Cone: A subset $ S \subseteq \mathbb{R}^n$ is a cone if $\alpha \geq 0 \implies \alpha S \subseteq S.$
Polar: A Polar $K^*$ of a cone $K$ is a closed convex cone such that $$K^*=\{y \in \mathbb{R}^n \mid x\in K \implies (x,y) \geq 0\}$$ where $(x,y)$ denote the standard inner product of vectors.
Problem If $K=\mathbb{R}^n_+ \cap R(A^T)$ then $K^*=\mathbb{R}^n_+ +N(A)$ where $A \in R^{m \times n}$ and $R(A),N(A)$ denote the range and null space of $A.$
Question How to approach this problem ? I know polar of a set is orthogonal to the original set and $N(A)$ and $R(A^T)$ is orthogonal for any matrix. Now how to reach to the desire result ?
Thank in advanced.
Edit: The dual of the intersection of the convex cones is the sum of their dual.
Since $(\mathbb{R}^n_+)^*=\mathbb{R}^n_+$ and $(R(A^T))^*=N(A)$ then $K^*=\mathbb{R}^n_+ +N(A)$ where $K=\mathbb{R}^n_+ \cap R(A^T)$.
Thanks copper.hat
Just for completion :-).
Let $K^+ = \{ x | \langle y, x \rangle \ge 0 \text{ for all } y \in K \}$. The key results are that if $K$ is a closed convex cone then $K^{++} = K$ and if $A \subset B$ then $B^+ \subset A^+$.
Suppose $A,B$ are closed convex cones. Then we will show that $(A \cap B)^+ = A^++B^+$.
Since $A\cap B \subset A$ we have $A^+ \subset (A \cap B)^+$ and similarly for $B$. Since $ (A \cap B)^+$ is a convex cone, we have $A^++B^+ \subset (A \cap B)^+$.
To show the other direction, it is sufficient to show that $(A^++B^+)^+ \subset (A \cap B)^{++} = A\cap B$.
Suppose $x \in (A^++B^+)^+$, then $\langle x, \alpha + \beta \rangle \ge 0$ for all $\alpha \in A^+, \beta \in B^+$. Setting $\beta =0$ shows that $x \in A^{++} = A$ and similarly for $B$ which shows that $x \in A \cap B$, the desired result.
Finally, it is straightforward to show that $(\mathbb{R}^n_+)^+ = \mathbb{R}^n_+$ and $({\cal R} A^T)^+ = ({\cal R} A^T)^\bot = \ker A$.