广义范德蒙德行列式计算

Generalized Vandermonde Matrix

http://10.12.0.10/www.hrpub.org/download/201309/ujam.2013.010209.pdf

柯斯特利金P101习题：

(扎哈洛夫——图拉, 1984)在平稳随机过程模型的研究中,出现了下述行列式:
$$\Delta _n\left( k_1,x_1;...;k_m,x_m \right) =\left| \begin{array}{c} M_{k_1}^{n}\left( x_1 \right)\\ M_{k_2}^{n}\left( x_2 \right)\\ \cdots \cdots\\ M_{k_m}^{n}\left( x_m \right) \end{array} \right|,$$
其中$x_1.x_2,\ldots,x_m$是未知量; $k_1,\ldots,k_m$是自然数, $k_1+k_2+\cdots+k_m=n$; $M_k^n(x)$是$k\times n$阶矩阵,形如
$$M_{k}^{n}\left( x \right) =\left( \begin{matrix} 1& x& x^2& \cdots& x^{n-1}\\ 0& 1& \binom 21x& \cdots& \binom{n-1}1x^{n-2}\\ 0& 0& 1& \cdots& \binom{n-1}2x^{n-3}\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ 0& 0& 0& \cdots& \binom{n-1}{k-1}x^{n-k} \end{matrix} \right).$$
证明
$$\Delta _n\left( k_1,x_1;...;k_m,x_m \right) =\prod_{1\leq j<i\leq m}{\left( x_i-x_j \right) ^{k_ik_j}}.$$
特别地,当$k_1=\cdots=k_m=1$时,即当$m=n$时,得到范德蒙德行列式.

---

证法一.由于$\binom nk-\binom{n-1}k=\binom{n-1}{k-1}$,我们对$\Delta_n$进行消法变换,将它的第$i-1$列乘以$-x_m$倍加到第$i$列,其中$i=n,n-1,\ldots,2$,则
$$M_{k_m}^{n}\left( x_m \right) \rightarrow M_{k_m,1}^{n}\left( x_m \right) =\left( \begin{matrix} 1& 0& 0& \cdots& 0\\ 0& 1& x_m& \cdots& x_{m}^{n-2}\\ 0& 0& 1& \cdots& \binom{n-2}1x_m^{n-3}\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ 0& 0& 0& \cdots& \binom{n-2}{k_m-2}x_m^{n-k_m} \end{matrix} \right),$$
而
$$\begin{align*} M_{k_i}^{n}\left( x_i \right) &\rightarrow M_{k_i,1}^{n}\left( x_i \right)\\ &=\left( \begin{matrix} 1& x_i-x_m& x_{i}^{2}-x_{m}x_i& \cdots& x_{i}^{n-1}-x_mx_{i}^{n-2}\\ 0& 1& \binom 21x_i-\binom 11x_m& \cdots& \binom{n-1}1x_{i}^{n-2}-\binom{n-2}1x_mx_{i}^{n-3}\\ 0& 0& 1& \cdots& \binom{n-1}2x_{i}^{n-3}-\binom{n-2}2x_mx_{i}^{n-4}\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ 0& 0& 0& \cdots& \binom{n-1}{k_i-1}x_{i}^{n-k_i}-\binom{n-2}{k_i-1}x_mx_{i}^{n-k_i-1} \end{matrix} \right). \end{align*}$$

