Showing linear independence using matrices. 
If $\mathbf{v}_1,...,\mathbf{v}_m \in F^n$ are written as rows of an $m\times n$ matrix $A$ and $B$ is the row-reduced echelon form of $A$, then $\{\mathbf{v}_1,...,\mathbf{v}_m\}$ is linearly independant if and only if $B$ has no all zero rows

I thought that a good way of showing this may be to show that linear dependence implies row of zeroes in reduced row echelon form and row of zeroes in reduced row echelon form implies linear dependence.
So I start from assuming that $B$ has a row of zeroes. I'm able to show that each row of $B$ is a linear combination of the rows of $A$ by considering the elementary row operations which reduce $A$ to $B$, and so since the rows of $A$ are $\mathbf{v}_1,...,\mathbf{v}_m$, I have that each row of $B$ is a linear combination of $\mathbf{v}_1,...,\mathbf{v}_m$.
So if $B$ has a row of zeroes, then we have that the zero vector is a linear combination of $\mathbf{v}_1,...,\mathbf{v}_m$. However, my issue is that I am unsure whether this linear combination could possibly be  $0\mathbf{v}_1+0\mathbf{v}_2+...+0\mathbf{v}_m = \mathbf{0}$ as in this case, having a row of zeroes wouldn't imply linear dependence of $\mathbf{v}_1,...,\mathbf{v}_m$. Also, I don't know how I would prove in the opposite direction: That if $\mathbf{v}_1,...,\mathbf{v}_m$ are linearly dependent then $B$ has a row of zeroes.
Can anyone help me with this? Is there a better way to approach this question?
 A: Nice problem :-)
This result is useful for to solve problem in linear algebra.  First, we will write an example of this statament amd then we will give a proof about this stament.
Example: Suppose that we need to analyze if the set $$S=\left\{\begin{pmatrix} 2 \\ 0 \\ 0\end{pmatrix}, \begin{pmatrix} 0 \\ 3 \\ 0\end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 4\end{pmatrix}   \right\}$$
I think it's clear that $S$ is independent linearly, but we will use that statement

Informally, your result says that if by putting my vectors as rows of a matrix and then reducing by rows, no row is transformed into a row filled with zeros, then I can state that the set I took to build the rows was linearly independent .

Let $$A=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{pmatrix}$$
using Gauss-Jordan's elimination we have $$A=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{pmatrix}\sim \cdots \sim \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}=B $$
since that no row in $B$ is filled with zeros, by the statement, we can see that the set is independent linearly.
Proof of the statament: Read here answer your question.
A: Note that $A$ and $B$ have the same row space, as the operations of Gauss-Jordan elimination leave the row space invariant.
So the span of $v_1,\dots,v_m$ is the same as the span of the rows of $B$. Further, the non-zero rows of $B$ are independent, from the construction of the echelon form.
So $B$ has no non-zero rows iff $A$ has rank $m$, i.e. iff $v_1,\dots,v_m$ are linearly independent.
