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I have the following problems when solving a linear equation.

Let $A=(a_{i,j})_{n \times n}$ be a non-negative matrix with $a_{i,j} \in (0,1)$, and let $0 be a scalar. Now we define a vector $x=(x_i)$ of length $n$ as follows:

(I) The first component of $x$ is 1, that is $x_1=1$.

(II) The other components of $x$ (except the first entry of $x$) satisfies the following equation:

$r \cdot Ax=x .$

Or equivalently, both (I) and (II) tell that $x$ satisfies the following equation: $max \{r \cdot Ax,e_1\} =x $ where $max$ is entry-wise maximum operator, and $e_1={(1,0,\cdots,0)}^T$ .

Based on such a defintion of $x$, I want study the relations between $x$ and the vector $y$ that satisfies $r \cdot Ay=y$ (including the first entry of $y$). In other words, can we compute $x$ from $y$ ?

I would really appreciate any suggestions.

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    John: Where are we at? If you consider that this question is a lost cause, then erase it. Otherwise, modify it, addressing the concerns raised in the comments.2012-09-09

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(This answer applies to the original version of the post, which might not reflect the question the OP has in mind. Some heroic tries to reach a proper formulation of the question are occurring right now...)


Both $x$ and $y$ are eigenvectors of $A$ for the eigenvalue $1/r$, hence, if the eigenvalue $1/r$ is simple, then $x$ and $y$ are proportional. Since $x_1=1$, this yields $y=y_1\cdot x$.

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    Still horrible. Here is a try: your (II) might in fact mean that there exists some real number $s$ with $rAx=x+se_1$. Do you confirm? Note that, even if so, (I) and (II) do not imply $\max\{rAx,e_1\}=x$ componentwise (call this (III)), since (III) says that $rAx=x+se_1$ for some **nonpositive** $s$. So, there is still some work to do before this post leaves the confines of NARQ...2012-09-01