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  • 文章目录矩阵分块法常用的分块法1) 按行分块2) 按列分块3) 分块对角矩阵(又称准对角矩阵)分块矩阵的运算分块矩阵的初等变换分块初等矩阵性质参考 矩阵分块法 定义1\large\color{magenta}{\boxed{\color{brown}{...


    矩阵线性代数笔记整理汇总,超全面

    1. 02矩阵01 ——概念、运算和基本矩阵、对角矩阵、方幂、数量矩阵、转置矩阵、对称矩阵、逆矩阵、奇异矩阵、三角矩阵、矩阵乘积的行列式与秩
    2. 03矩阵02——初等变换与高斯消元法、行阶梯形矩阵、行简化阶梯形矩阵、行阶梯形状与方程组解的关系、相抵
    3. 04矩阵03——逆矩阵、逆矩阵的求解、可逆矩阵的判别、伴随矩阵、以及性质、可逆矩阵的等价条件、克拉默法则的另一种推导法、矩阵乘积的秩的性质
    4. 05矩阵04——分块矩阵、分块矩阵的运算、分块矩阵的初等变换、分块初等矩阵的性质、按行分块、按列分块

    矩阵分块法

    定 义 1 \large\color{magenta}{\boxed{\color{brown}{定义1 } }} 1 对于行数和列数较高的矩阵A,运算时常采用分块法,使大矩阵的运算化成小矩阵的运算.

    我们将矩阵A用若干条纵线和横线分成许多个小矩阵,每一个小矩阵称为A的子块,以子块为元素的形式上的矩阵称为分块矩阵.

    例如将 3 × 4 3 \times 4 3×4 矩阵 A = ( a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 ) A=\left(\begin{array}{llll}a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34}\end{array}\right) A=a11a21a31a12a22a32a13a23a33a14a24a34 .

    分成子块的分法很多, 下面举出三种分块形式:

    image-20210103190403142 image-20210103190443336

    分法(1) 可记为 A = ( A 11 A 12 A 21 A 22 ) A=\left(\begin{array}{cc}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right) A=(A11A21A12A22).

    其中
    A 11 = ( a 11 a 12 a 21 a 22 ) , A 12 = ( a 13 a 14 a 23 a 24 )  ,  A 21 = ( a 31 a 32 ) , A 22 = ( a 33 a 34 ) A_{11}=\left(\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right), A_{12}=\left(\begin{array}{ll}a_{13} & a_{14} \\ a_{23} & a_{24}\end{array}\right) \text { , }\\ A_{21}=\left(\begin{array}{ll}a_{31} & a_{32}\end{array}\right), A_{22}=\left(\begin{array}{ll}a_{33} & a_{34}\end{array}\right) A11=(a11a21a12a22),A12=(a13a23a14a24) , A21=(a31a32),A22=(a33a34)
    A 11 , A 12 , A 21 , A 22 A_{11}, A_{12}, A_{21}, A_{22} A11,A12,A21,A22 A A A 的子块, 而 A A A 形式上成为以这些子块为元素的分块矩阵. 分法 (2) 及 (3) 的分块矩阵可类似写出, 这里略.

    常用的分块法

    设有 s × n s \times n s×n 矩阵 A = ( a i j ) s n , A=\left(a_{i j}\right)_{s n}, A=(aij)sn, 则对 A A A 有以下三种最常用的分块方法:

    1) 按行分块

    即把 A A A 的每一行当作一个子块, 这时每个子块为一行向量,也就是说,矩阵 A A A 是由一个行向量组组成 :
    A = ( α 1 α 2 ⋮ α s ) A=\left(\begin{array}{c} \alpha_{1} \\ \alpha_{2} \\ \vdots \\ \alpha_{s} \end{array}\right) A=α1α2αs

    2) 按列分块

    即把 A A A 的每一列当作一个子块,这时每个子块为一列向量,也就是说,矩阵 A A A 是由一个列向量组组成 :
    A = ( β 1 , β 2 , ⋯   , β n ) A=\left(\beta_{1}, \beta_{2}, \cdots, \beta_{n}\right) A=(β1,β2,,βn)
    B 1 , B 2 , ⋯   , B m B_{1}, B_{2}, \cdots, B_{m} B1,B2,,Bm 表示 B B B 的行向量,于是
    B = ( B 1 B 2 ⋮ B m ) B=\left(\begin{array}{c} B_{1} \\ B_{2} \\ \vdots \\ B_{m} \end{array}\right) B=B1B2Bm
    这就是 B B B 的一种分块.按分块相乘,就有
    A B = ( a 11 B 1 + a 12 B 2 + ⋯ + a 1 m B m a 21 B 1 + a 22 B 2 + ⋯ + a 2 m B m … … … … … a n 1 B 1 + a n 2 B 2 + ⋯ + a n m B m ) A B=\left(\begin{array}{c} a_{11} B_{1}+a_{12} B_{2}+\cdots+a_{1 m} B_{m} \\ a_{21} B_{1}+a_{22} B_{2}+\cdots+a_{2 m} B_{m} \\ \ldots \ldots \ldots \ldots \ldots \\ a_{n 1} B_{1}+a_{n 2} B_{2}+\cdots+a_{n m} B_{m} \end{array}\right) AB=a11B1+a12B2++a1mBma21B1+a22B2++a2mBman1B1+an2B2++anmBm
    用这个式子很容易看出 A B A B AB 的行向量是 B B B 的行向量的线性组合; 将 A B A B AB 进行另一种分块乘法,从结果中可以看出 A B A B AB 的列向量是 A A A 的列向量的线性组合.

