Find the spanning set of the range of the linear transformation $T(x)=Ax$. Let
$$
        A=
        \begin{bmatrix}
        -4 & -4 & 12 & 0 \\
        -4 & -4 & 12 & 0 \\
        4 & -2 & 0 &-6 \\
        1 &-4 &7 &-5 \\ 
        \end{bmatrix}
$$
Find the spanning set of the range of the linear transformation $T(x)=Ax$.
I have found the row reduced echelon form of A.
$$
        RREF(A)=
        \begin{bmatrix}
        1 & 0 & -1 & -1 \\
        0 & 1 & -2 & 1 \\
        0 & 0 & 0 & 0 \\
        0 & 0 & 0 & 0 \\ 
        \end{bmatrix}
$$
I don't know what to do with it after.

The range of $T$ is the column space of $A$. So the columns of $A$ already form a spanning set. If you want to find a linearly independent spanning set, you should find a column echelon form of $A$ instead of a row echelon form. I found that $\{(-4-4,4,1)^T,\,(0,0,-6,-5)^T\}$ is an answer, but depending on the column operations you perform, you may get a different answer.
Edit: for a starter,
$$
\begin{bmatrix}
-4 & -4 & 12 & 0 \\
-4 & -4 & 12 & 0 \\
4 & -2 & 0 &-6 \\
1 &-4 &7 &-5 \\ 
\end{bmatrix}
\stackrel{C_2-C_1,\, C_3+3C_1}{\longrightarrow}
\begin{bmatrix}
-4 & 0 & 0 & 0 \\
-4 & 0 & 0 & 0 \\
4 & -6 & 12 &-6 \\
1 &-5 &10 &-5 \\ 
\end{bmatrix}
\,\longrightarrow\cdots
\begin{bmatrix}
-4 & 0 & 0 & 0 \\
-4 & 0 & 0 & 0 \\
4 & -6 & 0 & 0 \\
1 &-5 & 0 & 0 \\ 
\end{bmatrix}
$$
