Let $x^1,\dots,x^m$ be $m$ vectors in $\mathbb{R}^n$ and let $K = \text{cone}(\{ x^1,\ldots, x^m \})$ be the cone generated by the set $\{ x^1,\ldots, x^m \}$. (Superscripts are indices not powers.)
Prove that the polar cone $K^\circ = \{v\in\mathbb{R}^n:Av\leq 0\}$ where $A$ is the matrix that has $x^i$ as its $i$-th row for ($i=1,\dots\,m$).
ATTEMPT
I use the following definitions:
The cone generated by $X$ is the set of all nonnegative combinations of $X$: $$ \text{cone}(X) = \bigg\{ \sum_{i\,=\,1}^{m} \lambda_i x_i:x_i\in X \text{ and } \lambda_i \geq0\bigg\}. $$ The polar cone of $X$ is defined by: $$ X^\circ=\big\{ v\in \mathbb{R}^n:\langle v,x \rangle \leq 0, \; \forall x \in X\big\}. $$
In this particular problem, $K$ contains all possible nonnegative combinations of $x^1,\dots,x^m$.
Its polar cone is $K^\circ = \big\{ v\in \mathbb{R}^n:\langle v,x \rangle \leq 0, \; \forall x \in K\big\}$.
Since each particular $x^{i_0}$ is a nonnegative combination by taking its corresponding $\lambda_{i_0}>0$ and all other $\lambda_i = 0$, then it is possible to write:
$$ \langle v,x^i \rangle \leq 0, \; \forall i \Longleftrightarrow \sum_{j\,=\,1}^{n} x_j^i v_j\leq 0, \;\forall i $$
This can be substituted by $Av \leq 0$ where:
$$ A = (x_{ij})_{1\leq i \leq m,\;1\leq j \leq n} = x^i_j \;\;\; \text{ and } \;\;\; v=(v_j)_{1 \leq j \leq n} $$
As desired.
I feel there is something wrong in my approach. It feels too straightforward.
Any insight is greatly appreciated.