第三章，矩阵，04-分块矩阵
第三章，矩阵，04-分块矩阵
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- 定义
- 运算
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- 加法
- 数乘
- 转置
- 乘法
- 分块对角阵
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- 分块对角阵乘法
- 分块对角阵行列式
- 按行列分块
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- 行向量
- 列向量
- 行列向量表示的矩阵乘法
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- 例
- 线性方程组分块表示
玩转线性代数(16)分块矩阵的笔记
定义
用一些纵、横线将矩阵A分割成若干小矩阵，以这些小矩阵为元素的矩阵称为分块矩阵，各个小矩阵称为A的子块
运算
对分块矩阵进行运算时，可以把每一个子块当作矩阵的一个元素来处理，但应保证矩阵运算的可行性。
加法
设矩阵A、B是两个同型矩阵，且分块方法一致，才能相加
A = ( A 11 A 12 ⋯ A 1 r A 21 A 22 ⋯ A 2 r ⋮ ⋮ ⋮ ⋮ A s 1 A s 2 ⋯ A s r ) , B = ( B 11 B 12 ⋯ B 1 r B 21 B 22 ⋯ B 2 r ⋮ ⋮ ⋮ ⋮ B s 1 B s 2 ⋯ B s r ) A=\begin{pmatrix} A_{11} & A_{12} & \cdots & A_{1r}\\ A_{21} & A_{22} & \cdots & A_{2r}\\ \vdots & \vdots & \vdots & \vdots \\ A_{s1} & A_{s2} & \cdots & A_{sr}\\ \end{pmatrix}, B=\begin{pmatrix} B_{11} & B_{12} & \cdots & B_{1r}\\ B_{21} & B_{22} & \cdots & B_{2r}\\ \vdots & \vdots & \vdots & \vdots \\ B_{s1} & B_{s2} & \cdots & B_{sr}\\ \end{pmatrix} A=⎝⎜⎜⎜⎛A11A21⋮As1A12A22⋮As2⋯⋯⋮⋯A1rA2r⋮Asr⎠⎟⎟⎟⎞,B=⎝⎜⎜⎜⎛B11B21⋮Bs1B12B22⋮Bs2⋯⋯⋮⋯B1rB2r⋮Bsr⎠⎟⎟⎟⎞
其中每一 A i j A_{ij} Aij与 B i j B_{ij} Bij的维数都对应相同，则规定加法为
A + B = ( A 11 + B 11 A 12 + B 12 ⋯ A 1 r + B 1 r A 21 + B 21 A 22 + B 22 ⋯ A 2 r + B 2 r ⋮ ⋮ ⋮ ⋮ A s 1 + B s 1 A s 2 + B s 2 ⋯ A s r + B s r ) , A+B=\begin{pmatrix} A_{11}+B_{11} & A_{12}+B_{12} & \cdots & A_{1r}+B_{1r}\\ A_{21}+B_{21} & A_{22}+B_{22} & \cdots & A_{2r}+B_{2r}\\ \vdots & \vdots & \vdots & \vdots \\ A_{s1}+B_{s1} & A_{s2}+B_{s2} & \cdots & A_{sr}+B_{sr}\\ \end{pmatrix}, A+B=⎝⎜⎜⎜⎛A11+B11A21+B21⋮As1+Bs1A12+B12A22+B22⋮As2+Bs2⋯⋯⋮⋯A1r+B1rA2r+B2r⋮Asr+Bsr⎠⎟⎟⎟⎞,
数乘
乘以常数 λ \lambda λ
