分块矩阵 对于行数和列数较高的矩阵 A A A A A A A A A
A = ( a 11 a 12 a 13 a 14 a 15 a 21 a 22 a 23 a 24 a 25 a 31 a 32 … … a 33 a 35 a 41 a 42 a 43 a 44 a 45 ) A=\left(\begin{array}{cc:ccc}
a_{11} & a_{12} & a_{13} & a_{14} & a_{15} \\
a_{21} & a_{22} & a_{23} & a_{24} & a_{25} \\
\hdashline a_{31} & a_{32} & \ldots \ldots & a_{33} & a_{35} \\
\hdashline a_{41} & a_{42} & a_{43} & a_{44} & a_{45}
\end{array}\right) A = a 11 a 21 a 31 a 41 a 12 a 22 a 32 a 42 a 13 a 23 …… a 43 a 14 a 24 a 33 a 44 a 15 a 25 a 35 a 45 记为
其中
A 11 = ( a 11 a 12 a 21 a 22 ) , A 12 = ( a 13 a 14 a 15 a 23 a 24 a 25 ) , A 21 = ( a 31 a 32 a 41 a 42 ) , A 22 = ( a 33 a 34 a 35 a 43 a 44 a 45 ) . A_{11}=\left(\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right), \quad A_{12}=\left(\begin{array}{lll}
a_{13} & a_{14} & a_{15} \\
a_{23} & a_{24} & a_{25}
\end{array}\right), \quad A_{21}=\left(\begin{array}{ll}
a_{31} & a_{32} \\
a_{41} & a_{42}
\end{array}\right), \quad A_{22}=\left(\begin{array}{lll}
a_{33} & a_{34} & a_{35} \\
a_{43} & a_{44} & a_{45}
\end{array}\right) . A 11 = ( a 11 a 21 a 12 a 22 ) , A 12 = ( a 13 a 23 a 14 a 24 a 15 a 25 ) , A 21 = ( a 31 a 41 a 32 a 42 ) , A 22 = ( a 33 a 43 a 34 a 44 a 35 a 45 ) . A = ( a 11 a 12 a 13 a 14 a 15 a 21 a 22 a 23 a 24 a 25 a 31 a 33 a 33 a 34 a 35 a 41 a 42 a 43 a 4 a 45 . ) A=\left(\begin{array}{cc:cc:c}
a_{11} & a_{12} & a_{13} & a_{14} & a_{15} \\
\hdashline a_{21} & a_{22} & a_{23} & a_{24} & a_{25} \\
a_{31} & a_{33} & a_{33} & a_{34} & a_{35} \\
\hdashline a_{41} & a_{42} & a_{43} & a_4 & a_{45} .
\end{array}\right) A = a 11 a 21 a 31 a 41 a 12 a 22 a 33 a 42 a 13 a 23 a 33 a 43 a 14 a 24 a 34 a 4 a 15 a 25 a 35 a 45 . 记为
A = ( A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ) A=\left(\begin{array}{lll}A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33}\end{array}\right) A = A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 A = ( A 11 , A 12 , A 13 , A 14 , A 15 ) , A=\left(A_{11}, A_{12}, A_{13}, A_{14}, A_{15}\right), A = ( A 11 , A 12 , A 13 , A 14 , A 15 ) ,
A 11 = ( a 11 , a 12 ) , A 12 = ( a 13 , a 14 ) , A 13 = a 15 , A 21 = ( a 21 a 22 a 31 a 32 ) , A 22 = ( a 23 a 24 a 33 a 34 ) , A 23 = ( a 25 a 35 ) A 31 = ( a 41 , a 42 ) , A 32 = ( a 43 , a 44 ) , A 33 = a 45 . \begin{aligned}
& \boldsymbol{A}_{11}=\left(a_{11}, a_{12}\right), \quad \boldsymbol{A}_{12}=\left(a_{13}, a_{14}\right), \quad \boldsymbol{A}_{13}=a_{15}, \\
& \boldsymbol{A}_{21}=\left(\begin{array}{ll}
a_{21} & a_{22} \\
a_{31} & a_{32}
\end{array}\right), \quad \boldsymbol{A}_{22}=\left(\begin{array}{ll}
a_{23} & a_{24} \\
a_{33} & a_{34}
\end{array}\right), \quad \boldsymbol{A}_{23}=\left(\begin{array}{l}
a_{25} \\
a_{35}
\end{array}\right) \\
& \boldsymbol{A}_{31}=\left(a_{41}, a_{42}\right), \quad \boldsymbol{A}_{32}=\left(a_{43}, a_{44}\right), \quad \boldsymbol{A}_{33}=a_{45} .
