The replacement argument doesn't convert one set into another set. What it does is takes a linear independent set and a spanning set and creates a basis (a linear independent spanning set). And as a corollary, it shows that every basis has the same size (the dimension of the vector space).
For instance, if you had
A = {(0,2,2)^T}
B= {(1,2,0)^T, (2,0,1)^T, (2,2,0)^T, (1,1,3)^T}
You would take the set
{(0,2,2)^T, (1,2,0)^T, (2,0,1)^T, (2,2,0)^T, (1,1,3)^T}
first remove (1,1,3)^T since it's a linear combination of the others
(1,1,3)^T = 11/10*(0,2,2)^T - 3/5*(1,2,0)^T + 4/5*(2,0,1)^T
So you're left with
{(0,2,2)^T, (1,2,0)^T, (2,0,1)^T, (2,2,0)^T}
Since {(0,2,2)^T, (1,2,0)^T} are linear independent (since one is not a multiple of the other), we must check them both against (2,0,1)^T
<br />
\left|<br />
\begin{array}{ccc}<br />
0 & 1 & 2\\<br />
2 & 2 & 0\\<br />
2 & 0 & 1<br />
\end{array}<br />
\right| = -4<br />
So these three vectors are linearly independent.
However, (2,2,0)^T is a linear combination of the others (you can check this)
so you can remove it, to get the basis
{(0,2,2)^T, (1,2,0)^T, (2,0,1)^T)}
I suppose if you want, you can use that technique to replace all of the vectors with the vectors from a basis, but it's really pretty useless since you already know that you have a basis. The theorem is really used (as far as I know, I'll be studying this more next year) to show that any linearly independent set of vectors of a vector space is less than or equal in size to any spanning set. This can also be used to show that all bases have the same size.
