why does the reduced row echelon form have the same null space as the original matrix? What is the proof for this and the intuitive explanation for why the reduced row echelon form have the same null space as the original matrix?

Say we have an $n$x$n$ matrix, $A$, and are going to row reduce it. Every time we do a row operation, it is the same as multiplying on the left side by an invertible matrix corresponding to the operation. So at the end of the process, we can conclude something like $B = L_1L_2...L_kA$, where $B$ is the row reduced matrix, and the $L_i$ are the matrices corresponding to the row operations. 
The null space of $A$ is the set $\left\{\vec{x} \in \mathbb{R}^n |\space A\vec{x} = 0\right\}$, so for any $\vec{x}$ in this set:
$B\vec{x} = L_1L_2...L_kA\vec{x} = L_1L_2...L_k\vec{0} = \vec{0}$. 
Conversely, if $x$ is in the null space of $B$ ($B\vec{x} = \vec{0}$) then
$A\vec{x} = L_k^{-1}...L_2^{-1}L_1^{-1}L_1L_2...L_kA\vec{x} = L_k^{-1}...L_2^{-1}L_1^{-1}B\vec{x} = L_k^{-1}...L_2^{-1}L_1^{-1}\vec{0} = \vec{0}$