\end{aligned} A 11 = ( a 11 , a 12 ) , A 12 = ( a 13 , a 14 ) , A 13 = a 15 , A 21 = ( a 21 a 31 a 22 a 32 ) , A 22 = ( a 23 a 33 a 24 a 34 ) , A 23 = ( a 25 a 35 ) A 31 = ( a 41 , a 42 ) , A 32 = ( a 43 , a 44 ) , A 33 = a 45 . 其中
A 11 = ( a 11 a 21 a 31 a 41 ) , A 12 = ( a 12 a 22 a 32 a 42 ) , A 13 = ( a 13 a 23 a 33 a 43 ) , A 14 = ( a 14 a 24 a 34 a 44 ) , A 15 = ( a 15 a 25 a 35 a 45 ) . \boldsymbol{A}_{11}=\left(\begin{array}{l}
a_{11} \\
a_{21} \\
a_{31} \\
a_{41}
\end{array}\right), \quad \boldsymbol{A}_{12}=\left(\begin{array}{l}
a_{12} \\
a_{22} \\
a_{32} \\
a_{42}
\end{array}\right), \quad \boldsymbol{A}_{13}=\left(\begin{array}{l}
a_{13} \\
a_{23} \\
a_{33} \\
a_{43}
\end{array}\right), \quad \boldsymbol{A}_{14}=\left(\begin{array}{l}
a_{14} \\
a_{24} \\
a_{34} \\
a_{44}
\end{array}\right), \quad \boldsymbol{A}_{15}=\left(\begin{array}{l}
a_{15} \\
a_{25} \\
a_{35} \\
a_{45}
\end{array}\right) .
A 11 = a 11 a 21 a 31 a 41 , A 12 = a 12 a 22 a 32 a 42 , A 13 = a 13 a 23 a 33 a 43 , A 14 = a 14 a 24 a 34 a 44 , A 15 = a 15 a 25 a 35 a 45 . 如果分块矩阵没有相应的运算，则 "分块矩阵" 这个概念是没有意义的。与矩阵的运算相对应，我们考虑分块矩阵的运算：加法、数乘、乘法、转置。当然基本要求是："分块前的运算结果与分块后的计算结果应该是一样的 ", 以矩阵加法为例, 设有矩阵 A , B \boldsymbol{A}, \boldsymbol{B} A , B A 1 , B 1 \boldsymbol{A}_1, \boldsymbol{B}_1 A 1 , B 1 A 1 + B 1 \boldsymbol{A}_1+\boldsymbol{B}_1 A 1 + B 1 A + B \boldsymbol{A}+\boldsymbol{B} A + B
分块矩阵的运算法则最好与通常矩阵的对应运算有类似法则。例如, 矩阵加法 A + B \boldsymbol{A}+\boldsymbol{B} A + B A B \boldsymbol{A} \boldsymbol{B} A B A B \boldsymbol{A} \boldsymbol{B} A B A \boldsymbol{A} A B \boldsymbol{B} B A B \boldsymbol{A} \boldsymbol{B} A B A \boldsymbol{A} A B \boldsymbol{B} B
分块矩阵的加法 A + B A+B A + B A + B = [ A 11 A 12 ⋯ A 1 s A 21 A 22 ⋯ A 2 s ⋮ ⋮ ⋮ A r 1 A r 2 ⋯ A r s ] + [ B 11 B 12 ⋯ B 1 s B 21 B 22 ⋯ B 2 s ⋮ ⋮ ⋮ B r 1 B r 2 ⋯ B r s ] = [ A 11 + B 11 A 12 + B 12 ⋯ A 1 s + B 1 s A 21 + B 21 A 22 + B 22 ⋯ A 2 s + B 2 s ⋮ ⋮ ⋮ A r 1 + B r 1 A r 2 + B r 2 ⋯ A r s + B r s ] = [ A i j + B i j ] r × s \begin{aligned}
& \boldsymbol{A}+\boldsymbol{B}=\left[\begin{array}{cccc}
\boldsymbol{A}_{11} & \boldsymbol{A}_{12} & \cdots & \boldsymbol{A}_{1 s} \\
