Linear algebra question on rank and null space This is an exercise from linear algebra and optimization by Gill, I do exercises to be prepared for my final exam and these are not homework questions! 

Exercise $\mathbf{6.1.}\,$ Consider the following matrix $A$ and vector $b$:
  $$
A=\begin{pmatrix}
2 & 4 \\
1 & 2  \\
1 & 2   \\
\end{pmatrix},\quad 
b=\begin{pmatrix}
3  \\
2   \\
1    \\
\end{pmatrix}.
$$
  $\textbf{(a)}$ What is the rank of $A$? Give a general form for any vector in the range of $A$.
$\textbf{(b)}$ Show that the dimension of the null space of $A^T$ is two, and display two linearly independent vectors $z_1$ and $z_2$ in $\operatorname{null}(A^T)$. Give a general form for every vector in $\operatorname{null}(A^T)$.
$\textbf{(c)}$ Find the vectors $b_R\in\operatorname{range}(A)$ and $b_N\in\operatorname{range}(A^T)$ such that $b=b_R+b_N$.
$\textbf{(d)}$ Give the general form of $b_A$ such that $b_R=Ab_A$. (Hint: consider all vectors $q$ such that $Aq=0$.)

For part (a), I think $rank(A)=1$ since all the columns are linear combination of each other. As for a general form for any vector in the range of $A$, when I write $Ax=b$ I get an over-determined system:
$$2x_1+4x_2=3$$
$$x_1+2x_2=2$$ 
$$x_1+2x_2=1$$ 
So I'm not sure about the general form. 
As for part (b), since the rank is two, dimension of $N(A^T)$ must be $3-1=1$, right?
And no idea about the rest. Any help would be greatly appreciated. 
 A: If you do row reduction, you find
$$
\begin{pmatrix}
2 & 4 \\
1 & 2 \\
1 & 2
\end{pmatrix}
\to
\begin{pmatrix}
1 & 2 \\
0 & 0 \\
0 & 0
\end{pmatrix}
$$
which means that the first column is a basis for the column space of $A$ (which is better terminology than “range of $A$”, in my opinion). So the general form of the vectors in the column space of $A$ is
$$
\begin{pmatrix}
2a \\
a \\
a
\end{pmatrix},\quad \text{$a$ any scalar}
$$
By the rank nullity theorem, the null space of $A$ has dimension $1$; the equation defining it is
$$
x_1+2x_2=0
$$
so a basis for it is the single vector
$$
\begin{pmatrix}
-2 \\
1
\end{pmatrix}
$$
The null space of $A^T$ has indeed dimension $2$; the row reduction on $A^T$ is
$$
\begin{pmatrix}
2 & 1 & 1 \\
4 & 2 & 2
\end{pmatrix}
\to
\begin{pmatrix}
2 & 1 & 1 \\
0 & 0 & 0
\end{pmatrix}
$$
so the equation defining the null space is $2x_1+x_2+x_3=0$ and a basis for it is
$$
\left\{
\begin{pmatrix}
-1 \\ 2 \\ 0
\end{pmatrix}
\,,
\begin{pmatrix}
-1 \\ 0 \\ 2
\end{pmatrix}
\right\}
$$
Writing $b=b_R+b_N$ should now be easy: the system to solve is
$$
\left(\begin{array}{ccc|c}
2 & -1 & -1 & 3 \\
1 & 2 & 0 & 2 \\
1 & 0 & 2 & 1
\end{array}\right)
$$
but you can as well find the orthogonal projection of $b$ on the column space of $A$:
$$
b_R=
\frac{(2\ 1\ 1)\begin{pmatrix}3\\2\\1\end{pmatrix}}
     {(2\ 1\ 1)\begin{pmatrix}2\\1\\1\end{pmatrix}}
\begin{pmatrix}2\\1\\1\end{pmatrix}
=\frac{9}{6}\begin{pmatrix}2\\1\\1\end{pmatrix}
=\begin{pmatrix}3\\3/2\\3/2\end{pmatrix}
$$
and $b_N=b-b_R$.
With this last idea it's easy to solve the last point.
