Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $m_i\in \mathbb{N}, 1\leq i \leq n $ such that wlog if $m_i < m_j\in \mathbb{Z^+}$ then $i < j$. Consider the Vandermonde matrix $$V= \begin{bmatrix} 1 & 1 & 1 & \ldots & 1 \\ x_1^{m_1} & x_2^{m_1} & x_3^{m_1} & \ldots & x_{n+1}^{m_1} \\ x_1^{m_2} & x_2^{m_2} & x_3^{m_2} & \ldots & x_{n+1}^{m_2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_1^{m_n} & x_2^{m_n} & x_3^{m_n} & \ldots & x_{n+1}^{m_n} \\ \end{bmatrix}$$ Under What conditions is the generalized Vandermonde Matrix Invertible? It is clear that if one lets $x_i=x_j$,then the determinant is Zero, and therefore the principle Vandermonde Determinant $V_p=\prod_{1 \leq i < j\leq n}^n(x_i-x_j)$ can be factored from the $det(V)$. What remains is the schur function, which is a homogeneous polynomial of degree d which depends on $n$, which symmetric over its variables $x_i,1\leq i\leq n$. There is a proof of the fact that the coefficients of the schur function are positive integers which was given by Mitchel long ago.

When one looks at the simplest case where $x_i \in \mathbb{C}, 1\leq i \leq n$ and $n=2$, $$V= \begin{bmatrix} 1 & 1\\ x_1^n & x_2^n\\ \end{bmatrix}$$ one sees that the Vandermonde determinant is invertible when $\frac{x_1}{x_2}$ is not an $n^{th}$ root of unity.

For $n=3$ things become difficult already as $$V_3= \begin{bmatrix} 1 & 1 & 1\\ x_1^{m_1} & x_2^{m_1} & x_3^{m_1}\\ x_1^{m_2} & x_2^{m_2} & x_3^{m_2} \end{bmatrix}$$ $det(V_3) = x_2^{m_2} x_1^{m_1}-x_3^{m_2} x_1^{m_1}-x_2^{m_1} x_1^{m_2}+x_3^{m_1} x_1^{m_2}-x_2^{m_2} x_3^{m_1}+x_2^{m_1} x_3^{m_2}$.

As far as I see (which is not very far), characterizing the zero set of the this polynomial is not at all simple in either $\mathbb{C^3}$ or $\mathbb{R^3}$ with exception of the trivial case we already mentioned where $x_i=x_j$ for some $i\neq j$ and perhaps some roots of unity. One can guess that over $\mathbb{C^n}$, if the $|x_i|$ are distinct and have no-non trivial relations, then the schur function is non-zero. Which brings me back to my question:

What are the conditions under which the generalized Vandermonde matrix is invertible over $\mathbb{R^n}$ or $\mathbb{C^n}$? Is there Hope?

Well yeah, a little bit.There are some results one could see from Mitchel's result about the positivity of the coefficients of the Schur function. If we are working over $\mathbb{R^n}$, then as a consequence of the positivity of the coefficients mentioned by Mitchel, one can deduce that if $x_i$ are distinct and positive for $1 \leq i \leq n$, then the schur function takes on positive values and therefore non-zero. This result is mentioned in this Paper.

Examples:

1)Let $V_3$ be a the following $3$x$3$ matrix: $$V_3= \begin{bmatrix} 1 & 1 & 1\\ x_1 & x_2 & x_3\\ x_1^{3} & x_2^{3} & x_3^{3} \end{bmatrix}$$ where $m_1 = 1, m_2 = 3$, then $det(V)=-(x_1-x_2) (x_1-x_3) (x_2-x_3) (x_1+x_2+x_3)$, so our schur function is the elementary symmetric function $x_1+x_2+x_3$ which is non-zero if $x_j \neq x_i \in \mathbb{R^+}$ $\forall i,j$

2) 1)Let $V_3$ be a the following $3$x$3$ matrix: $$V_3= \begin{bmatrix} 1 & 1 & 1\\ x_1^{4} & x_2^{4} & x_3^{4}\\ x_1^{6} & x_2^{6} & x_3^{6} \end{bmatrix}$$ where $m_1 = 4, m_2 = 6$, then

$det(V)=-x_2^4 x_1^6+x_3^4 x_1^6+x_2^6 x_1^4-x_3^6 x_1^4+x_2^4 x_3^6-x_2^6 x_3^4$

$det(V)=-(x_1-x_2) (x_1+x_2) (x_1-x_3) (x_2-x_3) (x_1+x_3) (x_2+x_3) (x_1^2 x_2^2+x_3^2 x_2^2+x_1^2 x_3^2)$, so our schur function is $ (x_1+x_2)(x_1+x_3) (x_2+x_3) (x_1^2 x_2^2+x_3^2 x_2^2+x_1^2 x_3^2)$ which is non-zero if $x_j \neq x_i \in \mathbb{R^+}$ $\forall i,j$

3) 1)Let $V_3$ be a the following $3$x$3$ matrix: $$V_3= \begin{bmatrix} 1 & 1 & 1\\ x_1^{4} & x_2^{4} & x_3^{4}\\ x_1^{7} & x_2^{7} & x_3^{7} \end{bmatrix}$$ where $m_1 = 4, m_2 = 7$, then

