﻿ 范德蒙行列式的矩阵形式推广及其应用

# 范德蒙行列式的矩阵形式推广及其应用The Popularization and Application of Matrix Form of Vandermonde’s Determinant

Abstract: The purpose of this article is to use the properties of block matrices and matrix direct products to generalize the elements of the Vandermonde determinant into a matrix form, so that the general-ized determinant still has a general solution formula similar to the Vandermonde determinant, and can solve more complicated problem of determinant evaluation.

1. 引言

2. 预备知识

1) n阶范德蒙行列式的通解公式

$d=|\begin{array}{ccccc}1& 1& 1& \cdots & 1\\ {a}_{1}& {a}_{2}& {a}_{3}& \cdots & {a}_{n}\\ {a}_{1}^{2}& {a}_{2}^{2}& {a}_{3}^{2}& \cdots & {a}_{n}^{2}\\ ⋮& ⋮& ⋮& & ⋮\\ {a}_{1}^{n-1}& {a}_{2}^{n-1}& {a}_{3}^{n-1}& \cdots & {a}_{n}^{n-1}\end{array}|=\underset{1\le j

2) 分块矩阵3点性质：

$A=\begin{array}{cc}& \begin{array}{cccc}{n}_{1}& {n}_{2}& \cdots & {n}_{l}\end{array}\\ \begin{array}{c}{s}_{1}\\ {s}_{2}\\ ⋮\\ {s}_{t}\end{array}& \left[\begin{array}{cccc}{A}_{11}& {A}_{12}& \cdots & {A}_{1l}\\ {A}_{21}& {A}_{22}& \cdots & {A}_{2l}\\ ⋮& ⋮& & ⋮\\ {A}_{t1}& {A}_{t2}& \cdots & {A}_{tl}\end{array}\right]\end{array}$$B=\begin{array}{cc}& \begin{array}{cccc}{m}_{1}& {m}_{2}& \cdots & {m}_{r}\end{array}\\ \begin{array}{c}{n}_{1}\\ {n}_{2}\\ ⋮\\ {n}_{l}\end{array}& \left[\begin{array}{cccc}{B}_{11}& {B}_{12}& \cdots & {B}_{1r}\\ {B}_{21}& {B}_{22}& \cdots & {B}_{2r}\\ ⋮& ⋮& & ⋮\\ {B}_{l1}& {B}_{l2}& \cdots & {B}_{lr}\end{array}\right]\end{array}$

$C=AB=\begin{array}{cc}& \begin{array}{cccc}{m}_{1}& {m}_{2}& \cdots & {m}_{r}\end{array}\\ \begin{array}{c}{s}_{1}\\ {s}_{2}\\ ⋮\\ {s}_{t}\end{array}& \left[\begin{array}{cccc}{C}_{11}& {C}_{12}& \cdots & {C}_{1r}\\ {C}_{21}& {C}_{22}& \cdots & {C}_{2r}\\ ⋮& ⋮& & ⋮\\ {C}_{t1}& {C}_{t2}& \cdots & {C}_{tr}\end{array}\right]\end{array}$

$|\begin{array}{cc}A& B\\ C& D\end{array}|=|A||D-C{A}^{-1}B|$

3) 矩阵直积的定义与性质：

$A\otimes B=\left(\begin{array}{cccc}{a}_{11}B& {a}_{12}B& \cdots & {a}_{1n}B\\ ⋮& ⋮& & ⋮\\ {a}_{m1}B& {a}_{m2}B& \cdots & {a}_{mn}B\end{array}\right)$

${P}^{-1}AP={J}_{A}$${Q}^{-1}BQ={J}_{B}$

${J}_{A}=\left(\begin{array}{cccc}{\lambda }_{1}& & & \\ {t}_{1}& {\lambda }_{2}& & \\ & \ddots & \ddots & \\ & & {t}_{m-1}& {\lambda }_{m}\end{array}\right)$${J}_{B}=\left(\begin{array}{cccc}{\mu }_{1}& & & \\ {s}_{1}& {\mu }_{2}& & \\ & \ddots & \ddots & \\ & & {s}_{n-1}& {\mu }_{n}\end{array}\right)$

${\left(P\otimes Q\right)}^{-1}\left(A\otimes B\right)\left(P\otimes Q\right)=\left({P}^{-1}AP\right)\otimes \left({Q}^{-1}BQ\right)={J}_{A}\otimes {J}_{B}$

$A\otimes B$ 的特征值即 ${J}_{A}\otimes {J}_{B}$ 的特征值是 ${\lambda }_{i}{\mu }_{j}$

$|A\otimes B|=\prod {\lambda }_{i}{\mu }_{j}={\left(\prod {\lambda }_{i}\right)}^{n}×{\left(\prod {u}_{j}\right)}^{m}={|A|}^{n}{|B|}^{m}$

