Date: Fri, 28 Mar 1997 13:33:49 -0500 (EST)
From: Prof William P Wardlaw
To: Torsten Sillke
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