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SL(n, Z)

Posted: Sun Feb 14, 2010 3:15 pm
by hopeful_statguy
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?

Re: SL(n, Z)

Posted: Sun Feb 14, 2010 4:41 pm
by ana3a
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.

Re: SL(n, Z)

Posted: Sun Feb 14, 2010 10:27 pm
by hopeful_statguy
Many thanks.