λ A = ( λ A 11 λ A 12 ⋯ λ A 1 r λ A 21 λ A 22 ⋯ λ A 2 r ⋮ ⋮ ⋮ ⋮ λ A s 1 λ A s 2 ⋯ λ A s r ) \lambda A=\begin{pmatrix} \lambda A_{11} & \lambda A_{12} & \cdots & \lambda A_{1r}\\ \lambda A_{21} & \lambda A_{22} & \cdots & \lambda A_{2r}\\ \vdots & \vdots & \vdots & \vdots \\ \lambda A_{s1} & \lambda A_{s2} & \cdots & \lambda A_{sr}\\ \end{pmatrix} λA=⎝⎜⎜⎜⎛λA11λA21⋮λAs1λA12λA22⋮λAs2⋯⋯⋮⋯λA1rλA2r⋮λAsr⎠⎟⎟⎟⎞
转置
子块行列互换的基础上还要对子块本身进行转置。
A T = ( A 11 T A 21 T ⋯ A s 1 T A 12 T A 22 ⋯ A s 2 T ⋮ ⋮ ⋮ ⋮ A 1 r T A 2 r T ⋯ A s r T ) A^T=\begin{pmatrix} A_{11}^T & A_{21}^T & \cdots & A_{s1}^T\\ A_{12}^T & A_{22} & \cdots & A_{s2}^T\\ \vdots & \vdots & \vdots & \vdots \\ A_{1r}^T & A_{2r}^T & \cdots & A_{sr}^T\\ \end{pmatrix} AT=⎝⎜⎜⎜⎛A11TA12T⋮A1rTA21TA22⋮A2rT⋯⋯⋮⋯As1TAs2T⋮AsrT⎠⎟⎟⎟⎞
乘法
设A是m×n矩阵，B是n×p矩阵，若将A分为r×s个子块 ( A k j ) r × s (A_{kj})_{r×s} (Akj)r×s，将B分为s×t个子块 ( B k j ) s × t (B_{kj})_{s×t} (Bkj)s×t，且A的列与B的行分块法一致，则规定A与B的乘法为：
( 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 t B 21 B 22 ⋯ B 2 t ⋮ ⋮ ⋮ ⋮ B s 1 B s 2 ⋯ B s t ) = ( C 11 C 12 ⋯ C 1 t C 21 C 22 ⋯ C 2 t ⋮ ⋮ ⋮ ⋮ C r 1 C r 2 ⋯ C r t ) \begin{pmatrix} A_{11} & A_{12} & \cdots & A_{1s}\\ A_{21} & A_{22} & \cdots & A_{2s}\\ \vdots & \vdots & \vdots & \vdots \\ A_{r1} & A_{r2} & \cdots & A_{rs}\\ \end{pmatrix} \begin{pmatrix} B_{11} & B_{12} & \cdots & B_{1t}\\ B_{21} & B_{22} & \cdots & B_{2t}\\ \vdots & \vdots & \vdots & \vdots \\ B_{s1} & B_{s2} & \cdots & B_{st}\\ \end{pmatrix}=\begin{pmatrix} C_{11} & C_{12} & \cdots & C_{1t}\\ C_{21} & C_{22} & \cdots & C_{2t}\\ \vdots & \vdots & \vdots & \vdots \\ C_{r1} & C_{r2} & \cdots & C_{rt}\\ \end{pmatrix} ⎝⎜⎜⎜⎛A11A21⋮Ar1A12A22⋮Ar2⋯⋯⋮⋯A1sA2s⋮Ars⎠⎟⎟⎟⎞⎝⎜⎜⎜⎛B11B21⋮Bs1B12B22⋮Bs2⋯⋯⋮⋯B1tB2t⋮Bst⎠⎟⎟⎟⎞=⎝⎜⎜⎜⎛C11C21⋮Cr1C12C22⋮Cr2⋯⋯⋮⋯C1tC2t⋮Crt⎠⎟⎟⎟⎞
其中 C i j = ∑ i = 1 s A i k B k j ， ( i = 1 , 2 , . . . , r ; j = 1 , 2 , . . . , t ) C_{ij}=\sum_{i=1}^s A_{ik}B_{kj}，(i=1,2,...,r;j=1,2,...,t) Cij=∑i=1sAikBkj，(i=1,2,...,r;j=1,2,...,t)
满足三点：
- 外能乘：在分块之前满足矩阵的乘法;
- 块能乘：即前矩阵分块后的行块数与后矩阵分块后列块数一致;
