对分块矩阵 M = [ A B C D ] (1) M=\left[\begin{matrix} A & B \\ C & D \end{matrix}\right]\tag{1} M=[ACBD](1)
有如下schur补和逆矩阵对比表：
| 可逆矩阵块 | Schur补 | 逆矩阵 |
|---|---|---|
| A | M / A = D − C A − 1 B M/A=D-CA^{-1}B M/A=D−CA−1B | [ A − 1 + A − 1 B ( M / A ) − 1 C A − 1 − A − 1 B ( M / A ) − 1 − ( M / A ) − 1 C A − 1 ( M / A ) − 1 ] \left[\begin{matrix}A^{-1}+A^{-1}B(M/A)^{-1}CA^{-1} & -A^{-1}B(M/A)^{-1}\\-(M/A)^{-1}CA^{-1} & (M/A)_{-1}\end{matrix}\right] [A−1+A−1B(M/A)−1CA−1−(M/A)−1CA−1−A−1B(M/A)−1(M/A)−1] |
| D | M / D = A − B D − 1 C M/D=A-BD^{-1}C M/D=A−BD−1C | [ ( M / D ) − 1 − ( M / D ) − 1 B D − 1 − D − 1 C ( M / D ) − 1 D − 1 + D − 1 C ( M / D ) − 1 B D − 1 ] \left[\begin{matrix} (M/D)^{-1} & -(M/D)^{-1}BD^{-1} \\ -D^{-1}C(M/D)^{-1} & D^{-1}+D^{-1}C(M/D)^{-1}BD^{-1}\end{matrix}\right] [(M/D)−1−D−1C(M/D)−1−(M/D)−1BD−1D−1+D−1C(M/D)−1BD−1] |
(注：shcur implementation和逆矩阵的快的位置可以按照顺时针记忆）
情况一：矩阵块D的Schur Complement
定义： 当矩阵块D可逆时，存在D的Schur Complement
S D = M / D = A − B D − 1 C (2) S_{D} = M/D = A - BD^{-1}C\tag{2} SD=M/D=A−BD−1C(2) 注：可以按照顺时针进行记忆
思想： 众所周知，如果能够将块矩阵进行如下变化构造有利于求逆
[ I X 1 0 I ] [ A B C D ] [ I 0 X 2 I ] = [ Y 1 0 0 Y 2 ] (3) \left[\begin{matrix} I & X_{1} \\ 0 & I \end{matrix}\right] \left[\begin{matrix} A & B \\ C & D \end{matrix}\right] \left[\begin{matrix} I & 0 \\ X_{2} & I \end{matrix}\right] =\left[\begin{matrix} Y_{1} & 0 \\ 0 & Y_{2} \end{matrix}\right]\tag{3} [I0X1I][ACBD][IX20I]=[Y100Y2](3)
所以根据如上思想进行如下构造;
[ I − B D − 1 0 I ] [ A B C D ] = [ A − B D − 1 C 0 C D ] (4) \left[\begin{matrix} I & {-BD^{-1}} \\ 0 & I \end{matrix}\right] \left[\begin{matrix} A & B \\ C & D \end{matrix}\right] =\left[\begin{matrix} A-BD^{-1}C & 0 \\ C & D \end{matrix}\right]\tag{4} [I0−BD−1I][ACBD]=[A−BD−1CC0D](4)
[ A − B D − 1 C 0 C D ] [ I 0 − D − 1 C I ] = [ A − B D − 1 C 0 0 D ] (5) \left[\begin{matrix} {A-BD^{-1}C} & 0 \\ C & D \end{matrix}\right] \left[\begin{matrix} {I} & {0} \\ {-D^{-1}C} & {I} \end{matrix}\right]= \left[\begin{matrix} {A-BD^{-1}C} & {0} \\ {0} & {D} \end{matrix}\right]\tag{5} [A−BD−1CC0D][I−D−1C0I]=[A−BD−1C00D](5)
则
