Giving a basis for the column space of A 
Let $A = \begin{bmatrix}3&3&3\\3&5&1\\-2&4&-8\\-2&-4&0\\4&9&-1\end{bmatrix}$
Give a basis for the column space of A

So what I've done so far is put it in RREF (which was a task itself) and got
$\begin{bmatrix}1&0&2\\0&1&-1\\0&0&0\\0&0&0\\0&0&0\end{bmatrix}$, but I'm not sure what to do next to give the "basis" of the column space of A


Fact. Let $A$ be a matrix. The nonzero rows of $\DeclareMathOperator{rref}{rref}\rref(A^\top)$ form a basis of the column space of $A$.

In our case we have
$$
A=
\left[\begin{array}{rrr}
3 & 3 & 3 \\
3 & 5 & 1 \\
-2 & 4 & -8 \\
-2 & -4 & 0 \\
4 & 9 & -1
\end{array}\right]
$$
Row-reducing $A^\top$ gives
$$
\rref(A^\top)=
\left[\begin{array}{rrrrr}
1 & 0 & -\frac{11}{3} & \frac{1}{3} & -\frac{7}{6} \\
0 & 1 & 3 & -1 & \frac{5}{2} \\
0 & 0 & 0 & 0 & 0
\end{array}\right]
$$
The fact above implies that
\begin{align*}
\langle1, 0, -{11}/{3}, {1}/{3}, -{7}/{6} \rangle
&&
\langle0,1, 3, -1, 5/2 \rangle
\end{align*}
forms a basis of the column space of $A$.