- 内能乘：每两个小块的乘法要满足矩阵的乘法。
分块对角阵
设A是r阶矩阵，若A的一个分块矩阵只有在主对角线上有非零子块，即 A = ( A 1 0 ⋯ ⋯ 0 0 A 2 0 ⋯ 0 0 0 ⋱ 0 0 0 ⋯ 0 A s − 1 0 0 0 ⋯ 0 A s ) A=\begin{pmatrix} A_{1} & 0 & \cdots & \cdots & 0\\ 0 & A_{2} & 0 & \cdots & 0\\ 0 & 0 & \ddots & 0 & 0\\ 0 & \cdots & 0 & A_{s-1} & 0\\ 0 & 0 & \cdots & 0 & A_{s}\\ \end{pmatrix} A=⎝⎜⎜⎜⎜⎛A100000A20⋯0⋯0⋱0⋯⋯⋯0As−100000As⎠⎟⎟⎟⎟⎞，
其中 A i A_i Ai是 r i r_i ri阶小方阵，i=1,2,…,s,
∑ i = 1 s r i = n \sum_{i=1}^sr_i=n ∑i=1sri=n，而其余非主对角子块都为零矩阵，那么称A为分块对角矩阵。
分块对角阵乘法
设有两个分块对角阵：
A = ( A 1 0 A 2 ⋱ 0 A k ) , B = ( B 1 0 B 2 ⋱ 0 B k ) A=\begin{pmatrix} A_1 & & & 0\\ & A_2& & \\ & &\ddots & \\ 0 & & & A_k\\ \end{pmatrix},B=\begin{pmatrix} B_1 & & & 0\\ & B_2& & \\ & &\ddots & \\ 0 & & & B_k\\ \end{pmatrix} A=⎝⎜⎜⎛A10A2⋱0Ak⎠⎟⎟⎞,B=⎝⎜⎜⎛B10B2⋱0Bk⎠⎟⎟⎞,
其中矩阵 A i A_i Ai与 B i B_i Bi都是 n i n_i ni阶方阵，因此 A i A_i Ai与 B i B_i Bi可以相乘，用分块矩阵乘法可得：
A B = ( A 1 B 1 0 A 2 B 2 ⋱ 0 A k B k ) AB=\begin{pmatrix} A_1B_1 & & & 0\\ & A_2B_2& & \\ & &\ddots & \\ 0 & & & A_kB_k\\ \end{pmatrix} AB=⎝⎜⎜⎛A1B10A2B2⋱0AkBk⎠⎟⎟⎞
即对应主对角线子块相乘即可。
分块对角阵行列式
根据拉普拉斯公式和递推法：
∣ A ∣ = ∣ A 1 0 A 2 ⋱ 0 A k ∣ = ∣ A 1 ∣ ∣ A 2 ∣ . . . ∣ A k ∣ |A|=\begin{vmatrix} A_1 & & & 0\\ & A_2& & \\ & &\ddots & \\ 0 & & & A_k\\ \end{vmatrix}=|A_1||A_2|...|A_k| ∣A∣=∣∣∣∣∣∣∣∣A10A2⋱0Ak∣∣∣∣∣∣∣∣=∣A1∣∣A2∣...∣Ak∣
按行列分块
矩阵 A = ( a i j ) m × n A=(a_{ij})_{m×n} A=(aij)m×n
行向量
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=\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \vdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn}\\ \end{pmatrix}=(a_1,a_2,...,a_n) A=⎝⎜⎜⎜⎛a11a21⋮am1a12a22⋮am2⋯⋯⋮⋯a1na2n⋮amn⎠⎟⎟⎟⎞=(a1,a2,...,an)
列向量
A = ( α 1 T α 2 T ⋮ α m T ) A=\begin{pmatrix} \alpha_1^T \\ \alpha_2^T \\ \vdots\\ \alpha_m^T \\ \end{pmatrix} A=⎝⎜⎜⎜⎛α1Tα2T⋮αmT⎠⎟⎟⎟⎞
其中:
