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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 a given positive vector, $A$ non 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}, 1 \leq i, j \leq n $ is positive.

Suppose $\Phi(\mathbf{1})=\mathbf{1}$, where $\mathbf{1}$ is the vector each entry being 1.

How can I prove that there cannot be two distinct vectors u, v such that u, v are different from the vector $\mathbf{1}$ and $\Phi(v)=v, \Phi(u)=u$,

$v, u$ are vectors with each entry positive and no greater than 1.

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Identical question posted to m.se, math.stackexchange.com/questions/221134/… --- voting to close. –  Gerry Myerson Oct 25 '12 at 22:02
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2 Answers

The property does not hold as stated. Consider the case $u=(\frac{1}{2},\dots,\frac{1}{2}), A=\frac{1}{4}I,$ and $Q(x)_i = \frac{1}{4}x_i^2$. Then $\Phi$ maps $(x_1,\dots,x_n)$ to $\frac{1}{4}(x_1^2+x_1+2,\dots,x_n^2+x_n+2)$, and all $x$ with $x_i\in\{1,2\}$ are fixed points.

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The OP was interested in vectors with entries in $[0,1]$. Nevertheless, it is easy to replace your quadratic with one with solutions $1$ and $\alpha<1$, and the counterexample does not change. –  Federico Poloni Oct 26 '12 at 12:56
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Are you sure that this holds in general? I have studied a very similar problem a few years ago (paper, arxiv), and there you can have up to $2^n$ solutions and you need a minimality property to characterize the physically meaningful solution. Maybe you can get something useful out of the paper.

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I forgot to mention that for the quadratic map $Q$, the diagonal entries $k_{ii}$ are zero. Any idea of the proof? –  youming Oct 26 '12 at 23:33
    
I still think that there will be counterexamples. Have you investigated what happens with $n=2$? I would start from that. –  Federico Poloni Oct 27 '12 at 8:09
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