Show that the extreme points of the unit simplex $\Delta_n$ are the unit vectors $e_1,\ldots ,e_n$

Question

Show that the extreme points of the unit simplex $\Delta_n$ are the unit vectors $e_1,\ldots ,e_n$.
I want to show that $ext(\Delta_n)=\{e_1,\ldots,e_n\}$
I've thought that proving $\rightarrow$ this part i would take some $x\in ext(\Delta_n)$ and assume that $x\neq e_i$ for all i but i got stuck.
any hints please?

Answer 1

So take some vector $x=\sum_{j=1}^n \alpha_je_j\in \Delta_n$ such that $x\not \in \{e_1,...,e_n\}$. Then, there exists $j_0$ such that $\alpha_{j_0}\in (0,1)$. Define $y=\frac{1}{\sum_{j\neq j_o}\alpha_j}(x-\alpha_{j_0} e_{j_0})$. Then, $y$ is well-defined since $\sum_{j\neq j_0} \alpha_j=1-\alpha_{j_0}\neq 0$ and $y\neq x$, since $\alpha_{j_0}\neq 0$. I'll leave it to you to check that $y$, indeed, lies in the simplex.

Furthermore,

$$ x= \alpha_{j_0} e_{j_0}+(1-\alpha_{j_0})y, $$ so $x$ is not an extreme point.

Now, let $1\leq i \leq n$ and assume that $e_i=tx+(1-t)y$ for some $x=\sum_{j=1}^n \alpha_j e_j,y=\sum_{j=1}^n \beta_j e_j\in \Delta_n$. Then, projecting on the $i$'th coordinate, we get that $1=t\alpha_i+(1-t) \beta_i$. But since $\alpha_i$ and $\beta_i$ are positive and smaller than $1$, this implies $\alpha_i=\beta_i=1$. Now, we simply observe that $\sum_{j=1}^n \alpha_j=\sum_{j=1}^n\beta_j=1,$ to get that $\alpha_j=\beta_j=0$ for all $j\neq i$. Thus, $x=y=e_i$. In conclusion, the canonical unit vectors are, indeed, extreme points of the $n$-simplex.