Re: How to prove this matrix identity?
- From: Angus Rodgers <twirlip@xxxxxxxxxxx>
- Date: Tue, 13 May 2008 04:02:34 +0100
On Mon, 12 May 2008 18:31:04 -0700 (PDT), Rotwang
<sg552@xxxxxxxxxxxxx> wrote:
Suppose that A is an invertible n x n matrix, D is an m x m matrix, B
is an n x m matrix and C is an m x n matrix. If the (n + m) x (n + m)
matrix M is given by
M = / A B \
\ C D /
then apparently
Det M = Det A * Det (D - C A^{-1} B).
Can anybody tell me how this is proved?
It's rather messy to write out, but one way to do it is first to
reduce to the case A = I:
/ A B \ / A^{-1} 0 \ = / I B \
\ C D / \ 0 I / \ CA^{-1} D /
therefore
(det M)*(det A)^{-1} = det(D - (CA^{-1})B)
- or else premultiply by same matrix, to get det(D - C(A^{-1}B)).
If A = I, we can reduce all the elements of B to 0 by subtracting
scalar multiples of the first n columns from the last m columns,
and this leaves I and C unchanged, and transforms D into D - CB.
(At least, that's how it looks in a few examples, but as I said,
it's messy to write out! ... and it's past my bedtime.)
--
Angus Rodgers
(twirlip@ eats spam; reply to angusrod@)
Contains mild peril
.
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