Column space of a 3x3 matrix Let $A$ be a $3 \times 3$ matrix $A$ with reduced row echelon form $
\left[ {\begin{array}{c}
 1 & 0 & 2 \\
 0 & 1 & 0 \\
 0 & 0 & 0
 \end{array} } \right]$.
Then
$\left[ {\begin{array}{c}
 1 \\
 0 \\
 0
 \end{array} } \right]$ and
$\left[ {\begin{array}{c}
 0 \\
 1 \\
 0
 \end{array} } \right]$ spans the column space of $A$.
I thought this would be a true statement but apparently it's not.
Isn't column space of A
$$span(\left[ {\begin{array}{c}
 1 \\
 0 \\
 0
 \end{array} } \right],
\left[ {\begin{array}{c}
 0 \\
 1 \\
 0
 \end{array} } \right],
\left[ {\begin{array}{c}
 2 \\
 0 \\
 0
 \end{array} } \right]) = span(\left[ {\begin{array}{c}
 1 \\
 0 \\
 0
 \end{array} } \right],
\left[ {\begin{array}{c}
 0 \\
 1 \\
 0
 \end{array} } \right]) $$  
Thank you in advance!

The column space of the original matrix is spanned by the columns of the original matrix which, when reduced to row echelon form, contain a pivot. So for a counterexample, consider:
$$
A = \begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 0 \\
0 & 1 & 0
\end{bmatrix}
$$
whose row reduced echelon form has the required form. Notice that:
$$
\text{Col}(A) = \text{span}\left\{
\begin{bmatrix}
1 \\ 0 \\ 0
\end{bmatrix},
\begin{bmatrix}
0 \\ 1 \\ 1
\end{bmatrix}
\right\}
\neq
\text{span}\left\{
\begin{bmatrix}
1 \\ 0 \\ 0
\end{bmatrix},
\begin{bmatrix}
0 \\ 1 \\ 0
\end{bmatrix}
\right\}
$$
