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Let $z_{1},\dots,z_{k}$ be distinct complex numbers with $\left|z_{j}\right|=1,\;j=1,\dots,k$. For any natural $N\geqslant k$ consider the rectangular Vandermonde matrix $$ V_{N}=\begin{pmatrix}1 & 1 & \dots & 1\\ z_{1} & z_{2} & & z_{k}\\ \vdots & \vdots & & \vdots\\ z_{1}^{N-1} & z_{2}^{N-1} & & z_{k}^{N-1} \end{pmatrix}. $$

Let $V_{N}^{*}$ denote the conjugate transpose of $V_{N}$. Since $V_{N}$ has full rank, the square $k\times k$ matrix $V_{N}^{*}V_N$ is nonsingular. We are interested in the quantity $$ D\left(N\right)=\det\left(V_{N}^{*}V_N\right). $$ For $N=k$ we have by the well-known explicit formula $$ D\left(k\right)=\prod_{i<j}\left|z_{i}-z_{j}\right|^{2}. $$

Question: Does there exist an explicit ``simple'' expression for $D\left(N\right)$ with arbitrary $N>k$?

Example of a simple expression in the special case $k=2$:

Without loss of generality $z_{1}=1$ and $z_{2}=\exp\left(\imath x\right)$ for $x\in\left[-\pi,\pi\right]\setminus\{0\}.$ An explicit computation gives the following: $$ D\left(N\right)=N^{2}-\frac{\sin^{2}\frac{N}{2}x}{\sin^{2}\frac{x}{2}}. $$

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This is a partial answer only - but there is a clear place to start.

The Cauchy-Binet theorem gives the answer as a sum of products of all maximal minors of the two matrices, where the minors are taken with matching sets of rows/columns. All of these minors are themselves ordinary, N by N Vandermonde determinants. So the answer is the symmetrization of the squared Vandermonde determinant, over all possible choices of N of the k variables.

This is probably not simple enough for you, but at least it's a formula. I haven't assumed anything about the z variables, nor attempted to do any simplification.

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  • $\begingroup$ Benjamin, thanks for the reference to Cauchy-Binet. However it does not seem to be helpful straight away. First notice that $N>k$ and thus the maximal minors would be $k\times k$. Second, these minors are definitely NOT the usual Vandermonde because they would involve non-consecutive powers of the $z$ variables. For instance, if $k=3$ and $N>5$ then there would be the minor $$\begin{pmatrix} z_1 & z_2 & z_3\\ z_1^3 & z_2^3 & z_3^3\\z_1^4 & z_2^4 & z_3^4 \end{pmatrix}.$$ $\endgroup$
    – dima
    Commented Jan 22, 2014 at 19:45
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    $\begingroup$ @dima: Determinants of that form are Schur polynomials times the Vandermonde determinant. $\endgroup$
    – Michael Joyce
    Commented Jan 22, 2014 at 22:18
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    $\begingroup$ @MichaelJoyce: So we get $D(k)$ times a sum of norm-squareds of Schur polynomials. Specifically, it is the sum of the norm-square of all Schur polynomials corresponding to partitions whose largest part is $<N-k$. Is there a simpler expression for this? $\endgroup$
    – Will Sawin
    Commented Jan 22, 2014 at 23:14
  • $\begingroup$ We can rewrite $|s_{\lambda}(z_1, z_2, \ldots, z_k)|^2$ as $s_{\lambda}(z_1, \ldots, z_l) s_{\lambda}(\bar{z}_1, \bar{z}_2, \ldots, \bar{z}_k)$. This recalls Cauchy's identity: $\sum_{\lambda} s_{\lambda}(x_1, x_2, \ldots, x_k) s_{\lambda}(y_1, y_2, \ldots, y_k) = \prod_{i,j} (1-x_i y_j)^{-1}$. But I don't know what happens when you bound the size of the top row. $\endgroup$
    – David E Speyer
    Commented Feb 21, 2014 at 20:58
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I guess that the determinant only factors if there is a stronger relation between the $z_j$. As example one can give a closed expression for $D_k(N)$ if $z_j = \exp(\iota (j-1) x)$. For $N$ even and $k=3$ one finds $$ D_k(N) = N^k -N \left[ 2 \frac{\sin^2\frac{Nx}{2}}{\sin^2\frac{x}{2}} + \frac{\sin^2 (Nx)}{\sin^2 x} \right] + 2\frac{\sin^2\frac{Nx}{2}}{\sin^2\frac{x}{2}} \frac{\sin(Nx)}{\sin x}. $$ I think the cases $k>3$ can also be derived. Note that for equidistant $z_j$ the Vandermonde becomes the discrete Fourier transform, at least if $N=k$.

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