# 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

## 2 Answers

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.

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.