Suppose T is strongly connected

**T is convergent if and only if it is aperiodic **

"strongly connected" is going to say that from every given individual, there exists a directed path from i to j.

$\exists$ directed path i to j $\forall i, j$

**T is convergent if and only if: lim $T^t = (1,1,...,1)^T s$ where s is the unique lhs(left hand side) eigenvector with eigenvalue 1**

Defn: T is **primitive** if $T_{ij}^t > 0, \forall i,j$ after some t

If T is strongly connected and stochastic then it is aperiodic if and only if it is primitive. (Perkins(1961))

If T is strongly connected and primitive then lim $T^t = (1,1,...,1)^Ts$

where s is the unique lhs eigenvector with eigenvalue "1"(e.g. see Meyer(2000))

So, strongly connected and aperiodic implies convergence.

Converse comes from showing:

- If T is strongly connected, stochastic and convergent, then it is primitive

Let $S = $ lim $T^t$ by convergence

Then $ST =$ lim $T^t T = S$

So each row is lhs eigenvector with eigenvalue "1": it is a positive vector by Perron-Frobenius theorem(an eigenvector of an irreducible non-negative matrix is strictly positive if and only if it is associated with its largest eigenvalue. This vector is unique if the matrix is primitive)

So since S is all positive, T is primitive

Since, T is primitive then Perron-Frobenius implies the eigenvector is unique, and all rows of S are the same s

- Aperiodicity is easy to satisfy:
- Have some agent weight him or herself
- Or have at least one communicating dyad and a transitive triple...