\boldsymbol{A}_{21} & \boldsymbol{A}_{22} & \cdots & \boldsymbol{A}_{2 s} \\
\vdots & \vdots & & \vdots \\
\boldsymbol{A}_{r 1} & \boldsymbol{A}_{r 2} & \cdots & \boldsymbol{A}_{r s}
\end{array}\right]+\left[\begin{array}{cccc}
\boldsymbol{B}_{11} & \boldsymbol{B}_{12} & \cdots & \boldsymbol{B}_{1 s} \\
\boldsymbol{B}_{21} & \boldsymbol{B}_{22} & \cdots & \boldsymbol{B}_{2 s} \\
\vdots & \vdots & & \vdots \\
\boldsymbol{B}_{r 1} & \boldsymbol{B}_{r 2} & \cdots & \boldsymbol{B}_{r s}
\end{array}\right] \\
& =\left[\begin{array}{cccc}
\boldsymbol{A}_{11}+\boldsymbol{B}_{11} & \boldsymbol{A}_{12}+\boldsymbol{B}_{12} & \cdots & \boldsymbol{A}_{1 s}+\boldsymbol{B}_{1 s} \\
\boldsymbol{A}_{21}+\boldsymbol{B}_{21} & \boldsymbol{A}_{22}+\boldsymbol{B}_{22} & \cdots & \boldsymbol{A}_{2 s}+\boldsymbol{B}_{2 s} \\
\vdots & \vdots & & \vdots \\
\boldsymbol{A}_{r 1}+\boldsymbol{B}_{r 1} & \boldsymbol{A}_{r 2}+\boldsymbol{B}_{r 2} & \cdots & \boldsymbol{A}_{r s}+\boldsymbol{B}_{r s}
\end{array}\right]=\left[\boldsymbol{A}_{i j}+\boldsymbol{B}_{i j}\right]_{r \times s}
\end{aligned} A + B = A 11 A 21 ⋮ A r 1 A 12 A 22 ⋮ A r 2 ⋯ ⋯ ⋯ A 1 s A 2 s ⋮ A rs + B 11 B 21 ⋮ B r 1 B 12 B 22 ⋮ B r 2 ⋯ ⋯ ⋯ B 1 s B 2 s ⋮ B rs = A 11 + B 11 A 21 + B 21 ⋮ A r 1 + B r 1 A 12 + B 12 A 22 + B 22 ⋮ A r 2 + B r 2 ⋯ ⋯ ⋯ A 1 s + B 1 s A 2 s + B 2 s ⋮ A rs + B rs = [ A ij + B ij ] r × s 分块矩阵的数乘 k A = [ k A 11 k A 12 ⋯ k A 1 s k A 21 k A 22 ⋯ k A 2 s ⋮ ⋮ ⋮ k A r 1 k A r 2 ⋯ k A n s ] = [ k A i j ] r × s \begin{aligned}
&k \boldsymbol{A}=\left[\begin{array}{cccc}
\boldsymbol{k} \boldsymbol{A}_{11} & k \boldsymbol{A}_{12} & \cdots & \boldsymbol{k} \boldsymbol{A}_{1 s} \\
k \boldsymbol{A}_{21} & k \boldsymbol{A}_{22} & \cdots & k \boldsymbol{A}_{2 s} \\
\vdots & \vdots & & \vdots \\
k \boldsymbol{A}_{r 1} & k \boldsymbol{A}_{r 2} & \cdots & k \boldsymbol{A}_{n s}
\end{array}\right]=\left[k \boldsymbol{A}_{i j}\right]_{r \times s}
\end{aligned} k A = k A 11 k A 21 ⋮ k A r 1 k A 12 k A 22 ⋮ k A r 2 ⋯ ⋯ ⋯ k A 1 s k A 2 s ⋮ k A n s = [ k A ij ] r × s 分块矩阵的转置 A T = [ A 11 T A 21 T ⋯ A r 1 T A 12 T A 22 T ⋯ A r 2 T ⋮ ⋮ ⋮ A 1 s T A 2 s T ⋯ A r s T ] \boldsymbol{A}^{\mathrm{T}}=\left[\begin{array}{cccc}
