After RREFing a matrix and finding the pivot columns, why can I go back to the original matrix and say the same columns are linearly independent? Suppose I have the following.
$$A=
 \begin{bmatrix}
    1 & -2 & 0 & 3 & 2 \\
    2 & -4 & 1 & 2 & 5 \\
1 & -2 & 1 & -1 & 3 \\
3 & -6 & 2 & 1 & 8 \\
  \end{bmatrix}\longrightarrow
  R=\begin{bmatrix}
    1 & -2 & 0 & 3 & 2 \\
    0 & 0 & 1 & -4 & 1 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
  \end{bmatrix}
.$$
I can see columns 1 and 3 of R must be linearly independent. But how does this imply that the first and third columns of $A$ are linearly independent if row operations don't preserve the column space?
I know that if we get $A^{T}$ into upper triangular form we obtain $$
 \begin{bmatrix}
    1 & 2 & 1 &3 \\
    0 & 1 & 1&2\\
     0 & 0 &0&0\\
     0 & 0 &0&0\\
     0 & 0 &0&0\\ 
  \end{bmatrix}
$$
which implies that cols 1 and 3 of A are independent. This is because row operations in the transpose of A preserve its row space, which is the original column space of A, but this doesn't show how RREF preserves linear dependence of the columns.
 A: If $R$ is the RREF of the matrix $A$, then you can write
$$
R=FA
$$
where $F$ is invertible. This is one of the main points in row reduction. Now let's write $A=\begin{bmatrix} a_1 & a_2 & \dots & a_n \end{bmatrix}$ and $R=\begin{bmatrix} r_1 & r_2 & \dots & r_n \end{bmatrix}$ ($a_i$ and $r_i$ the columns of $A$ and $R$). Therefore, by definition of matrix product,
$$
r_i=Fa_i \qquad (i=1,2,\dots,n)
$$
Suppose a column of $A$ can be written as a linear combination of other columns of $A$:
$$
a_j=\alpha_1a_{i_1}+\dots+\alpha_ka_{i_k}
$$
Then
$$
r_j=Fa_j=F(\alpha_1a_{i_1}+\dots+\alpha_ka_{i_k})=\alpha_1Fa_{i_1}+\dots+\alpha_kFa_{i_k}=\alpha_1r_{i_1}+\dots+\alpha_kr_{i_k}
$$
Similarly you can go from linear relations between columns of $R$ to the same linear relation between the corresponding columns of $A$, by using $F^{-1}$.
Since a set of vectors is linearly dependent if and only if one of the vectors is a linear combination of the others, it follows that a set of column in $A$ is linearly independent if and only if the corresponding set of columns of $R$ is linearly independent.
Since the pivot columns in $R$ form a maximal linearly independent subset, the same holds for the corresponding columns of $A$.
We have even more: the entries in a nonpivot column in $R$ allow us to write it as a linear combination of the pivot column using precisely those entries as coefficients, we also know how to express a column of $A$ as a linear combination of the columns corresponding to the pivot columns.
In your case, we have $r_4=3r_1-4r_3$, so also $a_4=3a_1-4a_3$ as you can verify directly.
