Rank of a rectangular Vandermonde Matrix to which weighted columns are added A Vandermonde matrix:
$\left(\begin{array}{ccc}
1 & \alpha_{0} & \dots & \alpha_{0}^{n} \\
1 & \alpha_{1} & \dots & \alpha_{1}^{n} \\
\vdots & \vdots & \ddots & \vdots \\
1 & \alpha_{2n} & \dots & \alpha_{2n}^{n}
\end{array}\right)$
has full rank $n+1$, provided $\alpha_{i}\neq \alpha_{j}$ for al $i\neq j$.
Can we say that a matrix of the form: $\left(\begin{array}{cccccc}
1 & \alpha_{0} & \dots & \alpha_{0}^{n} & v_{0} & v_{0}\alpha_{0} & \dots & v_{0}\alpha_{0}^{n} \\
1 & \alpha_{1} & \dots & \alpha_{1}^{n} & v_{1} & v_{1} \alpha_{1} & \dots & v_{1}\alpha_{1}^{n} \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\
1 & \alpha_{2n} & \dots & \alpha_{2n}^{n} & v_{2n} & v_{2n} \alpha_{2n} & \dots & v_{2n}\alpha_{2n}^{n}
\end{array}\right)$
where $v_{0}, \dots, v_{2n}$ are constants, $v_{i}\neq v_{j}$.
will continue to have rank $(2n+1)$ ? Do any more constrains on $\{v_{i}\}$ need to be assumed ?
I know that if we consider the above matrix as $(V | M)$ where $V$ is the initial Vandermonde matrix, both $V$ and $M$ will have rank $(2n+1)$.

Edit:
  Sorry, I meant row rank, in all cases. The context is that I want to use the matrix $(V|M)$ to describe a solution for $2n+1$ variables $\{x_{0}, \dots, x_{2n}\}$. If this variable vector is $\vec{x}$, then I want to solve for $\vec{x}$, in: $(V|M)\cdot \vec{x} = 0$. 
Hence I wanted to know whether the rows of $(V|M)$ specify linearly independent constraints on $\{x_{i}\}$.

 A: I can easily see one additional constraint. What if you have $v_i  = \alpha_i$? This then would give that the column of $v$ elements is the same as the column of $\alpha$ elements, which would mean that the column is not linearly independent within the new extended matrix $(V|M)$.
More generally speaking you have
$$(V|M) = (V|\operatorname{diag}(\vec{v})V)$$
Where $\operatorname{diag}(\vec{v})V$ represents the row scaling of the original $V$ matrix.
Your matrix $V$ is not "full rank" as you put it, since it has more rows than columns. Thus it does have a null space. In particular the columns do not and can not span the entire space. So there exists $y\ne 0$ such that
$$y^\top V = 0$$
Given such $y$ you need also that 
$$y^\top M = y^\top \operatorname{diag}(\vec{v})V\ne 0 $$
 otherwise the extended matrix still does not have a full spanning column space.
At the moment I cannot say what constraints could be put to $v$ to guarantee this for every such $y$.
You can also express this quantity as a Hadamard product
$$y^\top \operatorname{diag}(\vec{v}) =y^\top \circ v^\top $$
Not sure how it would help much, although I have not use the Hadamard product very much myself.