3. 主要结论

$A=\left(\begin{array}{ccccc}{A}_{11}& \cdots & {A}_{1j}& \cdots & {A}_{1n}\\ ⋮& \ddots & ⋮& \ddots & ⋮\\ {A}_{i1}& \cdots & {A}_{ij}& \cdots & {A}_{in}\\ ⋮& \ddots & ⋮& \ddots & ⋮\\ {A}_{n1}& \cdots & {A}_{nj}& \cdots & {A}_{nn}\end{array}\right)$

$B=\left(\begin{array}{ccccc}{A}_{11}& \cdots & M{A}_{1j}& \cdots & {A}_{1n}\\ ⋮& \ddots & ⋮& \ddots & ⋮\\ {A}_{i1}& \cdots & M{A}_{ij}& \cdots & {A}_{in}\\ ⋮& \ddots & ⋮& \ddots & ⋮\\ {A}_{n1}& \cdots & M{A}_{nj}& \cdots & {A}_{nn}\end{array}\right)$$C=\left(\begin{array}{ccccc}{A}_{11}& \cdots & {A}_{1j}& \cdots & {A}_{1n}\\ ⋮& \ddots & ⋮& \ddots & ⋮\\ M{A}_{i1}& \cdots & M{A}_{ij}& \cdots & M{A}_{in}\\ ⋮& \ddots & ⋮& \ddots & ⋮\\ {A}_{n1}& \cdots & {A}_{nj}& \cdots & {A}_{nn}\end{array}\right)$

$|D|=|\begin{array}{ccccc}E& E& E& \cdots & E\\ {D}_{1}& {D}_{2}& {D}_{3}& \cdots & {D}_{n}\\ {D}_{1}^{2}& {D}_{2}^{2}& {D}_{3}^{2}& \cdots & {D}_{n}^{2}\\ ⋮& ⋮& ⋮& & ⋮\\ {D}_{1}^{n-1}& {D}_{2}^{n-1}& {D}_{3}^{n-1}& \cdots & {D}_{n}^{n-1}\end{array}|$ (1)

$n=2$ 时由性质2.2，可得

$|\begin{array}{cc}E& E\\ {D}_{1}& {D}_{2}\end{array}|=|E||{D}_{2}-{D}_{1}EE|=|{D}_{2}-{D}_{1}|$

$\begin{array}{l}|\left(\begin{array}{cccc}E& & & \\ -{D}_{1}& E& & \\ & \ddots & \ddots & \\ & & -{D}_{1}& E\end{array}\right)\left(\begin{array}{cccc}E& E& \cdots & E\\ {D}_{1}& {D}_{2}& \cdots & {D}_{n}\\ ⋮& & \ddots & ⋮\\ {D}_{1}^{n-1}& {D}_{2}^{n-1}& \cdots & {D}_{n}^{n-1}\end{array}\right)|\\ =|\begin{array}{cccc}E& & & \\ -{D}_{1}& E& & \\ & \ddots & \ddots & \\ & & -{D}_{1}& E\end{array}|\cdot |\begin{array}{cccc}E& E& \cdots & E\\ {D}_{1}& {D}_{2}& \cdots & {D}_{n}\\ ⋮& & \ddots & ⋮\\ {D}_{1}^{n-1}& {D}_{2}^{n-1}& \cdots & {D}_{n}^{n-1}\end{array}|=|D|\end{array}$

$\begin{array}{c}|D|=|\begin{array}{cccc}E& E& \cdots & E\\ O& {D}_{2}-{D}_{1}& \cdots & {D}_{n}-{D}_{1}\\ ⋮& ⋮& & ⋮\\ O& {D}_{2}^{n-1}-{D}_{1}{D}_{2}^{n-2}& \cdots & {D}_{n}^{n-1}-{D}_{1}{D}_{n}^{n-2}\end{array}|\\ =|\begin{array}{cccc}{D}_{2}-{D}_{1}& {D}_{3}-{D}_{1}& \cdots & {D}_{n}-{D}_{1}\\ {D}_{2}^{2}-{D}_{1}{D}_{2}& {D}_{3}^{2}-{D}_{1}{D}_{3}& \cdots & {D}_{n}^{2}-{D}_{1}{D}_{n}\\ ⋮& ⋮& & ⋮\\ {D}_{2}^{n-1}-{D}_{1}{D}_{2}^{n-2}& {D}_{3}^{n-1}-{D}_{1}{D}_{3}^{n-2}& \cdots & {D}_{n}^{n-1}-{D}_{1}{D}_{n}^{n-2}\end{array}|\\ =|\begin{array}{cccc}{D}_{2}-{D}_{1}& {D}_{3}-{D}_{1}& \cdots & {D}_{n}-{D}_{1}\\ \left({D}_{2}-{D}_{1}\right){D}_{2}& \left({D}_{3}-{D}_{1}\right){D}_{3}& \cdots & \left({D}_{n}-{D}_{1}\right){D}_{n}\\ ⋮& ⋮& & ⋮\\ \left({D}_{2}-{D}_{1}\right){D}_{2}^{n-2}& \left({D}_{3}-{D}_{1}\right){D}_{3}^{n-2}& \cdots & \left({D}_{n}-{D}_{1}\right){D}_{n}^{n-2}\end{array}|\end{array}$

