Is there an $\alpha\in\mathbb{R}^m$, such that $\alpha_i > 0$ and $A\alpha\in S$?

Question

$A$ is a real $n\times m$ matrix and set $S\subseteq \mathbb{R}^n$ is defined as $$S = \{(x_1,\dots, x_n)\in \mathbb{R}^n\mid \forall(i,j)\in I.\; x_i< x_j\}\text{,}$$ where $I$ is a possibly empty index set of pairs of indices. The question is the following:

Is there an $\alpha\in\mathbb{R}^m$, such that $\alpha_i > 0$ and $A\alpha\in S$?

Background: there are $n$ films and $m$ persons. The number $A_{ij}$ tells us, how much does person $j$ like film $i$ (e.g., in the range from $1$ to $10$, $-5$ to $5$, or something similar: you can suggest additional assumptions that would help, I couldn't think of any).

Now, a new person comes, possibly with some preferences, e.g., film $1$ is better than $2$, film $7$ is better than $3$ and so on (hence $I = \{(2,1), (3,7), \dots\}$). The question is: can we aggregate the opinions of the first $m$ persons (create a weighted sum with the weights $\alpha_i$) in a way that would "agree" with the new person's preferences?

I really do not know, how to reduce this problem to linear system solving or something like that.

EDIT: As pointed out in the comments, the fact $A_{ij} \geq 0$ might help. Hence, from now on, we assume that $A_{ij}\geq 0$.

It seems that the answer depends heavily on the structure of $A$, and there might be no answer for some $A$, as pointed out in the answer below. I think this is because of the possible negative values of $A$. Things can be more interesting if $A$ becomes a nonnegative matrix, i.e., with all elements of $A$, non-negative. — Samrat Mukhopadhyay, Jun 28 '16 at 06:22
@SamratMukhopadhyay We can easily assume that $A$ is nonnegative, for example, let's say that the every person's opinion about a film is from the interval $[1,10]$. — Antoine, Jun 28 '16 at 06:25

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

2

This may have no solution. Consider $A=\begin{pmatrix}2&2\\1&2\end{pmatrix}$ with $I=\{(1,2)\}$. If $\alpha_1,\alpha_2>0$ and $x=A\alpha$ then $x_1=2\alpha_1+2\alpha_2>\alpha_1+2\alpha_2=x_2$. On the flip side there may be many solutions (even up to scalar multiple).

Determine whether a solution exists is essentially a linear programming problem. Let $$ I=\{(i_1,j_1),\ldots,(i_m,j_m)\} $$ and let $B$ denote the $m\times n$ matrix with $B_{kl}=\delta_{i_kl}-\delta_{j_kl}$. Then your problem is to find a vector $\alpha$ such that $$\begin{eqnarray*} \alpha>0,\\ BA\alpha>0. \end{eqnarray*}$$ By scaling $\alpha$, this has a solution if and only if $$\begin{eqnarray*} \alpha\geq1_n,\\ BA\alpha\geq1_m \end{eqnarray*}$$ has a solution, where $1_k$ is a vector of $k$ ones. This can be reduced to a standard linear programming problem (and solved using the simplex algorithm) as described in this answer.

edited Apr 13 '17 at 12:21

Community

1

answered Jun 28 '16 at 06:14

stewbasic

6,211

I know that this is not necessarily true. I would like to see an algorithm with $A$ as an input ant the answer as the ouput. – Antoine Jun 28 '16 at 06:27
I updated my answer with the assumption that $A_{ij}>0$. What would the algorithm do when there is no solution? – stewbasic Jun 28 '16 at 22:16
Algorithm should find a solution if it exists, or report that there is no solution. – Antoine Jun 29 '16 at 06:30
@Antoine see the update to my answer. – stewbasic Jun 29 '16 at 22:32
I do not like that solution very much (I hoped for a more "algebraic" one), but it is completely valid and the only one so far. – Antoine Jul 05 '16 at 05:35

Is there an $\alpha\in\mathbb{R}^m$, such that $\alpha_i > 0$ and $A\alpha\in S$?

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