第三节 分块矩阵
一、 相关概念
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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的子块，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 = \begin{bmatrix} 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 \\ \end{bmatrix} = \begin{bmatrix} A_{1} & A_{2} \\ \mathbf{O} & E_{3} \\ \end{bmatrix} A= 2100012000121000−3010−10001 =[A1OA2E3]
其中：
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A_{1} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \\ \end{bmatrix} ， A_{2} = \begin{bmatrix} 1 & 0 & -1 \\ 2 & -3 & 0 \\ \end{bmatrix} , \mathbf{O} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 0 \\ \end{bmatrix} , E_{3} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}
A1=[2112]，A2=[120−3−10],O=
000000
,E3=
100010001
- 按列分块矩阵、按行分块矩阵：
设
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m×n矩阵
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A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end{bmatrix}
A=
a11a21⋮am1a12a22⋮am2⋯⋯⋱⋯a1na2n⋮amn
- 把
A
A
A的每个列作为一个子块，即在列的方向分成
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n块，就得到
A
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A的按列分块矩阵，记为
A = [ α 1 , α 2 , ⋯ , α n ] A = \begin{bmatrix} \alpha_{1},\ \alpha_{2},\ \cdots,\ \alpha_{n} \end{bmatrix} A=[α1, α2, ⋯, αn]
其中， α j = [ a 1 j a 2 j ⋮ a m j ] , j = 1 , 2 , ⋯ , n . \alpha_{j} = \begin{bmatrix}a_{1j}\\ a_{2j}\\ \vdots \\a_{mj}\end{bmatrix}, j = 1,\ 2,\ \cdots,\ n. αj= a1ja2j⋮amj ,j=1, 2, ⋯, n.
- 类似的，把
A
A
A的每一行作为一个子块，就得到
A
A
A的按行分块矩阵，记为
A = [ β 1 β 2 ⋮ β m ] A = \begin{bmatrix} \beta_{1} \\ \beta_{2} \\ \vdots \\ \beta_{m} \\ \end{bmatrix} A= β1β2⋮βm
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
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分块矩阵的加法
设 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 = \begin{bmatrix} A_{11} & A_{12} & \cdots & A_{1t} \\ A_{21} & A_{22} & \cdots & A_{2t} \\ \vdots & \vdots & \ddots & \vdots \\ A_{s1} & A_{s2} & \cdots & A_{st} \\ \end{bmatrix} ， B = \begin{bmatrix} B_{11} & B_{12} & \cdots & B_{1t} \\ B_{21} & B_{22} & \cdots & B_{2t} \\ \vdots & \vdots & \ddots & \vdots \\ B_{s1} & B_{s2} & \cdots & B_{st} \\ \end{bmatrix} A= A11A21⋮As1A12A22⋮As2⋯⋯⋱⋯A1tA2t⋮Ast ，B= B11B21⋮Bs1B12B22⋮Bs2⋯⋯⋱⋯B1tB2t⋮Bst
其中
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A_{ij}
Aij与
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A+B = \begin{bmatrix} A_{11} + B_{11} & A_{12} + B_{12} & \cdots & A_{1t} + B_{1t} \\ A_{21} + B_{21} & A_{22} + B_{22} & \cdots & A_{2t} + B_{2t} \\ \vdots & \vdots & \ddots & \vdots \\ A_{s1} + B_{s} & A_{s2} + B_{s2} & \cdots & A_{st} + B_{st} \\ \end{bmatrix}
A+B=
A11+B11A21+B21⋮As1+BsA12+B12A22+B22⋮As2+Bs2⋯⋯⋱⋯A1t+B1tA2t+B2t⋮Ast+Bst
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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 = \begin{bmatrix} A_{11} & A_{12} & \cdots & A_{1t} \\ A_{21} & A_{22} & \cdots & A_{2t} \\ \vdots & \vdots & \ddots & \vdots \\ A_{s1} & A_{s2} & \cdots & A_{st} \\ \end{bmatrix} A= A11A21⋮As1A12A22⋮As2⋯⋯⋱⋯A1tA2t⋮Ast
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 = \begin{bmatrix} \lambda A_{11} & \lambda A_{12} & \cdots & \lambda A_{1t} \\ \lambda A_{21} & \lambda a_{22} & \cdots & \lambda A_{2t} \\ \vdots & \vdots & \ddots & \vdots \\ \lambda A_{s1} & \lambda A_{s2} & \cdots & \lambda A_{st} \\ \end{bmatrix} λA= λA11λA21⋮λAs1λA12λa22⋮λAs2⋯⋯⋱⋯λA1tλA2t⋮λAst -
分块矩阵的乘法
设 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 = \begin{bmatrix} A_{11} & A_{12} & \cdots & A_{1t} \\ A_{21} & A_{22} & \cdots & A_{2t} \\ \vdots & \vdots & \ddots & \vdots \\ A_{s1} & A_{s2} & \cdots & A_{st} \\ \end{bmatrix} , B = \begin{bmatrix} B_{11} & B_{12} & \cdots & B_{1r} \\ B_{21} & B_{22} & \cdots & B_{2r} \\ \vdots & \vdots & \ddots & \vdots \\ B_{t1} & B_{t2} & \cdots & B_{tr} \\ \end{bmatrix} A= A11A21⋮As1A12A22⋮As2⋯⋯⋱⋯A1tA2t⋮Ast ,B= B11B21⋮Bt1B12B22⋮Bt2⋯⋯⋱⋯B1rB2r⋮Btr
其中
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Ai1, Ai2, ⋯, Ait的列数分别等于
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AB = \begin{bmatrix} C_{11} & C_{12} & \cdots & C_{1r} \\ C_{21} & C_{22} & \cdots & C_{2r} \\ \vdots & \vdots & \ddots & \vdots \\ C_{s1} & C_{s2} & \cdots & C_{sr} \\ \end{bmatrix}
AB=
C11C21⋮Cs1C12C22⋮Cs2⋯⋯⋱⋯C1rC2r⋮Csr
其中，
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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).
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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 = \begin{bmatrix} A_{11} & A_{12} & \cdots & A_{1t} \\ A_{21} & A_{22} & \cdots & A_{2t} \\ \vdots & \vdots & \ddots & \vdots \\ A_{s1} & A_{s2} & \cdots & A_{st} \\ \end{bmatrix} ,\quad 则A_{T} = \begin{bmatrix} A_{11}^{T} & A_{21}^{T} & \cdots & A_{s1}^{T} \\ A_{12}^{T} & A_{22}^{T} & \cdots & A_{s2}^{T} \\ \vdots & \vdots & \ddots & \vdots \\ A_{1t}^{T} & A_{2t}^{T} & \cdots & A_{st}^{T} \\ \end{bmatrix} A= A11A21⋮As1A12A22⋮As2⋯⋯⋱⋯A1tA2t⋮Ast ,则AT= A11TA12T⋮A1tTA21TA22T⋮A2tT⋯⋯⋱⋯As1TAs2T⋮AstT