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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

(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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    I've retagged as it was in no way related to Combinatorics or Numerical Methods.. For the question: What exactly is the dimension of $x = (x_i)$ when $i \neq 1$? Note that it shouldn't be $(n-1) \times 1$ as you're multiplying it with an $n \times n$ matrix.2012-09-01
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    I think you should add "numeric" tag since it is hard to solve $x$ algebrically.2012-09-01
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    I still don't understand the question.. Can you give some example describing what's happening here?2012-09-01
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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

1 Answers 1

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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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    I don't think $x$ is an eigenvector of $A$. Actually, $x$ satisfies the following equation: $max \{Ax,e_1\} =x $, where $max$ is entry-wise maximum, and $e_1={(1,0,\cdots,0)}^T$ .2012-09-01
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    @JohnSmith How on Earth is any *entry-wise maximum* involved in your question?? You might wish to mention what $Ax$ is to you, for $A$ an $n\times n$ matrix and $x$ a vector of size $n$, if not the vector of size $n$ everybody understands. (And I seem to read *eigenvector* in your title, or am I dreaming?)2012-09-01
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    $x$ is a vector of $n$ size. The first entry of $x$ is always 1, and other entries of $x$ satisfy the equation: $r\cdot Ax=x$, as clearly described in my post. The title is eigenvector because $y$ is the eigenvector of $A$ that satisfies $r\cdot Ay=y$. So my goal is to compute $x$ from $y$!2012-09-01
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    Your post states that $rAx=x$, not $Ax=x$, please make up your mind. Anyway, if $Ax=x$ or if $rAx=x$, then $x$ is an eigenvector of $A$, whether its first coordinate $x_1$ is $1$ or not. Sorry but your *clearly described in my post* might be a tad too optimistic here.2012-09-01
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    It is $r\cdot Ax=x$ . The first coordinate $x_1$ is always 1, as I described in (I).2012-09-01
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    Here is a suggestion: rephrase (II), which at the moment makes no sense.2012-09-01
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    I rephrased (II). Hopefully, this time the post could be understood.2012-09-01
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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