From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: det(I-AB)=det(I-BA)?
Date: 14 Sep 2000 13:36:08 -0400
Newsgroups: sci.math
Summary: [missing]

In article <39C1032B.C042E38D@hotmail.com>,
du  <dax_urbszat@hotmail.com> wrote:

[next time, edit the overhead, please]

[...]
:Hi does anyone know whether it is generally true for A an m*n
:matrix, and B an n*m matrix that det(I-AB)=det(I-BA)? And if it is,
:what approach one might take to prove it? I have been able to show
:that AB and BA have the same set of eigenvalues and that for each
:eigenvalue the dimension of the corresponding eigenspaces are also
:the same for AB and BA. But wasn't able to go further. Thanks.

Consider the identity (fixed spacing required)

[I-AB   A] * [I   0]  =  [I   0] * [I  A   ]
[0      I]   [B   I]     [B   I]   [0  I-BA]

Cheers, ZVK(Slavek)