Consider a set of positive matrices $(P\_k)\_{k\in K}$ in $\mathbb R^{p\times p}\_{++}$ ($P\_k$ is positive in the sense that all entries of $P\_k$ are positive).


Let $X\_0\in\mathbb R\_{++}^p$ and define for every $n\ge0$ $X\_{n+1}=\max\_{k\in K}\{ P\_kX\_n\}$, where the max operator has to be understood element-wise: for all $i\in\{1,\ldots,p\}$, $[X\_{n+1}]\_i=\max\{k\in K:[P\_kX\_n]\_i\}$.


Do the following statements hold?


* There exists a positive matrix $P$ and $n\_0\ge0$ such that for all $n\ge n\_0$, $X\_{n+1}=PX\_n$.
* The matrix $P$ is constructed as follows. Consider the set of matrices, such that every row $i\in\{1,\ldots,p\}$ of a matrix $M$ in that set is the row $i$ of a matrix $P\_k$. (There are $(Card\ K)^p$ such matrices.) The matrix $P$ is the element of this set, which has the largest Perron-Frobenius eigenvalue, denoted $\lambda\_{PF}$.
* $(X\_n)\_n$ converges to the Perron-Frobenius eigenvector. More precisely, $\lim\_{n\rightarrow\infty} X\_n/||X\_n||$ exists and is denoted $x$. It is such that: $Px = \lambda\_{PF} x$.


I inferred the previous properties from the case $p=2$ but even in that simple case, I do manage to prove any of the above statements.


Thanks for any help or useful reference. 



Perron-Frobenius for a "maximum" of matrices
The answer is no, as the map you define is not linear. 


For example consider the matrices
$$ P\_1 = \begin{bmatrix} 2 & 1 \\ 2 & 1 \end {bmatrix},\quad P\_2 = \begin{bmatrix} 1 & 2 \\ 1 & 2 \end {bmatrix}.$$
For simplicity, I'll call the map in question $F(x) := \max\{P\_1x,P\_2x\}$, where as you say the maximum is taken componentwise, i.e. with respect to the order induced by the cone of nonegative vectors.


Then for the vectors 
$$ x\_1 = \begin{bmatrix} 2 \\ 1 \end {bmatrix},\quad x\_2 = \begin{bmatrix} 1 \\ 2 \end {bmatrix}$$
we have
$$ F(x\_1) = P\_1 x\_1 = \begin{bmatrix} 5 \\ 5 \end {bmatrix}, \quad F(x\_2) = P\_2 x\_2 = \begin{bmatrix} 5 \\ 5 \end {bmatrix}. $$
On the other hand
$$ F(x\_1+x\_2) = F\left( \begin{bmatrix} 3 \\ 3 \end {bmatrix}\right) = \begin{bmatrix} 9 \\ 9 \end {bmatrix},$$
which is clearly not equal to $F(x\_1) + F(x\_2)$. Now if there were a matrix $P$ as in your first question then the map $F$ would have to be linear. As the matrix $P$ does not exist your second and third question are no longer relevant. 


All is maybe not lost though. If what you are interested in is a Perron-Frobenius result, then you can use the nonlinear Perron-Frobenius theory. At least your map is homogeneous, i.e. $F(\alpha x) = \alpha F(x)$ for $\alpha > 0$. In addition, your map is monotone, i.e. if $x \leq y$, then $F(x) \leq F(y)$. Finally, the induced graph of your map is full (for details on that see the paper below). So Theorem 2 in


[S. Gaubert, J. Gunawardena, The Perron-Frobenius theorem for homogeneous, monotone functions](http://www.ams.org/journals/tran/2004-356-12/S0002-9947-04-03470-1/S0002-9947-04-03470-1.pdf)


shows that you have an eigenvector. For the convergence property that you want, I suggest to refer to the paper by Nussbaum cited in that paper.


