From: lcs Mixmaster Remailer <mix@anon.lcs.mit.edu>
Subject: Sums of Invertible Matrices
Date: 10 Jan 2000 01:20:37 -0000
Newsgroups: sci.math.research
Summary: [missing]

Suppose R is a commutative ring such that every element of R is
the sum of two units. Then the 1x1 matrix [a] is representable 
as the sum of [b] and [a-b] for two units b and (a-b) in the 
ring.  If R is the two element field, there is at least one 
matrix which is not the sum of two invertible matrices.

Let n > 1.  If an nxn matrix M has a 0 for one of its entries, 
say in position (i,j), then one can build A and B such that A 
has 1 in position (i,j), B has 0-1 in position (i,j), A otherwise 
has the same elements in row i that M does, and 0's in  column j,
B otherwise has the same elements in column j that M does, and 0's 
in row i, and the rest of A and B are filled by decomposing the 
(i,j) minor into the sum of two invertible matrices. Then A and B 
are invertible, and sum to M.

(In the case R has two elements, and n=2, one needs a different 
construction to find invertible A and B with A+B=M. But such a 
representation exists in these cases.)

Finally, if the nxn matrix has no nonzero entries, then one can 
break M into a lower triangular portion A which is 0 in entry 
(n,1), and M-A.  Then A is invertible if R is a field, and so is 
M-A. (Alternatively, split each diagonal element into a(i,i) and 
m(i,i) -a(i,i), two units; let A be lower triangular, then M-A is
upper triangular, and both are invertible.)

Thus among all fields and many rings R, there is exactly one 
matrix [1] which is not the sum of two invertible matrices over R,
and this happens when R is the two element field.

The bulk of this argument was provided by G. Paseman, and completed 
and streamlined by D. Radcliffe. A product of IRC Efnet #math.

G. Paseman, 2000.01.09
.