Choosing the columns of the reduced column form of the matrix as the basis elements? I was trying this tutorial problem Finding the basis and its dimension, I am trying to understand why she has not taken the first three columns of the reduced matrix as its basis elements rather she took the first three columns(that is the corresponding columns of the reduced matrix) as the basis elements as can be seen at slot 7:15.
 A: We are interested in a basis of the column space. Row operations does not preserve the column space.
As an example, $A_1 = \begin{bmatrix}0  & 0 \\ 1 & 0 \end{bmatrix}$ has its RREF being $R_1 = \begin{bmatrix}1  & 0 \\ 0 & 0 \end{bmatrix}$. Clearly $\left\{ \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right\}$ and $\left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}\right\}$ span different space.
Another example is , $A_2 = \begin{bmatrix}1  & 1 \\ 2 & 2 \end{bmatrix}$ has its RREF being $R_2=\begin{bmatrix}1  & 1 \\ 0 & 0 \end{bmatrix}$. Clearly $\left\{ \begin{bmatrix} 1 \\ 2 \end{bmatrix}\right\}$ and $\left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}\right\}$ span different space.
It is known that the pivot columns spans the column space. Hence, she took the corresponding column in the original matrix.
Edit:
The column space of $A_1$ can be described as $\operatorname{Span}\left\{ \begin{bmatrix} 0 \\ 1 \end{bmatrix},\begin{bmatrix} 0 \\ 0 \end{bmatrix} \right\}=\operatorname{Span}\left\{ \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right\}$
The column space of $R_1$ can be described as $\operatorname{Span}\left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix},\begin{bmatrix} 0 \\ 0 \end{bmatrix} \right\}=\operatorname{Span}\left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}\right\}$
