I need help to understand a problem in Linear Algebra about calculating a set. I'm trying to find the following set (some authors call it "column space"):
Given a matrix $A \in \mathbb{R} ^{m \times n}$, we define:
$$R(A)=\{b \in \mathbb{R}^{m} | \exists \space x\in \mathbb{R} ^n, b=Ax \}.$$
Let $A= \left[ \begin{array}{ccc}
2 & 2 & 0 & 0 \\
3 & 4 & -1 & 2 \\
-1 & 1 & -2 & 4 \end{array} \right]$. 
My attempt goes as follows:
I'm trying to find the elements of $R(A)$. Notice that the elements of $R(A)$ are vectors in $\mathbb{R} ^3 $. I want to know how those elements are. If I take a generic vector $b=(b_{1},b_{2},b_{3})^T \in \mathbb{R}^3$, then I'm trying to solve the following equation system:
  $$ \left[ \begin{array}{ccc}
2 & 2 & 0 & 0   &|& b_{1} \\
3 & 4 & -1 & 2  &|& b_{2}\\
-1 & 1 & -2 & 4 &|& b_{3}\end{array} \right]$$
I calculated this echelon form for A.
$$ \left[ \begin{array}{ccc}
1 & -1 & 2 & -1   &|& -b_{3} \\
0 & 7 & -7 & 14  &|& b_{2}+3b_{3}\\
0 & 0 & 0 & 0 &|& b_{1}-\frac{4}{7}b_{2}+\frac{2}{7}b_{3} \end{array} \right]$$.
(I think it is not necessary to put the steps to find the echelon form, but if you need it, let me know).
If I want the system $Ax=b$ to have solutions, it's clear that we need that $b_{1}-\frac{4}{7}b_{2}+\frac{2}{7}b_{3}=0$, which is equivalent to $b_{1}=\frac{4b_{2}-2b_{3}}{7}$.
Then my conclussion is that the elements of $R(A)$ must be all of the form $$\left[ \begin{array}{ccc}
b_{1}  \\
b_{2} \\
b_{3} \end{array} \right]=\left[ \begin{array}{ccc}
\frac{4b_{2}-2b_{3}}{7}  \\
b_{2} \\
b_{3} \end{array} \right]$$
because I need that condition if I want the system $Ax=b$ having solution.
The question is, is this the correct way for doing that?
If there's not, which is the steps that I need to follow?
I'm just learning Linear Algebra.
Thanks :)
 A: Note that you obtained the basis vectors as $\begin{bmatrix}4\\7\\0\end{bmatrix},\begin{bmatrix}-2\\0\\7\end{bmatrix}$ which is correct. An alternative method to find a basis of $R(A)$ would be to assume $(x,y,z,w)\in\Bbb R^4$ and find its image under $A$:
$\begin{bmatrix}
2 & 2 & 0 & 0 \\
3 & 4 & -1 & 2 \\
-1 & 1 & -2 & 4 \end{bmatrix}\begin{bmatrix}x\\y\\z\\w\end{bmatrix}=\begin{bmatrix}2x+2y\\3x+4y-z+2w\\-x+y-2z+4w\end{bmatrix}=x\begin{bmatrix}2\\3\\-1\end{bmatrix}+y\begin{bmatrix}2\\4\\1\end{bmatrix}+z\begin{bmatrix}0\\-1\\-2\end{bmatrix}+w\begin{bmatrix}0\\2\\4\end{bmatrix}$
$\therefore R(A)=\text{span}\Big\{\begin{bmatrix}2\\3\\-1\end{bmatrix},\begin{bmatrix}2\\4\\1\end{bmatrix},\begin{bmatrix}0\\-1\\-2\end{bmatrix},\begin{bmatrix}0\\2\\4\end{bmatrix}\Big\}$
Notice that these are just the column vectors of $A^1$. The basis of $R(A)$ is given by the linearly independent vectors in the set; for example, $B=\Big\{\begin{bmatrix}2\\4\\1\end{bmatrix},\begin{bmatrix}0\\-1\\-2\end{bmatrix}\Big\}$. Can you show that the basis you got and $B$ are equivalent?


$^1$Let $T:V^n\to W$ be a linear transformation. If
  $v_1,v_2,v_3,...,v_n$ are the basis vectors of $V$, then
  $R(T)\subseteq W$ is spanned by $T(v_1),T(v_2),T(v_3),...,T(v_n)$.

For $V=\Bbb R^4$, the most convenient choice of basis is the standard basis $B=\Big\{\begin{bmatrix}1\\0\\0\\0\end{bmatrix},\begin{bmatrix}0\\1\\0\\0\end{bmatrix},\begin{bmatrix}0\\0\\1\\0\end{bmatrix},\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\Big\}$.
This gives $T((1,0,0,0))=A\begin{bmatrix}1\\0\\0\\0\end{bmatrix}$ as the first column of the matrix $A$, $T((0,1,0,0))=A\begin{bmatrix}0\\1\\0\\0\end{bmatrix}$ as the second column and so on. Thus, the range of $T$ is the column-space of $A$ and is spanned by the column vectors of $A$.
