例题 5.2

例题 5.2.

证明 $n$ 阶 $(n \geq 2)$ 范德蒙德 (Vandermonde) 行列式 $V_{n} = 1 a_{1} a_{1}^{2} ⋮ a_{1}^{n - 2} a_{1}^{n - 1} 1 a_{2} a_{2}^{2} ⋮ a_{2}^{n - 2} a_{2}^{n - 1} 1 a_{3} a_{3}^{2} ⋮ a_{3}^{n - 2} a_{3}^{n - 1} \dots \dots \dots \dots \dots 1 a_{n - 1} a_{n - 1}^{2} ⋮ a_{n - 1}^{n - 2} a_{n - 1}^{n - 1} 1 a_{n} a_{n}^{2} ⋮ a_{n}^{n - 2} a_{n}^{n - 1} = 1 \leq j < i \leq n \prod (a_{i} - a_{j}) .$

\begin{proof} 对 $n$ 用数学归纳法.

当 $n = 2$ 时,有 $V_{2} = 1 a_{1} 1 a_{2} = a_{2} - a_{1}$ 结论成立.

假设对于 $n - 1$ 阶范德蒙德行列式 $V_{n - 1}$ 结论成立, 下证对于 $n$ 阶范德蒙德行列式结论成立.

在 $V_{n}$ 中,从第 $n$ 行开始逐行减去上一行的 $a_{n}$ 倍,得

$V_{n} \frac{r _{i} - a _{n} r _{i - 1}}{i = n , n - 1 , \dots , 2} 1 a_{1} - a_{n} a_{1} (a_{1} - a_{n}) ⋮ a_{1}^{n - 3} (a_{1} - a_{n}) a_{1}^{n - 2} (a_{1} - a_{n}) 1 a_{2} - a_{n} a_{2} (a_{2} - a_{n}) ⋮ a_{2}^{n - 3} (a_{2} - a_{n}) a_{2}^{n - 2} (a_{2} - a_{n}) \dots \dots \dots \dots \dots 1 a_{n - 1} - a_{n} a_{n - 1} (a_{n - 1} - a_{n}) ⋮ a_{n - 1}^{n - 3} (a_{n - 1} - a_{n}) a_{n - 1}^{n - 2} (a_{n - 1} - a_{n}) 100 ⋮ 00$

$按第 n 列展开 (- 1)^{1 + n} a_{1} - a_{n} a_{1} (a_{1} - a_{n}) ⋮ a_{1}^{n - 3} (a_{1} - a_{n}) a_{1}^{n - 2} (a_{1} - a_{n}) a_{2} - a_{n} a_{2} (a_{2} - a_{n}) ⋮ a_{2}^{n - 3} (a_{2} - a_{n}) a_{2}^{n - 2} (a_{2} - a_{n}) \dots \dots \dots \dots a_{n - 1} - a_{n} a_{n - 1} (a_{n - 1} - a_{n}) ⋮ a_{n - 1}^{n - 3} (a_{n - 1} - a_{n}) a_{n - 1}^{n - 2} (a_{n - 1} - a_{n})$

$\frac{c _{i} \div ( a _{i} - a _{n} )}{i = 1 , 2 , \dots , n - 1} (- 1)^{n + 1} 1 \leq j \leq n - 1 \prod (a_{j} - a_{n}) 1 a_{1} a_{1}^{2} ⋮ a_{1}^{n - 2} 1 a_{2} a_{2}^{2} ⋮ a_{2}^{n - 2} 1 a_{3} a_{3}^{2} ⋮ a_{3}^{n - 2} \dots \dots \dots \dots 1 a_{n - 1} a_{n - 1}^{2} ⋮ a_{n - 1}^{n - 2}$

$= 1 \leq j \leq n - 1 \prod (a_{n} - a_{j}) V_{n - 1}$

$= 1 \leq j \leq n - 1 \prod (a_{n} - a_{j}) 1 \leq j < i \leq n - 1 \prod (a_{i} - a_{j})$

$= 1 \leq j < i \leq n \prod (a_{i} - a_{j}) .$ \end{proof}

Youliang Zhong

例题 5.2

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