Let $\Phi: R^n \to R^n$ satisfy
$\Phi(x)=u+Ax+Q(x)$, with $x=(x_1, x_2,\ldots, x_n) \in R^n$. $u$ is positive vector, $A$ isnon negative matrix, and $Q(x)$ quadratic mapping with
$Q(x)_i=x_i(k_{i1}x_1+k_{i2}x_2+\ldots+k_{in}x_n)$, where all the $k_{ij}$ are nonnegative and at least one $k_{ij}$ is positive.
Suppose $\Phi(1)=1$, $1$ is the vector each entry being 1
How can I prove that there canot be two distinct vectors u, v such that u, v are different from the vector 1 and $\Phi(v)=v, \Phi(u)=u$,
$v, u$ are vectors with each entry positive and no greater than 1.
