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矩阵世界里的分块矩阵详解

最编程 2024-02-13 20:41:31
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22、分块矩阵.png

一、练习答案

1、已知n阶方阵A满足2A(A-E)=A^{3},求(E-A)^{-1}

解：由2A(A-E)=A^3,得：
A^3-2A^2+2A=0,
所以 \quad (A^3-E)-(2A^2-2A)=-E
(E-A^3)+(2A^2-2A)=E
A^2-A+E+A^2-A^3-AE=E
(A^2-A+E)(E-A)=E

即(E-A)^{-1}=A^2-A+E.

2、设A，B为n阶方阵，且E-AB与E-BA均可逆，
证明（E-BA）^{-1}=E+B（E-AB）^{-1}A.

证因为(E-BA)(E+B(E-AB)^{-1}A)
=E-BA+(E-BA)B(E-AB)^{-1}A
=E-BA+(B-BAB)(E-AB)^{-1}A
=E-BA+B(E-AB)(E-AB)^{-1}A
=E-BA+BA=E

故(E-BA)^{-1}=E+B(E-AB)^{-1}A

二、知识点

1、分块矩阵及求逆矩阵的公式

1.1分块矩阵的概念

将矩阵用若干纵横直线分成若干个小块，每一小块称为矩阵的子块（或子阵），以子块为元素形成的矩阵称为分块矩阵。
A=\left( \begin{array}{cc|cc} a_{11}&a_{12}&a_{13}&a_{14} \\ \hline a_{21}&a_{22}&a_{23}&a_{24} \\ a_{31}&a_{32}&a_{33}&a_{34} \\ \end{array} \right) = \left( \begin{array}{cccc} A_{11}&A_{12} \\ A_{21}&A_{22} \\ \end{array} \right)

A=\left( \begin{array}{c|cc|c} a_{11}&a_{12}&a_{13}&a_{14} \\ a_{21}&a_{22}&a_{23}&a_{24} \\ \hline a_{31}&a_{32}&a_{33}&a_{34} \\ \end{array} \right) = \left( \begin{array}{cccc} A_{11}&A_{12}&A_{13} \\ A_{21}&A_{22}&A_{23} \\ \end{array} \right)

1.2分块矩阵的运算

1.线性运算：加法与数乘
2.乘法运算：符合乘法的要求
3.转置运算：大块小块一起转

A^{T}= \left( \begin{array}{cccc} A_{11}&A_{12}&A_{13} \\ A_{21}&A_{22}&A_{23} \\ \end{array} \right) ^{T}= \left( \begin{array}{cccc} A_{11}^{T}&A_{21}^{T} \\ A_{12}^{T}&A_{22}^{T} \\ A_{13}^{T}&A_{23}^{T} \\ \end{array} \right)

1.3几种特殊的分块阵

1.3.1准对角阵(分块对角阵)

A=\left( \begin{array}{cccc} A_{1}&& \\ &A_{2}& \\ &&\ddots \\ &&&A_{s}\\ \end{array} \right) (A_{i}为方阵，i=1,2,\cdots,s)

A=\left( \begin{array}{cccc} A_{1}&& \\ &A_{2}& \\ &&\ddots \\ &&&A_{s}\\ \end{array} \right) \quad B=\left( \begin{array}{cccc} B_{1}&& \\ &B_{2}& \\ &&\ddots \\ &&&B_{s}\\ \end{array} \right)

(A_{i}，B(i)为同阶方阵，i=1,2,\cdots s)则有：

A\pm B=\left( \begin{array}{cccc} A_{1} \pm B_{1}&& \\ &A_{2} \pm B_{2}& \\ &&\ddots \\ &&&A_{s} \pm B_{s}\\ \end{array} \right)

kA=\left( \begin{array}{cccc} kA_{1}&& \\ &kA_{2}& \\ &&\ddots \\ &&&kA_{s}\\ \end{array} \right)

AB=\left( \begin{array}{cccc} A_{1}B_{1}&& \\ &A_{2}B_{2}& \\ &&\ddots \\ &&&A_{s}B_{s}\\ \end{array} \right)

