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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 $U = rref(A)$?

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

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  • $\begingroup$ Aren't the requirements for a basis: 1) Linearly independent vectors 2) Span the given subspace How do we know that the pivot columns span the subspace as well (in this case, the subspace is Range(A). I get the part on how they're linearly independent. $\endgroup$ – Jonathan Wang Dec 16 '17 at 3:38
  • $\begingroup$ Column of A span always a certain subspace with dimension n, to span that subspace you require numbers n of columns linearly independent. $\endgroup$ – user Dec 16 '17 at 3:50
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With regards to showing that the "pivot columns" span the range of $A$:

Can you show that the non-pivot columns can be written as linear combinations of the pivot columns?

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