Date: Fri, 28 Mar 1997 13:33:49 -0500 (EST)
From: Prof William P Wardlaw <wpw@nadn.navy.mil>
To: Torsten Sillke <sillke@Mathematik.Uni-Bielefeld.DE>
Subject: determinant identity

Dear Torsten,

        Here is my solution to your matrix problem:


        Let A, B, C, and  D  be  n x n  matrices over a commutative ring
R  such that   AC = CA.  Then

                |A B|  = |AD - CB|.
                |C D|

Proof:  Without loss of generality, assume  R  has an identity  1.  Then
AC = CA  implies that

                |A||A B| = | I 0| |A B|  = |  A    B   |
                   |C D|   |-C A| |C D|    |AC-CA AD-CD|

                         = | A    B   |  =  |A||AD-CB|.
                           | 0  AD-CB |

If  |A|  is not a divisor of zero in  R, we are done upon cancellation of
|A|.  Otherwise, let  A(t) = tI + A  be the n x n matrix over  R[t].
Clearly, A(t)C = CA(t) as before, and |A(t)|  is not a zero divisor in
R[t]  since it is a monic polynomial of degree  n.  Replacing  A  by  A(t)
in the above identity and cancelling  |A(t)|  gives

                |A(t)  B|  =  |A(t)D - CB|.
                | C    D|

Letting  t = 0  gives the desired identity.

[...]

Bill