For what values of $x = (x_0 x_1 \ldots x_n)$ the following inequality holds for all positive values of the vector $a = (a_0 a_1 \ldots a_n)$: $a x^T \ge 0$?
Simple-looking vector inequality
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inequality
vector-spaces
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1I answered your question thinking that the condition was that the inequality holds for all such a, but then I wasn't sure so I also said something about the case where $a$ is fixed beforehand. Could you please clarify which is intended? – 2010-12-01
1 Answers
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The condition is that $x_i\geq 0$ for all $i$. It is clear that $ax^T\geq0$ if this holds. On the other hand, if $x_i\lt0$ for some $i$, then choose $a$ such that $a_i=1$ and $|a_jx_j|\lt\frac{|x_i|}{n}$ for all $j\neq i$. Then $ax^T=x_i+(a_0x_0+\cdots+a_{i-1}x_{i-1}+a_{i+1}x_{i+1}+\cdots+a_nx_n) \lt x_i +n\frac{|x_i|}{n}=0$.
Or, if it was assumed that $a$ is fixed beforehand, then your inequality would describe a half-space whose boundary is the hyperplane described by $ax^T=0$, which would include those vectors $x$ such that $x_i\geq0$ for all $i$.
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0Awesome, Jonas, thanks! – 2010-12-01