﻿ 特殊矩阵行列式的计算

# 特殊矩阵行列式的计算Computation of the Determinants of Special Matrices

Abstract: Matrix and determinant are often encountered in science and engineering. The characteristic polynomial and eigenvalues of a matrix require computing its determinant. Aiming at some special matrices, this paper discusses the calculation problems of some determinants, including the characteristic polynomials of some special matrices.

1. 引言

2. 计算友矩阵特征多项式和行列式

${A}_{n}=\left(\begin{array}{ccccc}-{a}_{1}& -{a}_{2}& \cdots & -{a}_{n-1}& -{a}_{n}\\ 1& 0& \cdots & 0& 0\\ 0& 1& \cdots & 0& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& \dots & 1& 0\end{array}\right)$

${I}_{n}$ 是n阶单位阵。友矩阵的特征值多项式

${f}_{n}\left(s\right):=|s{I}_{n}-{A}_{n}|=|\begin{array}{ccccc}s+{a}_{1}& {a}_{2}& \cdots & {a}_{n-1}& {a}_{n}\\ -1& s& \cdots & 0& 0\\ 0& -1& \cdots & 0& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& \cdots & -1& s\end{array}|$

${f}_{2}\left(s\right)=|\begin{array}{cc}s+{a}_{1}& {a}_{2}\\ -1& s\end{array}|=s{f}_{1}\left(s\right)+{a}_{2}=s\left(s+{a}_{1}\right)+{a}_{2}$

${f}_{3}\left(s\right)=|\begin{array}{ccc}s+{a}_{1}& {a}_{2}& {a}_{3}\\ -1& s& 0\\ 0& -1& s\end{array}|=s{f}_{2}\left(s\right)-\left(-1\right)|\begin{array}{cc}s+{a}_{1}& {a}_{3}\\ -1& 0\end{array}|=s{f}_{2}\left(s\right)+{a}_{3}$

$\begin{array}{c}{f}_{4}\left(s\right)=|\begin{array}{cccc}s+{a}_{1}& {a}_{2}& {a}_{3}& {a}_{4}\\ -1& s& 0& 0\\ 0& -1& s& 0\\ 0& 0& -1& s\end{array}|=s{f}_{3}\left(s\right)-\left(-1\right)|\begin{array}{ccc}s+{a}_{1}& {a}_{2}& {a}_{4}\\ -1& s& 0\\ 0& -1& 0\end{array}|\\ =s{f}_{3}\left(s\right)+|\begin{array}{cc}s+{a}_{1}& {a}_{4}\\ -1& 0\end{array}|=s{f}_{3}\left(s\right)+{a}_{4}\end{array}$

$\begin{array}{l}{f}_{1}\left(s\right)=s+{a}_{1}\\ {f}_{2}\left(s\right)=s\left(s+{a}_{1}\right)+{a}_{2}={s}^{2}+s{a}_{1}+{a}_{2}\\ {f}_{3}\left(s\right)=s\left({s}^{2}+s{a}_{1}+{a}_{2}\right)+{a}_{3}={s}^{3}+{s}^{2}{a}_{1}+s{a}_{2}+{a}_{3}\\ {f}_{4}\left(s\right)=s\left({s}^{3}+{s}^{2}{a}_{1}+s{a}_{2}+{a}_{3}\right)+{a}_{4}={s}^{4}+{s}^{3}{a}_{1}+{s}^{2}{a}_{2}+s{a}_{3}+{a}_{4}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}⋮\\ {f}_{n}\left(s\right)={s}^{n}+{s}^{n-1}{a}_{1}+{s}^{n-2}{a}_{2}+{s}^{n-3}{a}_{3}+\cdots +{a}_{n}={s}^{n}+\underset{i=1}{\overset{n}{\sum }}{s}^{n-i}{a}_{i}\end{array}$

$\begin{array}{c}{f}_{n+1}\left(s\right)=s{f}_{n}\left(s\right)+\left(-1\right){\left(-1\right)}^{2n+1}|\begin{array}{cccccc}s+{a}_{1}& {a}_{2}& \cdots & {a}_{n-2}& {a}_{n-1}& {a}_{n+1}\\ -1& s& \cdots & 0& 0& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ 0& 0& \cdots & -1& s& 0\\ 0& 0& \cdots & 0& -1& 0\end{array}|\\ =s{f}_{n}\left(s\right)+\left(-1\right){\left(-1\right)}^{2n-1}|\begin{array}{cccccc}s+{a}_{1}& {a}_{2}& \cdots & {a}_{n-3}& {a}_{n-2}& {a}_{n+1}\\ -1& s& \cdots & 0& 0& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ 0& 0& \cdots & -1& s& 0\\ 0& 0& \cdots & 0& -1& 0\end{array}|\end{array}$