    3) 分块对角矩阵(又称准对角矩阵)

    n n n 级矩阵 A = ( a i j ) n A=\left(a_{i j}\right)_{n} A=(aij)n 中非零元素都集中在主对角线附近时,有时可将 A A A 分块成下面的分块对角矩阵(准对角矩阵).
    A = ( A 1 O A 2 ⋱ O A l ) A=\left(\begin{array}{llll} A_{1} & & & O \\ & A_{2} & & \\ & & \ddots & \\ O & & & A_{l} \end{array}\right) A=A1OA2OAl
    其中 A i A_{i} Ai n i n_{i} ni 级方阵 ( i = 1 , 2 , … , l ) (i=1,2, \ldots, l) (i=1,2,,l). 例如:

    image-20210103191152463

    例 1 \Large\color{violet}{例1 } 1 试用不同方法将 A m × n , B n × 1 A_{m \times n}, B_{n \times 1} Am×n,Bn×1 进行分块, 计算AB并写出 A B = 0 A B=0 AB=0 的充分必要条件.
    【解】 ( \left(\right. ( 方法1) 对于 m × n m \times n m×n 矩阵 A , A = ( α 1 α 2 ⋮ α m ) , A, A=\left(\begin{array}{c}\alpha_{1} \\ \alpha_{2} \\ \vdots \\ \alpha_{m}\end{array}\right), A,A=α1α2αm,
    A B = ( α 1 α 2 ⋮ α m ) B = ( α 1 B α 2 B ⋮ α m B ) . A B=\left(\begin{array}{c}\alpha_{1} \\ \alpha_{2} \\ \vdots \\ \alpha_{m}\end{array}\right) B=\left(\begin{array}{c}\alpha_{1} B \\ \alpha_{2} B \\ \vdots \\ \alpha_{m} B\end{array}\right) . AB=α1α2αmB=α1Bα2BαmB.
    这时 A B = 0 A B=0 AB=0 的充分必要条件是 α i B = 0 ( 1 ≤ i ≤ m ) . \alpha_{i} B=0(1 \leq i \leq m) . αiB=0(1im).
    解 (方法2) 对于 n × l n \times l n×l 矩阵 B , B = ( B 1 B 2 ⋯ B l ) B, B=\left(\begin{array}{llll}B_{1} & B_{2} & \cdots & B_{l}\end{array}\right) B,B=(B1B2Bl),

    A B = A ( B 1 B 2 ⋯ B l ) = ( A B 1 A B 2 ⋯ A B l ) A B=A\left(\begin{array}{llll}B_{1} & B_{2} & \cdots & B_{l}\end{array}\right)=\left(\begin{array}{llll}A B_{1} & A B_{2} & \cdots & A B_{l}\end{array}\right) AB=A(B1B2Bl)=(AB1AB2ABl)

    这时 A B = 0 A B=0 AB=0 的充分必要条件是 A B i = 0 ( 1 ≤ i ≤ l ) . A B_{i}=0(1 \leq i \leq l) . ABi=0(1il).

    分块矩阵的运算

    分块矩阵的运算规则与普通矩阵的运算规则相类似, 分别说明如下:

    1. 加法运算
      设矩阵 A A A B B B 的行数相同、列数相同, 采用相同的分块法, 有

    A = ( A 11 ⋯ A 1 r ⋮ ⋮ A s 1 ⋯ A s r ) , B = ( B 11 ⋯ B 1 r ⋮ ⋮ B s 1 ⋯ B s r ) A=\left(\begin{array}{ccc} A_{11} & \cdots & A_{1 r} \\ \vdots & & \vdots \\ A_{s 1} & \cdots & A_{s r} \end{array}\right), B=\left(\begin{array}{ccc} B_{11} & \cdots & B_{1 r} \\ \vdots & & \vdots \\ B_{s 1} & \cdots & B_{s r} \end{array}\right) A=A11As1A1rAsr,B=B11Bs1B1rBsr

    其中 A i j A_{i j} Aij B i j B_{i j} Bij 的行数相同、列数相同, 那么
    A + B = ( A 11 + B 11 ⋯ A 1 r + B 1 r ⋮ ⋮ A s 1 + B s 1 ⋯ A s r + B s r ) A+B=\left(\begin{array}{ccc} A_{11}+B_{11} & \cdots & A_{1 r}+B_{1 r} \\ \vdots & & \vdots \\ A_{s 1}+B_{s 1} & \cdots & A_{s r}+B_{s r} \end{array}\right) A+B=A11+B11As1+Bs1A1r+B1rAsr+Bsr

    1. 数乘运算

      A = ( A 11 ⋯ A 1 r ⋮ ⋮ A s 1 ⋯ A s r ) , λ A=\left(\begin{array}{ccc}A_{11} & \cdots & A_{1 r} \\ \vdots & & \vdots \\ A_{s 1} & \cdots & A_{s r}\end{array}\right), \quad \lambda A=A11As1A1rAsr,λ 为常数,那么
      λ A = ( λ A 11 ⋯ λ A 1 r ⋮ ⋮ λ A s 1 ⋯ λ A s r ) \lambda A=\left(\begin{array}{ccc} \lambda A_{11} & \cdots & \lambda A_{1 r} \\ \vdots & & \vdots \\ \lambda A_{s 1} & \cdots & \lambda A_{s r} \end{array}\right) λA=λA11λAs1λA1rλAsr

