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In the lecture notes I'm working through, it says that for a matrix A, the pivot columns of the matrix in reduced row echelon form are a basis for the column space of A. I've also seen some 'proofs' on the internet that support this.

But this doesn't seem correct: take for example $$ \begin{pmatrix} 2 & 1 & 4 \\ 2 & 1 & 4 \\ 2 & 1 & 4 \\ \end{pmatrix} $$

Then in reduced row echelon form this is :$$ \begin{pmatrix} 1 & 0.5 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} $$

and the pivot column is$$ \begin{pmatrix} 1 \\ 0 \\ 0 \\ \end{pmatrix} $$

But there's no way to write one of the columns of the original matrix say

$$ \begin{pmatrix} 1 \\ 1 \\ 1\\ \end{pmatrix} $$ as a linear combination of the pivot column.

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    $\begingroup$ The rows form a basis for the row space though. As you can see it's not true that the pivot columns form a basis for the column space unless you do column reduction. $\endgroup$ – Matt Samuel May 5 at 12:32
  • $\begingroup$ See math.stackexchange.com/q/1044230/265466. $\endgroup$ – amd May 6 at 2:24
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When you do row reduction, you are constructing a basis of the row space by eliminating dependent rows and ending up with a linearly independent set. It has little relation to the column space, and as you can see the columns do not necessarily form a basis for the column space. However, the column space has the same dimension as the space spanned by the pivot columns, so it is related by a change of basis.

The change of basis is described by the row operations that you perform, so it's not exactly arbitrary.

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Your lecture notes are wrong (or you’re misinterpreting them). The correct statement is that the columns of the original matrix that correspond to pivot columns in the RREF are a basis for the column space.

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