﻿ 分块矩阵的应用研究

# 分块矩阵的应用研究Research on the Application of Block Matrix

Abstract: In advanced mathematics, block matrix is a very important concept. It can make the representation of matrix more simple and clear, and simplify the operation of matrix. In this paper, through the simple analysis of the calculation and proof of block matrix, we discuss the method of matrix with n-order special attribute, and use block matrix to describe the solution of linear system and its related content.

1. 引言

${a}_{1}x+{b}_{1}y+{c}_{1}z={d}_{1},$ (1.1)

${a}_{2}x+{b}_{2}y+{c}_{2}z={d}_{2},$ (1.2)

${a}_{3}x+{b}_{3}y+{c}_{3}z={d}_{3},$ (1.3)

(1.4)

2. 分块矩阵在计算方面的应用

2.1. 逆矩阵方面的应用

$P=\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]$

B是r阶，C是k阶，当B与( $C-D{B}^{-1}A$ )两个方阵是可逆的，那么P也是可逆的，

1) 当，B与C都可逆时，有 ${P}^{-1}=\left[\begin{array}{cc}0& {C}^{-1}\\ {B}^{-1}& 0\end{array}\right]$

2) 当，B与C都可逆时，有 ${P}^{-1}=\left[\begin{array}{cc}-{C}^{-1}D{B}^{-1}& {C}^{-1}\\ {B}^{-1}& 0\end{array}\right]$

3) 当 $A\ne 0,D\ne 0$，B与C都可逆时，有 ${P}^{-1}=\left[\begin{array}{cc}0& {C}^{-1}\\ {B}^{-1}& -{B}^{-1}A{C}^{-1}\end{array}\right]$

$Q=\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]$

${Q}^{-1}={\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]}^{-1}=\left[\begin{array}{cc}{A}^{-1}+{A}^{-1}B{\left(D-C{A}^{-1}B\right)}^{-1}C{A}^{-1}& -{A}^{-1}B{\left(D-C{A}^{-1}B\right)}^{-1}\\ -{\left(D-C{A}^{-1}B\right)}^{-1}C{A}^{-1}& {\left(D-CAB\right)}^{-1}\end{array}\right]$

1) 当 $B=0,C=0$，A与D都可逆时，有 ${Q}^{-1}=\left[\begin{array}{cc}{A}^{-1}& 0\\ 0& {D}^{-1}\end{array}\right]$

2) 当，A与D都可逆时，有 ${Q}^{-1}=\left[\begin{array}{cc}{A}^{-1}& -{A}^{-1}B{D}^{-1}\\ 0& {D}^{-1}\end{array}\right]$

3) 当 $B=0,C\ne 0$，A与D都可逆时，有 ${Q}^{-1}=\left[\begin{array}{cc}{A}^{-1}& 0\\ -{D}^{-1}C{A}^{-1}& {D}^{-1}\end{array}\right]$

$M=\left(\begin{array}{cccccc}0& {m}_{1}& 0& \cdots & 0& 0\\ 0& 0& {m}_{2}& \cdots & 0& 0\\ ⋮& ⋮& ⋮& & ⋮& ⋮\\ 0& 0& 0& \cdots & 0& {m}_{n-1}\\ {m}_{n}& 0& 0& \cdots & 0& 0\end{array}\right)\left({m}_{i}\ne 0,i=1,\cdots ,n\right)$

$N=\left(\begin{array}{ccccc}{m}_{1}& 0& \cdots & 0& 0\\ 0& {m}_{2}& \cdots & 0& 0\\ ⋮& ⋮& & ⋮& ⋮\\ 0& 0& \cdots & 0& {m}_{n-1}\end{array}\right)$

${M}^{-1}=\left(\begin{array}{cc}0& {m}_{n}^{-1}\\ {N}^{-1}& 0\end{array}\right),{N}^{-1}=\left(\begin{array}{ccccc}{m}_{1}^{-1}& 0& \cdots & 0& 0\\ 0& {m}_{2}^{-1}& \cdots & 0& 0\\ ⋮& ⋮& & ⋮& ⋮\\ 0& 0& \cdots & 0& {m}_{n-1}^{-1}\end{array}\right)$

${M}^{-1}=\left(\begin{array}{cccccc}0& 0& 0& \cdots & 0& 1/{m}_{n}\\ 1/{m}_{1}& 0& 0& \cdots & 0& 0\\ ⋮& ⋮& ⋮& & ⋮& ⋮\\ 0& 0& 0& \cdots & 0& 0\\ 0& 0& 0& \cdots & 1/{m}_{n-1}& 0\end{array}\right)$

2.2. 行列式计算方面的应用

$H=\left[\begin{array}{cccc}{P}_{1}& & \cdots & P\\ 0& {P}_{2}& & ⋮\\ ⋮& ⋮& \ddots & \\ 0& 0& \cdots & {P}_{s}\end{array}\right]$

$|P|=\left[\begin{array}{ccccccc}x& 0& 0& \cdots & 0& 0& z\\ 0& x& 0& \cdots & 0& z& 0\\ 0& 0& x& \cdots & z& 0& 0\\ ⋮& ⋮& ⋮& & ⋮& ⋮& ⋮\\ 0& 0& z& \cdots & x& 0& 0\\ 0& z& 0& \cdots & 0& x& 0\\ z& 0& 0& \cdots & 0& 0& 0\end{array}\right]$

$A=\left[\begin{array}{cccc}x& & & \\ & x& & \\ & & \ddots & \\ & & & x\end{array}\right]=D,\text{\hspace{0.17em}}\text{\hspace{0.17em}}B=\left[\begin{array}{cccc}& & & z\\ & & z& \\ & ⋰& & \\ z& & & \end{array}\right]=C$

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

$|P|={x}^{n}{\left(x-{z}^{2}{x}^{-1}\right)}^{n}={\left({x}^{2}-{z}^{2}\right)}^{n}$

2.3. 矩阵特征值方面的应用

$\left[\begin{array}{cc}{E}_{r}& 0\\ 0& 0\end{array}\right]$

$\left[\begin{array}{cc}0& 0\\ 0& {E}_{r}\end{array}\right]$

1) 当 $R\left(A\right)=0$ 时，即，结论显然成立。

2) 设 $0，非零不可逆矩阵A，因为，所以有可逆矩阵P，

3. 分块矩阵在证明方面的应用

$R\left({A}_{s}\right)\ge R\left(A\right)+s-m$

$R\left(A+B\right)\le R\left(A\right)+R\left( B \right)$

$R\left[\begin{array}{c}0\\ B\end{array}\right]=R\left({A}_{s}\right)+m-s$

$T=P\left[\begin{array}{cc}{E}_{r}& 0\\ 0& 0\end{array}\right]Q=P\left[\begin{array}{c}{E}_{r}\\ 0\end{array}\right]\left(\begin{array}{cc}{E}_{r}& 0\end{array}\right)Q={T}_{1}{T}_{2}$

${T}_{1}=P\left[\begin{array}{c}{E}_{r}\\ 0\end{array}\right],\text{\hspace{0.17em}}{T}_{2}=\left(\begin{array}{cc}{E}_{r}& 0\end{array}\right)Q$

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