    2. 分块矩阵的乘法运算

      A A A m × l m \times l m×l 矩阵, B B B l × n l \times n l×n 矩阵, 分块成

    A = ( A 11 ⋯ A 1 t ⋮ ⋮ A s 1 ⋯ A s t ) , B = ( B 11 ⋯ B 1 r ⋮ ⋮ B t 1 ⋯ B t r ) A=\left(\begin{array}{ccc} A_{11} & \cdots & A_{1 t} \\ \vdots & & \vdots \\ A_{s 1} & \cdots & A_{s t} \end{array}\right), B=\left(\begin{array}{ccc} B_{11} & \cdots & B_{1 r} \\ \vdots & & \vdots \\ B_{t 1} & \cdots & B_{t r} \end{array}\right) A=A11As1A1tAst,B=B11Bt1B1rBtr

    其中 A i 1 , A i 2 , … , A i t A_{i 1}, A_{i 2}, \ldots, A_{i t} Ai1,Ai2,,Ait 的列数分别等于 B 1 j , B 2 j , … , B t j B_{1 j}, B_{2 j}, \ldots, B_{t j} B1j,B2j,,Btj 的行数,那么
    A B = ( C 11 ⋯ C 1 r ⋮ ⋮ C s 1 ⋯ C s r ) A B=\left(\begin{array}{ccc} C_{11} & \cdots & C_{1 r} \\ \vdots & & \vdots \\ C_{s 1} & \cdots & C_{s r} \end{array}\right) AB=C11Cs1C1rCsr
    其中 C i j = ∑ k = 1 t A i k B k j ( i = 1 , ⋯   , s ; j = 1 , ⋯   , r ) C_{i j}=\sum_{k=1}^{t} A_{i k} B_{k j}(i=1, \cdots, s ; j=1, \cdots, r) Cij=k=1tAikBkj(i=1,,s;j=1,,r)

    例 2 \Large\color{violet}{例2 } 2
    A = ( 1 0 0 0 0 1 0 0 − 1 2 1 0 1 1 0 1 ) , B = ( 1 0 1 0 − 1 2 0 1 1 0 4 1 − 1 − 1 2 0 ) A=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 2 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{array}\right), B=\left(\begin{array}{cccc} 1 & 0 & 1 & 0 \\ -1 & 2 & 0 & 1 \\ 1 & 0 & 4 & 1 \\ -1 & -1 & 2 & 0 \end{array}\right) A=1011012100100001,B=1111020110420110
    A B A B AB.

    【解】 把 A 、 B A 、 B AB 分块成

    image-20210103191831247


    A B = ( E O A 1 E ) ( B 11 E B 21 B 22 ) = ( B 11 E A 1 B 11 + B 21 A 1 + B 22 ) A B=\left(\begin{array}{cc}E & O \\ A_{1} & E\end{array}\right)\left(\begin{array}{cc}B_{11} & E \\ B_{21} & B_{22}\end{array}\right)=\left(\begin{array}{cc}B_{11} & E \\ A_{1} B_{11}+B_{21} & A_{1}+B_{22}\end{array}\right) AB=(EA1OE)(B11B21EB22)=(B11A1B11+B21EA1+B22)

     而  A 1 B 11 + B 21 = ( − 1 2 1 1 ) ( 1 0 − 1 2 ) + ( 1 0 − 1 − 1 ) = ( − 3 4 0 2 ) + ( 1 0 − 1 − 1 ) = ( − 2 4 − 1 1 ) A 1 + B 22 = ( − 1 2 1 1 ) + ( 4 1 2 0 ) = ( 3 3 3 1 ) \begin{aligned} \text { 而 } A_{1} B_{11}+B_{21} &=\left(\begin{array}{cc}-1 & 2 \\ 1 & 1\end{array}\right)\left(\begin{array}{cc}1 & 0 \\ -1 & 2\end{array}\right)+\left(\begin{array}{cc}1 & 0 \\ -1 & -1\end{array}\right) \\ &=\left(\begin{array}{cc}-3 & 4 \\ 0 & 2\end{array}\right)+\left(\begin{array}{cc}1 & 0 \\ -1 & -1\end{array}\right)=\left(\begin{array}{cc}-2 & 4 \\ -1 & 1\end{array}\right) \\ A_{1}+B_{22} &=\left(\begin{array}{cc}-1 & 2 \\ 1 & 1\end{array}\right)+\left(\begin{array}{cc}4 & 1 \\ 2 & 0\end{array}\right)=\left(\begin{array}{cc}3 & 3 \\ 3 & 1\end{array}\right) \end{aligned}   A1B11+B21A1+B22=(1121)(1102)+(1101)=(3042)+(1101)=(2141)=(1121)+(4210)=(3331)

    于是
    A B = ( 1 0 1 0 − 1 2 0 1 − 2 4 3 3 − 1 1 3 1 ) A B=\left(\begin{array}{cc|cc} 1 & 0 & 1 & 0 \\ -1 & 2 & 0 & 1 \\ -2 & 4 & 3 & 3 \\ -1 & 1 & 3 & 1 \end{array}\right) AB=1121024110330131

    1. 分块矩阵的转置

    A = ( A 11 ⋯ A 1 r ⋮ ⋮ A s 1 ⋯ A s r ) A=\left(\begin{array}{ccc}A_{11} & \cdots & A_{1 r} \\ \vdots & & \vdots \\ A_{s 1} & \cdots & A_{s r}\end{array}\right) A=A11As1A1rAsr

    A T = ( A 11 T ⋯ A 1 r T ⋮ ⋮ A s 1 T ⋯ A s r T ) A^{\mathrm{T}}=\left(\begin{array}{ccc}A_{11}^{\mathrm{T}} & \cdots & A_{1 r}^{\mathrm{T}} \\ \vdots & & \vdots \\ A_{s 1}^{\mathrm{T}} & \cdots & A_{s r}^{\mathrm{T}}\end{array}\right) AT=A11TAs1TA1rTAsrT

    1. 分块对角矩阵的运算

    A A A n n n 级矩阵,且 A A A 可分成如下分块对角矩阵.

