The second largest eigenvalue for Perron-Frobenius matrix The Perron-Frobenius theorem is about the largest eigenvalue and eigenvector of a non-negative (irreducible) matrix.
My question: Is there any estimation of the difference between the first and second largest eigenvalues, say an upper or a lower bound?
An general theory may be tough, so please feel free to add other conditions to limit our discussion.
 A: Well, if we limit our discussion to the weighted undirected graphs, there is Cheeger inequality which provides bounds for the 2nd eigenvalue in terms of other numerical characteristics of the graph: see e.g. here. 
The result may still hold for weighted graphs directed graphs that are symmetrizable: namely, there exists a positive vector $\mu$ such that $A(x,y)\mu(x) = A(y,x)\mu(y)$, see in this very accessible lecture notes. Without this reversible/symmetrizable structure you don't have Green's theorem, so results are pretty weak.
As Tim has mentioned, though, there are no non-trivial bounds in general since the 2nd eigenvalue may be as close to the 1st one as one can imagine. So anyways, it all depends on a particular case. I don't know of any interesting estimates for the non-reversible case.
A: There certainly isn't a useful one.  Consider the matrices
$$A_\lambda=\left(\begin{array}{cc} \tfrac{1+\lambda} 2&\tfrac{1-\lambda} 2 \\ \tfrac{1-\lambda} 2 & \tfrac{1+\lambda} 2\end{array}\right)$$
for $\lambda\in(0,1)$
Then $A_\lambda$ is irreducible and has eigenvalues $1$ and $\lambda$.
For an irreducible Markov chain the second eigenvalue represents the rate that the chain converges to its invariant distribution. This can be incredible fast, or incredibly slow. 
A: I think if you have Doeblin-Fortet (or Lasota-Yorke sometimes) inequalities satisfied you can do some estimates.