A^{T}=\left( \begin{array}{cccc} A_{1}^{T}&& \\ &A_{2}^{T}& \\ &&\ddots \\ &&&A_{s}^{T}\\ \end{array} \right)

A^{m}=\left( \begin{array}{cccc} A_{1}^{m}&& \\ &A_{2}^{m}& \\ &&\ddots \\ &&&A_{s}^{m}\\ \end{array} \right)

\left| A \right|=\left| A_{1} \right|\left| A_{2} \right| \cdots \left| A_{s} \right|

A可逆 \Leftrightarrow A_i 可逆（i=1,2, \cdots ,s）

A^{-1}\left( \begin{array}{cccc} A_{1}^{-1}&&& \\ &A_{2}^{-1}&& \\ &&\ddots& \\ &&&A_{s}^{-1} \\ \end{array} \right)

r(A)=r(A_1)+r(A_2)+\cdots+r(A_s)

例1. A=\left( \begin{array}{cccc} 1&2&0&0 \\ 0&1&0&0 \\ 0&0&2&3 \\ 0&0&1&3 \\ \end{array} \right)求A的行列式，秩及逆。

解：将矩阵分块A=\left( \begin{array}{cccc} A_1& \\ &A_2 \\ \end{array} \right) \Rightarrow \left | A \right|= \left | A_1 \right| \left | A_2 \right|=3,行列式不等于零，所以r(A)=4

A_1^{-1}=\frac{1}{1-0} \left( \begin{array}{cccc} 1&-2 \\ 0&1 \\ \end{array} \right)

A_2^{-1}=\frac{1}{6-3}\left( \begin{array}{cccc} 3&-3 \\ -1&2 \\ \end{array} \right) =\left( \begin{array}{cccc} 1&-1 \\ -\frac{1}{3}&\frac{2}{3} \\ \end{array} \right)

A^{-1}=\left( \begin{array}{cccc} A_1^{-1}& \\ &A_2^{-1} \\ \end{array} \right) = \left( \begin{array}{cccc} 1&2&0&0 \\ 0&1&0&0 \\ 0&0&1&-1 \\ 0&0&-\frac{1}{3}&\frac{2}{3} \\ \end{array} \right)

1.3.2分块三角阵

分块上三角阵或准上三角阵：

A=\left( \begin{array}{cccc} A_{11}&A_{12} \\ O&A_{22} \\ \end{array} \right) (A_{ii}为方阵，i=1,2.)

\left | A \right |=\left | A_{11} \right |\left | A_{22} \right |

A可逆\Leftrightarrow A_{ii}可逆 (i=1,2.)

A^{-1}=\left( \begin{array}{cccc} A_{11}^{-1}&-A_{11}^{-1}A_{12}A_{22}^{-1} \\ O&A_{22}^{-1}\\ \end{array} \right)

分块下三角阵：
A=\left( \begin{array}{cccc} A_{11}&O \\ A_{21}&A_{22} \\ \end{array} \right)

A^{-1}=\left( \begin{array}{cccc} A_{11}^{-1}&O\\ -A_{22}^{-1}A_{21}A_{11}^{-1} &A_{22}^{-1}\\ \end{array} \right)

证明分块上三角阵A^{-1}的公式：

设A^{-1}= \left( \begin{array}{cccc} X_{11}&X_{12} \\ X_{21}&X_{22} \\ \end{array} \right) 则AA^{-1}=\left( \begin{array}{cccc} A_{11}&A_{12} \\ O&A_{22} \\ \end{array} \right) \left( \begin{array}{cccc} X_{11}&X_{12} \\ X_{21}&X_{22} \\ \end{array} \right)

= \left( \begin{array}{cccc} A_{11}X_{11}+A_{12}X_{21}&A_{11}X_{12}+A_{12}X_{22} \\ A_{22}X_{21}&A_{22}X_{22} \\ \end{array} \right) = \left( \begin{array}{cccc} E&O \\ O&E \\ \end{array} \right)