    A = ( A 1 O A 2 ⋱ O A l ) A=\left(\begin{array}{cccc} A_{1} & & & O \\ & A_{2} & & \\ & & \ddots & \\ O & & & A_{l} \end{array}\right) A=A1OA2OAl

    分块对角矩阵的性质:

    1. ∣ A ∣ = ∣ A 1 ∣ ∣ A 2 ∣ … ∣ A l ∣ |A|=\left|A_{1}\right|\left|A_{2}\right| \ldots\left|A_{l}\right| A=A1A2Al

    2. ∣ A i ∣ ≠ 0 ( i = 1 , 2 , … , l ) , \left|A_{i}\right| \neq 0(i=1,2, \ldots, l), Ai=0(i=1,2,,l), ∣ A ∣ ≠ 0 , |A| \neq 0, A=0,
      A − 1 = ( A 1 − 1 O A 2 − 1 ⋱ O A l − 1 ) . A^{-1}=\left(\begin{array}{cccc} A_{1}^{-1} & & & O \\ & A_{2}^{-1} & & \\ & & \ddots & \\ O & & & A_{l}^{-1} \end{array}\right) . A1=A11OA21OAl1.
      A , B A, B A,B 是两个 n n n 级矩阵,且采用相同的分法可把它们都分成分块对角矩阵:
      A = ( A 1 O A 2 ⋱ O A l ) , B = ( B 1 O B 2 ⋱ O B l ) A=\left(\begin{array}{ccc} A_{1} & & O \\ & A_{2} & \\ & \ddots & \\ O & & & A_{l} \end{array}\right), B=\left(\begin{array}{cccc} B_{1} & & & O \\ & B_{2} & & \\ & & \ddots & \\ O & & & B_{l} \end{array}\right) A=A1OA2OAl,B=B1OB2OBl
      则有

    A B = ( A 1 B 1 O A 2 B 2 ⋱ O A l B l ) , A B=\left(\begin{array}{cccc}A_{1} B_{1} & & & O \\ & A_{2} B_{2} & & \\ & & \ddots & \\ O & & & A_{l} B_{l}\end{array}\right), AB=A1B1OA2B2OAlBl, A + B = ( A 1 + B 1 O A 2 + B 2 ⋱ O A l + B l ) A+B=\left(\begin{array}{cccc}A_{1}+B_{1} & & & O \\ & A_{2}+B_{2} & & \\ & & \ddots & \\ O & & & A_{l}+B_{l}\end{array}\right) A+B=A1+B1OA2+B2OAl+Bl

    例 3 \Large\color{violet}{例3 } 3 n n n 级矩阵 D D D 分块为
    D = ( A O C B ) D=\left(\begin{array}{ll} A & O \\ C & B \end{array}\right) D=(ACOB)
    其中 A , B A, B A,B 分别是 k k k 级和 r r r 级的可逆矩阵, C C C r × k r \times k r×k 级矩阵, O O O k × r k \times r k×r 级零知阵,求 D − 1 . D^{-1} . D1.

    【解】 \quad 因为 ∣ D ∣ = ∣ A ∣ ∣ B ∣ |D|=|A||B| D=AB, (行列式定理)所以当 A , B A,B AB 可逆时, D D D 也可逆。

    D − 1 = ( X 11 X 12 X 21 X 22 ) \quad D^{-1}=\left(\begin{array}{cc}X_{11} & X_{12} \\ X_{21} & X_{22}\end{array}\right) D1=(X11X21X12X22)
    于是 ( A O C B ) ( X 11 X 12 X 21 X 22 ) = ( A X 11 A X 12 C X 11 + B X 21 C X 12 + C X 22 ) \quad\left(\begin{array}{ll}A & O \\ C & B\end{array}\right)\left(\begin{array}{ll}X_{11} & X_{12} \\ X_{21} & X_{22}\end{array}\right)=\left(\begin{array}{cc}A X_{11} & A X_{12} \\ C X_{11}+B X_{21} & C X_{12}+C X_{22}\end{array}\right) (ACOB)(X11X21X12X22)=(AX11CX11+BX21AX12CX12+CX22)
    = ( E k O O E r ) =\left(\begin{array}{cc}E_{k} & O \\ O & E_{r}\end{array}\right) =(EkOOEr)

    由此可得

    image-20210103193310242

    所以 D − 1 = ( A − 1 O − B − 1 C A − 1 B − 1 ) \quad D^{-1}=\left(\begin{array}{cc}A^{-1} & O \\ -B^{-1} C A^{-1} & B^{-1}\end{array}\right) D1=(A1B1CA1OB1)

    例 4 \Large\color{violet}{例4 } 4
    A = ( 1 2 0 0 0 2 5 0 0 0 0 0 − 2 1 0 0 0 0 − 2 1 0 0 0 0 − 2 ) , B = ( 1 0 1 0 − 1 2 3 0 1 2 0 4 0 1 2 4 0 0 1 4 ) A=\left(\begin{array}{ccccc} 1 & 2 & 0 & 0 & 0 \\ 2 & 5 & 0 & 0 & 0 \\ 0 & 0 & -2 & 1 & 0 \\ 0 & 0 & 0 & -2 & 1 \\ 0 & 0 & 0 & 0 & -2 \end{array}\right), B=\left(\begin{array}{cccc} 1 & 0 & 1 & 0 \\ -1 & 2 & 3 & 0 \\ 1 & 2 & 0 & 4 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & 1 & 4 \end{array}\right) A=1200025000002000012000012,B=11100022101302100444
    用矩阵分块的方法 :(1) 计算 A 2 , A B ; A^{2}, A B ; A2,AB; (2) 求 A − 1 A^{-1} A1.

