If two coefficient matrices can be row reduced to be the same, then their columns have the same span. If two coefficient matrices can be row reduced to be the same, then their columns have the same span?
I said this statement is true because the columns in a matrix can be scaled up or down to look like a completely different matrix. However, the relation between each coordinate/variable (regardless of the vector's dimension) remains the same. For this reason, if two matrices row reduce to the same thing then their non-reduced versions are simply scales of each other.
Am I thinking correctly on this? Is the statement true?
Thanks.
 A: These matrices, $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$ have the same RREF. 
I don't think scaling can help in terms of making the column space the same.
A: Note that since the column operations don't change the colomn space, if you reduce to echelon form by column operations the statement is valid otherwise  if you reduce to echelon form by row operations the statement is uncorrect.
A: I was thinking the same, that perhaps information is lost after row reduction. But when you row reduce a matrix A, the position of the pivot columns tell you which columns you can pick to create the column space for  A. I'll show you an example:
\begin{equation}
A = \begin{pmatrix} 1&2&0 \\ 2&4&0\\1&2&1 \end{pmatrix} , \:\; B = \begin{pmatrix} 0&1&2 \\ 0&2&4\\1&1&2 \end{pmatrix}
\end{equation}
Notice that $\text{Col}(A) = \text{Col}(B)$, and that for $A$ and $B$ one of the columns is two times larger than the other. When we row reduce both, we get:
\begin{equation}
\texttt{rref(A)} = \begin{pmatrix} 1&2&0 \\ 0&0&1\\0&0&0 \end{pmatrix} , \:\; \texttt{rref(B)} = \begin{pmatrix} 1&0&0 \\ 0&1&2\\0&0&0 \end{pmatrix}
\end{equation}
We know $A$ and $B$ span the same space, but their row reduced forms are different. The pivot columns for A are in the first and third column. That means that I can remove the second column in A. The pivot columns for B are in the first and second, so the first two columns of B span the column space. Notice that when we rowreduce two different matrices and end up with the same matrix, and that the $n$-th column is not a pivot column, we can remove the $n$-th column from both matrices. The dimension would be the same, BUT the spans could be completely different.
