= Solution
The independent-column representation gives $T(x)=c_1(x_1+3x_3)+c_2(x_2+2x_4)$. Thus a <vector> lies in the <kernel of a linear map> represented by this <matrix> precisely when $x_1=-3x_3$ and $x_2=-2x_4$. Writing the two free coordinates separately gives
$$
x=x_3\begin{pmatrix}-3\\0\\1\\0\end{pmatrix}
+x_4\begin{pmatrix}0\\-2\\0\\1\end{pmatrix}.
$$
The two displayed <vectors> are <linearly independent>, so \b[they form a <basis> of the kernel]. Its dimension two also agrees with the <rank-nullity theorem>, $4=2+2$.
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