矩阵求逆引理

 首先必须记住的是可逆矩阵A+BCD的逆可以表示成A-1+X,其中X为未知矩阵

故有(A+BCD)(A-1+X)=E

E+AX+BCDA-1+BCDX=E;

(A+BCD)X+BCDA-1=0

X=-(A+BCD)-1BCDA-1

X=-[B(B-1A+CD)]-1BCDA-1

X=-(B-1A+CD)-1B-1B CDA-1

X=-(B-1A+CD)-1CDA-1 

X=-[C(C-1B-1A+D)]-1CDA-1

X=-(C-1B-1A+D)-1DA-1

X=-[(C-1B-1+DA-1)A]-1DA-1

X=-A-1(C-1B-1+DA-1)-1DA-1

X=-A-1[(C-1+DA-1B)B-1]-1DA-1 

X=-A-1B(C-1+DA-1B)-1DA-1

将其带入A-1+X,就得到了引理的结果:

(A+BCD)-1=A-1-A-1B(C-1+DA-1B)-1DA-1

 

法二来源https://people.cs.umu.se/myshao/teaching/notes/SMW.pdf

 

posted @ 2020-05-17 22:37  柠檬味呀  阅读(1431)  评论(0编辑  收藏  举报