TITLE: Reduced row echelon form and linear independence
QUESTION [8 upvotes]: 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.

REPLY [7 votes]: 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.