Boyd's book, my understanding of cone and dual cone for 2-space is: If we think of a circle in $\mathbb{R}^2$ space, cone $K$ and dual cone $K^*$ would be like this:

enter image description here

Here, $K^* = y | x^Ty \geq 0 \text{ for all } x \in K$

Now question is: How do you draw cone and dual cone for these:

  1. $K = \{ (a_1, a_2) \in \mathbb{R} | |a_1| \leq a_2 \}$

  2. $ K = \{ Ax | x \geq 0 \}$

I am just trying to get an intuitive idea about the geometry of the cones and dual cones.

  • $\begingroup$ Usually, cones extend to infinity and are not bounded by a circle. Also, the definition that I typically use for dual cone results in a different region than what you have highlighted. Perhaps you could provide some additional background. $\endgroup$ – Michael Burr Feb 14 '17 at 15:53

Your understanding is wrong. Consider the vector $x=(1,1)$ and the vector $y=(-1,-1)$. $x$ is clearly in $K$. You claim that $y$ is in $K^{*}$, but $x^{T}y=-2<0$. In fact, the dual of the positive orthant $R^{n}_{+}$ is $R^{n}_{+}$.

  • $\begingroup$ I have updated the picture. Do you think it's right now? $\endgroup$ – jhon_wick Feb 14 '17 at 16:13
  • $\begingroup$ No, it's still not correct. Keep in mind that $K^{*}$ is the set of $y$ such that $x^{T}y\geq 0$ for all $x$ in $K$. That means that $y$ must be within 90 degrees of every vector $x$ in $K$. $\endgroup$ – Brian Borchers Feb 14 '17 at 16:26
  • $\begingroup$ my understanding is, we can find hyperplanes which supports cone $K$ and those hyperpalnes goes through the 2nd and 4th quardent.(?). Could you please sketch the correct one? What about the other two? $\endgroup$ – jhon_wick Feb 14 '17 at 16:31

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