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I encountered this question: Find the dual cone of $K = \{(x,y)|(x+y=0)\}$. To find the dual ($K^* = \{y \mid x'y \geq 0 \text{ for all }x \in K \}$), I did the following: \begin{align} (x,-x)(y_1,y_2)'& \geq 0 \\ xy_1-xy_2 & \geq 0 \\ x(y_1-y_2) & \geq 0 \\ y_1 & \geq y_2 \end{align}

But the Dual of this cone is $\{(y_1,y_2) \mid y_1=y_2 \}$. Can you find out where is the mistake? Thanks!

littleO
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Jyothi Jain
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    It would help if you elaborated on how you went from $x(y_1 - y_2) \geq 0$ to $y_1 \geq y_2$. – littleO Jul 14 '17 at 10:26

1 Answers1

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You have correctly deduced that $y_1 \geq y_2$. You can also deduce that $y_1 \leq y_2$. It follows that $y_1 = y_2$. So, the dual cone of $K$ is $$ K^* = \{ (y_1,y_2) \mid y_1 = y_2\}. $$

By the way, here is a useful fact: if $K$ is a subspace of $\mathbb R^n$, the dual cone of $K$ is the subspace orthogonal to $K$.

littleO
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