$det(V)=-(x_1-x_2) (x_1-x_3) (x_2-x_3)$ *

$(x_2^3 x_1^5+x_3^3 x_1^5+x_2 x_3^2 x_1^5+x_2^2 x_3 x_1^5+x_2^4 x_1^4+x_3^4 x_1^4+$ $2 x_2 x_3^3 x_1^4+2 x_2^2 x_3^2 x_1^4+2 x_2^3 x_3 x_1^4+x_2^5 x_1^3+x_3^5 x_1^3+$ $2 x_2 x_3^4 x_1^3+3 x_2^2 x_3^3 x_1^3+3 x_2^3 x_3^2 x_1^3+2 x_2^4 x_3 x_1^3+x_2 x_3^5$ $x_1^2+2 x_2^2 x_3^4 x_1^2+3 x_2^3 x_3^3 x_1^2+2 x_2^4 x_3^2 x_1^2+x_2^5 x_3 x_1^2+$ $x_2^2 x_3^5 x_1+2 x_2^3 x_3^4 x_1+2 x_2^4 x_3^3 x_1+x_2^5 x_3^2 x_1+x_2^3 x_3^5+$ $x_2^4 x_3^4+x_2^5 x_3^3)$

So our schur function is everything that we have after dividing out by $V_p$. This function is non-zero if $x_j \neq x_i \in \mathbb{R^+}$ $\forall i,j$

My goal is to really answer the following interesting question, Let ${m_i}$, $i \in \mathbb{N}$ be an increasing sequence of distinct positive integers, then consider the schur function obtained by dividing out the following Vandermonde Determinant by the principle Vandermonde:

$$V= \begin{bmatrix} 1 & 1 & 1 & \ldots & 1 \\ x_1^{m_{i_1}} & x_2^{m_{i_1}} & x_3^{m_{i_1}} & \ldots & x_{n+1}^{m_{i_1}} \\ x_1^{m_{i_2}} & x_2^{m_{i_2}} & x_3^{m_{i_2}} & \ldots & x_{n+1}^{m_{i_2}} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_1^{m_{i_n}} & x_2^{m_{i_n}} & x_3^{m_{i_n}} & \ldots & x_{n+1}^{m_{i_n}} \\ \end{bmatrix}$$

where $m_{i_j}$ are terms in the sequence ${m_i}$, $i \in \mathbb{N}$ such that $m_{i_{j_1}}< m_{i_{j_2}}$ implies $j_1 < j_2$. Let $\Gamma$ ={all possible $m_{i_j}, 1 \leq j \leq n $}, then each element of $\Gamma$ corresponds to a unique schur function. Let $S_{\Gamma}$ ={schur functions induced by $\Gamma$}. Now Let $s \in S_{\Gamma}$ and $Z_s$ ={zeros of $s$}, then I would like to show that $\cap_{s \in S_{\Gamma} }Z_s$ is empty, provided that there are no non-trivial relations between the variables of the schur function, i.e. the $n$ variables of the schur function form a free abelian group of rank $n$, i.e. if the product of the coordinates of a zero of the schur function does not satisfy that condition then you should exclude it.

As pointed out by Emmanuel Briand in the comments, if the sequence $m_i$ contains $n$ consecutive integers, then the zeros of the corresponding schur function are trivial as they are of the form $(0,\cdots , 1, \cdots , 0)$ where 1 is located in the $i^{th}$ position. As a special case, you can see that if $m_i$ contains $1,2,\cdots , n-1$, then the answer to the question above is positive, i.e. the $\cap_{s \in S_{\Gamma} }Z_s$ is empty.

share|improve this question
1  
I don't get your questions: I am not sure what kind of results you want to know about the zeros. As for the group question, you point out yourself that the symmetric group acts on the zeros, I am not sure what else you could be looking for. I see that you've been editing your question quite a bit, and your editing improved it, but there is still room for clarification. –  Thierry Zell Aug 27 '11 at 14:06
    
By the way, this question mathoverflow.net/questions/60938/… discussed Vandermonde determinants and Schur functions, from a slightly different angle. The answers were pretty detailed, so maybe there is something there for you. –  Thierry Zell Aug 27 '11 at 14:10
    
I hope that clarifies things a bit. –  Ahmed Roman Aug 29 '11 at 8:51
    
Like Thierry, I do not understand well the kind of answer you expect. The condition of non-invertibility of a generalized Vandermonde matrix is already expressed as an equation. Are you looking for a geometrical study of the corresponding algebraic variety ? (Such a study is addressed in the following draft: www2.math.su.se/~shapiro/Articles/SML.pdf) Of its real points ? Are you interested in sufficient conditions for invertibility (like: all $x_i$ positive and pairwise distinct that you give) ? –  Emmanuel Briand Aug 29 '11 at 22:28
1  
I have seen your edit but there is still a technical problem: your $\Gamma$ seems to include the sequence $m=(1,2,\ldots,n-1)$, whose corresponding Schur function is $1$. Trivially the corresponding $Z_s$ is empty. Maybe the correct question is different. Then probably the correct answer will be obtained by considering the fact that the Schur polynomials are a linear basis for the symmetric polynomials. –  Emmanuel Briand Aug 31 '11 at 14:02

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.