\boldsymbol{A}_{11}^{\mathrm{T}} & \boldsymbol{A}_{21}^{\mathrm{T}} & \cdots & \boldsymbol{A}_{r 1}^{\mathrm{T}} \\
\boldsymbol{A}_{12}^{\mathrm{T}} & \boldsymbol{A}_{22}^{\mathrm{T}} & \cdots & \boldsymbol{A}_{r 2}^{\mathrm{T}} \\
\vdots & \vdots & & \vdots \\
\boldsymbol{A}_{1 s}^{\mathrm{T}} & \boldsymbol{A}_{2 s}^{\mathrm{T}} & \cdots & \boldsymbol{A}_{r s}^{\mathrm{T}}
\end{array}\right] A T = A 11 T A 12 T ⋮ A 1 s T A 21 T A 22 T ⋮ A 2 s T ⋯ ⋯ ⋯ A r 1 T A r 2 T ⋮ A rs T 分块矩阵的乘法 根据分块矩阵运算的原则, 母矩阵 A , B \boldsymbol{A}, \boldsymbol{B} A , B A \boldsymbol{A} A B \boldsymbol{B} B
A = [ A 11 A 12 ⋯ A 1 s A 21 A 22 ⋯ A 2 s ⋮ ⋮ ⋮ A r 1 A r 2 ⋯ A r s ] = [ A i j ] r × s \boldsymbol{A}=\left[\begin{array}{cccc}
\boldsymbol{A}_{11} & \boldsymbol{A}_{12} & \cdots & \boldsymbol{A}_{1 s} \\
\boldsymbol{A}_{21} & \boldsymbol{A}_{22} & \cdots & \boldsymbol{A}_{2 s} \\
\vdots & \vdots & & \vdots \\
\boldsymbol{A}_{r 1} & \boldsymbol{A}_{r 2} & \cdots & \boldsymbol{A}_{r s}
\end{array}\right]=\left[\boldsymbol{A}_{i j}\right]_{r \times s} A = A 11 A 21 ⋮ A r 1 A 12 A 22 ⋮ A r 2 ⋯ ⋯ ⋯ A 1 s A 2 s ⋮ A rs = [ A ij ] r × s 则分块矩阵 B \boldsymbol{B} B
B = [ B 11 B 12 ⋯ B 1 k B 21 B 22 ⋯ B 2 k ⋮ ⋮ ⋮ B s 1 B s 2 ⋯ B s k ] = [ B i j ] s × k \boldsymbol{B}=\left[\begin{array}{cccc}
\boldsymbol{B}_{11} & \boldsymbol{B}_{12} & \cdots & \boldsymbol{B}_{1 k} \\
\boldsymbol{B}_{21} & \boldsymbol{B}_{22} & \cdots & \boldsymbol{B}_{2 k} \\
\vdots & \vdots & & \vdots \\
\boldsymbol{B}_{s 1} & \boldsymbol{B}_{s 2} & \cdots & \boldsymbol{B}_{s k}
\end{array}\right]=\left[\boldsymbol{B}_{i j}\right]_{s \times k} B = B 11 B 21 ⋮ B s 1 B 12 B 22 ⋮ B s 2 ⋯ ⋯ ⋯ B 1 k B 2 k ⋮ B s k = [ B ij ] s × k 其结果是
C i j = ∑ t = 1 s A i t B t j \boldsymbol{C}_{i j}=\sum_{t=1}^s \boldsymbol{A}_{i t} \boldsymbol{B}_{t j} C ij = t = 1 ∑ s A i t B t j 例如假设A
A = [ 5 − 2 − 2 1 3 4 − 3 − 2 0 2 − 2 2 3 1 6 1 4 2 3 − 2 ] = [ A 11 A 12 A 21 A 22 ] \boldsymbol{A}=\left[\begin{array}{cc:ccc}
5 & -2 & -2 & 1 & 3 \\
4 & -3 & -2 & 0 & 2 \\
\hdashline-2 & 2 & 3 & 1 & 6 \\
1 & 4 & 2 & 3 & -2
