Re: How to prove this matrix identity?
- From: rob@xxxxxxxxxxxxxx (Rob Johnson)
- Date: Tue, 13 May 2008 12:13:38 GMT
In article <d1aaf241-4c74-4598-84ca-cd2366a84fd9@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
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?
If A is invertible, let Aa = I. Then
[ A B ] [ I -aB ] [ A 0 ]
[ ] [ ] = [ ] [1]
[ C D ] [ 0 I ] [ C D-CaB ]
and we get
[ A B ]
det [ ] = det(A) det(D-CaB) [2]
[ C D ]
This proves your claim. However, more can be shown.
If D is invertible, let Dd = I. Then
[ A B ] [ I 0 ] [ A-BdC B ]
[ ] [ ] = [ ] [3]
[ C D ] [ -dC I ] [ 0 D ]
and we get
[ A B ]
det [ ] = det(A-BdC) det(D) [4]
[ C D ]
Equation [4] is analagous to [2]. Now, let A = D = I, so that
a = d = I, and we get from [2] that
[ I B ]
det [ ] = det(I-CB) [5]
[ C I ]
and from [4] that
[ I B ]
det [ ] = det(I-BC) [6]
[ C I ]
Thus, we have shown that det(I-BC) = det(I-CB), which is discussed in
the thread rooted at <http://tinyurl.com/5zrjv4>, where, eventually,
we get to the following:
Expanding on this a bit, suppose A is mxn and B is nxm where n >= m.
Using this result, it is pretty simple to show that
n-m
det(Ix - BA) = x det(Ix - AB)
That is, the characteristic polynomial of BA is x^{n-m} times that of
AB.
Rob Johnson <rob@xxxxxxxxxxxxxx>
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- How to prove this matrix identity?
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