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.
Best Answer
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.