$|{D}_{2}-{D}_{1}|×|{D}_{3}-{D}_{1}|×\cdots ×|{D}_{n}-{D}_{1}|×|\begin{array}{cccc}E& E& \cdots & E\\ {D}_{2}& {D}_{3}& \cdots & {D}_{n}\\ ⋮& ⋮& & ⋮\\ {D}_{2}^{n-2}& {D}_{3}^{n-2}& \cdots & {D}_{n}^{n-2}\end{array}|$

$|D|=|\begin{array}{ccccc}B& B& B& \cdots & B\\ {a}_{1}B& {a}_{2}B& {a}_{3}B& \cdots & {a}_{m}B\\ {a}_{1}^{2}B& {a}_{2}^{2}B& {a}_{3}^{2}B& \cdots & {a}_{m}^{2}B\\ ⋮& ⋮& ⋮& & ⋮\\ {a}_{1}^{m-1}B& {a}_{2}^{m-1}B& {a}_{3}^{m-1}B& \cdots & {a}_{m}^{m-1}B\end{array}|$

$|D|={\left(\underset{1\le j

$A=\left(\begin{array}{ccccc}1& 1& 1& \cdots & 1\\ {a}_{1}& {a}_{2}& {a}_{3}& \cdots & {a}_{n}\\ {a}_{1}^{2}& {a}_{2}^{2}& {a}_{3}^{2}& \cdots & {a}_{n}^{2}\\ ⋮& ⋮& ⋮& & ⋮\\ {a}_{1}^{m-1}& {a}_{2}^{m-1}& {a}_{3}^{m-1}& \cdots & {a}_{m}^{m-1}\end{array}\right)$

$|D|=|A\otimes B|={|A|}^{n}{|B|}^{m}$

$|A|=\underset{1\le j

$|D|={\left(\underset{1\le j

4. 应用举例

1) 计算 $|D|=|\begin{array}{cccccc}1& 0& 1& 0& 1& 0\\ 0& 1& 0& 1& 0& 1\\ 1& 2& 1& 3& 1& 4\\ 2& 3& 3& 4& 4& 5\\ 5& 8& 10& 15& 17& 24\\ 8& 13& 15& 25& 24& 41\end{array}|$ 的值。

${D}_{1}=|\begin{array}{cc}1& 2\\ 2& 3\end{array}|$${D}_{2}=|\begin{array}{cc}1& 3\\ 3& 4\end{array}|$${D}_{3}=|\begin{array}{cc}1& 4\\ 4& 5\end{array}|$

$|D|=|{D}_{2}-{D}_{1}||{D}_{3}-{D}_{1}||{D}_{3}-{D}_{2}|=|\begin{array}{cc}0& 1\\ 1& 1\end{array}|×|\begin{array}{cc}0& 2\\ 2& 2\end{array}|×|\begin{array}{cc}0& 1\\ 1& 1\end{array}|=\left(-1\right)×\left(-4\right)×\left(-1\right)=-4.$

2) 计算 $|D|=|\begin{array}{cccccc}1& 2& 1& 2& 1& 2\\ 3& 4& 3& 4& 3& 4\\ x& 2x& y& 2y& z& 2z\\ 3x& 4x& 3y& 4y& 3z& 4z\\ {x}^{2}& 2{x}^{2}& {y}^{2}& 2{y}^{2}& z{}^{2}& 2z{}^{2}\\ 3{x}^{2}& 4{x}^{2}& 3{y}^{2}& 4{y}^{2}& 3z{}^{2}& 4z{}^{2}\end{array}|$ 的值。

$|D|=|\left(\begin{array}{ccc}1& 1& 1\\ x& y& z\\ {x}^{2}& {y}^{2}& {z}^{2}\end{array}\right)\otimes \left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)|$

$|D|={|\begin{array}{ccc}1& 1& 1\\ x& y& z\\ {x}^{2}& {y}^{2}& {z}^{2}\end{array}|}^{2}×{|\begin{array}{cc}1& 2\\ 3& 4\end{array}|}^{3}={\left(\left(z-y\right)\left(z-x\right)\left(y-x\right)\right)}^{2}{|\begin{array}{cc}1& 2\\ 3& 4\end{array}|}^{3}=-8{\left(\left(z-y\right)\left(z-x\right)\left(y-x\right)\right)}^{2}$

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