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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 $0<r<1$ 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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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. –  Rijul Saini Sep 1 '12 at 10:31
    
I think you should add "numeric" tag since it is hard to solve $x$ algebrically. –  John Smith Sep 1 '12 at 10:44
    
I still don't understand the question.. Can you give some example describing what's happening here? –  Rijul Saini Sep 1 '12 at 10:58
    
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. –  Did Sep 9 '12 at 15:32
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1 Answer

(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 $e_1={(1,0,\cdots,0)}^T$ . –  John Smith Sep 1 '12 at 10:41
    
@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?) –  Did Sep 1 '12 at 10:46
    
$x$ 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 $y$! –  John Smith Sep 1 '12 at 10:48
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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. –  Did Sep 1 '12 at 10:51
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Here is a suggestion: rephrase (II), which at the moment makes no sense. –  Did Sep 1 '12 at 10:58
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