Deriving left nullspace of matrix from $EA=R$ Let $A$ be an $m\times n$ matrix, $R$ be its row-reduced echelon form, and $E$ be the sequence of matrices $E_k\dots E_1$ used to bring $A$ to $R$, such that $EA = R$. In one of the books on linear algebra it is said that we can use the fact that $EA=R$ to find the basis for the left nullspace of $A$, without the need to bring $A^T$ to the row-reduced echelon form. Is my understanding correct that all we need to do is, since $(EA)^T=A^T$, just row-reduce $R^T$? Or, even better, just read the basis off from $R^T$ without even row-reducing it?
 A: Here are two detailed examples:
Find base and dimension of given subspace 
Given a matrix and its reduced row echelon form, resolve the image and the kernel.

Fundamental Theorem of Linear Algebra
Given $\mathbf{A}\in\mathbb{C}^{m\times n}$, the four fundamental subspaces are
$$
\begin{align}
%
  \mathbf{C}^{n} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus
    \color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\
%
  \mathbf{C}^{m} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus
    \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)}
%
\end{align}
$$

Column space
$$
\begin{align}
\left[
\begin{array}{c|c}
 \mathbf{A} & \mathbf{I}_{4} \\
\end{array}
\right]
& \mapsto
\left[
\begin{array}{c|c}
 \mathbf{E_{A}} & \mathbf{R} \\
\end{array}
\right]  \\
%
\left[
\begin{array}{rrr|cccc}
 1 & 1 & -1 & 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 & 1 & 0 & 0 \\
 -1 & 0 & 1 & 0 & 0 & 1 & 0 \\
 -2 & 1 & 2 & 0 & 0 & 0 & 1 \\
\end{array}
\right]
& \mapsto
\left[
\begin{array}{ccr|rrcc}
 \boxed{1} & 0 & -1 & 1 & -1 & 0 & 0 \\
 0 & \boxed{1} & 0 & 0 & 1 & 0 & 0 \\\hline
 0 & 0 & 0 & \color{red}{1} & \color{red}{-1} & \color{red}{1} & \color{red}{0} \\
 0 & 0 & 0 & \color{red}{2} & \color{red}{-3} & \color{red}{0} & \color{red}{1} \\
\end{array}
\right]
\tag{1}
\end{align}
$$
The boxed pivot entries identify the fundamental columns of the $\color{blue}{range}$ space. The $\color{red}{red}$ row vectors form a span for the $\color{red}{null}$ space.
The column space is now resolved.
$$
\begin{align}
%
  \mathbf{C}^{m} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus
    \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)}
%
% range
&=
\text{span } \left\{ \,
\color{blue}{\left[
\begin{array}{r}
  1 \\
  0 \\
 -1 \\
 -2 \\
\end{array}
\right]}, \,
\color{blue}{\left[
\begin{array}{r}
  1 \\
  1 \\
  0 \\
  1 \\
\end{array}
\right]},
\, \right\} 
% null
\oplus
\text{span } \left\{ \,
\color{red}{
\left[
\begin{array}{r}
  1 \\
 -1 \\
  1 \\
  0 \\
\end{array}
\right], 
\,
\left[
\begin{array}{r}
  2 \\
 -3 \\
  0 \\
  1 \\
\end{array}
\right]}
\, \right\}%
%
\end{align}
$$

