Let $A$ be a matrix, and let $B$ be the result of doing a row operation to $A$. Show that the null space of $A$ is the same as the null space of $B$. Let $A$ be a matrix, and let $B$ be the result of doing a row operation to $A$. Show that the null space of $A$ is the same as the null space of $B$.  
Any ideas of what I can do to show this? If the null spaces have to be equal then the vector $x$ in $Ax=0$ and $Bx=0$ have to be equal, but how can I show that $B$ is the result of doing a row operation to $A$ and that $A$ and $B$ are equal?
 A: Applying a row operation simply amounts to left-multiplying $A$ by an elementary matrix $E$. The crucial point is that $E$ is invertible, so $B = EA$ and $A = E^{-1}B$. From this is follows that $\ker A = \ker B$.
A: Doing a row operation on $A$ can be obtained by multiplying $A$ by an invertible matrix $E$. So $B=EA$ for a suitable invertible matrix $E$. If $Av=0$, then $Bv=EAv=0$. Since $A=E^{-1}B$, the converse inclusion also holds.
The matrix $E$ is obtained from the identity $m\times m$ matrix (if $A$ is $m\times n$) by applying the same elementary row operation.
A: Let $P$ be the matrix corresponding to the row operation performed.
We define the null space as the set L= $\left\{\vec{x} \in \mathbb{F}^n |\space A\vec{x} = 0\right\}$
 So, for any vector $\vec{x}\in L $ for $B$ we have: $B\vec{x}=PA\vec{x}=\vec{0}$
We work in the exact same manner for a vector in the null space of $B$, say $L'$ to show $L=L'$.
A: Any row operation is achievable by left-multiplying $A$ by some regular matrix. So we have $B=EA$, where $R$ is the (regular) matrix obtained by multiplying some finitely many (regular) matrices corresponding to the elementary row operations. Since multiplying by a regular matrix does not change the rank (there are many possible justifications for this), the result follows.
A: There exists an elementary matrix $E$ s.t.
$$B=EA$$
Does $B$ have the same nullspace as $A$?
$$Bx=0$$
$$\iff EAx = 0$$
$$\stackrel{?}{\iff} Ax = 0$$
$\leftarrow$ is obvious.
What about $\rightarrow$? Might all elementary matrices be invertible?
