Q:

What is the implication of Perron Frobenius Theorem?

The Perron Frobenius theorem states:

Any square matrix $A$ with positive entries has a unique eigenvector with
  positive entries (up to a multiplication by a positive scalar), and
  the corresponding eigenvalue has multiplicity one and is strictly
  greater than the absolute value of any other eigenvalue.

So I tempted fate using this matrix:
$$ A =\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$
I find my eigenvalues to be $\lambda_1 = 0, \lambda_2 = 2$
Now I find my eigenvector, taking $v_{11} = 1$, $v_{21} = 1$
I find 
$v_1 = \begin{bmatrix} v_{11} & v_{12} \end{bmatrix}$ = $\begin{bmatrix} 1 & -1 \end{bmatrix}$
$v_1 = \begin{bmatrix} v_{21} & v_{22} \end{bmatrix}$ = $\begin{bmatrix} 1 & 1 \end{bmatrix}$
This verifies the Perron-Frobenius theorem.

Now what is the great implication that every positive square matrix has a real eigenvector with an eigenvalue that is the largest of all eigenvalues?
Can someone show me an application of this theorem?

A:

The Perron-Frobenius theorem is used in Google's Pagerank. It's one of the things that make sure that the algorithm works. Here it's explained why the the theorem is useful, you have a lot of information with easy explanations on Google.

A:

For example, there are important applications in the theory of finite Markov chains.  Among other things, it implies that a Markov chain with strictly positive transition probabilities has a unique stationary distribution.

A:

First of all, the conclusion is more interesting if you also try an example where the matrix entries are still real, but not positive.  For example, if $A$ is the $2\times2$ rotation matrix
$$A=\pmatrix{\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta}$$
then you can check that the eigenvectors are no longer real, and there's no longer a unique eigenvalue of maximum modulus: the two eigenvalues are $e^{i\theta}$ and $e^{-i\theta}$.
I second Robert Israel's nomination of Markov chains as a great application.  I'll add the following, though.  (I'll remark that you don't actually need the matrix to have positive entries - just for some power of it to have positive entries.)  Suppose you have a finite connected graph (or strongly connected digraph) such that the gcd of the cycle lengths is $1$.  Then if $A$ is the adjacency matrix for the graph, some power $A^r$ will have positive entries so Perron-Frobenius applies.  Since the entries of $A^n$ count paths in the graph, we conclude that for any pair of vertices in the graph, the number of paths of length $n$ between them is $c\lambda_1^n(1+o(1))$, where $\lambda_1$ is the Perron-Frobenius eigenvalue and $c$ is a computable constant.  Applications of this particular consequence of Perron-Frobenius include asymptotic growth rate results for "regular languages," which are defined in terms of paths on graphs.
In particular, there is a cute formula giving the $n$-th Fibonacci number as
$$F_n=\left\langle \frac1{\sqrt5}\phi^n\right\rangle$$
where $\phi$ is the "golden ratio" $(1+\sqrt5)/2$ and $\langle\cdot\rangle$ denotes "closest integer."  This formula, and many others like it, become less surprising
once you check that
$$\pmatrix{1&1\\1&0}^n=\pmatrix{F_{n+1}&F_n\\ F_n&F_{n-1}}$$
and note that $\phi$ is the Perron-Frobenius eigenvalue (and the other eigenvalue is smaller than $1$ in absolute value.)