按$M_{k_m,1}^{n}\left( x_m \right)$的第一行展开,并$M_{k_i,1}^{n}\left( x_i \right)$第一行中提取$x_i-x_m$,在我们有
$$\begin{align*} M_{k_m,1}^{n}\left( x_m \right) \rightarrow M_{k_m,2}^{n}\left( x_m \right)=\left( \begin{matrix} 1& x_m& x_{m}^{2}& \cdots& x_{m}^{n-2}\\ 0& 1& \binom 21x_m& \cdots& \binom{n-2}1x_{m}^{n-3}\\ 0& 0& 1& \cdots& \binom{n-2}2x_{m}^{n-4}\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ 0& 0& 0& \cdots& \binom{n-2}{k_m-2}x_{m}^{n-k_m} \end{matrix} \right). \end{align*}$$
而
$$\begin{align*}M_{k_i,1}^{n}\left( x_i \right) &\rightarrow M_{k_i,2}^{n}\left( x_i \right)\\ &=\left( \begin{matrix} 1& x_i& x_{i}^{2}& \cdots& x_{i}^{n-2}\\ 1& \binom 21x_i-\binom 11x_m& \binom 31x_{i}^{2}-\binom 21x_mx_i& \cdots& \binom{n-1}1x_{i}^{n-2}-\binom{n-2}1x_mx_{i}^{n-3}\\ 0& 1& \binom 32x_{i}-\binom 22x_m& \cdots& \binom{n-1}2x_{i}^{n-3}-\binom{n-2}2x_mx_{i}^{n-4}\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ 0& 0& 0& \cdots& \binom{n-1}{k_i-1}x_{i}^{n-k_i}-\binom{n-2}{k_i-1}x_mx_{i}^{n-k_i-1} \end{matrix} \right).\end{align*}$$
再次利用$\binom nk-\binom{n-1}k=\binom{n-1}{k-1}$,将新行列式的第$i-1$列乘以$-x_m$倍加到第$i$列,其中$i=n,n-1,\ldots,2$,则
$$M_{k_m,2}^{n}\left( x_m \right) \rightarrow M_{k_m,3}^{n}\left( x_m \right) =\left( \begin{matrix} 1& 0& 0& \cdots& 0\\ 0& 1& x_m& \cdots& x_{m}^{n-3}\\ 0& 0& 1& \cdots& \binom{n-3}1x_{m}^{n-3}\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ 0& 0& 0& \cdots& \binom{n-3}{k_m-3}x_{m}^{n-k_m} \end{matrix} \right),$$
而
$$\begin{align*} &M_{k_i,2}^{n}\left( x_i \right) \rightarrow M_{k_i,3}^{n}\left( x_i \right)\\ &=\left( \begin{matrix} 1& x_i-x_m& x_i\left( x_i-x_m \right)& \cdots& x_{i}^{n-3}\left( x_i-x_m \right)\\ 1& \binom 21\left( x_i-x_m \right)& \left( \binom 31 x_i-x_m \right) \left( x_i-x_m \right)& \cdots& \left( \binom{n-1}1x_i-\binom{n-3}1x_m \right) \left( x_i-x_m \right) x_{i}^{n-4}\\ 0& 1& \binom 32x_i-2\binom 22x_m& \cdots& \binom{n-1}2x_{i}^{n-3}-2\binom{n-2}2x_mx_{i}^{n-4}+\binom{n-3}2x_{m}^{2}x_{i}^{n-5}\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ 0& 0& 0& \cdots& \binom{n-1}{k_i-1}x_{i}^{n-k_i}-2\binom{n-2}{k_i-1}x_mx_{i}^{n-k_i-1}+\binom{n-3}{k_i-1}x_{m}^{2}x_{i}^{n-k_i-2} \end{matrix} \right). \end{align*}$$
按$M_{k_m,1}^{n}\left( x_m \right)$的第一行展开,并$M_{k_i,1}^{n}\left( x_i \right)$第一行,第二行中提取$x_i-x_m$,在我们有
$$M_{k_m,3}^{n}\left( x_m \right) \rightarrow M_{k_m,4}^{n}\left( x_m \right) =\left( \begin{matrix} 1& x_m& x_m^2& \cdots& x_{m}^{n-3}\\ 0& 1& \binom 21x_m& \cdots& \binom{n-3}1x_{m}^{n-4}\\ 0& 0& 1& \cdots& \binom{n-3}2x_{m}^{n-5}\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ 0& 0& 0& \cdots& \binom{n-3}{k_m-3}x_{m}^{n-k_m} \end{matrix} \right),$$
而
$$\begin{align*} &M_{k_i,2}^{n}\left( x_i \right) \rightarrow M_{k_i,3}^{n}\left( x_i \right)\\ &=\left( \begin{matrix} 1& x_i& x_i^2& \cdots& x_{i}^{n-3}\\ \binom 21& \left( \binom 31 x_i-x_m \right) & \left( \binom 41 x_i-\binom 21 x_m \right) x_i& \cdots& \left( \binom{n-1}1x_i-\binom{n-3}1x_m \right) x_{i}^{n-4}\\ 1& \binom 32x_i-2\binom 22x_m & \binom 42x_{i}^2-2\binom{3}2x_mx_{i}+\binom{2}2x_{m}^{2} & \cdots& \binom{n-1}2x_{i}^{n-3}-2\binom{n-2}2x_mx_{i}^{n-4}+\binom{n-3}2x_{m}^{2}x_{i}^{n-5}\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ 0& 0& 0& \cdots& \binom{n-1}{k_i-1}x_{i}^{n-k_i}-2\binom{n-2}{k_i-1}x_mx_{i}^{n-k_i-1}+\binom{n-3}{k_i-1}x_{m}^{2}x_{i}^{n-k_i-2} \end{matrix} \right). \end{align*}$$
类似地进行下去,我们得到
$$\Delta _n\left( k_1,x_1;...;k_m,x_m \right) =\prod_{j=1}^{m-1}{\left( x_m-x_j \right) ^{k_mk_j}}\Delta _{n-k_m}\left( k_1,x_1;...;k_{m-1},x_{m-1} \right),$$
因此
$$\Delta _n\left( k_1,x_1;...;k_m,x_m \right) =\prod_{1\leq j<i\leq m}{\left( x_i-x_j \right) ^{k_ik_j}}.$$

证法二：参考普丰山，陈军《广义范德蒙行列式及其应用》.