out $(I+T)^n$ you get terms involving $T^k$ for every $k$ from $0$ to $n$. If $n$ is large enough, this includes all the $k(i,j)$ from the definition of irreducible. Since all terms are non-negative, if one is positive the sum is positive.