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Sherman–Morrison 公式
对可逆矩阵A及向量u,v,有
(A+uv′)−1=A−1−1+v′A−1uA−1uv′A−1.
证明1:
注意到u+uv′A−1u=u(1+v′A−1u)=(A+uv′)A−1u
有
(A+uv′)−1u=1+v′A−1uA−1u.
故
A−1=(A+uv′)−1(A+uv′)A−1=(A+uv′)−1(I+uv′A)=(A+uv′)−1+(A+uv′)−1uv′A=(A+uv′)−1+1+v′A−1uA−1uv′A.A−1=(A+uv′)−1(A+uv′)A−1=(A+uv′)−1(I+uv′A)=(A+uv′)−1+(A+uv′)−1uv′A=(A+uv′)−1+1+v′A−1uA−1uv′A.
证明2:
考虑分块矩阵方程
(Av′u−1)(Av′u−1)(XY)(XY)=(I0)(I0).
此即矩阵方程组
{AX+uY=I,v′X−Y=0.{AX+uY=I,v′X−Y=0.
有AX+uv′X=I,从而
X=(A+uv′)−1,
所以只要求X.
而由方程组第一个方程又有
X=A−1(I−uY).
代入第二个方程,
v′A−1(I−uY)=Y,v′A−1=(1+v′A−1u)Y,Y=1+v′A−1uv′A−1.v′A−1(I−uY)=Y,v′A−1=(1+v′A−1u)Y,Y=1+v′A−1uv′A−1.
再代回第一个方程:
AX+1+v′A−1uuv′A−1=IX=A−1−1+v′A−1uA−1uv′A−1.AX+1+v′A−1uuv′A−1=IX=A−1−1+v′A−1uA−1uv′A−1.
证明3:
由LDU分解:
(Av′u1)(Av′u1)=(Iv′A−101)(Iv′A−101)(A1−v′A−1u)(A1−v′A−1u)(I0′A−1u1)(I0′A−1u1),
求逆得
(Av′u1)(Av′u1)−1=(I0′A−1u1)(I0′A−1u1)−1(A1−v′A−1u)(A1−v′A−1u)−1(Iv′A−101)(Iv′A−101)−1
即
(Av′u1)(Av′u1)−1=(I0′−A−1u1)(I0′−A−1u1)(A−1(1−v′A−1u)−1)(A−1(1−v′A−1u)−1)(I−v′A−101)(I−v′A−101)
右边计算结果的1行1列元素:
(I−A−1u)(I−A−1u)(A−1(1−v′A−1u)−1)(A−1(1−v′A−1u)−1)(I−v′A−1)(I−v′A−1)=A−1+1+v′A−1uA−1uv′A−1.
原矩阵又可LDU分解:
(Av′u1)(Av′u1)=(I0′u1)(I0′u1)(A−uv′1)(A−uv′1)(Iv′01)(Iv′01),
求逆
(Av′u1)(Av′u1)−1=(I−v′01)(I−v′01)((A−uv′)−11)((A−uv′)−11)(I0′−u1)(I0′−u1),
第1行1列元素为
(I0)(I0)((A−uv′)−11)((A−uv′)−11)(I0′)(I0′)=(A−uv′)−1,
综合两方面结果就有
(A−uv′)−1=A−1+1+v′A−1uA−1uv′A−1.
与原式当然是等价的。
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