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If $Ax\geq b$, under what conditions on $K$ can I premultiply it and preserve the sign? i.e

What is a sufficient condition on $K$ for: $$KAx\geq Kb$$

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You're going to have to elaborate on what you mean by "$\ge$". – EuYu Nov 9 '12 at 21:35
Component wise. – Inquest Nov 9 '12 at 21:40
Are you sure you want a sufficient condition? One sufficient condition is that $K$ is the identity. – joriki Nov 9 '12 at 21:52
up vote 1 down vote accepted

Suppose you have two vectors which satisfies the component-wise inequality (denoted $\ge_c$) $$\mathbb{x} \ge_c \mathbb{y}$$ If you want a matrix such that $$A\mathbb{x} \ge_c A\mathbb{y}$$ then that is equivalent to requiring $$\mathbb{a_i}\cdot \mathbb{x} \ge \mathbb{a_i}\cdot \mathbb{y}$$ for each row vector $\mathbb{a_i}$ of $A$. It is sufficient (and probably necessary) that you require $A$ to be non-negative.

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Let $y = Ax-b$

Assume $A$ is full rank. Then your question is the condition for matrix $K$, so that $Ky ≥_c 0,\ \forall y \geq_c 0 $

Necessary and sufficient condition for this, is non negativitity of $A$.

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