<s>[INST] 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.
 [/INST] 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 $$T(v)=
        \begin{pmatrix}
        1 & 0 & 0 & 0 \\
        0 & 1 & 0 & 0 \\
        0 & 0 & 0 & 0 \\
        0 & 0 & 0 & 0 \\ 
        \end{pmatrix}\begin{pmatrix}
        x  \\
        y  \\
        z  \\
        t  \\ 
        \end{pmatrix}=\begin{pmatrix}
        x  \\
        y  \\
       0  \\
        0  \\ 
        \end{pmatrix}=x\begin{pmatrix}
        1  \\
        0  \\
       0  \\
        0  \\ 
        \end{pmatrix}+y\begin{pmatrix}
        0  \\
        1  \\
       0  \\
        0  \\ 
        \end{pmatrix}$$
</s>