Let $$A=\left[\begin{array}{cccc} 6 &-8 &-7 &1\cr -2 &6 &4 &-2\cr 2 &-6 &-4 &2\cr 2 &-6 &-4 &2 \end{array}\right].$$
Find a basis for the image of $A$ (or, equivalently, for the linear transformation $T(x)=Ax$).

A basis for the image of $A$ is $\Bigg\lbrace$ $\left[\Rule{0pt}{4.8em}{0pt}\right.$$\left]\Rule{0pt}{4.8em}{0pt}\right.$, $\left[\Rule{0pt}{4.8em}{0pt}\right.$$\left]\Rule{0pt}{4.8em}{0pt}\right.$ $\Bigg\rbrace.$