M = [ A B C D ] = [ I − B D − 1 0 I ] − 1 [ A − B D − 1 C 0 0 D ] [ I 0 − D − 1 C I ] − 1 = [ I B D − 1 0 I ] [ A − B D − 1 C 0 0 D ] [ I 0 D − 1 C I ] ( 注： [ I X 0 I ] − 1 = [ I − X 0 I ] , [ I 0 X I ] − 1 = [ I 0 − X I ] − 1 ) M=\left[\begin{matrix} {A} & {B} \\ {C} & {D} \end{matrix}\right]\\= \left[\begin{matrix} I & {-BD^{-1}} \\ 0 & I \end{matrix}\right]^{-1} \left[\begin{matrix} {A-BD^{-1}C} & {0} \\ {0} & {D} \end{matrix}\right] \left[\begin{matrix} {I} & {0} \\ {-D^{-1}C} & {I} \end{matrix}\right]^{-1} \\\\= \left[\begin{matrix} I & {BD^{-1}} \\ 0 & I \end{matrix}\right] \left[\begin{matrix} {A-BD^{-1}C} & {0} \\ {0} & {D} \end{matrix}\right] \left[\begin{matrix} {I} & {0} \\ {D^{-1}C} & {I} \end{matrix}\right] \\ (注：\left[\begin{matrix} {I} & {X} \\ {0} & {I}\end{matrix}\right]^{-1}=\left[\begin{matrix} {I} & {-X} \\ {0} & {I}\end{matrix}\right], \left[\begin{matrix} {I} & {0} \\ {X} & {I}\end{matrix}\right]^{-1}=\left[\begin{matrix} {I} & {0} \\ {-X} & {I}\end{matrix}\right]^{-1}) M=[ACBD]=[I0−BD−1I]−1[A−BD−1C00D][I−D−1C0I]−1=[I0BD−1I][A−BD−1C00D][ID−1C0I](注：[I0XI]−1=[I0−XI],[IX0I]−1=[I−X0I]−1)
则
M − 1 = [ A B C D ] − 1 = ( [ I B D − 1 0 I ] [ A − B D − 1 C 0 0 D ] [ I 0 D − 1 C I ] ) − 1 = [ I 0 − D − 1 C I ] [ ( A − B D − 1 C ) − 1 0 0 D − 1 ] [ I − B D − 1 0 I ] = [ ( A − B D − 1 C ) − 1 − ( A − B D − 1 C ) − 1 B D − 1 − D − 1 C ( A − B D − 1 C ) − 1 D − 1 + D − 1 C ( A − B D − 1 C ) − 1 B D − 1 ] ( 注： [ A 0 0 D ] − 1 = [ A − 1 0 0 D − 1 ] ) M^{-1}=\left[\begin{matrix} {A} & {B} \\ {C} & {D} \end{matrix}\right]^{-1}\\= \left(\left[\begin{matrix} I & {BD^{-1}} \\ 0 & I \end{matrix}\right] \left[\begin{matrix} {A-BD^{-1}C} & {0} \\ {0} & {D} \end{matrix}\right] \left[\begin{matrix} {I} & {0} \\ {D^{-1}C} & {I} \end{matrix}\right]\right)^{-1} \\= \left[\begin{matrix} {I} & {0}\\ {-D^{-1}C} & {I} \end{matrix}\right] \left[\begin{matrix} {(A-BD^{-1}C)^{-1}} & 0 \\ {0} & {D^{-1}} \end{matrix}\right] \left[\begin{matrix} {I} & {-BD^{-1}} \\ {0} & {I} \end{matrix}\right]\\= \left[\begin{matrix} (A-BD^{-1}C)^{-1} & -(A-BD^{-1}C)^{-1}BD^{-1} \\ -D^{-1}C(A-BD^{-1}C)^{-1} & D^{-1}+D^{-1}C(A-BD^{-1}C)^{-1}BD^{-1} \end{matrix}\right]\\ \\(注：\left[\begin{matrix} A & 0 \\ 0 & D \end{matrix}\right]^{-1}=\left[\begin{matrix} A^{-1} & 0 \\ 0 & D^{-1} \end{matrix}\right]) M−1=[ACBD]−1=([I0BD−1I][A−BD−1C00D][ID−1C0I])−1=[I−D−1C0I][(A−BD−1C)−100D−1][I0−BD−1I]=[(A−BD−1C)−1−D−1C(A−BD−1C)−1−(A−BD−1C)−1BD−1D−1+D−1C(A−BD−1C)−1BD−1](注：[A00D]−1=[A−100D−1])
当我们将矩阵块D的shcur complement引入后可以得到
M − 1 = [ A B C D ] − 1 = [ ( M / D ) − 1 − ( M / D ) − 1 B D − 1 − D − 1 C ( M / D ) − 1 D − 1 + D − 1 C ( M / D ) − 1 B D − 1 ] M^{-1}=\left[\begin{matrix}A & B \\ C & D \end{matrix}\right]^{-1}=\left[\begin{matrix} (M/D)^{-1} & -(M/D)^{-1}BD^{-1} \\ -D^{-1}C(M/D)^{-1} & D^{-1}+D^{-1}C(M/D)^{-1}BD^{-1} \end{matrix}\right] M−1=[ACBD]−1=[(M/D)−1−D−1C(M/D)−1−(M/D)−1BD−1D−1+D−1C(M/D)−1BD−1]
情况二：矩阵块A的Schur Complement
定义： 当矩阵块A可逆时，存在D的Schur Complement
S A = M / A = D − C A − 1 B S_{A} = M/A = D - CA^{-1}B SA=M/A=D−CA−1B
则有
M = [ A B C D ] = [ I 0 C A − 1 I ] [ A 0 0 D − C A − 1 B ] [ I A − 1 B 0 I ] M=\left[\begin{matrix} {A} & {B} \\ {C} & {D} \end{matrix}\right]\\= \left[\begin{matrix} I & 0 \\ CA^{-1} & I \end{matrix}\right] \left[\begin{matrix} {A} & {0} \\ {0} & {D-CA^{-1}B} \end{matrix}\right] \left[\begin{matrix} {I} & {A^{-1}B} \\ {0} & {I} \end{matrix}\right] M=[ACBD]=[ICA−10I][A00D−CA−1B][I0A−1BI]
求逆可得：
M − 1 = [ A B C D ] − 1 = [ A − 1 + A − 1 B ( D − C A − 1 B ) − 1 C A − 1 − A − 1 B ( D − C A − 1 B ) − 1 ( D − C A − 1 B ) − 1 C A − 1 ( D − C A − 1 B ) − 1 ] = [ A − 1 + A − 1 B ( M / A ) − 1 C A − 1 − A − 1 B ( M / A ) − 1 − ( M / A ) − 1 C A − 1 ( M / A ) − 1 ] M^{-1}=\left[\begin{matrix}A & B \\ C & D \end{matrix}\right]^{-1} \\=\left[\begin{matrix} A^{-1} + A^{-1}B(D- CA^{-1}B)^{-1}CA^{-1} & -A^{-1}B(D- CA^{-1}B)^{-1} \\ (D- CA^{-1}B)^{-1}CA^{-1} &(D- CA^{-1}B)^{-1} \end{matrix}\right] \\=\left[\begin{matrix} A^{-1}+A^{-1}B(M/A)^{-1}CA^{-1} & -A^{-1}B(M/A)^{-1} \\ -(M/A)^{-1}CA^{-1} & (M/A)^{-1} \end{matrix}\right] M−1=[ACBD]−1=[A−1+A−1B(D−CA−1B)−1CA−1(D−CA−1B)−1CA−1−A−1B(D−CA−1B)−1(D−CA−1B)−1]=[A−1+A−1B(M/A)−1CA−1−(M/A)−1CA−1−A−1B(M/A)−1(M/A)−1]
