Starting with a Vandermonde matrix:
$$V = \begin{bmatrix} 1^1 & 1^2 & 1^3 & \dots & 1^n \\ 2^1 & 2^2 & 2^3 & \dots & 2^n \\ 3^1 & 3^2 & 3^3 & \dots & 3^n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n^1 & n^2 & n^3 & \dots & n^n \\ \end{bmatrix} $$
...which, according to this question, always has a solution in reduced row-echelon form, we can extend the matrix to include additional columns:
$$V(a) = \begin{bmatrix} (a_0)^1 & (a_0)^2 & (a_0)^3 & \dots & (a_0)^n \\ (a_1)^1 & (a_1)^2 & (a_1)^3 & \dots & (a_1)^n \\ (a_2)^1 & (a_2)^2 & (a_2)^3 & \dots & (a_2)^n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ (a_n)^1 & (a_n)^2 & (a_n)^3 & \dots & (a_n)^n \\ (a_{n+1})^1 & (a_{n+1})^2 & (a_{n+1})^3 & \dots & (a_{n+1})^n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ (a_m)^1 & (a_m)^2 & (a_m)^3 & \dots & (a_m)^n \end{bmatrix} $$
The idea is that we can add two extended matrices together, like so:
$$A=\begin{bmatrix} V(a) \, | \, 0 \end{bmatrix}$$ $$B=\begin{bmatrix} 0 \, | \, V(b) \end{bmatrix}$$ $$A+B = \begin{bmatrix} V(a) \, | \, V(b) \end{bmatrix}$$
...We should be able to reduce this to the identity. My question is, what if the columns overlap? In other words, if we have:
$$A=\begin{bmatrix} V(a) \, | \, V(a_2) \, | \, 0 \end{bmatrix}$$ $$B=\begin{bmatrix} 0 \, | \, V(b) \, | \, V(b_2) \end{bmatrix}$$ $$A+B = \begin{bmatrix} V(a) \, | \, V(a_2) + V(b) \, | \, V(b_2) \end{bmatrix}$$
I believe that this should still reduce to the identity. Am I correct? Could someone possibly prove this? Or, if a proof is elusive, can someone point me in the right direction?