<s>[INST] Finding more than one basis for a column space Let's say the matrix A is defined as the following:
$$A=\begin{pmatrix}6 & 3 & -1 & 0 \\ 1 & 1 & 0 & 4 \\ -2 & 5 & 0 & 2\end{pmatrix}$$
Whose RRE form is:
$$A_{RRE}=\begin{pmatrix}1 & 0 & 0 & * \\ 0 & 1 & 0 & * \\ 0 & 0 & 1 & *\end{pmatrix}$$
Then I can write a basis for the column space of matrix A using the three linearly independent vectors as such:
$$B_1=\left\{\begin{pmatrix}6\\ 1 \\ -2\end{pmatrix},\begin{pmatrix}3\\ 1 \\ 5\end{pmatrix},\begin{pmatrix}-1\\ 0 \\ 0\end{pmatrix}\right\}$$
However, what if I want to find a different basis for colA besides the one above, such that no vector in the basis is a scalar multiple of any of the vectors in B1? Is this method valid?
1. Re-arrange columns of matrix, which won't change the column span:
$$A'=\begin{pmatrix}0 & 6 & 3 & -1 \\ 4 & 1 & 1 & 0 \\ 2 & -2 & 5 & 0\end{pmatrix}$$
2. Put into RRE form:
$$A'_{RRE}=\begin{pmatrix}1 & 0 & 0 & * \\ 0 & 1 & 0 & * \\ 0 & 0 & 1 & *\end{pmatrix}$$
3. Preliminary basis:
$$B'_2=\left\{\begin{pmatrix}0\\ 4 \\ 2\end{pmatrix},\begin{pmatrix}6\\ 1 \\ -2\end{pmatrix},\begin{pmatrix}3\\ 1 \\ 5\end{pmatrix}\right\}$$
4. Take linear combinations of above basis:
$$B_2=\left\{\begin{pmatrix}0\\ 4 \\ 2\end{pmatrix},\begin{pmatrix}6\\ 5 \\ 0\end{pmatrix},\begin{pmatrix}9\\ 2 \\ 3\end{pmatrix}\right\}$$
I know a better method would be to just take linear combinations of the existing basis B1, but I want to know if this method works and if not, why.
 [/INST] Yes it works fine and all the basis you have found are correct as any triple of linearly independent vectors. 
A simpler choice would be: $v_1=(1,1,0)$, $v_2=(0,1,0)$, $v_3=(0,0,1)$.
</s>