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?

The operations (elementary row operations) that occur in Gauss-Jordan elimination for the purposes of row reduction are mathematically equivalent to left multiplication by invertible matrices known as elementary matrices; these in fact generate the general linear group of $n\times n$ invertible real matrices.
Because elementary matrices are invertible, it follows that left multiplication does not change the kernel (also known as the null space). In other words, $$\ker EA=\ker A$$ where $E$ is an elementary matrix for all suitable matrices $A$.
Since no single elementary matrix changes the kernel, it follows that a product of any finite number of elementary matrices will not change the kernel, either.
