Reduced row echelon form and linear independence Let's say I have the set of vectors $S = \{v_1, v_2, ..., v_n\}$ where $v_j \in R^m$, $v_j = (a_{1j}, a_{2j}, ..., a_{mj})$.
If the matrix formed by each of the vectors $A=[v_1, v_2, ..., v_n]$ looks like this (I believe), which is not a square matrix:
$$A = \begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots & a_{mn}
\end{pmatrix}$$
Then does A's reduced row echelon form help me determine whether the vectors of $S$ are linear dependent or independent? If so, how?
I hope I got all the indices, notation and terminology right, since I am a beginner in linear algebra, and English is not my native language.

Yes, but since you are considering the column vectors, you know that they are all independent if each column has a leading one in the reduced row-echelon form.
If we look at an explicit example:
$A = \begin{bmatrix}
    1       & 3 & -1 & 0 \\
    4       & 1 & 7 & 11 \\
    0       & 4 & -4 & -4 \\
    2       & 0 & 4 & 6 
\end{bmatrix}$
Then we have
$RREF(A)=\begin{bmatrix}
    1       & 0 & 2 & 3 \\
    0       & 1 & -1 & -1 \\
    0       & 0 & 0 & 0 \\
    0       & 0 & 0 & 0 
\end{bmatrix}$
Notice that since columns $3$ and $4$ in $RREF(A)$ do not have leading $1$'s, they are dependent on the first two columns. Specifically:
$\begin{bmatrix}
    -1        \\
   7 \\
-4 \\
4
\end{bmatrix}=2\begin{bmatrix}
    1        \\
   4 \\
0 \\
2
\end{bmatrix}-\begin{bmatrix}
    3        \\
   1 \\
4 \\
0
\end{bmatrix}$
$\begin{bmatrix}
    0        \\
   11 \\
-4 \\
6
\end{bmatrix}=3\begin{bmatrix}
    1        \\
   4 \\
0 \\
2
\end{bmatrix}-\begin{bmatrix}
    3        \\
   1 \\
4 \\
0
\end{bmatrix}$
Notice that the entries in column $3$ of $RREF(A)$ respectively correspond to the scalars for columns $1$ and $2$, such that column $3$ can be written as a linear combination of those columns. And the same goes for column $4$.
But if each column of $RREF(A)$ has a leading $1$, then each column is linearly independent.
Sidenote: The number of rows with all zeros in $RREF(A)$ tell only how many rows are independent/dependent (unless it is a square matrix, then the number of (in)dependent rows/columns will be the same) Looking at the leading 1's is much more helpful in determining linear dependence.
