= Solution
<Gaussian elimination> gives
$$
\operatorname{rref}(A)=\begin{pmatrix}1&0&2&1\\0&1&1&2\\0&0&0&0\end{pmatrix}.
$$
The first two original columns form an <image> <basis>:
$$
\boxed{\left\{\begin{pmatrix}1\\2\\1\end{pmatrix},\begin{pmatrix}-1\\1\\0\end{pmatrix}\right\}.}
$$
Their independence follows already from their third and first coordinates. The third column is twice the first plus the second, and the fourth is the first plus twice the second, so they span every output of $T$.
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