\end{array}\right]=\left[\begin{array}{ll}
\boldsymbol{A}_{11} & \boldsymbol{A}_{12} \\
\boldsymbol{A}_{21} & \boldsymbol{A}_{22}
\end{array}\right] A = 5 4 − 2 1 − 2 − 3 2 4 − 2 − 2 3 2 1 0 1 3 3 2 6 − 2 = [ A 11 A 21 A 12 A 22 ] 那么 B \boldsymbol{B} B B \boldsymbol{B} B A \boldsymbol{A} A
B = [ 5 − 2 − 2 4 − 3 − 2 − 2 2 3 1 4 2 2 1 4 ] , B = [ 5 − 2 − 2 4 − 3 − 2 − 2 2 3 1 4 2 2 1 4 ] \boldsymbol{B}=\left[\begin{array}{c:cc}
5 & -2 & -2 \\
4 & -3 & -2 \\
\hdashline-2 & 2 & 3 \\
1 & 4 & 2 \\
2 & 1 & 4
\end{array}\right], \quad \boldsymbol{B}=\left[\begin{array}{ccc}
5 & -2 & -2 \\
4 & -3 & -2 \\
\hdashline-2 & 2 & 3 \\
1 & 4 & 2 \\
2 & 1 & 4
\end{array}\right] B = 5 4 − 2 1 2 − 2 − 3 2 4 1 − 2 − 2 3 2 4 , B = 5 4 − 2 1 2 − 2 − 3 2 4 1 − 2 − 2 3 2 4 都是可接受的。
分块矩阵求逆矩阵 分块矩阵求逆没有普遍的规律，但是对于以下情形则比较简单。例如
A B = [ A 1 B 11 A 1 B 12 ⋯ A 1 B 1 s A 2 B 21 A 2 B 22 ⋯ A 2 B 2 s ⋮ ⋮ ⋮ A s B s 1 A s B s 2 ⋯ A s B s s ] = [ E 1 E 2 ⋱ E n ] 由于 A 1 , A 2 , ⋯ , A s 都可逆, 所以有 A − 1 = [ A 1 − 1 A 2 − 1 ⋱ A s − 1 ] \begin{aligned}
&\boldsymbol{A} \boldsymbol{B}=\left[\begin{array}{cccc}
\boldsymbol{A}_1 \boldsymbol{B}_{11} & \boldsymbol{A}_1 \boldsymbol{B}_{12} & \cdots & \boldsymbol{A}_1 \boldsymbol{B}_{1 s} \\
\boldsymbol{A}_2 \boldsymbol{B}_{21} & \boldsymbol{A}_2 \boldsymbol{B}_{22} & \cdots & \boldsymbol{A}_2 \boldsymbol{B}_{2 s} \\
\vdots & \vdots & & \vdots \\
\boldsymbol{A}_s \boldsymbol{B}_{s 1} & \boldsymbol{A}_s \boldsymbol{B}_{s 2} & \cdots & \boldsymbol{A}_s \boldsymbol{B}_{s s}
\end{array}\right]=\left[\begin{array}{llll}
\boldsymbol{E}_1 & & & \\
& \boldsymbol{E}_2 & & \\
& & \ddots & \\
& & & \boldsymbol{E}_n
\end{array}\right]\\
&\text { 由于 } A_1, A_2, \cdots, A_s \text { 都可逆, 所以有 }\\
&\boldsymbol{A}^{-1}=\left[\begin{array}{llll}
\boldsymbol{A}_1^{-1} & & & \\
& \boldsymbol{A}_2^{-1} & & \\
& & \ddots & \\
& & & \boldsymbol{A}_s^{-1}
\end{array}\right]
\end{aligned} A B = A 1 B 11 A 2 B 21 ⋮ A s B s 1 A 1 B 12 A 2 B 22 ⋮ A s B s 2 ⋯ ⋯ ⋯ A 1 B 1 s A 2 B 2 s ⋮ A s B ss = E 1 E 2 ⋱ E n 由于 A 1 , A 2 , ⋯ , A s 都可逆 , 所以有 A − 1 = A 1 − 1 A 2 − 1 ⋱ A s − 1 分块矩阵乘法例题 例设有矩阵