α i = ( a i 1 a i 2 ⋮ a i n ) , α i T = ( a i 1 a i 2 ⋯ a i n ) \alpha_i=\begin{pmatrix} a_{i1} \\ a_{i2} \\ \vdots\\ a_{in} \\ \end{pmatrix},\alpha_i^T=\begin{pmatrix}a_{i1} & a_{i2}&\cdots & a_{in}\end{pmatrix} αi=⎝⎜⎜⎜⎛ai1ai2⋮ain⎠⎟⎟⎟⎞,αiT=(ai1ai2⋯ain)
行列向量表示的矩阵乘法
A = ( a i j ) m × s , B = ( b i j ) s × n A=(a_{ij})_{m×s},B=(b_{ij})_{s×n} A=(aij)m×s,B=(bij)s×n，则A与B的乘积 C = ( c i j ) m × n C=(c_{ij})_{m×n} C=(cij)m×n,把A按行、B按列分块后有：
A B = ( α 1 T α 2 T ⋮ α m T ) ( b 1 b 2 ⋯ b n ) = ( α 1 T b 1 α 1 T b 2 ⋯ α 1 T b n α 2 T b 1 α 2 T b 2 ⋯ α 2 T b n ⋮ ⋮ ⋮ ⋮ α m T b 1 α m T b 2 ⋯ α m T b n ) AB=\begin{pmatrix} \alpha_1^T \\ \alpha_2^T \\ \vdots\\ \alpha_m^T \end{pmatrix} \begin{pmatrix}b_{1} & b_{2}&\cdots & b_{n}\end{pmatrix} =\begin{pmatrix} \alpha_1^T b_1 & \alpha_1^T b_2 & \cdots & \alpha_1^T b_n\\ \alpha_2^T b_1 & \alpha_2^T b_2 & \cdots & \alpha_2^T b_n\\ \vdots & \vdots & \vdots & \vdots \\ \alpha_m^T b_1 & \alpha_m^T b_2 & \cdots & \alpha_m^T b_n \end{pmatrix} AB=⎝⎜⎜⎜⎛α1Tα2T⋮αmT⎠⎟⎟⎟⎞(b1b2⋯bn)=⎝⎜⎜⎜⎛α1Tb1α2Tb1⋮αmTb1α1Tb2α2Tb2⋮αmTb2⋯⋯⋮⋯α1Tbnα2Tbn⋮αmTbn⎠⎟⎟⎟⎞
例
设 A T A = 0 A^TA=0 ATA=0，证明A=0.
证明：设 A = ( a i j ) m × n A=(a_{ij})_{m×n} A=(aij)m×n,把A按列分块有 A = ( a 1 , a 2 , . . . , a n ) A=(a_1,a_2,...,a_n) A=(a1,a2,...,an)，则
A T A = ( a 1 T a 2 T ⋮ a m T ) ( a 1 a 2 ⋯ a n ) = ( a 1 T a 1 a 1 T a 2 ⋯ a 1 T a n a 2 T a 1 a 2 T a 2 ⋯ a 2 T a n ⋮ ⋮ ⋮ ⋮ a m T a 1 a m T a 2 ⋯ a n T a n ) A^TA=\begin{pmatrix} a_1^T \\ a_2^T \\ \vdots\\ a_m^T \end{pmatrix} \begin{pmatrix}a_{1} & a_{2}&\cdots & a_{n}\end{pmatrix} =\begin{pmatrix} a_1^T a_1 & a_1^T a_2 & \cdots & a_1^T a_n\\ a_2^T a_1 & a_2^T a_2 & \cdots & a_2^T a_n\\ \vdots & \vdots & \vdots & \vdots \\ a_m^T a_1 & a_m^T a_2 & \cdots & a_n^T a_n \end{pmatrix} ATA=⎝⎜⎜⎜⎛a1Ta2T⋮amT⎠⎟⎟⎟⎞(a1a2⋯an)=⎝⎜⎜⎜⎛a1Ta1a2Ta1⋮amTa1a1Ta2a2Ta2⋮amTa2⋯⋯⋮⋯a1Tana2Tan⋮anTan⎠⎟⎟⎟⎞
即 A T A A^TA ATA的(i,j)元为 a i T a j a_i^Ta_j aiTaj，因 A T A = 0 A^TA=0 ATA=0,故 a i T a j = 0 ( i , j = 1 , 2 , . . . , n ) a_i^Ta_j=0(i,j=1,2,...,n) aiTaj=0(i,j=1,2,...,n),特殊地，有 a i T a i = 0 ( i = 1 , 2 , . . . , n ) a_i^Ta_i=0(i=1,2,...,n) aiTai=0(i=1,2,...,n)
