# Find the polar set of a convex cone

I am stuck with the following question:

Given the convex cone $$K = \{x \in \mathbb R^n : x_1 \geq x_2 \geq \cdots \geq x_n \geq 0\}$$ determine the polar set of $$K$$, which we denote by $$K^{*}$$.

I know that $$K$$ is a convex cone, which is easy to proof. I also know from our lecture, that $$K^{*}$$ is again a convex cone and
$$K^{*}=\{y\in \mathbb{R^n}: \langle x,y\rangle \leq 0\ \forall x\in K\},$$ where $$\langle\cdot,\cdot\rangle$$ is the inner product.

It's quite obvious, that all $$y\in \mathbb{R}^n$$ where $$y_i\leq0\ \forall i\in[n]$$ are in $$K^{*}$$. But there have to be some more inequalities or easier ways to determine $$K^{*}$$. Do you have any hints/suggestions?

• $K$ is actually just $A \cdot \mathbb{R}^n_+$, where $A$ is a certain invertible matrix. Does this help? Commented Nov 9, 2015 at 19:18
• If I would find out more about A, this would help a lot! Thank you so much :-)
– D1Fu
Commented Nov 9, 2015 at 19:51
• Do you have any hints for A? It might be too obvious for me...
– D1Fu
Commented Nov 9, 2015 at 20:03
• Think I got it :-)
– D1Fu
Commented Nov 9, 2015 at 20:48
• Good. Then post your answer, and get points! :-) Commented Nov 9, 2015 at 21:17

Ok, with Michael Grant's hint, the solution should be the following: $Let A\in \mathbb{R^{n\times n}}$ be the upper triangular matrix consisting of $1$'s. Then $K=A\cdot \mathbb{R^{n}_+}$. A is invertiable, and the following proposition can be easily shown:
$$(MX)^* = M^{-T}X^*\hspace{1mm} where \hspace{1mm} M\in \mathbb{R^{n\times n}} \hspace{1mm} invertiable$$
So for the solution of our problem, we just have to find out $A^{-T}$ and determine what ($\mathbb{R^{n}_+})^*$ is, which is not too difficult, since for $y\in R^{n}:{<y,x>}\leq 0 \forall \hspace{1mm} x\in \mathbb{R^{n}_+}$, so $y\in\mathbb{R^{n}_-}$.