Is there a closed form expression for the determinant of the $n\times n$ Vandermonde-type matrix


$$A = \left(\begin{array}{}
1&g\_1 & x\_1&g\_1 x\_1 & x\_1^2&g\_1 x\_1^2 & \cdots & x\_1^{n/2-1} & g\_1 x\_1^{n/2-1} \\
1&g\_2 & x\_2&g\_2 x\_2 & x\_2^2&g\_2 x\_2^2 & \cdots & x\_2^{n/2-1} & g\_2 x\_2^{n/2-1} \\
\vdots &\vdots &\vdots &\vdots &\vdots & \vdots & \ddots & \vdots & \vdots \\
1&g\_n & x\_n&g\_n x\_n & x\_n^2&g\_n x\_n^2 & \cdots & x\_n^{n/2-1} & g\_n x\_n^{n/2-1}
\end{array}\right), $$


with $n$ even?
The $g\_i$ can be assumed to be functions of $x\_i$: $g\_i\equiv g(x\_i)$.
Furthermore, it might be helpful to assume the $x\_i$ to be the complex roots of a known polynomial of order $n$.



Determinant of a certain Vandermonde matrix
In Appendix B of <https://arxiv.org/abs/2103.10776> (J. Phys. A: Math. Theor. 54, 375201 (2021)), I derived a transformation of the block Vandermonde determinant above to a Hankel matrix, which in my case dramatically simplified the problem. This is a block-matrix generalization of the well known "Vandermonde factorization of a Hankel matrix":


Let $\mathbf V\_{\!\vec{x}}$ be a general Vandermonde matrix
\begin{equation}
\mathbf V\_{\!\vec{x}} \equiv \left[\vphantom{A\_{2}^{2}}x\_{\mu}^{j}\right]\_{\mu=1,\,j=0}^{M\hspace{1.5ex}M-1}\label{eq:V\_x}\tag{1}
\end{equation}
and let
\begin{equation}
P\_{\vec{x}}(x)=\prod\_{\mu=1}^{M}(x-x\_{\mu})=\sum\_{n=0}^{M}b\_{n}x^{n},\label{eq:P\_x(x)}\tag{2}
\end{equation}
be the related characteristic polynomial in $x$,
where $b\_{M}=1$ by construction. If we further let
\begin{align}
\mathbf D\_{\vec{x}} & \equiv \left[\frac{\delta\_{\mu\nu}}{P\_{\vec{x}}'(x\_{\mu})}\right]\_{\mu,\nu=1}^{M}, & \mathbf H\_{\vec{x}}^{-1} & \equiv\left[\vphantom{A\_{2}^{2}}b\_{i+j+1}\right]\_{i,j=0}^{M-1},\tag{3}
\end{align}
then
\begin{align}
\mathbf V^T\_{\!\vec{x}}\,\mathbf D\_{\vec{x}}\mathbf V\_{\!\vec{x}} & = \mathbf H\_{\vec{x}}, & {\det}^{2}\,\mathbf V\_{\!\vec{x}}\det\mathbf D\_{\vec{x}} & =1.\label{eq:Vandermonde\_Hankel\_Identity}\tag{4}
\end{align}
While $\mathbf H\_{\vec{x}}^{-1}$ is upper anti-triangular, $\mathbf H\_{\vec{x}}$
itself is lower anti-triangular, e.g. for $M=4$,
\begin{equation}
\mathbf H\_{\vec{x}}^{-1}=\begin{pmatrix}b\_{1} & b\_{2} & b\_{3} & 1\\
b\_{2} & b\_{3} & 1 & 0\\
b\_{3} & 1 & 0 & 0\\
1 & 0 & 0 & 0
\end{pmatrix}\quad\Rightarrow\quad\mathbf H\_{\vec{x}}=\begin{pmatrix}0 & 0 & 0 & 1\\
0 & 0 & 1 & \tilde{b}\_{5}\\
0 & 1 & \tilde{b}\_{5} & \tilde{b}\_{6}\\
1 & \tilde{b}\_{5} & \tilde{b}\_{6} & \tilde{b}\_{7}
\end{pmatrix},\label{eq:H\_x\_example}\tag{5}
\end{equation}
with $\tilde{b}\_{n}=\sum\_{\mu}x\_{\mu}^{n-1}/P\_{\vec{x}}'(x\_{\mu})$.


As a side note, both $\tilde{b}\_{n}=s\_{(n-M)}(\boldsymbol{x})$ and
$b\_{n}=(-1)^{n}s\_{\boldsymbol{1}\_{M-n}}(\boldsymbol{x})$ are Schur
polynomials.


