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.
 A: The "row rank" of $A$ is the number of linearly independent rows it has, and the "column rank" the number of its linearly independent columns.  The key facts are (for any matrix $A$) that:


*

*The row rank is equal to the column rank.

*The row (equiv. column) rank is unchanged by elementary row operations.
Therefore you can get the rank of $A$ (as we say for simplicity) by counting the leading ones of its row echelon form.  Since $S$ has $n$ vectors, we need the rank of $A$ to be $n$ (it cannot be more) in order for $S$ to be a linearly independent set.
A: Yes, if you can convert the matrix into reduced row echelon form(or even just row echelon form) without a row of $0$s,then the vectors are linearly independent.
A: 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.
