# 浅谈范德蒙德(Vandermonde)方阵的逆矩阵与拉格朗日(Lagrange)插值的关系以及快速傅里叶变换(FFT)中IDFT的原理

$V(x_0,x_1,\cdots ,x_{n-1})=\begin{bmatrix} {1}&{1}&{\cdots}&{1}\\ {x_{0}}&{x_{1}}&{\cdots}&{x_{n-1}}\\ {x_{0}^2}&{x_{1}^2}&{\cdots}&{x_{n-1}^2}\\ {\vdots}&{\vdots}&{}&{\vdots}\\ {x_{0}^{n-1}}&{x_{1}^{n-1}}&{\cdots}&{x_{n-1}^{n-1}}\\ \end{bmatrix}$

$V(x_0,x_1,\cdots ,x_{n-1})=\prod _{n > i > j \geq 0}(x _{i}-x _{j})$

$V=\begin{pmatrix} {1}&{1}&{\cdots}&{1}\\ {x_{0}}&{x_{1}}&{\cdots}&{x_{n-1}}\\ {x_{0}^2}&{x_{1}^2}&{\cdots}&{x_{n-1}^2}\\ {\vdots}&{\vdots}&{}&{\vdots}\\ {x_{0}^{n-1}}&{x_{1}^{n-1}}&{\cdots}&{x_{n-1}^{n-1}}\\ \end{pmatrix}$

$(V^*)_{ij}=c_{ij}$

$V^{-1}={V* \over det(V)}$

$(V^{-1})_{ij}=(-1)^{j+1}{ \sum\limits_{0 \leq p_1<\cdots < p_{n-j} < n;\ p_1,p_2,\cdots p_{n-j} \ne i} x_{p_1} x_{p_2} \cdots x_{p_{n-j}} \over \prod\limits_{0 \leq k < n;\ k\ne i} (x_k-x_i)}$

${\begin{pmatrix} {a_0}\\ {a_1}\\ {a_2}\\ {\vdots}\\ {a_{n-1}}\\ \end{pmatrix} }^T\begin{pmatrix} {1}&{1}&{\cdots}&{1}\\ {x_{0}}&{x_{1}}&{\cdots}&{x_{n-1}}\\ {x_{0}^2}&{x_{1}^2}&{\cdots}&{x_{n-1}^2}\\ {\vdots}&{\vdots}&{}&{\vdots}\\ {x_{0}^{n-1}}&{x_{1}^{n-1}}&{\cdots}&{x_{n-1}^{n-1}}\\ \end{pmatrix}={ \begin{pmatrix} {y_0}\\ {y_1}\\ {y_2}\\ {\vdots}\\ {y_{n-1}} \end{pmatrix}}^T$

${\begin{pmatrix} {a_0}\\ {a_1}\\ {a_2}\\ {\vdots}\\ {a_{n-1}}\\ \end{pmatrix}}^T ={\begin{pmatrix} {y_0}\\ {y_1}\\ {y_2}\\ {\vdots}\\ {y_{n-1}} \end{pmatrix}}^T \begin{pmatrix} {1}&{1}&{\cdots}&{1}\\ {x_{0}}&{x_{1}}&{\cdots}&{x_{n-1}}\\ {x_{0}^2}&{x_{1}^2}&{\cdots}&{x_{n-1}^2}\\ {\vdots}&{\vdots}&{}&{\vdots}\\ {x_{0}^{n-1}}&{x_{1}^{n-1}}&{\cdots}&{x_{n-1}^{n-1}}\\ \end{pmatrix}^{-1}$

$f(x)=\sum_{i}y_i\prod_{j\ne i} {x-x_j \over x_i-x_j}$

$1$$n$次复根$w$,如果没有特别说明，以下本文中的$w$都为$e^{2\pi \over n}$

$\prod\limits_{j\ne i} {(x-w^j) \over (w^i-w^j)}={\prod\limits_{j\ne i} (x-w^j) \over \prod\limits_{j\ne i} (w^i-w^j)}$

$w^{0},w^1,\cdots,w^{n-1}$都是1的n次复根，根据代数基本定理，显然有$$G(x)=x^n-1$$

$\lim _{x \to w^i}{G(x) \over {x-w^i}}=\lim _{x \to w^i} G'(x)=n \times w^{i(n-1)}=n \times w^{-i}$

${\prod\limits_{j\ne i} (x-w^j)}={G(x) \over {x-w^i}}={1-x^n \over {w^i-x}}\\ =w^{-i} \times \begin{pmatrix}{1 \over 1- x w^{-i}}-{x^n \over 1-xw^{-i}}\end{pmatrix}\\ =w^{-i} \times \begin{pmatrix}{\sum_{j=0}^{\infty} w^{-ij}x^j -\sum_{j=n}^{\infty} w^{-i(j-n)}x^j} \end{pmatrix}\\ =w^{-i} \times \sum_{j=0}^{n-1} w^{-ij} x^j$

$原式={w^{-i} \times \sum\limits_{j=0}^{n-1} w^{-ij} x^j \over n \times w^{-i}}\\ =\sum_{j=0}^{n-1} {w^{-ij} \over n}x^j$

$\begin{pmatrix} {1}&{1}&{\cdots}&{1}\\ {1}&{w^{1}}&{\cdots}&{w^{n-1}}\\ {1}&{w^{2}}&{\cdots}&{w^{(n-1)2}}\\ {\vdots}&{\vdots}&{}&{\vdots}\\ {1}&{w^{n-1}}&{\cdots}&{w^{(n-1)(n-1)}}\\ \end{pmatrix}^{-1} ={1 \over n}\begin{pmatrix} {1}&{1}&{\cdots}&{1}\\ {1}&{w^{-1}}&{\cdots}&{w^{-(n-1)}}\\ {1}&{w^{-2}}&{\cdots}&{w^{-(n-1)2}}\\ {\vdots}&{\vdots}&{}&{\vdots}\\ {1}&{w^{-(n-1)}}&{\cdots}&{w^{-(n-1)(n-1)}}\\ \end{pmatrix}$

posted @ 2018-10-04 11:58  Deadecho  阅读(9861)  评论(6编辑  收藏  举报