# Determinant of special generalized Vandermonde matrix

Good evening!

I have a generalized Vandermonde matrix of special form: $\left( \begin{array}{ccccc} a_{0,0} & a_{0,1} \cdot x_0 & a_{0,2} \cdot x_0^2 & \ldots & a_{0,m-1} \cdot x_0^{m-1} \newline a_{1,0} & a_{1,1} \cdot x_1 & a_{1,2} \cdot x_1^2 & \ldots & a_{1,m-1} \cdot x_1^{m-1} \newline & & \vdots & & \newline a_{m-1,0} & a_{m-1,1} \cdot x_{m-1} & a_{m-1,2} \cdot x_{m-1}^2 & \ldots & a_{m-1,m-1} \cdot x_{m-1}^{m-1} \end{array}\right)$

This is a Hadamard product of arbitrary matrix $A$ and Vandermonde matrix $V$. What is determinant of this? I observe fact that if $\sum_i a_i$ is a determinant of $A$ and $\sum_i v_i$ is determinant of $V$ than determinant of Hadamard product $A \circ B$ is $\sum_i (a_i \cdot b_i)$. Main question is what constraints $A$ should satisfy that $A \circ V$ is non-degenerate in case of $det{V} \not= 0$.

Thank you!

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Since the entries of $A$ are completely general, so are the entries of $A\circ V$ when each $x_i\neq 0$. Therefore the problem seems equivalent to asking when an arbitrary matrix in non-degenerate. – Richard Stanley Mar 28 '11 at 17:24
I don't know what it means for $\sum_ia_i$ to be a determinant of $A$. In any event, I'd suggest the same thing I suggested in response to another question you asked about Hadamard products and Vandermonde matrices; have you written out the $2\times2$ case in detail to see what happens there? – Gerry Myerson Mar 28 '11 at 22:50