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- Date: Tue, 5 Jan 1993 18:03:50 GMT
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- From: "Randall D. Tobias" <sasrdt@UNX.SAS.COM>
- Subject: Re: Matrix problem
-
- I'm posting this solution rather than just mailing, in order to
- advertise my primary source.
-
- In article <STAT-L%93010511153853@VM1.MCGILL.CA>, Just Jensen
- <LOFJUST@VM.UNI-C.DK> writes:
- |> During this procedure I have to prove a matrix identity that I think is
- |> well known but i do'nt seem to be able to prove it.
- |>
- |> Let A be a symetric P.d. matrix.
- |>
- |> Partition A as A = ! A11 A12 !
- |> ! A21 A22 !
- |>
- |> Let C=Inv(A)= ! C11 C12 !
- |> ! c21 C22 !
- |>
- |> where C is partitioned in the same way as A and inv means inverse.
- |>
- |> Then the following relationship exists:
- |>
- |> Inv(A11)*A12 = -C12*Inv(C22)
- |>
- |> what I need is the proof! (any hints?? or better full proofs)
-
- I just happen to be looking at the same sort of partitioned inverses
- right now myself, in connection with block designs. The appendices in
- Mardia, Kent, and Bibby (_Multivariate_Analysis_, Academic Press, 1979)
- are my standard reference for all sorts of useful matrix identities
- like this. Formula (A.2.4g) there gives
-
- C12 = -Inv(A11 - A12*Inv(A22)*A21)*A12*Inv(A22)
- C22 = Inv(A22 - A21*Inv(A11)*A12)
-
- Thus,
-
- -C12*Inv(C22) = Inv(A11 - A12*Inv(A22)*A21)
- * A12*Inv(A22)*(A22 - A21*Inv(A11)*A12)
- = Inv(A11 - A12*Inv(A22)*A21)
- * (A12 - A12*Inv(A22)*A21*Inv(A11)*A12)
- = Inv(A11 - A12*Inv(A22)*A21)
- * (A11 - A12*Inv(A22)*A21)*Inv(A11)*A12
- = Inv(A11)*A12
-
- as required.
-
- Question: Is non-singularity of all the relevent matrices really
- required? I doubt it.
-
- |> Thanks in advance.
-
- Glad to help!
- --
-
- Randy Tobias SAS Institute Inc. sasrdt@unx.sas.com
- (919) 677-8000 x7933 SAS Campus Dr. 72450.2545@compuserve.com
- (919) 677-8123 (Fax) Cary, NC 27512-8000 norat@aol.com
-
- ... just my $(-exp(2*sqrt(-1)*arcos(0))/(((2**(2 + 1)) - 1)**2 + 1)).
-