$\begin{array}{c}{f}_{n+1}\left(s\right)=s{f}_{n}\left(s\right)+|\begin{array}{cc}s+{a}_{1}& {a}_{n+1}\\ -1& 0\end{array}|=s{f}_{n}\left(s\right)+{a}_{n+1}\\ =s\left({s}^{n}+\underset{i=1}{\overset{n}{\sum }}{a}_{i}{s}^{n-i}\right)+{a}_{n+1}={s}^{n+1}+\underset{i=1}{\overset{n+1}{\sum }}{a}_{i}{s}^{n+1-i}\end{array}$

3. 计算十字叉形壹矩阵的特征多项式

n阶十字叉形壹矩阵具有下列形式：

${A}_{n}={\left(\begin{array}{ccccc}1& & & & 1\\ & 1& & 1& \\ & & \ddots & & \\ & 1& & 1& \\ 1& & & & 1\end{array}\right)}_{n×n}$ (未写出部分的值均为0)，

${A}_{n}$ 的特征多项式为

${f}_{n}\left(s\right)=|s{I}_{n}-{A}_{n}|=|\begin{array}{ccccc}s-1& & & & -1\\ & s-1& & -1& \\ & & \ddots & & \\ & -1& & s-1& \\ -1& & & & s-1\end{array}|$

① 当n为奇数时，

$\begin{array}{c}{f}_{n}\left(s\right)=\left(s-1\right)|\begin{array}{cccc}s-1& \cdots & -1& 0\\ ⋮& \ddots & ⋮& ⋮\\ -1& \cdots & s-1& 0\\ 0& \cdots & 0& s-1\end{array}|+\left(-1\right)|\begin{array}{cccc}0& \cdots & 0& -1\\ s-1& \cdots & -1& 0\\ ⋮& \ddots & ⋮& ⋮\\ -1& \cdots & s-1& 0\end{array}|\\ ={\left(s-1\right)}^{2}{f}_{n-2}\left(s\right)+\left(-1\right)\left(-1\right){\left(-1\right)}^{n}{f}_{n-2}\left(s\right)\\ =\left({s}^{2}-2s\right){f}_{n-2}\left(s\right)\end{array}$

② 当n为偶数时，

$\begin{array}{c}{f}_{n}\left(s\right)=\left(s-1\right)|\begin{array}{cccc}s-1& \cdots & -1& 0\\ ⋮& \ddots & ⋮& ⋮\\ -1& \cdots & s-1& 0\\ 0& \cdots & 0& s-1\end{array}|-\left(-1\right)|\begin{array}{cccc}0& \cdots & 0& -1\\ s-1& \cdots & -1& 0\\ ⋮& \ddots & ⋮& ⋮\\ -1& \cdots & s-1& 0\end{array}|\\ ={\left(s-1\right)}^{2}{f}_{n-2}\left(s\right)-\left(-1\right)\left(-1\right){\left(-1\right)}^{n}{f}_{n-2}\left(s\right)\\ =\left({s}^{2}-2s\right){f}_{n-2}\left(s\right)\end{array}$

${f}_{n}\left(s\right)=\left\{\begin{array}{l}\left(s-1\right){\left({s}^{2}-2s\right)}^{\frac{n-1}{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\text{ }为奇数\\ {\left({s}^{2}-2s\right)}^{\frac{n}{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }n\text{ }为偶数\end{array}$

4. 计算块单位阵的特征值多项式

2n阶块单位阵 ${A}_{2n}={\left[\begin{array}{cc}{I}_{n}& {I}_{n}\\ {I}_{n}& {I}_{n}\end{array}\right]}_{2n×2n}$ 的特征多项式为

${f}_{2n}\left(s\right):=|s{I}_{2n}-{A}_{2n}|=|\begin{array}{cccccc}s-1& & & -1& & \\ & \ddots & & & \ddots & \\ & & s-1& & & -1\\ -1& & & s-1& & \\ & \ddots & & & \ddots & \\ & & -1& & & s-1\end{array}|$

