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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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    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 $A_1 \neq A_2$.2012-08-31

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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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    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.2012-09-23