If both and have integer entries, show that and are in . I can't figure this out .
ps this is pretty urgent if anyone can give me a pointer real quick
Thanks.
Induction Proof.
Base Case: Let such that a,b,c,d are integers.
, and since the set of integers is closed under addition and multiplication, we know we will be getting an integer determinant.
Now, let , such that a,b,c,d,e,f,g,h,i are integers.
Using cofactor expansion, we can write
as the sum of three 2x2 determinants multiplied by integer coefficients.
Now, since we showed that a 2x2 integer matrix has an integer determinant, and again, the set of integers is closed under multiplication and addition, is too clearly an integer.
Induction Step:
Consider an NxN integer matrix. Assume that any square integer matrix smaller than NxN has an integer determinant. The determinant of the NxN integer matrix is the sum of N (N-1)x(N-1) integer determinants times their integer coefficients. Again, since the set of integers is closed under addition and multiplication, the NxN integer matrix has an integer determinant, for N>=2.
Yeah it is obvious, since a determinant is just a polynomial of its terms so if all terms are integers, then obviously the determinant will be an integer. I wasn't thinking straight last night
It becomes clear when seeing the Leibniz formula for the determinant. We have that for an -matrix , then
where is the symmetric group of elements.