From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Inverse over Z
Date: 9 Aug 2000 19:31:17 -0400
Newsgroups: sci.math
Summary: [missing]

In article <8ms5r1$1c8$1@nnrp1.deja.com>,
Alejandro Rivero  <arivero@my-deja.com> wrote:
:Hello,
:
:Does anybody knows how to build or clasify the set
:of integer matrices which have also an integer
:matrix as inverse?

A characterization "from above" (given an integer matrix, decide if it has
an integer inverse) has been given by others: if and only if the
determinant has magnitude 1.

There is a characterization "from below" (find building blocks and rules
for building): actually many such characterization. I will bring two:

Assume n > 1.

(1) Take the n-by-n identity matrix and replace the zero in the
    (1,2) position with 1. Call the resulting matrix S.
    Form all finite products of finite ordered lists containing 
    S, S^(-1), and permutation matrices. The resulting set is the set of
    all integer matrices with integer inverses.

(2) First, build all integer matrices with determinant +1. To do this,
    take all integer upper triangular matrices with unit diagonal, and all
    transposes of the above. Then products of finite lists of these form
    the set of all integer matrices with determinant +1.
    To get determinant -1, just switch the sign of the first column of a
    suitable matrix from the previous set.

Proof is done by rather tedious induction, unless someone knows a clever
trick to avoid it.

(In a course on elementary group theory, this can make an exercise on
identifying normal subgroups, generators, commutators, etc.; more fun
comes from allowing Gaussian integers.)

(Also, you can show that given an n-by-k integer matrix with n>k>0, it can
be augmented by (n-k) integer columns to a matrix with integer inverse if
and only if the gcd of all its k-by-k subdeterminants is 1. This relates
to solvability of systems of linear diophantine equations.)

Cheers, ZVK(Slavek).