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
$\endgroup$1 Answer
$\begingroup$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 $K$).
$\endgroup$ 2
