[Question 2014a04] Answer

[Question 2014a04] Answer

(1) The condition can be \ (AB + BA = 0 \), that is, \ (AB =-Ba \). Therefore, \ [AB = a ^ 2b = a (AB) = a (-BA) =-(AB) A =-(-BA) A = BA ^ 2 = BA, \] And \ (AB = BA = 0 \).

(2) The condition can be \ (0 = B (AB) ^ Ka = (BA) ^ {k + 1} \), SO \ [(I _n-BA) \ big (I _n + BA + \ cdots + (BA) ^ k \ big) = I _n, \] SO \ (I _n-BA \) reversible.

(3) We provide three solutions to this question.

Solution 1 (join factor method)

The join factor method is the inverse array of \ (A-BD ^ {-1} c \), the key of the method is constant deformation, generate the \ (A-BD ^ {-1} c \) factor. we will illustrate this typical example in several steps.

First set \ (H = (D-CA ^ {-1} B) ^ {-1} \), then \ [(D-CA ^ {-1} B) H = I _n. \ cdots (1) \] (1) is our starting point, and then we start to deform. our goal is to generate \ (A-BD ^ {-1} c \), so what is needed is \ (d ^ {-1} \) instead of \ (d \), so (1) the two sides at the same time left multiplication \ (d ^ {-1} \) can get \ [(I _n-D ^ {-1} Ca ^ {-1} B) H = d ^ {-1 }. \ cdots (2) \] In order to generate \ (A-BD ^ {-1} c \), left multiplication \ (B \) on both sides of the (2) Formula \) right multiplication \ (C \) Available \ [BHC-BD ^ {-1} Ca ^ {-1} BHC = BD ^ {-1} C. \ cdots (3) \] (3) formula raised on the left of the Public factor \ (a ^ {-1} BHC \), \ (BD ^ {-1} c \) on the Right \) move to the left, and add \ (A \) on both sides to make \ (A-BD ^ {-1} c \), get \ [(A-BD ^ {-1} C) A ^ {-1} BHC + (A-BD ^ {-1} c) =. \ cdots (4) \] raised the (4) formula left public factor \ (A-BD ^ {-1} c, and right multiplication \ (a ^ {-1} \) on both sides at the same time can be \ [(A-BD ^ {-1} C) \ big (I _n + A ^ {-1} BHC \ big) a ^ {-1} = I _n. \ cdots (5) \] Get \ [(A-BD ^ {-1} c) by (5) ^ {-1} = a ^ {-1} + A ^ {-1} bhca ^ {-1} = a ^ {-1} + A ^ {-1} B (D-CA ^ {-1} B) ^ {-1} Ca ^ {-1 }. \, \, \ Box \]

Solution 2 (using certified conclusions)

In class, I proved the following conclusions:

If \ (I _n-AB \) is reversible, \ (I _n-BA \) is also reversible and \ (I _n-BA) ^ {-1} = I _n + B (I _n-AB) ^ {-1} \).

At that time, I used the methods of factor pooling and power series expansion + verification to prove the above conclusions, and this conclusion is also a special case of this question. the downgrading formula is easily proved \ (| A-BD ^ {-1} c | \ NEQ 0 \), SO \ (A-BD ^ {-1} c \) is not different. we perform the following Deformation:

\ [(A-BD ^ {-1} c) ^ {-1} = \ big (A (I _n-A ^ {-1} BD ^ {-1} c) \ big) ^ {-1} = (I _n-A ^ {-1} BD ^ {-1} c) ^ {-1} a ^ {-1 }. \] regard \ (a ^ {-1} B \) and \ (d ^ {-1} c \) as two groups respectively, and use the above conclusions to obtain

\ [(A-BD ^ {-1} C) ^ {-1 }=\ big (I _n + A ^ {-1} B (I _n-D ^ {-1} Ca ^ {-1} B) ^ {-1} d ^ {-1} c \ big) a ^ {-1} \]

\ [= \ Big (I _n + A ^ {-1} B (D-CA ^ {-1} B) ^ {-1} c \ big) A ^ {-1} = a ^ {-1} + A ^ {-1} B (D-CA ^ {-1} B) ^ {-1} Ca ^ {-1 }. \, \, \ Box \]

Solution 3 (partition Elementary Transformation)

Block Matrix \ (\ begin {bmatrix} A & B \ C & D \ end {bmatrix }\) block elementary transformation can be changed to Block Diagonal arrays \ (\ begin {bmatrix} A & 0 \ 0 & D-CA ^ {-1} B \ end {bmatrix }\) and \ (\ begin {bmatrix} A-BD ^ {-1} C & 0 \ 0 & D \ end {bmatrix }\). to rewrite the above process with the multiplication of the block primary array

\ [\ Begin {bmatrix} I _n & 0 \-Ca ^ {-1} & I _n \ end {bmatrix} \ begin {bmatrix} A & B \ C & D \ end {bmatrix} \ begin {bmatrix} I _n &-a ^ {-1} B \ 0 & I _n \ end {bmatrix} = \ begin {bmatrix} A & 0 \\ 0 & D-CA ^ {-1} B \ end {bmatrix }, \]

\ [\ Begin {bmatrix} I _n &-BD ^ {-1} \ 0 & I _n \ end {bmatrix} \ begin {bmatrix} A & B \ C & D \ end {bmatrix} \ begin {bmatrix} I _n & 0 \-d ^ {-1} C & I _n \ end {bmatrix} = \ begin {bmatrix} A-BD ^ {-1} C & 0 \ 0 & D \ end {bmatrix }. \]

Therefore, we have

\ [\ Begin {bmatrix} (A-BD ^ {-1} C) ^ {-1} & 0 \ 0 & d ^ {-1} \ end {bmatrix} = \ begin {bmatrix} I _n & 0 \ d ^ {-1} c & I _n \ end {bmatrix} \ begin {bmatrix} A & B \ C & D \ end {bmatrix} ^ {-1} \ begin {bmatrix} I _n & BD ^ {-1} \ 0 & I _n \ end {bmatrix} \]

\ [= \ Begin {bmatrix} I _n & 0 \ d ^ {-1} C & I _n \ end {bmatrix} \ begin {bmatrix} I _n &-a ^ {-1} B \ 0 & I _n \ end {bmatrix} \ begin {bmatrix} a ^ {-1} & 0 \ 0 & (D-CA ^ {-1} B) ^ {-1} \ end {bmatrix} \ begin {bmatrix} I _n & 0 \-Ca ^ {-1} & I _n \ end {bmatrix} \ begin {bmatrix} I _n & BD ^ {-1} \ 0 & I _n \ end {bmatrix} \]

\ [= \ Begin {bmatrix} a ^ {-1} + A ^ {-1} B (D-CA ^ {-1} B) ^ {-1} Ca ^ {-1} & 0 \ 0 & d ^ {-1} \ end {bmatrix}, \]

Thus \ (A-BD ^ {-1} C) ^ {-1} = a ^ {-1} + A ^ {-1} B (D-CA ^ {-1} B) ^ {-1} Ca ^ {-1 }\). \ (\, \, \ Box \)

[Question 2014a04] Answer