「高等代数」浅谈缺项 Vandermonde 行列式
前置知识
Vandermonde 行列式定义: 形如下列的行列式称为 Vandermonde 行列式.
\[ V_n = \begin{vmatrix} 1 & 1 & \cdots & 1 \\ a_1 & a_2 & \cdots & a_n \\ a_1^2 & a_2^2 & \cdots & a_n^2 \\ \vdots & \vdots & & \vdots \\ a_1^{n - 2} & a_2^{n - 2} & \cdots & a_n^{n - 2} \\ a_1^{n - 1} & a_2^{n - 1} & \cdots & a_n^{n - 1} \end{vmatrix} = \prod\limits_{1 \le i < j \le n}{(a_j - a_i)} \]
证明可用经典的数学归纳法和隔行相消证明.
易证当 \(n = 2\) 时成立. 设当 \(n = k - 1\) 时成立, 则 \(n = k\) 时:
\[ \begin{align*} V_k &= \begin{vmatrix} 1 & 1 & \cdots & 1 \\ a_1 & a_2 & \cdots & a_k \\ a_1^2 & a_2^2 & \cdots & a_k^2 \\ \vdots & \vdots & & \vdots \\ a_1^{k - 2} & a_2^{k - 2} & \cdots & a_k^{k - 2} \\ a_1^{k - 1} & a_2^{k - 1} & \cdots & a_k^{k - 1} \end{vmatrix} \xlongequal[j = n, n - 1, \cdots, 2]{r_j - ar_{j - 1}} \begin{vmatrix} 1 & 1 & \cdots & 1 \\ 0 & a_2 - a_1 & \cdots & a_k - a_1 \\ 0 & (a_2 - a_1)a_2 & \cdots & (a_k - a_1)a_k \\ \vdots & \vdots & & \vdots \\ 0 & (a_2 - a_1)a_2^{k - 3} & \cdots & (a_k - a_1)a_k^{k - 3} \\ 0 & (a_2 - a_1)a_2^{k - 2} & \cdots & (a_k - a_1)a_k^{k - 2} \end{vmatrix} \\ &= (a_2 - a_1)(a_3 - a_1) \cdots (a_k - a_1) \begin{vmatrix} 1 & 1 & \cdots & 1 \\ a_2 & a_3 & \cdots & a_k \\ \vdots & \vdots & & \vdots \\ a_2^{k - 2} & a_3^{k - 2} & \cdots & a_k^{k - 2} \end{vmatrix} = \prod\limits_{1 \le i < j \le k}{(a_j - a_i)} \end{align*} \]
故 \(n = k\) 成立. 由数学归纳法得证.
缺项 Vandermonde 行列式
例: 求下列 \(n\) 阶行列式的值:
\[ \begin{vmatrix} 1 & x_1 & \cdots & x_1^{k - 1} & x_1^{k + 1} & \cdots & x_1^n \\ 1 & x_2 & \cdots & x_2^{k - 1} & x_2^{k + 1} & \cdots & x_2^n \\ \vdots & \vdots & & \vdots & \vdots & & \vdots\\ 1 & x_{n - 1} & \cdots & x_{n - 1}^{k - 1} & x_{n - 1}^{k + 1} & \cdots & x_{n - 1}^n \\ 1 & x_n & \cdots & x_n^{k - 1} & x_n^{k + 1} & \cdots & x_n^n \end{vmatrix} (2 \le k \le n) \]
可见该行列式可看作 Vandermonde 行列式缺了一列.
设上述行列式为 \(D\). 将其加边转化为 Vandermonde 行列式. 取行列式:
\[ V = \begin{vmatrix} 1 & x_1 & \cdots & x_1^{k - 1} & x_1^k & x_1^{k + 1} & \cdots & x_1^n \\ 1 & x_2 & \cdots & x_2^{k - 1} & x_2^k & x_2^{k + 1} & \cdots & x_2^n \\ \vdots & \vdots & & \vdots & \vdots & \vdots & & \vdots\\ 1 & x_{n - 1} & \cdots & x_{n - 1}^{k - 1} & x_{n - 1}^k & x_{n - 1}^{k + 1} & \cdots & x_{n - 1}^n \\ 1 & x_n & \cdots & x_n^{k - 1} & x_n^k & x_n^{k + 1} & \cdots & x_n^n \\ 1 & y & \cdots & y^{k - 1} & y^k & y^{k + 1} & \cdots & y^n \end{vmatrix} \]
对第 \(n + 1\) 行展开, 有:
\[ V = (-1)^{n + 2} M_{1 \ n + 1} + (-1)^{n + 3} y M_{2 \ n + 1} + (-1)^{n + 4} y^2 M_{3 \ n + 1} + \cdots + (-1)^{2n + 2} y^n M_{n+1 \ n + 1} \]
又有 \(M_{k \ n + 1} = D\). 即取 \(V\) 中 \(y^k\) 的系数 \(X = (-1)^{n + k + 2}M_{k \ n + 1} \Rightarrow D = (-1)^{n + k}X\).
又由 Vandermonde 行列式得:
\[ V = \left[\prod\limits_{1 \le i \le j < n}{(x_j - x_i)}\right]\left[\prod\limits_{i = 1}^n{(y - x_i)}\right] \]
故由排列组合得:
\[ X = \left[\prod\limits_{1 \le i \le j < n}{(x_j - x_i)}\right]\left(-1\right)^{n - k}\left(\sum\limits_{1 \le j_1 < j_2 < \cdots < j_{n - k} \le n}{\prod\limits_{i = 1}^{n - k}{x_{j_i}}}\right) \]
故带入得:
\[ D = \left[\prod\limits_{1 \le i \le j < n}{(x_j - x_i)}\right]\left(\sum\limits_{1 \le j_1 < j_2 < \cdots < j_{n - k} \le n}{\prod\limits_{i = 1}^{n - k}{x_{j_i}}}\right) \]