$n=1$ 时， ${f}_{2}\left(s\right)=|\begin{array}{cc}s-1& -1\\ -1& s-1\end{array}|=s\left(s-2\right)$ 。注意到此行列式为中心对称行列式，可以进行操作： ${r}_{1}+{r}_{2n}$${r}_{2}+{r}_{2n-1}$$\cdots$${r}_{n}+{r}_{n+1}$ ( ${r}_{i}$ 表示第i行)，即把最后一行的值加到第1行上，把倒数第2行的值加到第2行上，依次类推，一直加到最中间的那两行。但此时也需要对n进行奇偶讨论。

① 当n为奇数时，

${f}_{2n}\left(s\right)=|\begin{array}{cccccccccc}s-1& & & & -1& -1& & & & s-1\\ & \ddots & & ⋰& & & \ddots & & ⋰& \\ & & s-2& & & & & s-2& & \\ & ⋰& & \ddots & & & ⋰& & \ddots & \\ -1& & & & s-1& s-1& & & & -1\\ -1& & & & & s-1& & & & \\ & \ddots & & & & & \ddots & & & \\ & & -1& & & & & s-1& & \\ & & & \ddots & & & & & \ddots & \\ & & & & -1& & & & & s-1\end{array}|$

② 当n为偶数时，

${f}_{2n}\left(s\right)=|\begin{array}{cccccccccccc}s-1& & & & & -1& -1& & & & & s-1\\ & \ddots & & & ⋰& & & \ddots & & & ⋰& \\ & & s-1& -1& & & & & -1& s-1& & \\ & & -1& s-1& & & & & s-1& -1& & \\ & ⋰& & & \ddots & & & ⋰& & & \ddots & \\ -1& & & & & s-1& s-1& & & & & -1\\ & & -{I}_{n}& & & & & & \left(s-1\right){I}_{n}& & & \end{array}|$

① 当n为奇数时，

$\begin{array}{c}{f}_{2n}\left(s\right)=|\begin{array}{cccccccccc}s-1& & & & -1& & & & & \\ & \ddots & & ⋰& & & & & & \\ & & s-2& & & & & 0& & \\ & ⋰& & \ddots & & & & & & \\ -1& & & & s-1& & & & & \\ -1& & & & & s-1& & & & 1\\ & \ddots & & & & & \ddots & & ⋰& \\ & & -1& & & & & s& & \\ & & & \ddots & & & ⋰& & \ddots & \\ & & & & -1& 1& & & & s-1\end{array}|\\ =|\begin{array}{ccccc}s-1& & & & -1\\ & \ddots & & ⋰& \\ & & s-2& & \\ & ⋰& & \ddots & \\ -1& & & & s-1\end{array}|\cdot |\begin{array}{ccccc}s-1& & & & 1\\ & \ddots & & ⋰& \\ & & s& & \\ & ⋰& & \ddots & \\ 1& & & & s-1\end{array}|\end{array}$

$\begin{array}{c}{f}_{n}\left(s\right)=\left(s-2\right){\left({s}^{2}-2s\right)}^{\frac{n-1}{2}}\cdot s{\left({s}^{2}-2s\right)}^{\frac{n-1}{2}}\\ ={\left({s}^{2}-2s\right)}^{n}\end{array}$

② 当n为偶数时，

$\begin{array}{c}{f}_{2n}\left(s\right)=|\begin{array}{cccccccccccc}s-1& & & & & -1& & & & & & \\ & \ddots & & & ⋰& & & & & & & \\ & & s-1& -1& & & & & \begin{array}{l}\\ 0\end{array}& & & \\ & & -1& s-1& & & & & & & & \\ & ⋰& & & \ddots & & & & & & & \\ -1& & & & & s-1& & & & & & \\ -1& & & & & & s-1& & & & & 1\\ & \ddots & & & & & & \ddots & & & ⋰& \\ & & -1& & & & & & s-1& 1& & \\ & & & -1& & & & & 1& s-1& & \\ & & & & \ddots & & & ⋰& & & \ddots & \\ & & & & & -1& 1& & & & & s-1\end{array}|\\ =|\begin{array}{cccccc}s-1& & & & & -1\\ & \ddots & & & ⋰& \\ & & s-1& -1& & \\ & & -1& s-1& & \\ & ⋰& & & \ddots & \\ -1& & & & & s-1\end{array}|\cdot |\begin{array}{cccccc}s-1& & & & & 1\\ & \ddots & & & ⋰& \\ & & s-1& 1& & \\ & & 1& s-1& & \\ & ⋰& & & \ddots & \\ 1& & & & & s-1\end{array}|\\ =\left({s}^{2}-2s\right){\left({s}^{2}-2s\right)}^{\frac{n-2}{2}}\cdot \left({s}^{2}-2s\right){\left({s}^{2}-2s\right)}^{\frac{n-2}{2}}\\ ={\left({s}^{2}-2s\right)}^{n}\end{array}$

