vandermonde_det.md

$$ D_n= \begin{vmatrix} 1&1&\cdots&1\ x_1&x_2&\cdots&x_n\ x_1^2&x_2^2&\cdots&x_n^2\ \vdots&\vdots&&\vdots\ x_1^{n-1}&x_2^{n-1}&\cdots&x_n^{n-1} \end{vmatrix}

\prod_{1\le i\lt j\le n}(x_j-x_i) $$

证明

对$Vandermonde$行列式的阶数$n$应用归纳法。

当$n=2$时

$$ D_2= \begin{vmatrix} 1&1\\ x_1&x_2 \end{vmatrix}= (x_2-x_1)= \prod_{1\le i\lt j\le 2}(x_j-x_i) $$

成立

假设原式对$n-1$阶$Vandermonde$行列式成立，即

$$ D_{n-1}= \begin{vmatrix} 1&1&\cdots&1\\ x_2&x_3&\cdots&x_n\\ x_2^2&x_3^2&\cdots&x_n^2\\ \vdots&\vdots&&\vdots\\ x_2^{n-2}&x_3^{n-2}&\cdots&x_n^{n-2} \end{vmatrix}= \prod_{2\le i\lt j\le n}(x_j-x_i) $$

只需证原式对$n$阶$Vandermonde$行列式成立即可

为此，将$D_n$的第$n-1$行的$-x_1$倍加到第$n$行，第$n-2$行$-x_1$倍加到第$n-1$行，如此作下去，直到第$1$行$-x_1$倍加到第$2$行，得

$$ \begin{align} D_n&= \begin{vmatrix} 1&1&1&\cdots&1\\ 0&x_2-x_1&x_3-x_1&\cdots&x_n-x_1\\ 0&x_2(x_2-x_1)&x_3(x_3-x_1)&\cdots&x_n(x_n-x_1)\\ \vdots&\vdots&\vdots&&\vdots\\ 0&x_2^{n-2}(x_2-x_1)&x_3^{n-2}(x_3-x_1)&\cdots&x_n^{n-2}(x_n-x_1) \end{vmatrix}\\\\ &= 1\cdot(-1)^{1+1} \begin{vmatrix} x_2-x_1&x_3-x_1&\cdots&x_n-x_1\\ x_2(x_2-x_1)&x_3(x_3-x_1)&\cdots&x_n(x_n-x_1)\\ \vdots&\vdots&&\vdots\\ x_2^{n-2}(x_2-x_1)&x_3^{n-2}(x_3-x_1)&\cdots&x_n^{n-2}(x_n-x_1) \end{vmatrix}\\\\ &=(x_2-x_1)(x_3-x_1)\cdots(x_n-x_1) \begin{vmatrix} 1&1&\cdots&1\\ x_2&x_3&\cdots&x_n\\ x_2^2&x_3^2&\cdots&x_n^2\\ \vdots&\vdots&&\vdots\\ x_2^{n-2}&x_3^{n-2}&\cdots&x_n^{n-2} \end{vmatrix}\\\\ &=(x_2-x_1)(x_3-x_1)\cdots(x_n-x_1)\prod_{2\le i\lt j\le n}(x_j-x_i)\\ &=\prod_{1\le i\lt j\le n}(x_j-x_i) \end{align} $$

根据数学归纳法，证得$Vandermonde$行列式

$$ D_n=\prod_{1\le i\lt j\le n}(x_j-x_i) $$

对任意的正整数$n(n\ge 2)$都成立