Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\{x_{i}\}$.

share|improve this question
    
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
4  
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.