# Do the pivot columns of a matrix in reduced row echelon form form a basis for the column space of the matrix?

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.

• 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. – Matt Samuel May 5 at 12:32
• – amd May 6 at 2:24