Question: 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.
> 
> 
> 

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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.**

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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?

Answer: 
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.
One of the nice things about it is that for $v$ a vector with all positive entries, if we let $v\_k = A^k v$, then $\lim\_{k\to\infty} \frac{v^k}{|| v\_k ||}$ exists and is an eigen vector for the Perron-Frobenius eigen value.