Why are pivot columns the basis of A?

Why is it that when we determine the pivot columns of an m x n matrix $A$, the pivot columns form a basis for the $Range(A)$? I understand that the pivot columns are linearly independent (the reason why we chose them as the pivot columns), but how do we know that they also span the range of $A$?

Also, why is that we're choosing the columns of $A$ to be the vectors that form a basis for $Range(A)$ rather than the columns of $U = rref(A)$?

• Basically a duplicate of math.stackexchange.com/q/2521737/265466. – amd Dec 16 '17 at 3:54
• @Jonathan Wang If you are ok, you can set as solved. Thanks! – user Dec 17 '17 at 8:00

The pivot columns form a basis for the $Range(A)$ since they are linearly independent.
We're choosing the columns of $A$ to be the vectors that form a basis for $Range(A)$ rather than the columns of $U = rref(A)$ because $U$ is generally obtained by row operations which modify the column space.
With regards to showing that the "pivot columns" span the range of $A$: