Let extreme points be the set X={x greater or equal to 0 given Ax=b for vector x and b} and a point x is extreme if for all y,z in X, x=(1-a)y+az for a=[0,1]
Basic Feasible solution x is if for A be matrix m*n, there are at most m non-zero entries.
I would like to see why it is equivalent (the proof) as i dont understand some notes given online (http://www.statslab.cam.ac.uk/~qb204/teaching/optim_2019.pdf, theorem 3.1) Thank you!
Let $x\in X$ be a feasible point and $I\subset\{1,2,...,n\}$ be the subset of indices so that $x_{i}=0$ if and only if $i\in I$. Then if $$|I|+m<n$$ the expanded system $$ Av=0,~v_i=0~~\forall i\in I $$ has a non-zero solution by simple rank considerations, the solution space of that system has dimension at least $n-(m+|I|)>0$.
Take such a vector $v$ and observe that $x+tv\in X$ for small positive and negative $t$. Take an $\varepsilon$ so that $x\pmεv\in X$, then $x=\frac12(x+εv)+\frac12(x-εv)$ is not an extremal point of the set.
Thus in any extremal point of $X$, there are at least $n-m$ components that are zero, or in other words, at most $m$ non-zero entries.