# 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 $\{x_{i}\}$.

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The rank is the same as the number of linearly independent columns. Adding columns won't change that. This is true independent of the $v_i$. – copper.hat Nov 15 '12 at 18:22
Your very first statement that the $(2n+1)\times(n+1)$ retangular matrix "has full rank $(2n+1)$, provided $\alpha_i\not=\alpha_j$" is already wrong. Yes, that matrix is of full rank, but its rank is $n+1$, not $2n+1$. – user1551 Nov 16 '12 at 6:28

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.

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