5. 计算壹矩阵的特征值多项式

n阶壹矩阵具有下列形式：

$A={\left(\begin{array}{ccccc}1& 1& \cdots & 1& 1\\ 1& 1& \cdots & 1& 1\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 1& 1& \cdots & 1& 1\\ 1& 1& \cdots & 1& 1\end{array}\right)}_{n×n}$

${f}_{n}\left(s\right)=|\begin{array}{ccccc}s-n& -1& \cdots & -1& -1\\ s-n& s-1& \cdots & -1& -1\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ s-n& -1& \cdots & s-1& -1\\ s-n& -1& \cdots & -1& s-1\end{array}|=\left(s-n\right)\cdot |\begin{array}{ccccc}1& -1& \cdots & -1& -1\\ 1& s-1& \cdots & -1& -1\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 1& -1& \cdots & s-1& -1\\ 1& -1& \cdots & -1& s-1\end{array}|$

${f}_{n}\left(s\right)=\left(s-n\right)|\begin{array}{ccccc}1& 0& \cdots & 0& 0\\ 1& s& \cdots & 0& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 1& 0& \cdots & s& 0\\ 1& 0& \cdots & 0& s\end{array}|=\left(s-n\right){s}^{n-1}$

6. 计算斜对称矩阵的行列式

${A}_{n}={|\begin{array}{ccccc}a& b& b& \cdots & b\\ c& a& b& \cdots & b\\ c& c& a& \cdots & b\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ c& c& c& \cdots & a\end{array}|}_{n×n}$

$\begin{array}{c}{A}_{n}=|\begin{array}{cccccc}a& b& b& \cdots & b& b\\ c& a& b& \cdots & b& b\\ c& c& a& \cdots & b& b\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ c& c& c& \cdots & a& b\\ c& c& c& \cdots & c& a\end{array}|\begin{array}{c}\underset{_}{\underset{_}{\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}}\\ \end{array}|\begin{array}{cccccc}a& b& b& \cdots & b& b\\ c-a& a-b& 0& \cdots & 0& 0\\ 0& c-a& a-b& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& 0& \cdots & a-b& 0\\ 0& 0& 0& \cdots & c-a& a-b\end{array}|\\ =a{\left(a-b\right)}^{n-1}+\left(a-c\right)\cdot b|\begin{array}{cccccc}1& 1& 1& \cdots & 1& 1\\ c-a& a-b& 0& \cdots & 0& 0\\ 0& c-a& a-b& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& 0& \cdots & a-b& 0\\ 0& 0& 0& \cdots & c-a& a-b\end{array}|\end{array}$

${x}_{n-1}={\left(a-b\right)}^{n-2}+\left(a-c\right){x}_{n-2}$

$\begin{array}{l}{x}_{1}=1,\\ {x}_{2}=\left(a-b\right)+\left(a-c\right),\\ {x}_{3}={\left(a-b\right)}^{2}+\left(a-b\right)\left(a-c\right)+{\left(a-c\right)}^{2},\\ {x}_{4}={\left(a-b\right)}^{3}+{\left(a-b\right)}^{2}\left(a-c\right)+\left(a-b\right){\left(a-c\right)}^{2}+{\left(a-c\right)}^{3},\cdots \end{array}$

$\begin{array}{c}{x}_{n-1}={\left(a-b\right)}^{n-2}+{\left(a-b\right)}^{n-3}\left(a-c\right)+\cdots +\left(a-b\right){\left(a-c\right)}^{n-3}+{\left(a-c\right)}^{n-2}\\ =\frac{{\left(a-b\right)}^{n-1}-{\left(a-c\right)}^{n-1}}{\left(a-b\right)-\left(a-c\right)}=\frac{{\left(a-b\right)}^{n-1}-{\left(a-c\right)}^{n-1}}{c-b}\end{array}$

$\begin{array}{c}{x}_{n}=a{\left(a-b\right)}^{n-1}+b\left(a-c\right)\cdot \frac{{\left(a-b\right)}^{n-1}-{\left(a-c\right)}^{n-1}}{c-b}\\ =\frac{a\left(c-b\right){\left(a-b\right)}^{n-1}+b\left(a-c\right){\left(a-b\right)}^{n-1}-b{\left(a-c\right)}^{n}}{c-b}\\ =\frac{c{\left(a-b\right)}^{n}-b{\left(a-c\right)}^{n}}{c-b}\end{array}$