而
a j T a j = 0 = ( a 1 j , a 2 j , . . . , a m j ) ( a 1 j a 2 j ⋮ a m j ) = a 1 j 2 + a 2 j 2 + . . . + a m j 2 a_j^Ta_j=0=(a_{1j},a_{2j},...,a_{mj}) \begin{pmatrix} a_{1j} \\ a_{2j} \\ \vdots \\ a_{mj} \end{pmatrix}=a_{1j}^2+a_{2j}^2+...+a_{mj}^2 ajTaj=0=(a1j,a2j,...,amj)⎝⎜⎜⎜⎛a1ja2j⋮amj⎠⎟⎟⎟⎞=a1j2+a2j2+...+amj2
由 a 1 j 2 + a 2 j 2 + . . . + a m j 2 = 0 a_{1j}^2+a_{2j}^2+...+a_{mj}^2=0 a1j2+a2j2+...+amj2=0,得
a 1 j = a 2 j = . . . = a m j = 0 , ( j = 1 , 2 , . . . , n ) a_{1j}=a_{2j}=...=a_{mj}=0,(j=1,2,...,n) a1j=a2j=...=amj=0,(j=1,2,...,n),
即A=0
线性方程组分块表示
将线性方程组
{ a 11 x 1 + a 12 x 2 + a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + a 2 n x n = b 2 ⋯ a m 1 x 1 + a m 2 x 2 + a m n x n = b m \begin{cases} a_{11}x_1+ a_{12}x_2 + a_{1n}x_n =b_1\\ a_{21}x_1+ a_{22}x_2 + a_{2n}x_n =b_2\\ \cdots\\ a_{m1}x_1+ a_{m2}x_2 + a_{mn}x_n =b_m\\ \end{cases} ⎩⎪⎪⎪⎨⎪⎪⎪⎧a11x1+a12x2+a1nxn=b1a21x1+a22x2+a2nxn=b2⋯am1x1+am2x2+amnxn=bm
简记为 A X = b AX=b AX=b
其中 A = ( a 11 ⋯ a 1 n ⋮ ⋱ ⋮ a m 1 ⋯ a m n ) , X = ( x 1 ⋮ x n ) , b = ( b 1 ⋮ b m ) A=\begin{pmatrix} a_{11} & \cdots & a_{1n}\\ \vdots & \ddots & \vdots\\ a_{m1} & \cdots & a_{mn} \end{pmatrix}, X=\begin{pmatrix}x_1\\ \vdots \\ x_n \end{pmatrix}, b=\begin{pmatrix}b_1\\ \vdots \\ b_m \end{pmatrix} A=⎝⎜⎛a11⋮am1⋯⋱⋯a1n⋮amn⎠⎟⎞,X=⎝⎜⎛x1⋮xn⎠⎟⎞,b=⎝⎜⎛b1⋮bm⎠⎟⎞
列分块
( a 1 , a 2 , . . . , a n ) ( x 1 x 2 ⋮ x n ) = b (a_1,a_2,...,a_n)\begin{pmatrix}x_1\\ x_2 \\ \vdots \\ x_n \end{pmatrix}=b (a1,a2,...,an)⎝⎜⎜⎜⎛x1x2⋮xn⎠⎟⎟⎟⎞=b
即 x 1 a 1 + x 2 a 2 + . . . + x n a n = b x_1a_1+x_2a_2+...+x_na_n=b x1a1+x2a2+...+xnan=b
行分块
{ a 1 T x = b 1 a 2 T x = b 2 ⋯ a m T x = b m \begin{cases} a_1^Tx =b_1\\ a_2^Tx =b_2\\ \cdots\\ a_m^Tx =b_m\\ \end{cases} ⎩⎪⎪⎪⎨⎪⎪⎪⎧a1Tx=b1a2Tx=b2⋯amTx=bm