    【解】把矩阵A,B进行如下分块

    image-20210103193814714

    并令
    A = ( A 1 O O A 2 ) , B = ( B 11 B 12 B 21 B 22 ) A=\left(\begin{array}{ll} A_{1} & O \\ O & A_{2} \end{array}\right), B=\left(\begin{array}{ll} B_{11} & B_{12} \\ B_{21} & B_{22} \end{array}\right) A=(A1OOA2),B=(B11B21B12B22)
    其中
    A 1 = ( 1 2 2 5 ) , A 2 = ( − 2 1 0 0 − 2 1 0 0 − 2 ) B 11 = ( 1 0 − 1 2 ) , B 12 = ( 1 0 3 0 ) B 21 = ( 1 2 0 1 0 0 ) , B 22 = ( 0 4 2 4 1 4 ) \begin{array}{l} A_{1}=\left(\begin{array}{ll} 1 & 2 \\ 2 & 5 \end{array}\right), A_{2}=\left(\begin{array}{ccc} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -2 \end{array}\right) \\ B_{11}=\left(\begin{array}{cc} 1 & 0 \\ -1 & 2 \end{array}\right), B_{12}=\left(\begin{array}{cc} 1 & 0 \\ 3 & 0 \end{array}\right) \\ B_{21}=\left(\begin{array}{cc} 1 & 2 \\ 0 & 1 \\ 0 & 0 \end{array}\right), B_{22}=\left(\begin{array}{cc} 0 & 4 \\ 2 & 4 \\ 1 & 4 \end{array}\right) \end{array} A1=(1225),A2=200120012B11=(1102),B12=(1300)B21=100210,B22=021444
    则1)
    A 2 = ( A 1 O O A 2 ) ( A 1 O O A 2 ) = ( A 1 2 O O A 2 2 ) A^{2}=\left(\begin{array}{ll}A_{1} & O \\ O & A_{2}\end{array}\right)\left(\begin{array}{ll}A_{1} & O \\ O & A_{2}\end{array}\right)=\left(\begin{array}{ll}A_{1}^{2} & O \\ O & A_{2}^{2}\end{array}\right) A2=(A1OOA2)(A1OOA2)=(A12OOA22)
    其中
    A 1 2 = ( 1 2 2 5 ) ( 1 2 2 5 ) = ( 5 12 12 29 ) A 2 2 = ( − 2 1 0 0 − 2 1 0 0 − 2 ) ( − 2 1 0 0 − 2 1 0 0 − 2 ) = ( 4 − 4 1 0 4 − 4 0 0 4 ) \quad A_{1}^{2}=\left(\begin{array}{ll}1 & 2 \\ 2 & 5\end{array}\right)\left(\begin{array}{ll}1 & 2 \\ 2 & 5\end{array}\right)=\left(\begin{array}{cc}5 & 12 \\ 12 & 29\end{array}\right)\\ A_{2}^{2}=\left(\begin{array}{ccc}-2 & 1 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -2\end{array}\right)\left(\begin{array}{ccc}-2 & 1 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -2\end{array}\right)=\left(\begin{array}{ccc}4 & -4 & 1 \\ 0 & 4 & -4 \\ 0 & 0 & 4\end{array}\right) A12=(1225)(1225)=(5121229)A22=200120012200120012=400440144

    A B = ( A 1 O O A 2 ) ( B 11 B 12 B 21 B 22 ) = ( A 1 B 11 A 1 B 12 A 2 B 21 A 2 B 22 ) A B=\left(\begin{array}{cc}A_{1} & O \\ O & A_{2}\end{array}\right)\left(\begin{array}{cc}B_{11} & B_{12} \\ B_{21} & B_{22}\end{array}\right)=\left(\begin{array}{cc}A_{1} B_{11} & A_{1} B_{12} \\ A_{2} B_{21} & A_{2} B_{22}\end{array}\right) AB=(A1OOA2)(B11B21B12B22)=(A1B11A2B21A1B12A2B22)

    其中
    A 1 B 11 = ( 1 2 2 5 ) ( 1 0 − 1 2 ) = ( − 1 4 − 3 10 ) \quad A_{1} B_{11}=\left(\begin{array}{cc}1 & 2 \\ 2 & 5\end{array}\right)\left(\begin{array}{cc}1 & 0 \\ -1 & 2\end{array}\right)=\left(\begin{array}{cc}-1 & 4 \\ -3 & 10\end{array}\right) A1B11=(1225)(1102)=(13410)

    A 1 B 12 = ( 1 2 2 5 ) ( 1 0 3 0 ) = ( 7 0 17 0 ) A_{1} B_{12}=\left(\begin{array}{ccc}1 & 2 \\ 2 & 5\end{array}\right)\left(\begin{array}{cc}1 & 0 \\ 3 & 0\end{array}\right)=\left(\begin{array}{cc}7 & 0 \\ 17 & 0\end{array}\right) A1B12=(1225)(1300)=(71700)

    A 2 B 21 = ( − 2 1 0 0 − 2 1 0 0 − 2 ) ( 1 2 0 1 0 0 ) = ( − 2 − 3 0 − 2 0 0 ) A_{2} B_{21}=\left(\begin{array}{ccc}-2 & 1 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -2\end{array}\right)\left(\begin{array}{cc}1 & 2 \\ 0 & 1 \\ 0 & 0\end{array}\right)=\left(\begin{array}{cc}-2 & -3 \\ 0 & -2 \\ 0 & 0\end{array}\right) A2B21=200120012100210=200320

    A 2 B 22 = ( − 2 1 0 0 − 2 1 0 0 − 2 ) ( 0 4 2 4 1 4 ) = ( 2 − 4 − 3 − 4 − 2 − 8 )  所以  A B = ( − 1 4 7 0 − 3 10 17 0 − 2 − 3 2 − 4 0 − 2 − 3 − 4 0 0 − 2 − 8 ) \begin{array}{l}A_{2} B_{22}=\left(\begin{array}{ccc}-2 & 1 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -2\end{array}\right)\left(\begin{array}{rr}0 & 4 \\ 2 & 4 \\ 1 & 4\end{array}\right)=\left(\begin{array}{cc}2 & -4 \\ -3 & -4 \\ -2 & -8\end{array}\right) \\ \text { 所以 } \quad A B= \left(\begin{array}{cccc}-1 & 4 & 7 & 0 \\ -3 & 10 & 17 & 0 \\ -2 & -3 & 2 & -4 \\ 0 & -2 & -3 & -4 \\ 0 & 0 & -2 & -8\end{array}\right)\end{array} A2B22=200120012021444=232448 所以 AB=1320041032071723200448

    1. 因为 A = ( A 1 O O A 2 ) , A=\left(\begin{array}{cc}A_{1} & O \\ O & A_{2}\end{array}\right), A=(A1OOA2), 所以 A − 1 = ( A 1 − 1 O O A 2 − 1 ) A^{-1}=\left(\begin{array}{cc}A_{1}^{-1} & O \\ O & A_{2}^{-1}\end{array}\right) A1=(A11OOA21),
      A 1 = ( 1 2 2 5 ) \quad A_{1}=\left(\begin{array}{ll}1 & 2 \\ 2 & 5\end{array}\right) A1=(1225) 得到 A 1 − 1 = ( 5 − 2 − 2 1 ) A_{1}^{-1}=\left(\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right) A11=(5221)
      A 2 = ( − 2 1 0 0 − 2 1 0 0 − 2 ) \quad A_{2}=\left(\begin{array}{ccc}-2 & 1 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -2\end{array}\right) A2=200120012,得到 A 2 − 1 = − 1 8 ( 4 2 1 0 4 2 0 0 4 ) A_{2 }^{-1}=-\frac{1}{8}\left(\begin{array}{lll}4 & 2 & 1 \\ 0 & 4 & 2 \\ 0 & 0 & 4\end{array}\right) A21=81400240124

    所以 A − 1 = ( 5 − 2 0 0 0 − 2 1 0 0 0 0 0 − 1 2 − 1 4 − 1 8 0 0 0 − 1 2 − 1 4 0 0 0 0 − 1 2 ) . \quad A^{-1}=\left(\begin{array}{ccccc}5 & -2 & 0 & 0 & 0 \\ -2 & 1 & 0 & 0 & 0 \\ 0 & 0 & -\frac{1}{2} & -\frac{1}{4} & -\frac{1}{8} \\ 0 & 0 & 0 & -\frac{1}{2} & -\frac{1}{4} \\ 0 & 0 & 0 & 0 & -\frac{1}{2}\end{array}\right) . A1=5200021000002100004121000814121.

    例 5 \Large\color{violet}{例5 } 5 已知 A = ( 0 1 0 0 0 1 0 0 0 ) , A=\left(\begin{array}{ccc}\mathbf{0} & \mathbf{1} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & \mathbf{1} \\ \mathbf{0} & \mathbf{0} & \mathbf{0}\end{array}\right), A=000100010, A n A^{n} An

    A 2 = A ( 0 1 0 0 0 1 0 0 0 ) = A ( 0 ε 1 ε 2 ) = ( O A ε 1 A ε 2 ) = ( 0 0 1 0 0 0 0 0 0 ) \begin{aligned} A^{2} &=A\left(\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right)=A\left(\begin{array}{lll} 0 & \varepsilon_{1} & \varepsilon_{2} \end{array}\right) \\ &=\left(\begin{array}{lll} \boldsymbol{O} & A \varepsilon_{1} & A \varepsilon_{2} \end{array}\right) \\ &=\left(\begin{array}{lll} \mathbf{0} & \mathbf{0} & \mathbf{1} \\ \mathbf{0} & \mathbf{0} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & \mathbf{0} \end{array}\right) \end{aligned} A2=A000100010=A(0ε1ε2)=(OAε1Aε2)=000000100

    A 3 = A 2 ( 0 1 0 0 0 1 0 0 0 ) = A 2 ( O ε 1 ε 2 ) = ( O A 2 ε 1 A 2 ε 2 ) = ( 0 0 0 0 0 0 0 0 0 ) \begin{aligned} A^{3} &=A^{2}\left(\begin{array}{ccc} \mathbf{0} & \mathbf{1} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & \mathbf{1} \\ \mathbf{0} & \mathbf{0} & \mathbf{0} \end{array}\right)=\boldsymbol{A}^{2}\left(\begin{array}{lll} \boldsymbol{O} & \boldsymbol{\varepsilon}_{1} & \boldsymbol{\varepsilon}_{2} \end{array}\right) \\ &=\left(\begin{array}{lll} \boldsymbol{O} & \boldsymbol{A}^{2} \boldsymbol{\varepsilon}_{1} & \boldsymbol{A}^{2} \boldsymbol{\varepsilon}_{2} \end{array}\right) \\ &=\left(\begin{array}{ccc} \mathbf{0} & \mathbf{0} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0} \end{array}\right) \end{aligned} A3=A2000100010=A2(Oε1ε2)=(OA2ε1A2ε2)=000000000

    从而 当 n ≥ 3 n \geq 3 n3 , A n = ( 0 0 0 0 0 0 0 0 0 ) , A^{n}=\left(\begin{array}{ccc}\mathbf{0} & \mathbf{0} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & \mathbf{0}\end{array}\right) ,An=000000000

