B is obtained from A by elementary row operations, where A,B are 3x3 matrices in the reals then prove they have the same rank? $A,B \in M_{3}(\mathbb{R})$ if a series of ``Elementary Row Operations" are performed on them how do we show they have the same rank? Where I believe rank is the dimension of the span of the rows?
 A: By construction/definition, ERO's preserve the solution set of a system Ax=b (trivial to see for exchanges and for scaling, a little harder for shears ), so that, in particular, ERO's would preserve the set of solutions to Ax=0, i.e., the kernel, so, by rank-nullity, they would preserve the column space, and therefore its dimension, i.e., the rank of A.
A: Another way to think about the problem: elementary row operations are linear combinations of row vectors that do not change the span of the rows.
For instance, let
$$
A = 
\begin{pmatrix}
a^1_1    & \cdots & a^1_n   \\
\vdots   & \ddots & \vdots   \\
a^m_1    & \cdots & a^m_n
\end{pmatrix}
=
\begin{pmatrix}
a^1 \\
\vdots \\
a^m
\end{pmatrix}
$$
Then, if you interchange rows $i$ and $j$, obviously the span of the row vectors $[a^1 , \dots , a^m]$ doesn't change:
$$
[\dots , a^i , \dots , a^j, \dots] = [\dots , a^j , \dots , a^i, \dots]  \ .
$$
The same happens if you multiply one row by some $\lambda \neq 0$:
$$
[\dots , \lambda a^i , \dots ] = [\dots , a^i , \dots]
$$
because every linear combination you can do with the vectors on the right is also a linear combination with the vectors on the left and vice versa:
$$
\dots + \alpha_i (\lambda a^i ) + \dots = \dots + (\alpha_i \lambda ) a^i  + \dots
$$
and
$$
\dots + \alpha_i  a^i  + \dots = \dots + \frac{\alpha_i}{\lambda} (\lambda a^i ) + \dots
$$
Finally, if you add two rows, you also have
$$
[\dots , a^i , \dots , a^j, \dots ] = [\dots , a^i , \dots , a^i + a^j, \dots ] \ .
$$
Exercise: why?