7. 计算对称矩阵的行列式

1) 计算n阶行列式 [5] ：

${D}_{n}=|\begin{array}{cccc}1+{a}_{1}& 1& \cdots & 1\\ 1& 1+{a}_{2}& \cdots & 1\\ ⋮& ⋮& \ddots & ⋮\\ 1& 1& \cdots & 1+{a}_{n}\end{array}|$ ，其中 ${a}_{1}{a}_{2}\cdots {a}_{n}\ne 0$

2) ${F}_{n}=|\begin{array}{cccc}1+{a}_{1}& {a}_{1}& \cdots & {a}_{1}\\ {a}_{2}& 1+{a}_{2}& \cdots & {a}_{2}\\ ⋮& ⋮& \ddots & ⋮\\ {a}_{n}& {a}_{n}& \cdots & 1+{a}_{n}\end{array}|$

$|\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ ⋮& ⋮& & ⋮\\ {b}_{1}+{c}_{1}& {b}_{2}+{c}_{2}& \cdots & {b}_{n}+{c}_{n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}|=|\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ ⋮& ⋮& & ⋮\\ {b}_{1}& {b}_{2}& \cdots & {b}_{n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}|+|\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ ⋮& ⋮& & ⋮\\ {c}_{1}& {c}_{2}& \cdots & {c}_{n}\\ ⋮& ⋮& & ⋮\\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}|$

$\begin{array}{c}{D}_{n}=\left({a}_{2}{a}_{3}\cdots {a}_{n}+{a}_{1}{a}_{3}{a}_{4}\cdots {a}_{n}+{a}_{1}{a}_{2}{a}_{4}{a}_{5}\cdots {a}_{n}+\cdots +{a}_{1}{a}_{2}\cdots {a}_{n-1}\right)+{a}_{1}{a}_{2}\cdots {a}_{n}\\ ={a}_{1}{a}_{2}\cdots {a}_{n}\left(\frac{1}{{a}_{1}}+\frac{1}{{a}_{2}}+\cdots +\frac{1}{{a}_{n}}+1\right)\\ ={a}_{1}{a}_{2}\cdots {a}_{n}\left(1+\underset{i=1}{\overset{n}{\sum }}\frac{1}{{a}_{i}}\right)\end{array}$

$A=\left(\begin{array}{cccc}1& & & \\ & 1& & \\ & & \ddots & \\ & & & 1\end{array}\right)$$B=\left(\begin{array}{cccc}{a}_{1}& a{}_{1}& \cdots & {a}_{1}\\ {a}_{2}& {a}_{2}& \cdots & {a}_{2}\\ ⋮& ⋮& \ddots & ⋮\\ {a}_{n}& {a}_{n}& \cdots & {a}_{n}\end{array}\right)$

${F}_{n}=|A+B|$ 。同样对于展开之后的 ${2}^{n}$ 个子行列式来说，其中值不为零的行列式只包含两种情况：

① 子行列式中的每一列均来自于 $A$

② 子行列式中有且只有一列来自于 $B$

$|\begin{array}{cccccc}1& & & {a}_{1}& & \\ ⋮& 1& & {a}_{2}& & \\ ⋮& & \ddots & ⋮& & \\ 0& \cdots & \cdots & {a}_{i}& & \\ & & & ⋮& \ddots & \\ & & & {a}_{n}& & 1\end{array}|\begin{array}{c}\underset{_}{\underset{_}{{c}_{1}↔{c}_{i}}}\\ \end{array}-|\begin{array}{cccccc}{a}_{1}& & & 1& & \\ {a}_{2}& 1& & ⋮& & \\ ⋮& & \ddots & ⋮& & \\ {a}_{i}& \cdots & \cdots & 0& & \\ ⋮& & & & \ddots & \\ {a}_{n}& & & & & 1\end{array}|\begin{array}{c}\underset{_}{\underset{_}{{r}_{1}↔{r}_{i}}}\\ \end{array}|\begin{array}{cccccc}{a}_{i}& & & & & \\ {a}_{2}& 1& & & & \\ ⋮& & \ddots & & & \\ {a}_{1}& \cdots & \cdots & 1& & \\ ⋮& & & & \ddots & \\ {a}_{n}& & & & & 1\end{array}|={a}_{i}$

