Quote:
Originally Posted by smcdonn2
BTW this is not homework, I am just trying to get an understanding
So for example if I want to show that do this one:
determine whether the given set of invertible nxn matrices with real number entries is a subgroup of GL(n,r)
the given set is nxn matrice with determinant 1 or -1
Let the given set be H.
Is H closed under multiplication? Yes, because the determinant of the product of two matrices is equal to the product of their two determinants, so this will be +/- 1.
Is the identity element in H? The identity element of GL(n,r) has determinant 1, so it is in H.
Does every element of H have an inverse in H? A square matrix is invertible if it does not have determinant zero, so all elements of H have an inverse in G. For any element h of H, we have h * h^-1 = e, and so |h| * |h^-1| = |e|, so |h^-1| = 1 / |h|. Thus h^-1 has determinant 1 (if h has determinant 1), or -1 (if h has determinant -1). Thus h^-1 is in H as well, and we have shown that H forms a subgroup.