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The similar question is here , but I do not see the desired answer.

Assume a cone $K=\{(x,y) |\ x+y=0\}$, find the dual cone of $K$.

The definition of dual cone is here: $K^*=\{y|x^{T}y\geq0, \forall x\in K \}$

And the given answer is $K^∗=\{(x,y)| \ x-y=0 \}$

I am very confused. Where is the $\geq$ in the definition formula?? Why it disappears? How could I derive the dual cone from the definition ?

Any help is appreciable . Thanks!

Edit: Another question: find the dual cone of $K=\{(x,y)∣|x|≤y\} $, in the question, we could use the inequality

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stander Qiu
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1 Answers1

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If the cone $K$ is a subspace, then the dual cone is the orthogonal complement of $K$.

This is because $x^T y \geq 0$ for all $x$ in $K$ implies that $x^T y = 0$ for all $x$ in $K$ (when $K$ is a subspace).

More detail: Assume that $K$ is a subspace let $y \in K^*$ (the dual cone for $K$). If $x \in K$, then $-x \in K$ also (because $K$ is a subspace), and so it follows from the definition of $K^*$ that $y^Tx \geq 0$ and $y^T(-x) \geq 0$. In other words, $y^T x = 0$. Hence, if $x \in K^*$, then $x \in K^\perp$ (the orthogonal complement of $K$).

littleO
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  • Thanks for your reply! I know what you mean. But when does the inequality come to effect? I have edited my question ( add another question) Thanks! – stander Qiu Aug 21 '16 at 03:49
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    For the new question, if you visualize $K$, you will see that it looks a bit like an ice cream cone. And if you think it $y^Tx \geq 0$ as meaning that the angle between $x$ and $y$ is at most 90 degrees, then you can visualize what the dual cone must look like. – littleO Aug 21 '16 at 04:08