# Uniqueness of the unbounded solution for linear equation

Let $\mathbb X$ be the set of all vectors $x$ such that $x_i\in(-\infty,\infty]$ for all $1\leq i \leq n$. Let $A,b$ be a matrix and a vector with non-negative real entries (bounded) and consider the equation $$x = b+Ax\quad (1)$$ where the solution $\hat x$ has to belong to $\mathbb X$. We say that an equation $(1)$ has a unique solution if for any $x,y$ which satisfies $(1)$ it holds that either $x_i=y_i\in (-\infty,\infty)$ or $x_i = \infty$ and $y_i=\infty$. For the multiplication the rule is $$a\cdot\infty=\begin{cases} \infty,&\text{ if }a>0, \\ 0,&\text{ if }a=0, \\ -\infty,&\text{ if }a<0 \end{cases}$$ and $a+\infty = \infty$, $a-\infty = -\infty$ for all bounded $a$.

Did anybody read about the uniqueness of such equations? The difference with the bounded case is the following: if the solution has to be bounded then non-unique solution of $x=Ax$ leads to the non-unique solution of $(1)$ provided that the latter exists. Indeed, if there is non-zero solution $x'$ of $x=Ax$ and at least one solution $x''$ of $(1)$ then $x'+\alpha x'$ is a solution of $(1)$ different from $x''$ for all $\alpha\neq 0$,

As an example: $A = I_2$ is an identity matrix of the dimension $2$ and $b = (0 \quad1)^T$. The equation $(1)$ does not admit any bounded solutions but $\hat x = (0\quad \infty)^T$ is a unique unbounded solution, though any vector solves the homogeneous equation $x=Ax$.

This question is motivated by Expectation of a stopping time uniquely determined by a function.

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$A$ and $b$ have non-negative entries? Or just $b$? – André Caldas Oct 17 '11 at 19:05
@AndréCaldas You can assume that both $A$ and $b$ has non-negative entries. – Ilya Oct 17 '11 at 19:09