Row space
$$
\begin{align}
%
\left[
\begin{array}{c|c}
 \mathbf{A}^{T} & \mathbf{I}_{4} \\
\end{array}
\right]
  & \mapsto 
\left[
\begin{array}{c|c}
 \mathbf{E_{A^{T}}} & \mathbf{R} \\
\end{array}
\right] \\
%
\left[
\begin{array}{rcrr|ccc}
  1 & 0 & -1 & -2 & 1 & 0 & 0 \\
  1 & 1 &  0 & 1 & 0 & 1 & 0 \\
 -1 & 0 &  1 & 2 & 0 & 0 & 1 \\
\end{array}
\right] 
  &\mapsto
\left[
\begin{array}{ccrr|rcc}
 \boxed{1} & 0 & -1 & -2 & 1 & 0 & 0 \\
 0 & \boxed{1} & 1 & 3 & -1 & 1 & 0 \\\hline
 0 & 0 & 0 & 0 & \color{red}{1} & \color{red}{0} & \color{red}{1} \\
\end{array}
\right]
\tag{2}
%
\end{align}
$$
The row space is now resolved:
$$
\begin{align}
%
  \mathbf{C}^{n} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus
    \color{red} {\mathcal{N} \left( \mathbf{A} \right)}
%
% range
&=
\text{span } \left\{ \,
\color{blue}{\left[
\begin{array}{r}
  1 \\
  1 \\
 -1 \\
\end{array}
\right]}, \,
\color{blue}{\left[
\begin{array}{r}
  0 \\
  1 \\
  0 \\
\end{array}
\right]},
\right\} 
% null
\oplus
\text{span } \left\{ \,
\color{red}{
\left[
\begin{array}{r}
  1 \\
  0 \\
  1 \\
\end{array}
\right]}
\, \right\}
%
\end{align}
$$

Challenge
As an example for this question, the task is to use the blue row vectors in (1) 
$$
\left[
\begin{array}{ccr|rrcc}
 \color{blue}{1} & \color{blue}{0} & \color{blue}{-1} & 1 & -1 & 0 & 0 \\
 \color{blue}{0} & \color{blue}{1} & \color{blue}{0} & 0 & 1 & 0 & 0 \\\hline
 0 & 0 & 0 & \color{red}{1} & \color{red}{-1} & \color{red}{1} & \color{red}{0} \\
 0 & 0 & 0 & \color{red}{2} & \color{red}{-3} & \color{red}{0} & \color{red}{1} \\
\end{array}
\right],
\tag{3}
$$
to find a vector in 
$$
\text{span } \left\{ \,
\color{red}{
\left[
\begin{array}{r}
  1 \\
  0 \\
  1 \\
\end{array}
\right]}
\, \right\}.
\tag{4}
$$
What the reduction in (1) provides is another span
$$
\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)}
=
\text{span } \left\{ \,
\color{blue}{\left[
\begin{array}{r}
  1 \\
  1 \\
 -1 \\
\end{array}
\right]}, \,
\color{blue}{\left[
\begin{array}{r}
  0 \\
  1 \\
  0 \\
\end{array}
\right]}
\, \right\}.
$$
And yes, you could look at that span and conclude (4). But this is not "reading" the vectors directly as in the red terms in (1) and (2). For example, can you look at (3) and guess the spans for the column space?
Conclusion
To summarize: There is no general method to resolve the row space from the reduction matrix $\left[
\begin{array}{c|c}
 \mathbf{E_{A}} & \mathbf{R} \\
\end{array}
\right]$.
The row vectors in the upper left quadrant of $\mathbf{E_{A}}$ are in the span of $\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)}$ and in some simple cases you will be able to deduce a span for $\color{red}{\mathcal{N} \left( \mathbf{A} \right)}$.
A: $EA = R$
It is trivial to find the vectors the vectors that span the null space of R.
for each vector your basis for the left null-space of R
i.e. (0,0,0,0,1) , (0,0,0,1,0) if R has 2 zero rows.
$\mathbf xEA = \mathbf xR = 0$
$\mathbf x E$ is in the null-space of A.
For however many zero rows are in $R$ that many of the bottom rows of $E$ span the left null space of $A.$ 
A: It is given that $EA=R$. Since $E$ is product of elementary matrix, it is invertible. So we can write $A=E^{-1}R$. Then take transpose on both sides which gives, $A^T=R^T(E^{-1})^T$. So to find the left nullspace of matrix $A$, we need to solve the system $A^T X=R^T(E^{-1})^TX=0$. Thus we don't need to calculate the row reduced form of $A^T$ to get the left nullspace of $A$.
