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?

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.