所以：求得X_{11}，X_{12}，X_{21}，X_{22}

  • A_{22}X_{22}=E \Rightarrow X_{22}=A_{22}^{-1}

  • A_{22}X_{21}=0 \Rightarrow A_{22}^{-1}A_{22}X_{21}=0 \times A_{22}^{-1}

\Rightarrow X_{21}=O

  • A_{11}X_{11}+A_{12}X_{21}=E \Rightarrow X_{11}=A_{11}^{-1}

  • A_{11}X_{12}+A_{12}X_{22}=O \Rightarrow A_{11}X_{12}=-A_{12}X_{22}

A_{11}^{-1}A_{11}X_{12}=-A_{11}^{-1}A_{12}X_{22} \Rightarrow

X_{12}=-A_{11}^{-1}A_{12}X_{22}=-A_{11}^{-1}A_{12}A_{22}^{-1}

例2.求矩阵的逆 A=\left( \begin{array}{cccc} 2&1&3&4 \\ 0&2&1&3 \\ 0&0&2&1 \\ 0&0&0&2 \\ \end{array} \right)

解：将矩阵分块

  • A=\left( \begin{array}{cccc} A_{11}&A_{12} \\ O&A_{22} \\ \end{array} \right) \ A_{11}^{-1}=\left( \begin{array}{cccc} \frac{1}{2}&-\frac{1}{4} \\ 0&\frac{1}{2} \\ \end{array} \right)=A_{22}^{-1}

-A_{11}^{-1}A_{12}A_{22}^{-1} =-\left( \begin{array}{cccc} \frac{1}{2}&-\frac{1}{4} \\ 0&\frac{1}{2} \\ \end{array} \right) \left( \begin{array}{cccc} 3&4 \\ 1&3 \\ \end{array} \right) \left( \begin{array}{cccc} \frac{1}{2}&-\frac{1}{4} \\ 0&\frac{1}{2} \\ \end{array} \right)

=-\left( \begin{array}{cccc} \frac{5}{4}&\frac{5}{4} \\ \frac{1}{2}&\frac{3}{2} \\ \end{array} \right)\left( \begin{array}{cccc} \frac{1}{2}&-\frac{1}{4} \\ 0&\frac{1}{2} \\ \end{array} \right) =\left( \begin{array}{cccc} -\frac{5}{8}&-\frac{5}{16} \\ -\frac{1}{4}&-\frac{5}{8} \\ \end{array} \right)

  • A^{-1}=\left( \begin{array}{cccc} A_{11}^{-1}&-A_{11}^{-1}A_{12}A_{22}^{-1} \\ O&A_{22}^{-1}\\ \end{array} \right)

=\left( \begin{array}{cccc} \frac{1}{2}&-\frac{1}{4}&-\frac{5}{8}&-\frac{5}{16} \\ 0&\frac{1}{2}&-\frac{1}{4}&-\frac{5}{8} \\ 0&0&\frac{1}{2}&-\frac{1}{4} \\ 0&0&0&\frac{1}{2} \\ \end{array} \right)

1.3.3分块斜对角阵

M= \left( \begin{array}{cccc} O&A \\ B&O \\ \end{array} \right) \quad M可逆 \Leftrightarrow A,B可逆

M^{-1}= \left( \begin{array}{cccc} O&B^{-1} \\ A^{-1}&O \\ \end{array} \right)

例3.求矩阵的逆M= \left( \begin{array}{cccc} 0&0&3&5 \\ 0&0&1&2 \\ 1&2&0&0 \\ 0&3&0&0 \\ \end{array} \right)

解：将矩阵分块

M= \left( \begin{array}{cccc} O&A \\ B&O \\ \end{array} \right) ，M^{-1}= \left( \begin{array}{cccc} O&B^{-1} \\ A^{-1}&O \\ \end{array} \right) = \left( \begin{array}{cccc} 0&0&1&-\frac{2}{3} \\ 0&0&0&\frac{1}{3} \\ 2&-5&0&0 \\ -1&3&0&0 \\ \end{array} \right)

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