Prove that $conv\{e_1,e_2,-e_1,-e_2\} = \{ x \in \mathbb{R}^2 : |x_1| + |x_2| \leq 1 \} $ where $e_1 = (1,0)^T, e_2 = (0,1)^T$ where $conv\{ \}$ is the convex hull of the points.
I am new to convex optimization and I can see why the convex hull of the above points is given diamond shape $|x_1| + |x_2| \leq 1$
But how to prove that analytically?
Why not show that the convex hull of $\pm e_i$, $i=1, \ldots, n$ equals $\{ (x_i)\ | \sum_{i=1}^n |x_i|\le 1\}$. Clearly $\operatorname{conv} (\pm e_i)$ is contained in the set, since $(x_i) \mapsto \sum |x_i|$ is a norm, and the we have equality for each $\pm e_i$. Conversely, let $x=(x_i)$, with $\sum |x_i|\le 1$. Consider $\epsilon_i = \operatorname{sign} x_i$. Let's show that $x$ is in the convex hull of $\epsilon_i e_i$ and $0$. Indeed, we have $$x= \sum_{i=1}^n |x_i| \cdot \epsilon_i e_i + (1-\sum |x_i|) \cdot 0$$
Now, note that $0$ is a convex combination of $e_i$ and $-e_i$ ( $i$ arbitrary).