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Consider vectors $w_1, w_2, w_3, w_4 \in \mathbb{R}^{n}$, $n \in \mathbb{Z}_{\geq 1}$.

Assume the following statement. For all $(c_1,c_2) \in (\mathbb{R}_{\geq 0} \times \mathbb{R}_{\geq 0}) \setminus\{(0,0)\}$ there exists $x \in \mathbb{R}^n$ such that

$$ \left( c_1 w_1 + c_2 w_2 \right)^\top x < 0 \quad \text{ and } \quad \left( c_1 w_3 + c_2 w_4 \right)^\top x < 0. $$

Find a case in which the following statement is false. There exists $y \in \mathbb{R}^n$ such that

$$ w_i^\top y < 0 \quad \forall i \in \{1,2,3,4\}. $$

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Maybe you could simplify your hypotheses just assuming that you're given vectors $w_1, w_2, w_3, w_4$? I mean, if there is no restriction on matrices $A_1, A_2$, I cannot find any restriction on vectors $w_i$ either. – a.r. Aug 31 '12 at 12:40
@Adam: it would help if you showed what work you had already done, or why you couldn't do any. – Ben Millwood Aug 31 '12 at 14:22
Ok, I'll just use $w_i$s. I already found that if $A_1 = A_2$, then the first statement implies the second. Therefore I'm claiming that this implication becomes false whenever $A_1 \neq A_2$. – Adam Aug 31 '12 at 14:53

Smallest example: $n=2$, $w_1=(1, 0)^T$, $w_2=(1, 1)^T$, $w_3=(-1, 1)^T$, $w_4=(-1, 0)^T$.

To show that the condition holds for this, try $x=(1, -2)^T$ or $x=(-1, -1)^T$. At least one of these will work (check that! You'll obtain $-c_1$ or $-c_1$ as product).

But $w_1=-w_4$ implies that for each $y$ we have $w_1^T y\ge 0$ or $w_4^T y\ge 0$.

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Neither $x_1 := (1,-2)^\top$ nor $x_2 := (-1,-1)^\top$ can work. In fact $(c_1 w_1 + c_2 w_2)^\top x_1 = c_1 - c_2$ that is not always negative. Analogously, $(c_1 w_3 + c_2 w_4)^\top x_2 = c_2$ that is non negative. – Adam Sep 1 '12 at 10:34
I think that there is a typo in the answer: if you just take $w_3 = w_2$ then you find that $x_1$ is good whenever $c_2 > c_1$, and $x_2$ is good whenever $2 c_1 > c_2$. At least one of those options will hold. – Jakub Konieczny Sep 23 '12 at 13:03

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