We now turn to the block matrix version:


Let $\boldsymbol{\mathcal{V}}\_{\vec{g},\vec{x}}$ be the generalized $1\times B$ block Vandermonde matrix with $M\times N$ blocks
\begin{equation}
\boldsymbol{\mathcal{V}}\_{\vec{g},\vec{x}} \equiv \left[\vphantom{A\_{2}^{2}}g\_{\mu}^{b}\,x\_{\mu}^{n}\right]\_{b=0;\,\,\mu=1,\,n=0}^{B-1\,M\hspace{2ex}N-1} = \left[\vphantom{A\_{2}^{2}}\mathbf G^{b}\,\mathbf V\_{\!\vec{x}}\right]\_{b=0}^{B-1},\tag{6}
\end{equation}
with
\begin{equation}
\mathbf V\_{\!\vec{x}} \equiv \left[\vphantom{A\_{2}^{2}}x\_{\mu}^{n}\right]\_{\mu=1,\,n=0}^{M\hspace{1.5ex}N-1},\qquad \mathbf G \equiv \left[\vphantom{A\_{2}^{2}}\delta\_{\mu\nu}\,g\_{\mu}\right]\_{\mu,\nu=1}^{M}.\tag{7}
\end{equation}
As an example, for $M=4$, $B=3$ and $N=3$ we have
\begin{equation}
\boldsymbol{\mathcal{V}}\_{\vec{g},\vec{x}} = 
\left[\begin{array}{ccc|ccc|ccc}
1 & x\_{1} & x\_{1}^{2} & g\_{1} & g\_{1}x\_{1} & g\_{1}x\_{1}^{2} & g\_{1}^{2} & g\_{1}^{2}x\_{1} & g\_{1}^{2}x\_{1}^{2}\\
1 & x\_{2} & x\_{2}^{2} & g\_{2} & g\_{2}x\_{2} & g\_{2}x\_{2}^{2} & g\_{2}^{2} & g\_{2}^{2}x\_{2} & g\_{2}^{2}x\_{2}^{2}\\
1 & x\_{3} & x\_{3}^{2} & g\_{3} & g\_{3}x\_{3} & g\_{3}x\_{3}^{2} & g\_{3}^{2} & g\_{3}^{2}x\_{3} & g\_{3}^{2}x\_{3}^{2}\\
1 & x\_{4} & x\_{4}^{2} & g\_{4} & g\_{4}x\_{4} & g\_{4}x\_{4}^{2} & g\_{4}^{2} & g\_{4}^{2}x\_{4} & g\_{4}^{2}x\_{4}^{2}
\end{array}\right].
\end{equation}
Furthermore, let $\mathbf D$ be an arbitrary $M\times M$ diagonal matrix.
Then, the $B\times B$ block Hankel matrix with $N\times N$ blocks
\begin{equation}
\boldsymbol{\mathcal{H}}\_{\vec{g},\vec{x}} \equiv \left[\sum\_{\mu=1}^{M}d\_{\mu}g\_{\mu}^{a+b}\,x\_{\mu}^{m+n}\right]\_{a,b=0;\,m,n=0}^{B-1\;\;N-1} = \left[\mathbf V^T\_{\!\vec{x}}\,\mathbf D\,\mathbf G^{a+b}\,\mathbf V\_{\!\vec{x}}\right]\_{a,b=0}^{B-1}\tag{8}
\end{equation}
trivially fulfills the identity
\begin{equation}
\boldsymbol{\mathcal{H}}\_{\vec{g},\vec{x}} = \boldsymbol{\mathcal{V}}^T\_{\vec{g},\vec{x}} \, \mathbf D \, \boldsymbol{\mathcal{V}}\_{\vec{g},\vec{x}}.\tag{9}
\end{equation}
Note that the upper left block $\hat{\mathbf H}\_{0} \equiv (\boldsymbol{\mathcal{H}}\_{\vec{g},\vec{x}})\_{0,0} = \mathbf V^T\_{\!\vec{x}}\,\mathbf D\,\mathbf V\_{\!\vec{x}}$
is free of $g\_{\mu}$ and is therefore a usual Vandermonde product
in $x\_{\mu}$. Consequently, if the matrix $\mathbf D$ is set to the diagonal matrix $\mathbf D\_{\vec{x}}$ containing the reciprocal first derivatives of the characteristic polynomial $P\_{\vec{x}}(x)=\prod\_{\mu=1}^M(x-x\_\mu)$,
\begin{equation}
\mathbf D\_{\vec{x}} \equiv \left[\frac{\delta\_{\mu\nu}}{P\_{\vec{x}}'(x\_{\mu})}\right]\_{\mu,\nu=1}^{M},\tag{10}
\end{equation}
and if additionally $M\geq2N$, then $\hat{\mathbf H}\_{0}=\mathbf 0$
vanishes identically.


For the problem above, $B=2$ and $N=M/2$, such that
\begin{equation}
\mathbf A^T \mathbf D\_{\vec{x}}\mathbf A = 
\boldsymbol{\mathcal{H}}\_{\vec{g},\vec{x}} =
\begin{bmatrix}
\mathbf 0 & \hat{\mathbf H}\_{1} \\
\hat{\mathbf H}\_{1} & \hat{\mathbf H}\_{2}
\end{bmatrix}.\tag{11}
\end{equation}


The desired determinant $\det \mathbf A$ therefore fulfills
\begin{equation}
\det(\mathbf A^T \mathbf D\_{\vec{x}} \mathbf A)
= {\det}^{2}\,\mathbf A\det\mathbf D\_{\vec{x}}
= \det \boldsymbol{\mathcal{H}}\_{\vec{g},\vec{x}}
= {\det}^{2}\,\hat{\mathbf H}\_{1}.\tag{12}
\end{equation}


For further details, see <https://arxiv.org/abs/2103.10776>.


