Prove row space of RREF equals row space of matrix 
Let $X$ be a $m\times n$ matrix, Let $R = \text{RREF } (X)$. Show that $\text{row } X = \text{row } R$

Hints only please
I don't see how to approach this, I know the definition of row space is
$\text{row } X = \{X^{T}\overrightarrow{x}| \overrightarrow{x} \in \mathbb{R}^m\}$
So basically we are trying to show:
$$X^T x = R^T x$$ for any $x \in \mathbb{R}^n$, but I have no clue why this is even true or how this true.
It is not in general true that $X^T = R^T$ is it?
 A: You are correct to say that in general $X^T \neq R^T$. Since you asked for hints only I'll just say
1) If you multiply two matrices $A\cdot B=C$, then by just writing out the definition of $A\cdot B$ shows each row of $C$ is in the rowspace of $B$.
2) There is a sequence of elementary row operation matricies $E_i$ such that 
$$
E_n \cdots E_i X = R
$$
A: Interpret ${}^\mathrm{t\!}X$ as the matrix of a linear map $f\colon\mathbf R^m\to\mathbf R^n$ and the row-rank of $X$ is the rank of this linear map.
We perform on $X$ a succession of elementary row operations, which correspond to left-multiplying $X$ by invertible  $m\times m$ matrices of elementary operations $A_k$. Transposing,  ${}^\mathrm{t\!}X$ is right-multiplied  by the transposed matrices ${}^\mathrm{t\!}A_k$ in the reverse order. Eventually,  ${}^\mathrm{t\!}X$ is right-multiplied by the  invertible $m\times m$ matrix 
$$B={}^\mathrm{t}\Bigl(\prod_k A_k\Bigr).$$
Now, $B$ is the matrix of an automorphism $u$ of $\mathbf R^m$, and  $ \;{}^\mathrm{t\!}XB$ is the matrix of the linear map $f\circ u$. The rank of a linear map is the same after composition  on either side by an isomorphism. Thus
$$\operatorname{rank}(f\circ u)=\operatorname{rank}(f)\iff \operatorname{rank}({}^\mathrm{t\!}XB)=\operatorname{rank}({}^\mathrm{t\!}X).$$
