范德蒙德行列式证明

范德蒙德行列式:

111x1x2xnx12x22xn2x1n1x2n1xnn1=ni>j1(xixj) \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 & \ddots & \vdots \\ x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \end{vmatrix} = \prod_{n \ge i > j \ge 1}(x_i-x_j)

证明如下:

从第 nn 行开始,将行列式中的后一行减去前一行的 x1x_1 倍,得到下式:

1110x2x1xnx10x2(x2x1)xn(xnx1)0x2n2(x2x1)xnn2(xnx1)  =1110x2x1xnx10x2(x2x1)xn(xnx1)0x2n2(x2x1)xnn2(xnx1) \begin{align*} \begin{vmatrix} 1 & 1 & \cdots & 1 \\ 0 & x_2-x_1 & \cdots & x_n-x_1 \\ 0 & x_2(x_2-x_1) & \cdots & x_n(x_n-x_1) \\ \vdots & \vdots & \ddots & \vdots \\ 0 & x_2^{n-2}(x_2-x_1) & \cdots & x_n^{n-2}(x_n-x_1) \end{vmatrix}\\ \ \\ \ \\ = \begin{vmatrix} 1 & 1 & \cdots & 1 \\ 0 & x_2-x_1 & \cdots & x_n-x_1 \\ 0 & x_2(x_2-x_1) & \cdots & x_n(x_n-x_1) \\ \vdots & \vdots & \ddots & \vdots \\ 0 & x_2^{n-2}(x_2-x_1) & \cdots & x_n^{n-2}(x_n-x_1) \end{vmatrix} \end{align*}

将得到的行列式按第 11 行展开,得到

x2x1x3x1xnx1x2(x2x1)x3(x3x1)xn(xnx1)x2n2(x2x1)x3n2(x3x1)xnn2(xnx1)  =(x2x1)(x3x1)(xnx1)111x2x3xnx2n2x3n2xnn2 \begin{align*} &\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 & \ddots & \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 \\ \vdots & \vdots & \ddots & \vdots \\ x_2^{n-2} & x_3^{n-2} & \cdots & x_n^{n-2} \end{vmatrix} \end{align*}

此时该式右端为一个 n1n-1 阶的范德蒙德行列式,可以写成:

(x3x2)(x4x2)(xnx2)111x3x4xnx3n3x4n3x3n3 (x_3-x_2)(x_4-x_2) \cdots (x_n-x_2) \begin{vmatrix} 1 & 1 & \cdots & 1 \\ x_3 & x_4 & \cdots & x_n \\ \vdots & \vdots & \ddots & \vdots \\ x_3^{n-3} & x_4^{n-3} & \cdots & x_3^{n-3} \end{vmatrix}

根据数学归纳法,可以把范德蒙德行列式写作:

(x2x1)(x3x1)(xnx1)(x3x2)(x4x2)(xnxn1)  =ni>j1(xixj) \begin{align*} &(x_2-x_1)(x_3-x_1) \cdots (x_n-x_1)(x_3-x_2)(x_4-x_2) \cdots (x_n-x_{n-1}) \\ \ \\ \ \\ &= \prod_{n \ge i > j \ge 1}(x_i-x_j) \end{align*}

证毕。

翻译: