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《线性代数》第一章第三节“分块矩阵”笔记_分块矩阵笔记

分块矩阵笔记

第三节 分块矩阵

一、 相关概念
  • 将矩阵 A A A通过若干条纵线和横线分成许多个小矩阵,每个小矩阵称为 A A A子块,以子块为元素的矩阵称为分块矩阵,如:

    A = [ 2 1 1 0 − 1 1 2 2 − 3 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 ] = [ A 1 A 2 O E 3 ] A = [2110112230001000001000001] = [A1A2OE3] A= 2100012000121000301010001 =[A1OA2E3]

其中:
A 1 = [ 2 1 1 2 ] , A 2 = [ 1 0 − 1 2 − 3 0 ] , O = [ 0 0 0 0 0 0 ] , E 3 = [ 1 0 0 0 1 0 0 0 1 ] A_{1} = [2112] , A_{2} = [101230] , \mathbf{O} = [000000] , E_{3} = [100010001] A1=[2112]A2=[120310],O= 000000 ,E3= 100010001

  • 按列分块矩阵、按行分块矩阵

m × n m \times n m×n矩阵
A = [ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋱ ⋮ a m 1 a m 2 ⋯ a m n ] A = [a11a12a1na21a22a2nam1am2amn] A= a11a21am1a12a22am2a1na2namn

  1. A A A的每个列作为一个子块,即在列的方向分成 n n n块,就得到 A A A按列分块矩阵,记为
    A = [ α 1 ,   α 2 ,   ⋯   ,   α n ] A =
    [α1, α2, , αn]
    A=[α1, α2, , αn]

其中, α j = [ a 1 j a 2 j ⋮ a m j ] , j = 1 ,   2 ,   ⋯   ,   n . \alpha_{j} = [a1ja2jamj], j = 1,\ 2,\ \cdots,\ n. αj= a1ja2jamj ,j=1, 2, , n.

  1. 类似的,把 A A A的每一行作为一个子块,就得到 A A A按行分块矩阵,记为
    A = [ β 1 β 2 ⋮ β m ] A =
    [β1β2βm]
    A= β1β2βm

其中, β i = [ a i 1 ,   a i 2 ,   ⋯   ,   a i m ] , i = 1 ,   2 ,   ⋯   ,   m . \beta_{i} = [a_{i1},\ a_{i2},\ \cdots,\ a_{im}], i = 1,\ 2,\ \cdots,\ m. βi=[ai1, ai2, , aim],i=1, 2, , m.

二、分块矩阵的运算法则
  1. 分块矩阵的加法

    A = [ a i j ] m × n ,   B = [ b i j ] m × n A = [a_{ij}]_{m \times n},\ B = [b_{ij}]_{m \times n} A=[aij]m×n, B=[bij]m×n,采用相同的分块法(每个对应子块都是同型矩阵),有
    A = [ A 11 A 12 ⋯ A 1 t A 21 A 22 ⋯ A 2 t ⋮ ⋮ ⋱ ⋮ A s 1 A s 2 ⋯ A s t ] , B = [ B 11 B 12 ⋯ B 1 t B 21 B 22 ⋯ B 2 t ⋮ ⋮ ⋱ ⋮ B s 1 B s 2 ⋯ B s t ] A = [A11A12A1tA21A22A2tAs1As2Ast] , B = [B11B12B1tB21B22B2tBs1Bs2Bst] A= A11A21As1A12A22As2A1tA2tAst B= B11B21Bs1B12B22Bs2B1tB2tBst

其中 A i j A_{ij} Aij B i j B_{ij} Bij的行相同,列数也相同,则有:
A + B = [ A 11 + B 11 A 12 + B 12 ⋯ A 1 t + B 1 t A 21 + B 21 A 22 + B 22 ⋯ A 2 t + B 2 t ⋮ ⋮ ⋱ ⋮ A s 1 + B s A s 2 + B s 2 ⋯ A s t + B s t ] A+B = [A11+B11A12+B12A1t+B1tA21+B21A22+B22A2t+B2tAs1+BsAs2+Bs2Ast+Bst] A+B= A11+B11A21+B21As1+BsA12+B12A22+B22As2+Bs2A1t+B1tA2t+B2tAst+Bst

