矩阵求逆引理(Matrix inversion lemma)推导

  矩阵求逆引理(Matrix inversion lemma):

  现有矩阵 A A 可以写为如下分块矩阵形式:

A=[A11A21A12A22](m+n)×(m+n) A = [ A 11 A 12 A 21 A 22 ] ( m + n ) × ( m + n )

  矩阵 A A (m+n) ( m + n ) 阶方阵,其中 A11 A 11 n n 阶非奇异方阵, A22 A 22 m m 阶非奇异方阵。那么可以得到: (A11A12A122A21) ( A 11 − A 12 A 22 − 1 A 21 ) (A22A21A111A12) ( A 22 − A 21 A 11 − 1 A 12 ) 都是非奇异矩阵。

  引理结论:

A1=[A111+A111A12(A22A21A111A12)1A21A111(A22A21A111A12)1A21A111A111A12(A22A21A111A12)1(A22A21A111A12)1] A − 1 = [ A 11 − 1 + A 11 − 1 A 12 ( A 22 − A 21 A 11 − 1 A 12 ) − 1 A 21 A 11 − 1 − A 11 − 1 A 12 ( A 22 − A 21 A 11 − 1 A 12 ) − 1 − ( A 22 − A 21 A 11 − 1 A 12 ) − 1 A 21 A 11 − 1 ( A 22 − A 21 A 11 − 1 A 12 ) − 1 ]

A1=[(A11A12A122A21)1A122A21(A11A12A122A21)1(A11A12A122A21)1A12A122A122+A122A21(A11A12A122A21)1A12A122] A − 1 = [ ( A 11 − A 12 A 22 − 1 A 21 ) − 1 − ( A 11 − A 12 A 22 − 1 A 21 ) − 1 A 12 A 22 − 1 − A 22 − 1 A 21 ( A 11 − A 12 A 22 − 1 A 21 ) − 1 A 22 − 1 + A 22 − 1 A 21 ( A 11 − A 12 A 22 − 1 A 21 ) − 1 A 12 A 22 − 1 ]

  以上为矩阵 A1 A − 1 的两种等价结果,其中对应位置的分块矩阵分别相等,由此可以分别得到4个推广等式,其中对角线上的两组等式又是相互等价的,最终得到2个等式:

(A11A12A122A21)1=A111+A111A12(A22A21A111A12)1A21A111 ( A 11 − A 12 A 22 − 1 A 21 ) − 1 = A 11 − 1 + A 11 − 1 A 12 ( A 22 − A 21 A 11 − 1 A 12 ) − 1 A 21 A 11 − 1

A111A12(A22A21A111A12)1=(A11A12A122A21)1A12A122 A 11 − 1 A 12 ( A 22 − A 21 A 11 − 1 A 12 ) − 1 = ( A 11 − A 12 A 22 − 1 A 21 ) − 1 A 12 A 22 − 1


证明:

  对矩阵 A A 以块为单位进行LU分解。得到以下两种分解结果:

A=[InA21A1110Im][A110A12A22A21A111A12] A = [ I n 0 A 21 A 11 − 1 I m ] [ A 11 A 12 0 A 22 − A 21 A 11 − 1 A 12 ]

A=[In0A12A122Im][A11A12A122A21A210A22] A = [ I n A 12 A 22 − 1 0 I m ] [ A 11 − A 12 A 22 − 1 A 21 0 A 21 A 22 ]

  对以上分解得到的四个矩阵分别求逆(分块矩阵求逆方法):

[InA21A1110Im]1=[InA21A1110Im] [ I n 0 A 21 A 11 − 1 I m ] − 1 = [ I n 0 − A 21 A 11 − 1 I m ]

[A110A12A22A21A111A12]1=[A110A111A12(A22A21A111A12)1(A22A21A111A12)1] [ A 11 A 12 0 A 22 − A 21 A 11 − 1 A 12 ] − 1 = [ A 11 − A 11 − 1 A 12 ( A 22 − A 21 A 11 − 1 A 12 ) − 1 0 ( A 22 − A 21 A 11 − 1 A 12 ) − 1 ]

[In0A12A122Im]1=[In0A12A122Im] [ I n A 12 A 22 − 1 0 I m ] − 1 = [ I n − A 12 A 22 − 1 0 I m ]

[A11A12A122A21A210A22]1=[(A11A12A122A21)1A122A21(A11A12A122A21)10A122] [ A 11 − A 12 A 22 − 1 A 21 0 A 21 A 22 ] − 1 = [ ( A 11 − A 12 A 22 − 1 A 21 ) − 1 0 A 22 − 1 A 21 ( A 11 − A 12 A 22 − 1 A 21 ) − 1 A 22 − 1 ]

  由LU分解的结果,矩阵 A1 A − 1 可以写为:

A1=[A110A12A22A21A111A12]1[InA21A1110Im]1 A − 1 = [ A 11 A 12 0 A 22 − A 21 A 11 − 1 A 12 ] − 1 [ I n 0 A 21 A 11 − 1 I m ] − 1

A1=[A11A12A122A21A210A22]1[In0A12A122Im]1 A − 1 = [ A 11 − A 12 A 22 − 1 A 21 0 A 21 A 22 ] − 1 [ I n A 12 A 22 − 1 0 I m ] − 1

  带入分块矩阵求逆的结果得到:

A1=[A111+A111A12(A22A21A111A12)1A21A111(A22A21A111A12)1A21A111A111A12(A22A21A111A12)1(A22A21A111A12)1] A − 1 = [ A 11 − 1 + A 11 − 1 A 12 ( A 22 − A 21 A 11 − 1 A 12 ) − 1 A 21 A 11 − 1 − A 11 − 1 A 12 ( A 22 − A 21 A 11 − 1 A 12 ) − 1 − ( A 22 − A 21 A 11 − 1 A 12 ) − 1 A 21 A 11 − 1 ( A 22 − A 21 A 11 − 1 A 12 ) − 1 ]

A1=[(A11A12A122A21)1A122A21(A11A12A122A21)1(A11A12A122A21)1A12A122A122+A122A21(A11A12A122A21)1A12A122] A − 1 = [ ( A 11 − A 12 A 22 − 1 A 21 ) − 1 − ( A 11 − A 12 A 22 − 1 A 21 ) − 1 A 12 A 22 − 1 − A 22 − 1 A 21 ( A 11 − A 12 A 22 − 1 A 21 ) − 1 A 22 − 1 + A 22 − 1 A 21 ( A 11 − A 12 A 22 − 1 A 21 ) − 1 A 12 A 22 − 1 ]

  根据两个矩阵对应位置的分块矩阵相同,得到结论:

(A11A12A122A21)1=A111+A111A12(A22A21A111A12)1A21A111 ( A 11 − A 12 A 22 − 1 A 21 ) − 1 = A 11 − 1 + A 11 − 1 A 12 ( A 22 − A 21 A 11 − 1 A 12 ) − 1 A 21 A 11 − 1

A111A12(A22A21A111A12)1=(A11A12A122A21)1A12A122 A 11 − 1 A 12 ( A 22 − A 21 A 11 − 1 A 12 ) − 1 = ( A 11 − A 12 A 22 − 1 A 21 ) − 1 A 12 A 22 − 1

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