${F}_{n}=1+{a}_{1}+{a}_{2}+\cdots +{a}_{n}$

8. 计算带状矩阵的行列式

1) 试证： $|\begin{array}{cccccc}a+b& ab& 0& \cdots & 0& 0\\ 1& a+b& ab& \cdots & 0& 0\\ 0& 1& a+b& \cdots & 0& 0\\ ⋮& ⋮& ⋮& & ⋮& ⋮\\ 0& 0& 0& \cdots & 1& a+b\end{array}|=\frac{{a}^{n+1}-{b}^{n+1}}{a-b}$ $\left(a\ne b\right)$

$n=1$ 时，原式 $=a+b$ ；当 $n=2$ 时，原式 $={a}^{2}+ab+{b}^{2}$ ，等式均成立。

$\begin{array}{c}原式=\left(a+b\right)\cdot |\begin{array}{cccccc}a+b& ab& 0& \cdots & 0& 0\\ 1& a+b& ab& \cdots & 0& 0\\ 0& 1& a+b& \cdots & 0& 0\\ ⋮& ⋮& ⋮& & ⋮& ⋮\\ 0& 0& 0& \cdots & 1& a+b\end{array}|-|\begin{array}{cccccc}ab& 0& 0& \cdots & 0& 0\\ 1& a+b& ab& \cdots & 0& 0\\ 0& 1& a+b& \cdots & 0& 0\\ ⋮& ⋮& ⋮& & ⋮& ⋮\\ 0& 0& 0& \cdots & 1& a+b\end{array}|\\ =\left(a+b\right)\cdot \frac{{a}^{n+1}-{b}^{n+1}}{a-b}-ab\cdot \frac{{a}^{n}-{b}^{n}}{a-b}=\frac{{a}^{n+2}-{b}^{n+2}}{a-b}\end{array}$

2) 计算下列n阶带状矩阵的行列式：

${X}_{n}=|\begin{array}{cccccc}a& b& 0& \cdots & 0& 0\\ c& a& b& \cdots & 0& 0\\ 0& c& a& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& 0& \cdots & a& b\\ 0& 0& 0& \cdots & c& a\end{array}|$

${y}_{1}=\frac{1}{2}a+\frac{1}{2}\sqrt{{a}^{2}-4bc}$${y}_{2}=\frac{1}{2}-\frac{1}{2}\sqrt{{a}^{2}-4bc}$

① 当 ${a}^{2}-4bc>0$ 时， ${y}_{1}$${y}_{2}$ 是两个不同的实根，则 ${X}_{n}={C}_{1}{y}_{1}^{n}+{C}_{2}{y}_{2}^{n}$ ( ${C}_{1}$${C}_{2}$ 为待定系数，它们的值取决于整个数列的前两项的值，即 ${X}_{1}$${X}_{2}$ 的值，下同)。分别令 $n=1$ 和2，得到

$\left\{\begin{array}{l}{C}_{1}{y}_{1}+{C}_{2}{y}_{2}={X}_{1}\\ {C}_{1}{y}_{1}^{2}+{C}_{2}{y}_{2}^{2}={X}_{2}\end{array}$ ，解得 ${C}_{1}=\frac{{X}_{2}-{X}_{1}{y}_{2}}{{y}_{1}\left({y}_{1}-{y}_{2}\right)}$${C}_{2}=\frac{{X}_{2}-{X}_{1}{y}_{1}}{{y}_{2}\left({y}_{2}-{y}_{1}\right)}$

② 当 ${a}^{2}-4bc<0$ 时， ${y}_{1}$${y}_{2}$ 是两个相同的实根， ${y}_{1}={y}_{2}=\frac{1}{2}a$ ，则 ${X}_{n}=\left({C}_{1}+{C}_{2}n\right){y}_{1}^{n}$ 。代入初值有

$\left\{\begin{array}{l}\left({C}_{1}+{C}_{2}\right){y}_{1}={X}_{1}\\ \left({C}_{1}+2{C}_{2}\right){y}_{1}^{2}={X}_{2}\end{array}$ ，解得 ${C}_{1}=\frac{2{X}_{1}{y}_{1}-{X}_{2}}{{y}_{1}^{2}}$${C}_{2}=\frac{{X}_{2}-{X}_{1}{y}_{1}}{{y}_{1}^{2}}$

③ 当 ${a}^{2}-4bc<0$ 时，和是两个不同的复根 $r\left(\mathrm{cos}\theta ±i\mathrm{sin}\theta \right)$ ，则

