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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 $U = rref(A)$?

Jonathan Wang
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2 Answers2

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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.

user
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  • 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.

     – Jonathan Wang Dec 16 '17 at 03:38
  • Column of A span always a certain subspace with dimension n, to span that subspace you require numbers n of columns linearly independent. – user Dec 16 '17 at 03:50
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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?

Anne
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