I have a simplex, n + 1 points in $\mathbb{R}^n$, 
which may have rank $r < n$.  

Is there a cheap way of "inflating" it to rank $n$,
changing a few, all but $r$, of the points ? 


The points are also ordered, rows 1 2 3 $\dots$ in a matrix, and
I'd like to keep as many of the leading rows as possible unchanged.



Inflate a simplex, change rows to make the rank n
You are given a matrix with $ n + 1 $ rows (one for each point) and $ n $ columns. Without loss of generality we can say the first point is the origin, so the first row is all zeros and we delete it from the matrix to get a square $ n $ by $ n $ matrix $ A $.


Now you want to modify as few rows as possible of $ A $ so that the span of the rows of $ A $ is $ \mathbb R^n $. But the fundamental theorem of linear algebra says $$ \text{nulspace}(A) = (\text{span of rows}(A))^\perp = \text{image}(A^T)^\perp $$


So find a basis for the nulspace of $ A $, and for each vector you get replace one of the linearly dependent rows of $ A $ with that vector.