${X}_{n}={r}^{n}\left({C}_{1}\mathrm{cos}n\theta +{C}_{2}\mathrm{sin}n\theta \right)$

$\begin{array}{l}{X}_{1}=a\\ {X}_{2}={a}^{2}-bc\\ {X}_{3}={a}^{3}-2abc\\ {X}_{4}={a}^{4}-3{a}^{2}bc+{b}^{2}{c}^{2}\\ {X}_{5}={a}^{5}-4{a}^{3}bc+3{a}^{2}{b}^{2}{c}^{2}\end{array}$

$\begin{array}{l}{X}_{6}={a}^{6}-5{a}^{4}bc+6{a}^{2}{b}^{2}{c}^{2}-{b}^{3}{c}^{3}\\ {X}_{7}={a}^{7}-6{a}^{5}bc+10{a}^{3}{b}^{2}{c}^{2}-4a{b}^{3}{c}^{3}\\ {X}_{8}={a}^{8}-7{a}^{6}bc+15{a}^{4}{b}^{2}{c}^{2}-10{a}^{2}{b}^{3}{c}^{3}+{b}^{4}{c}^{4}\\ {X}_{9}=\cdots \end{array}$

1 1

1 1 2

1 2 3

1 3 1 5

1 4 3 8

15 6 1 13

1 6 104 21

1 7 15101 34

1 8 21205 55

1 9 2835151 89

${A}_{n}={X}_{n-1}={C}_{n-1}^{0}+{C}_{n-2}^{1}+{C}_{n-3}^{2}+\cdots \left(n\ge 2\right)$

$n=1$ 时， ${A}_{n}={C}_{0}^{0}=1$ ，刚好也满足于上述的通项公式。这便是斐波那契数列的通项公式的另一种表现形式： ${A}_{n}={C}_{n-1}^{0}+{C}_{n-2}^{1}+{C}_{n-3}^{2}+\cdots$ ，有别于用特征方程所得到的通项公式

${A}_{n}=\frac{\sqrt{5}}{5}\left[{\left(\frac{1+\sqrt{5}}{2}\right)}^{n}-{\left(\frac{1-\sqrt{5}}{2}\right)}^{n}\right]$

9. 计算范德蒙德矩阵的行列式

${x}_{n}=|\begin{array}{ccccc}1& 1& 1& \cdots & 1\\ {a}_{1}& {a}_{2}& {a}_{3}& \cdots & {a}_{n}\\ {a}_{1}^{2}& {a}_{2}^{2}& {a}_{3}^{2}& \cdots & {a}_{n}^{2}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ {a}_{1}^{n-1}& {a}_{2}^{n-1}& {a}_{3}^{n-1}& \cdots & {a}_{n}^{n-1}\end{array}|$

${x}_{n}=|\begin{array}{cccccc}1& 1& 1& \cdots & 1& 1\\ 0& {a}_{2}-{a}_{1}& {a}_{3}-{a}_{1}& \cdots & {a}_{n-1}-{a}_{1}& {a}_{n}-{a}_{1}\\ 0& {a}_{2}^{2}-{a}_{2}{a}_{1}& {a}_{3}^{2}-{a}_{3}{a}_{1}& \cdots & {a}_{n-1}^{2}-{a}_{n-1}{a}_{1}& {a}_{n}^{2}-{a}_{n}{a}_{1}\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& {a}_{2}^{n-2}-{a}_{2}^{n-3}{a}_{1}& {a}_{3}^{n-2}-{a}_{3}^{n-3}{a}_{1}& \cdots & {a}_{n-1}^{n-2}-{a}_{n-1}^{n-3}{a}_{1}& {a}_{n}^{n-2}-{a}_{n}^{n-3}{a}_{1}\\ 0& {a}_{2}^{n-1}-{a}_{2}^{n-2}{a}_{1}& {a}_{3}^{n-1}-{a}_{3}^{n-2}{a}_{1}& \cdots & {a}_{n-1}^{n-1}-{a}_{n-1}^{n-2}{a}_{1}& {a}_{n}^{n-1}-{a}_{n}^{n-2}{a}_{1}\end{array}|$

