SL(n, Z)

Forum for the GRE subject test in mathematics.
hopeful_statguy
Posts: 7
Joined: Sun Sep 13, 2009 5:45 pm

SL(n, Z)

Postby hopeful_statguy » Sun Feb 14, 2010 3:15 pm

On page 229 of Cracking the GRE Math Subject test, it's written:
"Notice that if we replace R by Z in GL(n,R), we get a monoid but not a group, because the inverse of a nonsingular matrix with integer entries may not have integer entries. However, if the determinant of such a matrix is equal to 1, then the entries of the inverse will also be integers..."

Why is it the case that if the determinant of such a matrix is 1, then the entries of the inverse will also be integers?

ana3a
Posts: 26
Joined: Fri Mar 13, 2009 6:12 am

Re: SL(n, Z)

Postby ana3a » Sun Feb 14, 2010 4:41 pm

A * adj(A) = adj(A) * A = det(A) * I

If A only contains integers, then the adjugate matrix also only contains integers (arises from entries of A without division). So we see A^(-1) exists in GL(n,Z), in which case it equals adj(A), if and only if det(A) = 1.

hopeful_statguy
Posts: 7
Joined: Sun Sep 13, 2009 5:45 pm

Re: SL(n, Z)

Postby hopeful_statguy » Sun Feb 14, 2010 10:27 pm

Many thanks.




Return to “Mathematics GRE Forum: The GRE Subject Test in Mathematics”



Who is online

Users browsing this forum: No registered users and 5 guests