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, 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.
