Perron-Frobenius for a "maximum" of matrices 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. 
 A: 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
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.