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

${x}_{n}=\left({a}_{2}-{a}_{1}\right)\left({a}_{3}-{a}_{1}\right)\cdots \left({a}_{n}-{a}_{1}\right)\cdot \left({a}_{3}-{a}_{2}\right)\left({a}_{4}-{a}_{2}\right)\cdots \left({a}_{n}-{a}_{2}\right)|\begin{array}{ccccc}1& 1& \cdots & 1& 1\\ {a}_{3}& {a}_{4}& \cdots & {a}_{n-1}& {a}_{n}\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ {a}_{3}^{n-4}& {a}_{4}^{n-4}& \cdots & {a}_{n-1}^{n-4}& {a}_{n}^{n-4}\\ {a}_{3}^{n-3}& {a}_{4}^{n-3}& \cdots & {a}_{n-1}^{n-3}& {a}_{n}^{n-3}\end{array}|$

${x}_{n}=\underset{1\le i

10. 范德蒙德矩阵的行列式的扩展

$|\begin{array}{ccccc}1& 1& 1& 1& 1\\ {a}_{1}& {a}_{2}& {a}_{3}& {a}_{4}& x\\ {a}_{1}^{2}& {a}_{2}^{2}& {a}_{3}^{2}& {a}_{4}^{2}& {x}^{2}\\ {a}_{1}^{3}& {a}_{2}^{3}& {a}_{3}^{3}& {a}_{4}^{3}& {x}^{3}\\ {a}_{1}^{4}& {a}_{2}^{4}& {a}_{3}^{4}& {a}_{4}^{4}& {x}^{4}\end{array}|$

$|\begin{array}{ccccc}1& 1& 1& 1& 1\\ {a}_{1}& {a}_{2}& {a}_{3}& {a}_{4}& x\\ {a}_{1}^{2}& {a}_{2}^{2}& {a}_{3}^{2}& {a}_{4}^{2}& {x}^{2}\\ {a}_{1}^{3}& {a}_{2}^{3}& {a}_{3}^{3}& {a}_{4}^{3}& {x}^{3}\\ {a}_{1}^{4}& {a}_{2}^{4}& {a}_{3}^{4}& {a}_{4}^{4}& {x}^{4}\end{array}|=\left(x-{a}_{1}\right)\left(x-{a}_{2}\right)\left(x-{a}_{3}\right)\left(x-{a}_{4}\right)\underset{1\le i

$m=\left({a}_{1}+{a}_{2}+{a}_{3}+{a}_{4}\right)\underset{1\le i

$|\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ {a}_{1}& {a}_{2}& {a}_{3}& {a}_{4}& x& y\\ {a}_{1}^{2}& {a}_{2}^{2}& {a}_{3}^{2}& {a}_{4}^{2}& {x}^{2}& {y}^{2}\\ {a}_{1}^{3}& {a}_{2}^{3}& {a}_{3}^{3}& {a}_{4}^{3}& {x}^{3}& {y}^{3}\\ {a}_{1}^{4}& {a}_{2}^{4}& {a}_{3}^{4}& {a}_{4}^{4}& {x}^{4}& {y}^{4}\\ {a}_{1}^{5}& {a}_{2}^{5}& {a}_{3}^{5}& {a}_{4}^{5}& {x}^{5}& {y}^{5}\end{array}|$

$\left(x-{a}_{1}\right)\left(x-{a}_{2}\right)\left(x-{a}_{3}\right)\left(x-{a}_{4}\right)\left(y-{a}_{1}\right)\left(y-{a}_{2}\right)\left(y-{a}_{3}\right)\left(y-{a}_{4}\right)\underset{_}{\left(y-x\right)}\underset{1\le i

$-\left(\underset{1\le i

$\left({a}_{1}{a}_{2}{a}_{3}+{a}_{1}{a}_{2}{a}_{4}+{a}_{1}{a}_{3}{a}_{4}+{a}_{2}{a}_{3}{a}_{4}\right)\underset{1\le i

$\underset{1\le i

$\underset{1\le i

11. 结语

[1] Ding, F. (2010) Transformations between Some Special Matrices. Computers & Mathematics with Applications, 59, 2676-2695.
https://doi.org/10.1016/j.camwa.2010.01.036

[2] Ding, F. (2013) Computation of Matrix Exponentials of Special Matrices. Applied Mathematics and Computation, 223, 311-326.
https://doi.org/10.1016/j.amc.2013.07.079

[3] 严坤妹. 一类矩阵的特征值多项式的计算[J]. 莆田学报, 2007, 14(5): 19-21.

[4] 冯纯伯. 矩阵多项式特征值的计算[J]. 控制与决策, 1999, 14(6): 653-657.

[5] 同济大学数学系, 编. 工程数学线性代数[M]. 第六版. 北京: 科学出版社, 2014.

[6] 王中良, 编. 线性代数与解析几何[M]. 北京: 科学出版社, 2017.

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