= Solution
Since $L$ is an <invertible matrix>, $Av=0$ is equivalent to $Uv=0$. The last nontrivial row gives $v_3=v_4$; the second gives $v_2=v_4$; and the first gives $2v_1=v_2+v_4$, so $v_1=v_4$. A <basis> of the <null space> is therefore
$$
\boxed{\mathcal B_{\ker A}=\{(1,1,1,1)^T\}.}
$$
Its single <basis vector> has <Euclidean norm> $2$, so the unit <null space> <basis vector> is $(1,1,1,1)^T/2$.
The <pivot elements> of $U$ lie in columns $1,2,3$. Take these columns from the \b[original] <matrix> $A$ to obtain a <basis> of its <column space>:
$$
\boxed{\mathcal B_{\operatorname{Col} A}=
\left\{\begin{pmatrix}-2\\2\\0\\0\end{pmatrix},
\begin{pmatrix}1\\-2\\1\\0\end{pmatrix},
\begin{pmatrix}0\\0\\-2\\2\end{pmatrix}\right\}.}
$$
The fourth column is minus the sum of these three columns, and the leading three-by-three <matrix minor> has nonzero <determinant> $-4$, confirming <linear independence>.
<Elementary row operations> preserve the <row space>. Thus the nonzero rows of $U$ give a <basis> of the <row space> of $A$:
$$
\boxed{\mathcal B_{\operatorname{Row} A}=
\{(-2,1,0,1),\ (0,-1,0,1),\ (0,0,-2,2)\}.}
$$
These <basis vectors> need not have unit <Euclidean norm> or be <orthogonal>. Their <Euclidean norms> are $\sqrt6,\sqrt2,2\sqrt2$, respectively; the three column <basis vectors> have <Euclidean norms> $2\sqrt2,\sqrt6,2\sqrt2$.
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