Compute basis for nullspace by inversion and transpose A known technique to compute a basis for the nullspace of a matrix $A^T$ is to extend $A$ with arbitrary columns $V$ so that $B = [A V]$ is nonsingular. After computing $B^{⁻T} = [Y Z]$, $Z$ is a basis for the nullspace of $A^T$.
Does this hold also when $B$ is not as clearly partitioned into $A$ and $V$?
Example: Let $A = \begin{bmatrix}a_1 & a_2 & a_3\end{bmatrix}, V = \begin{bmatrix}v_1 & v_2\end{bmatrix}$ with $a_i, v_i \in \mathbb{R}⁵$. Let $B = \begin{bmatrix}a_1 & v_1 & a_2 & v_2 & a_3\end{bmatrix}$. Is a basis for the nullspace of $A^T$ given by the 2nd and 4th column of $B^{⁻T}$?
Example: 
$A^T = \begin{bmatrix} 2 & 3 & 4\\ 1 & 0 & 1\end{bmatrix}$. 
Choose $V = \begin{bmatrix}1\\1\\1\\ \end{bmatrix}$.
Then $B = \begin{bmatrix} 2 & 1 & 1\\ 3 & 0 & 1\\ 4 & 1 & 1\end{bmatrix}$, $B^{-T} = \begin{bmatrix}-0.5 & 0.5 & 1.5\\ 0 & -1 & 1\\ 0.5 & 0.5 & -1.5\end{bmatrix}$, and $Z = \begin{bmatrix}1.5\\ 1\\ -1.5\end{bmatrix}$ is a valid basis for the nullspace of $A^T$.
If instead $V$ is injected inbetween the two colums of $A$, $B = \begin{bmatrix} 2 & 1 & 1\\ 3 & 1 & 0\\ 4 & 1 & 1\end{bmatrix}$, $B^{-T} = \begin{bmatrix}-0.5 & 1.5 & 0.5\\ 0 & 1 & -1\\ 0.5 & -1.5 & 0.5\end{bmatrix}$. The second column, corresponding to $V$, is now a valid basis of the nullspace of $A^T$. Is this a general result?
 A: To begin with, let’s deal with some unstated assumptions in your question. Obviously, $A$ must have more rows than columns. In addition, it must have full rank. If the columns of $A$ are not linearly independent, then there’s no way to fill it out so that the resulting square matrix is nonsingular. So, we’ll assume that $A$ is an $m\times n$ matrix with $m\gt n$ and $\operatorname{rank}(A)=n$.  
With that, let’s examine why the method works. The null space of a matrix is the orthogonal complement of its row space, or equivalently, the orthogonal complement of the column space of its transpose. Now, for any appropriately-shaped matrices $M$ and $N$, the $ij$-th element of the product $MN$ is the dot product of the $i$th row of $M$ with the $j$th column of $N$. When $MN$ is a diagonal matrix, this implies that for every column $[N]_j$ of $N$, all of the rows $[M]_i$ of $M$ with $i\ne j$ are orthogonal to $[N]_j$. The columns of $B$, and so also the rows of $B^{-1}$, are linearly independent, which means that for each column $[B]_j$, the $m-1$ rows $[B^{-1}]_i$ $(i\ne j)$ form a basis for the orthogonal complement of $[B]_j$. If we take a subset of columns of $B$, it should be evident that a basis for the orthogonal complement of their span is formed by the complementary rows of $B^{-1}$—this is just a matter of intersecting the individual complements. Thus, if we choose the columns that correspond to those of $A$, then the complementary rows of $B^{-1}$, i.e., those with the same indices as the “extra” columns in $B$, are a basis for the null space of $A^T$.  
Nowhere in the preceding paragraph did we require that the columns of $A$ appear in any particular place in $B$, so, as you’ve conjectured, this method still works if the additional columns are placed anywhere in $B$.
