Null space basis Let $V\in\mathbb{R}^{a\times b}$
  be a matrix such that it is not a full column rank. Then there will be a nonsingular matrix $H$ such that $$VH=\left[\begin{array}{cc} V_{1} & 0_{a\times q}\end{array}\right]$$
 where $q$ is the nullity of $V$. Now let $W\in\mathbb{R}^{b\times b}$ be some arbitrary square matrix such that $\left[\begin{array}{c}
V\\
W
\end{array}\right]$ is full column rank. Then $$\left[\begin{array}{c}
V\\
W
\end{array}\right]H=\left[\begin{array}{cc}
V_{1} & 0_{a\times q}\\
W_{1} & W_{2}
\end{array}\right]$$
where $W_{2}$ is full column rank. 


*

*Do the columns of $W_{2}$ form the basis for the null space of the matrix $V$? 

*If yes, how do we prove that?

*If no, what do the columns of $W_2$ signify?

 A: Since $H$ is nonsingular, the equation
$$VH = \left[V_1\; 0_{a\times q}\right]$$
means that the last $q$ columns of $H$ form a basis of the null space of $V$. Splitting $H = \left[H_1\; H_2\right]$, we then have
$$W_1 = W\cdot H_1\quad \text{ and } \quad W_2 = W\cdot H_2.$$
If we consider the map $T_W\colon x \mapsto W\cdot x$, we see that the columns of $W_2$ span the space $T_W(\ker T_V)$, and since $W_2$ has full column rank, they form a basis of that space (i.e. $\ker T_V \cap \ker T_W = \{0\}$).
In general, the null space of $V$ is not invariant under $T_W$, so the answer to the first question

Do the columns of $W_2$ form the basis for the null space of the matrix $V$?

is "maybe, but probably not". They can be a basis of any subspace with the same dimension as the null space of $V$.
The answer to the third question

If no, what do the columns of $W_2$ signify?

is then accordingly "nothing". The columns of $W_2$ can be any $q$-tuple of linearly independent vectors in $\mathbb{R}^b$. The matrix with significance is $H$, in particular the part $H_2$, whose columns are a basis of the null space of $V$.