  1. 数与分块矩阵的乘法

    设分块矩阵
    A = [ A 11 A 12 ⋯ A 1 t A 21 A 22 ⋯ A 2 t ⋮ ⋮ ⋱ ⋮ A s 1 A s 2 ⋯ A s t ] A = [A11A12A1tA21A22A2tAs1As2Ast] A= A11A21As1A12A22As2A1tA2tAst
    λ \lambda λ为数,则有:
    λ A = [ λ A 11 λ A 12 ⋯ λ A 1 t λ A 21 λ a 22 ⋯ λ A 2 t ⋮ ⋮ ⋱ ⋮ λ A s 1 λ A s 2 ⋯ λ A s t ] \lambda A = [λA11λA12λA1tλA21λa22λA2tλAs1λAs2λAst] λA= λA11λA21λAs1λA12λa22λAs2λA1tλA2tλAst

  2. 分块矩阵的乘法

    A = [ a i j ] m × n ,   B = [ b i j ] m × n A = [a_{ij}]_{m \times n},\ B = [b_{ij}]_{m \times n} A=[aij]m×n, B=[bij]m×n,将 A ,   B A,\ B A, B分块成
    A = [ A 11 A 12 ⋯ A 1 t A 21 A 22 ⋯ A 2 t ⋮ ⋮ ⋱ ⋮ A s 1 A s 2 ⋯ A s t ] , B = [ B 11 B 12 ⋯ B 1 r B 21 B 22 ⋯ B 2 r ⋮ ⋮ ⋱ ⋮ B t 1 B t 2 ⋯ B t r ] A = [A11A12A1tA21A22A2tAs1As2Ast] , B = [B11B12B1rB21B22B2rBt1Bt2Btr] A= A11A21As1A12A22As2A1tA2tAst ,B= B11B21Bt1B12B22Bt2B1rB2rBtr

其中 A i 1 ,   A i 2 ,   ⋯   ,   A i t A_{i1},\ A_{i2},\ \cdots,\ A_{it} Ai1, Ai2, , Ait的列数分别等于 B i j ,   B 2 j ,   ⋯   ,   B t j B_{ij},\ B_{2j},\ \cdots,\ B_{tj} Bij, B2j, , Btj的行数,则有
A B = [ C 11 C 12 ⋯ C 1 r C 21 C 22 ⋯ C 2 r ⋮ ⋮ ⋱ ⋮ C s 1 C s 2 ⋯ C s r ] AB = [C11C12C1rC21C22C2rCs1Cs2Csr] AB= C11C21Cs1C12C22Cs2C1rC2rCsr
其中, C i j = ∑ k = 1 t ( A i k B k j ) ( i = 1 ,   2 ,   ⋯   ,   s ;   j = 1 ,   2 ,   ⋯   ,   r ) C_{ij} = \sum_{k = 1}^{t}(A_{ik}B_{kj})(i = 1,\ 2,\ \cdots,\ s;\ j = 1,\ 2,\ \cdots,\ r) Cij=k=1t(AikBkj)(i=1, 2, , s; j=1, 2, , r).

  1. 分块矩阵的转置

    设分块矩阵
    A = [ A 11 A 12 ⋯ A 1 t A 21 A 22 ⋯ A 2 t ⋮ ⋮ ⋱ ⋮ A s 1 A s 2 ⋯ A s t ] , 则 A T = [ A 11 T A 21 T ⋯ A s 1 T A 12 T A 22 T ⋯ A s 2 T ⋮ ⋮ ⋱ ⋮ A 1 t T A 2 t T ⋯ A s t T ] A = [A11A12A1tA21A22A2tAs1As2Ast] ,\quad 则A_{T} = [AT11AT21ATs1AT12AT22ATs2AT1tAT2tATst] A= A11A21As1A12A22As2A1tA2tAst ,AT= A11TA12TA1tTA21TA22TA2tTAs1TAs2TAstT

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