I need to find vector \(f\) such that \((I\rho A)f\) is an increasing vector (i.e. for \(i > j, g_i \ge g_j\) for some \(g \in R^N\). Some of the properties of the different parameters are:
I can find a \(f\) based on the power series representation of \((I\rho A)^{1}\). However, I am looking at some alternate methods for constructing vector \(f\). Any trivial/non trivial construction would suffice. asked 26 Jan '14, 12:20 anon123 
Assuming that \(\rho\) and \(A\) are specified, this seems like a rather straightforward LP feasibility model. You could try the following in any LP solver: \[ \begin{array}{ccccc} \min & g_N\\ \textrm{s.t.} & \left(I\rho A\right)f & = & g\\ & g_{i} + \epsilon & \le & g_{i+1} & \forall i\in\left\{ 1,\dots,N1\right\}\\ & g & \ge & 0 \end{array} \] answered 27 Jan '14, 11:59 Paul Rubin ♦♦ I am aware of solving an LP to check for feasibility. I was looking for some sort of theoretical construction which would illustrate that the LP always have a solution. Or perhaps show that above LP will always have a solution.
(27 Jan '14, 12:48)
anon123
Is there a result in LP which says that if \(x\) is feasible for an LP then all the points in the neighborhood of \(x\) will also be feasible. For my problem \(x\) is a constant vector that is all entries of vector \(x\) are for example one.
(27 Jan '14, 13:13)
anon123
I have no idea what \(x\) is in relation to your original question. Also, the term "neighborhood" implies a topology. If you mean euclidean neighborhood (all points within some distance of \(x\)), then no, not all nearby points need be feasible. The feasible region will be a polyhedron, and \(x\) could be on the boundary.
(27 Jan '14, 14:28)
Paul Rubin ♦♦
My fault for being flawed and imprecise. The question that I was trying to pose concern the LP that you wrote. My question was that does there exist a feasible f and g for the given LP. f being a constant vector is always feasible as the corresponding g are all constant too (check f = [1,...,1] yields g = [1p,..,1p]). I was looking to construct f such that g is strictly increasing.
(27 Jan '14, 15:04)
anon123
I changed the model (bounded \(g_N\) and gave it a nontrivial objective function) in order to avoid the trivial solution \(f = 0 = g\), which was feasible in the original formulation.
(27 Jan '14, 15:29)
Paul Rubin ♦♦
Do you require \(f\) (and therefore \(g\)) to be nonnegative? Without that, I'm not positive my modified model is bounded, although I suspect it is. With nonnegativity, it's definitely bounded.
(27 Jan '14, 15:31)
Paul Rubin ♦♦
Instead of adding an objective you could have set g_n=1 to normalize the solution. The problem is homogeneous so that should be ok. Btw if \((I\rho{}A)\) is nonsingular then it has a solution I think.
(28 Jan '14, 01:34)
Erling_MOSEK
@Erling: You are correct about normalization, provided that we can assume \(g\ge 0\). The OP never said that, and I've been trying to avoid that assumption.
(28 Jan '14, 10:41)
Paul Rubin ♦♦
I just changed the model again (sigh), since the problem description changed again. Apparently \(g\) needs to be strictly increasing. I've also constrained \(g\ge 0\), although I'm not sure that is warranted. The LP has a solution iff the original problem does, and gives a mechanism to construct a solution when one exists, but does not prove a solution exists (unless the dual is provably bounded).
(28 Jan '14, 10:47)
Paul Rubin ♦♦
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By "increasing vector" do you mean \(i\gt j \implies f_i\ge f_j\)? Also,do you need \(f\ge 0\)?
Yes I added the definition in the question too