    例 6 \Large\color{violet}{例6 } 6 A = ( 1 0 0 0 0 1 0 0 0 0 − 1 1 0 0 1 − 1 ) , A=\left(\begin{array}{cccc}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & -1\end{array}\right), A=1000010000110011, A 2016 A^{2016} A2016.
    解 记 M = ( − 1 1 1 − 1 ) , M=\left(\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right), M=(1111), A = ( E O o M ) A=\left(\begin{array}{cc}E & O \\ o & M\end{array}\right) A=(EoOM)
    于是 A 2016 = ( E 2016 0 0 M 2016 ) = ( E O 0 M 2016 ) A^{2016}=\left(\begin{array}{cc}E^{2016} & 0 \\ 0 & M^{2016}\end{array}\right)=\left(\begin{array}{cc}E & O \\ 0 & M^{2016}\end{array}\right) A2016=(E201600M2016)=(E0OM2016)
    α = ( − 1 1 ) , \alpha=\left(\begin{array}{c}-1 \\ 1\end{array}\right), α=(11), 由于 M = ( α − α ) = α ( 1 − 1 ) M=\left(\begin{array}{cc}\alpha & -\alpha\end{array}\right)=\alpha\left(\begin{array}{ll}1 & -1\end{array}\right) M=(αα)=α(11)

    image-20210105112248757

    如此继续得 M 2016 = ( − 2 ) 2015 M . M^{2016}=(-2)^{2015} M . M2016=(2)2015M.
    因此 A 2016 = ( 1 0 0 0 0 1 0 0 0 0 2 2015 − 2 2015 0 0 − 2 2015 2 2015 ) . A^{2016}=\left(\begin{array}{cccc}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2^{2015} & -2^{2015} \\ 0 & 0 & -2^{2015} & 2^{2015}\end{array}\right) . A2016=10000100002201522015002201522015.

    分块矩阵的初等变换

    定 义 2 \large\color{magenta}{\boxed{\color{brown}{定义2 } }} 2 把单位矩阵 E E E 如下进行分块:
    E = ( E m O O E n ) E=\left(\begin{array}{cc} E_{m} & O \\ O & E_{n} \end{array}\right) E=(EmOOEn)
    对它进行三种初等变换所得到的矩阵称为 分 块 初 等 矩 阵 \large\color{#70f3ff}{\boxed{\color{green}{分块初等矩阵}}} .

    分块初等矩阵有以下三种:

     1) 分块对换矩阵   对换两行(列)所得到  \begin{array}{ll}\color{red}{\text { 1) 分块对换矩阵 } }& \text { 对换两行(列)所得到 }\end{array}  1) 分块对换矩阵  对换两行()所得到 
    ( O E n E m O ) \left(\begin{array}{cc} O & E_{n} \\ E_{m} & O \end{array}\right) (OEmEnO)
     2) 分块倍乘矩阵   某一行(列)左乘(右乘)一个矩阵  P  所得到  \begin{array}{ll}\color{red}{\text { 2) 分块倍乘矩阵 } }& \text { 某一行(列)左乘(右乘)一个矩阵 } P \text { 所得到 }\end{array}  2) 分块倍乘矩阵  某一行()左乘(右乘)一个矩阵 P 所得到 
    ( P O O E n ) , ( E m O O P ) \left(\begin{array}{ll} P & O \\ O & E_{n} \end{array}\right),\left(\begin{array}{cc} E_{m} & O \\ O & P \end{array}\right) (POOEn),(EmOOP)
     3) 分块倍加矩阵   一行(列)加上另一行(列)的  P  (矩阵)倍数所得到  \begin{array}{ll}\color{red}{\text { 3) 分块倍加矩阵 } }& \text { 一行(列)加上另一行(列)的 } P \text { (矩阵)倍数所得到 }\end{array}  3) 分块倍加矩阵  一行()加上另一行() P (矩阵)倍数所得到 
    ( E m P O E n ) , ( E m O P E n ) \left(\begin{array}{cc} E_{m} & P \\ O & E_{n} \end{array}\right),\left(\begin{array}{cc} E_{m} & O \\ P & E_{n} \end{array}\right) (EmOPEn),(EmPOEn)

    分块初等矩阵的性质

    和初等矩阵与初等变换的关系一样,分块初等矩阵有与初等矩阵类似的性质:

    用分块初等矩阵左乘分块矩阵A,在保证可乘的情况下,其作用相当于对分块矩阵A进行一次相应的初等行变换;用分块初等矩阵右乘分块矩阵A,其作用相当于对分块矩阵A进行一次相应的初等列变换.

    例如,设有如下分块矩阵
    ( A B C D ) \left(\begin{array}{ll} A & B \\ C & D \end{array}\right) (ACBD)
    分别用三种分块初等矩阵左乘它,其结果如下:
    ( O E n E m O ) ( A B C D ) = ( C D A B ) ( P O O E n ) ( A B C D ) = ( P A P B C D ) ( E m O P E n ) ( A B C D ) = ( A B C + P A D + P B ) \begin{array}{l} \left(\begin{array}{cc} O & E_{n} \\ E_{m} & O \end{array}\right)\left(\begin{array}{cc} A & B \\ C & D \end{array}\right)=\left(\begin{array}{cc} C & D \\ A & B \end{array}\right) \\ \left(\begin{array}{cc} P & O \\ O & E_{n} \end{array}\right)\left(\begin{array}{cc} A & B \\ C & D \end{array}\right)=\left(\begin{array}{cc} P A & P B \\ C & D \end{array}\right) \\ \left(\begin{array}{cc} E_{m} & O \\ P & E_{n} \end{array}\right)\left(\begin{array}{cc} A & B \\ C & D \end{array}\right)=\left(\begin{array}{cc} A & B \\ C+P A & D+P B \end{array}\right) \end{array} (OEmEnO)(ACBD)=(CADB)(POOEn)(ACBD)=(PACPBD)(EmPOEn)(ACBD)=(AC+PABD+PB)

