Rate of convergence for the Perron–Frobenius theorem The Perron–Frobenius Theorem states the following.

Let $A = (a_{ij})$ be an $n \times n$ irreducible, non-negative matrix ($a_{ij} \geq 0, \forall i,j: 1\leq i,j \leq n$). Then the following statements are true.


$A$ has a real eigenvalue $c \geq 0$ such that $c &gt; |c'|$ for all other eigenvalues $c'$.
There is an eigenvector $v$ with non-negative real components corresponding to the largest
eigenvalue $c: Av = cv, v_i \ge 0, 1 \leq i \leq n$, and $v$ is unique up to multiplication by a constant.
If the largest eigenvalue $c$ is equal to $1$, then for any starting vector $x^{\langle 0\rangle} \neq 0$ with non-negative components, the sequence of vectors $A^k x^{\langle 0\rangle}$ converge to a vector in the direction of $v$ as $k \rightarrow \infty$.


But the theorem does not say how fast the sequence of vectors $A^k x^{\langle 0\rangle}$ will converge. Are there any known results on the rate of convergence? What are some good, polynomial-time algorithms to compute this limiting vector?
See the power method for computing eigenvectors:
http://en.wikipedia.org/wiki/Power_iteration

Convergence is exponential (geometric in the ratio of the top two eigenvalues).