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.
 A: 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,
$$
\begin{bmatrix}
-4 & -4 & 12 & 0 \\
-4 & -4 & 12 & 0 \\
4 & -2 & 0 &-6 \\
1 &-4 &7 &-5 \\ 
\end{bmatrix}
\stackrel{C_2-C_1,\, C_3+3C_1}{\longrightarrow}
\begin{bmatrix}
-4 & 0 & 0 & 0 \\
-4 & 0 & 0 & 0 \\
4 & -6 & 12 &-6 \\
1 &-5 &10 &-5 \\ 
\end{bmatrix}
\,\longrightarrow\cdots
\begin{bmatrix}
-4 & 0 & 0 & 0 \\
-4 & 0 & 0 & 0 \\
4 & -6 & 0 & 0 \\
1 &-5 & 0 & 0 \\ 
\end{bmatrix}
$$
A: It $T(v)=Av,~~~v\in\mathbb K^4$, then $$
        T(v)=
        \begin{pmatrix}
        1 & 0 & -1 & -1 \\
        0 & 1 & -2 & 1 \\
        0 & 0 & 0 & 0 \\
        0 & 0 & 0 & 0 \\ 
        \end{pmatrix}\begin{pmatrix}
        x  \\
        y  \\
        z  \\
        t  \\ 
        \end{pmatrix}=\begin{pmatrix}
        x-z-t  \\
        y-2z+t  \\
       0  \\
        0  \\ 
        \end{pmatrix}
$$ but $$\begin{pmatrix}
        1 & 0 & -1 & -1 \\
        0 & 1 & -2 & 1 \\
        0 & 0 & 0 & 0 \\
        0 & 0 & 0 & 0 \\ 
        \end{pmatrix}\approx\begin{pmatrix}
        1 & 0 & 0 & 0 \\
        0 & 1 & 0 & 0 \\
        0 & 0 & 0 & 0 \\
        0 & 0 & 0 & 0 \\ 
        \end{pmatrix}  $$ so $$T(v)=
        \begin{pmatrix}
        1 & 0 & 0 & 0 \\
        0 & 1 & 0 & 0 \\
        0 & 0 & 0 & 0 \\
        0 & 0 & 0 & 0 \\ 
        \end{pmatrix}\begin{pmatrix}
        x  \\
        y  \\
        z  \\
        t  \\ 
        \end{pmatrix}=\begin{pmatrix}
        x  \\
        y  \\
       0  \\
        0  \\ 
        \end{pmatrix}=x\begin{pmatrix}
        1  \\
        0  \\
       0  \\
        0  \\ 
        \end{pmatrix}+y\begin{pmatrix}
        0  \\
        1  \\
       0  \\
        0  \\ 
        \end{pmatrix}$$
A: There is no need to use row echelon form here...the question is only asking for a spanning set.
Consider a general vector $v = (x,y,z,w)^T$, then 
$Av = (-4x-4y+12z,-4x-4y+12z,4x-2y-6w,x-4y+7z-5w)$
$= x(-4,-4,4,1) + y(-4,-4,-2,-4) + z(12,12,0,7) + w(0,0,-6,-5)$
so those four vectors form a spanning set for the range. Notice that they are the columns of $A$.
Of course this isn't the BEST choice for a spanning set since they aren't linearly independent...but that would be your aim if you were finding a basis.