      ( E m O P E n ) ( A B C D ) = ( A B C + P A D + P B ) \text { }\left(\begin{array}{cc} E_{m} & O \\ P & E_{n} \end{array}\right)\left(\begin{array}{cc} A & B \\ C & D \end{array}\right)=\left(\begin{array}{cc} A & B \\ C+P A & D+P B \end{array}\right)  (EmPOEn)(ACBD)=(AC+PABD+PB)
    中,适当选择 P , P, P, 可使 C + P A = O . C+P A=O . \quad C+PA=O. 例如 A A A 可逆时, 选 P = − C A − 1 P=-C A^{-1} P=CA1,则 C + P A = O . C+P A=O . C+PA=O. 于是上式右端成为
    ( A B O D − C A − 1 B ) \left(\begin{array}{cc} A & B \\ O & D-C A^{-1} B \end{array}\right) (AOBDCA1B)
    这种形状的矩阵在求行列式、逆矩阵和解决其他问题时是比较方便的,因此这种运算非常有用.

    例 1 \Large\color{violet}{例1 } 1 T = ( A O C D ) T=\left(\begin{array}{ll} A & O \\ C & D \end{array}\right) T=(ACOD),其中 A , D A, D A,D 可逆,求 T − 1 T^{-1} T1.

    【解 】 \quad 因为
    ( E m O − C A − 1 E n ) ( A O C D ) = ( A O O D ) \left(\begin{array}{cc} E_{m} & O \\ -C A^{-1} & E_{n} \end{array}\right)\left(\begin{array}{ll} A & O \\ C & D \end{array}\right)=\left(\begin{array}{ll} A & O \\ O & D \end{array}\right) (EmCA1OEn)(ACOD)=(AOOD)
    ( A O O D ) − 1 = ( A − 1 O O D − 1 ) \left(\begin{array}{ll}A & O \\ O & D\end{array}\right)^{-1}=\left(\begin{array}{cc}A^{-1} & O \\ O & D^{-1}\end{array}\right) (AOOD)1=(A1OOD1) ,所以
    T − 1 = ( A − 1 O O D − 1 ) ( E m O − C A − 1 E n ) = ( A − 1 O − D − 1 C A − 1 D − 1 ) \begin{aligned}\quad T^{-1} &=\left(\begin{array}{cc}A^{-1} & O \\ O & D^{-1}\end{array}\right)\left(\begin{array}{cc}E_{m} & O \\ -C A^{-1} & E_{n}\end{array}\right) \\ &=\left(\begin{array}{cc}A^{-1} & O \\ -D^{-1} C A^{-1} & D^{-1}\end{array}\right) \end{aligned} T1=(A1OOD1)(EmCA1OEn)=(A1D1CA1OD1)

    例 2 \Large\color{violet}{例2 } 2 T 1 = ( A B C D ) T_{1}=\left(\begin{array}{cc} A & B \\ C & D \end{array}\right) T1=(ACBD),其中 T 1 , D T_{1}, D T1,D 可逆,试证 ( A − B D − 1 C ) − 1 \left(A-B D^{-1} C\right)^{-1} (ABD1C)1 存在,并求 T 1 − 1 T_{1}^{-1} T11.

    【证明 】 \quad 因为
    ( E m − B D − 1 O E n ) ( A B C D ) = ( A − B D − 1 C O C D ) \begin{aligned} &\left(\begin{array}{cc} E_{m} & -B D^{-1} \\ O & E_{n} \end{array}\right)\left(\begin{array}{ll} A & B \\ C & D \end{array}\right) \\ &=\left(\begin{array}{cc} A-B D^{-1} C & O \\ C & D \end{array}\right) \end{aligned} (EmOBD1En)(ACBD)=(ABD1CCOD)

    因为 T 1 T_{1} T1 可逆,对它进行初等变换后仍可逆,即
    ( A − B D − 1 C O C D ) \left(\begin{array}{cc} A-B D^{-1} C & O \\ C & D \end{array}\right) (ABD1CCOD)
    可逆,故 ( A − B D − 1 C ) − 1 \left(A-B D^{-1} C\right)^{-1} (ABD1C)1 存在. 由
    ( E m − B D − 1 O E n ) T 1 = ( A − B D − 1 C O C D ) \left(\begin{array}{cc} E_{m} & -B D^{-1} \\ O & E_{n} \end{array}\right) T_{1}=\left(\begin{array}{cc} A-B D^{-1} C & O \\ C & D \end{array}\right) (EmOBD1En)T1=(ABD1CCOD)
    解得 T 1 − 1 = ( A − B D − 1 C O C D ) − 1 ( E m − B D − 1 O E n ) T_{1}^{-1}=\left(\begin{array}{cc}A-B D^{-1} C & O \\ C & D\end{array}\right)^{-1}\left(\begin{array}{cc}E_{m} & -B D^{-1} \\ O & E_{n}\end{array}\right) T11=(ABD1CCOD)1(EmOBD1En)

    再由例 1,得
    T 1 − 1 = ( A − B D − 1 C O C D ) − 1 ( E m − B D − 1 O E n ) = ( ( A − B D − 1 C ) − 1 O − D − 1 C ( A − B D − 1 C ) − 1 D − 1 ) ( E m − B D − 1 O E n ) = ( ( A −