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 ] \boldsymbol{A}=\left[\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
-1 & 2 & 1 & 0 \\
1 & 1 & 0 & 1
\end{array}\right], \quad \boldsymbol{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 = 1 0 − 1 1 0 1 2 1 0 0 1 0 0 0 0 1 , B = 1 − 1 1 − 1 0 2 0 − 1 1 0 4 2 0 1 1 0 计算 A B \boldsymbol{A B} AB
A = [ 1 0 0 0 0 1 0 0 − 1 2 1 0 1 1 0 1 ] = [ E O A 1 E ] A 1 = [ − 1 2 1 1 ] A=\left[\begin{array}{cc:cc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
\hdashline-1 & 2 & 1 & 0 \\
1 & 1 & 0 & 1
\end{array}\right]=\left[\begin{array}{ll}
E & O \\
A_1 & E
\end{array}\right] \quad A_1=\left[\begin{array}{cc}
-1 & 2 \\
1 & 1
\end{array}\right] A = 1 0 − 1 1 0 1 2 1 0 0 1 0 0 0 0 1 = [ E A 1 O E ] A 1 = [ − 1 1 2 1 ] B = [ 1 0 1 0 − 1 2 0 1 1 0 4 1 − 1 − 1 2 0 ] = [ B 11 E B 21 B 22 ] \boldsymbol{B}=\left[\begin{array}{cc:cc}
1 & 0 & 1 & 0 \\
-1 & 2 & 0 & 1 \\
\hdashline 1 & 0 & 4 & 1 \\
-1 & -1 & 2 & 0
\end{array}\right]=\left[\begin{array}{cc}
B_{11} & \boldsymbol{E} \\
\boldsymbol{B}_{21} & \boldsymbol{B}_{22}
\end{array}\right] B = 1 − 1 1 − 1 0 2 0 − 1 1 0 4 2 0 1 1 0 = [ B 11 B 21 E B 22 ] B 11 = [ 1 0 − 1 2 ] , B 21 = [ 1 0 − 1 − 1 ] , B 22 = [ 4 1 2 0 ] 根据分块矩阵乘法的运算规则, 有 A B = [ E O A 1 E ] [ B 11 E B 21 B 22 ] = [ E B 11 + O B 21 E × E + O B 22 A 1 B 11 + B 21 A 1 E + I B 21 ] = [ B 11 E A 1 B 11 + B 21 A 1 + B 21 ] \begin{aligned}
&\boldsymbol{B}_{11}=\left[\begin{array}{cc}
1 & 0 \\
-1 & 2
\end{array}\right], \quad \boldsymbol{B}_{21}=\left[\begin{array}{cc}
1 & 0 \\
-1 & -1
\end{array}\right], \quad \boldsymbol{B}_{22}=\left[\begin{array}{ll}
4 & 1 \\
2 & 0
\end{array}\right]\\
&\text { 根据分块矩阵乘法的运算规则, 有 }\\
&\begin{aligned}
& \boldsymbol{A} \boldsymbol{B}=\left[\begin{array}{ll}
\boldsymbol{E} & \boldsymbol{O} \\
\boldsymbol{A}_1 & \boldsymbol{E}
\end{array}\right]\left[\begin{array}{cc}
\boldsymbol{B}_{11} & \boldsymbol{E} \\
\boldsymbol{B}_{21} & \boldsymbol{B}_{22}
\end{array}\right]=\left[\begin{array}{cc}
\boldsymbol{E} \boldsymbol{B}_{11}+\boldsymbol{O} \boldsymbol{B}_{21} & \boldsymbol{E} \times \boldsymbol{E}+\boldsymbol{O} \boldsymbol{B}_{22} \\
