有四個矩陣,分別為A: n x n, B: n x k, C: k x k, D: k x n,
其中n,k是正整數,假設(A + BCD)是可逆的,
證明: (A + BCD)-1 = A-1 - A-1B(C-1 + DA-1B)-1DA-1
我是直接拿(A + BCD)(A-1 - A-1B(C-1 + DA-1B)-1DA-1) 下去求I,
假設:X: n x n, Y: n x n, Z: n x n, W: n x n,
那麼 XYZ - XWZ = X(Y - W)Z 是對的嗎?
其中 X(Y - W)Z 的 Y - W 在什麼情況下要變成 Y + W 還是根本不會有此情況?
I - XZ - XYZ = I - X(I - Y)Z 是對的嗎?
Thank for your respondence.
It seems that parts of my concepts are correct.
But I just can't get identity matrix I by direct product.
Could you show that (A + BCD)(A -1 - A -1 B(C -1 + DA -1 B) -1 DA -1 ) = ... = I in great detail?
- no nicknameLv 51 0 年前最佳解答
First, (A + BCD)-1 = A-1 - A-1B(C-1 + DA-1B)-1DA-1
is proved by direct product. Notice that : the multiplication is not commutative. I mean AxB not equals BxA . However, you still can product them tern by tern. Just computation problems.
Second, Matrix multiplication satisfies "distribution laws" so, A(B+C) = AB + AC is always true. so, XYZ - XWZ = X(Y - W)Z is always true.
Third, I - XZ - XYZ = I - X(I - Y)Z is always true.
I think you can check any linear algebra book, the matrix multiplication satisfies "distribution law", and I think you need a good reference book for linear algebra. Ask your friends in mathematics department, you will know which books are good.
you can check NTU math website.
2008-12-31 05:02:29 補充：
Come on man. You make a stupid and obviously wrong -A -B = - (A+B) not -(A-B) . The fifth row should be I - B (I + CDA^(-1)B) (C.....)
So, your computation are totally right except here. ^_^