\boldsymbol{A}_1 \boldsymbol{B}_{11}+\boldsymbol{B}_{21} & \boldsymbol{A}_1 \boldsymbol{E}+\boldsymbol{I} \boldsymbol{B}_{21}
\end{array}\right] \\
& =\left[\begin{array}{cc}
\boldsymbol{B}_{11} & \boldsymbol{E} \\
\boldsymbol{A}_1 \boldsymbol{B}_{11}+\boldsymbol{B}_{21} & \boldsymbol{A}_1+\boldsymbol{B}_{21}
\end{array}\right]
\end{aligned}
\end{aligned} B 11 = [ 1 − 1 0 2 ] , B 21 = [ 1 − 1 0 − 1 ] , B 22 = [ 4 2 1 0 ] 根据分块矩阵乘法的运算规则 , 有 A B = [ E A 1 O E ] [ B 11 B 21 E B 22 ] = [ E B 11 + O B 21 A 1 B 11 + B 21 E × E + O B 22 A 1 E + I B 21 ] = [ B 11 A 1 B 11 + B 21 E A 1 + B 21 ] 从上式可知, 只需要做两个二阶方阵的乘法。因此根据上式计算 A B \boldsymbol{A B} AB A B \boldsymbol{A B} AB
注意: 用分块矩阵简化矩阵运算, 只有在矩阵乘法中才有可能达到, 并且需要矩阵经过适当分块后有较多的 0 块、单位矩阵子块。
分块矩阵的作用 对于线性方程组，使用分块
A m × n X n × 1 = b m × 1 \boldsymbol{A}_{m \times n} \boldsymbol{X}_{n \times 1}=\boldsymbol{b}_{m \times 1} A m × n X n × 1 = b m × 1 其中 A \boldsymbol{A} A X \boldsymbol{X} X b \boldsymbol{b} b
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 1 A 2 ⋯ A n ] A=\left[\begin{array}{c:c:c:c}
a_{11} & a_{12} & \cdots & a_{1 n} \\
a_{21} & a_{22} & \cdots & a_{2 n} \\
\vdots & \vdots & & \vdots \\
a_{m 1} & a_{m 2} & \cdots & a_{m n}
\end{array}\right]=\left[\begin{array}{llll}
A_1 & A_2 & \cdots & A_n
\end{array}\right] A = a 11 a 21 ⋮ a m 1 a 12 a 22 ⋮ a m 2 ⋯ ⋯ ⋯ a 1 n a 2 n ⋮ a mn = [ A 1 A 2 ⋯ A n ] 则 A X = b \boldsymbol{A} \boldsymbol{X}=\boldsymbol{b} A X = b
x 1 A 1 + x 2 A 2 + ⋯ + x n A n = b x_1 \boldsymbol{A}_1+x_2 \boldsymbol{A}_2+\cdots+x_n \boldsymbol{A}_n=\boldsymbol{b} x 1 A 1 + x 2 A 2 + ⋯ + x n A n = b 也就是
x 1 [ a 11 a 21 ⋮ a m 1 ] + x 2 [ a 12 a 22 ⋮ a m 2 ] + ⋯ + x n [ a 1 n a 2 n ⋮ a m n ] = [ b 1 b 2 ⋮ b m ] x_1\left[\begin{array}{c}
a_{11} \\
a_{21} \\
\vdots \\
a_{m 1}
\end{array}\right]+x_2\left[\begin{array}{c}
a_{12} \\
a_{22} \\
\vdots \\
a_{m 2}
\end{array}\right]+\cdots+x_n\left[\begin{array}{c}
a_{1 n} \\
a_{2 n} \\
\vdots \\
a_{m n}
\end{array}\right]=\left[\begin{array}{c}
b_1 \\
b_2 \\
\vdots \\
b_m
\end{array}\right] x 1 a 11 a 21 ⋮ a m 1 + x 2 a 12 a 22 ⋮ a m 2 + ⋯ + x n a 1 n a 2 n ⋮ a mn = b 1 b 2 ⋮ b m 该式称为线性方程组的向量表示。这在后面向量空间里经常使用。