Prove these facts a if U is upper triangular and invertible

Prove these facts

(a) if U is upper triangular and invertible then U^-1 is upper triangular.

(b) The inverse of a unit lower triangular matrix is unit lower triangular

(c) The product of two upper or (two lower triangular) matrices is upper or (lower) triangular

Solution

By definition, (i, j) entry of AB = (row i of A) · (column j of B) = [a_i1 , a_i2 , . . . , a_in ] · [b_1j , b_2j , . . . , b_nj ] = a_i1 b_1j + a _i2 b_2j + · · ·+ a_in b_nj. This is a sum of terms of the form a_ik b_kj ( with i and j are as before) , and where k denotes the coordinates being currently multiplied. This sum turns out to be zero as long as (i, j) is an entry below the diagonal, because every single entry in this sum is zero. Let us assume that i > j. We will show that the (i, j) entry of AB is 0. It would be enough to show that a_ikb_kj = 0 for every single value of k.Thus, we need to show that either a_ik or b_kj is 0 for all k. Let us consider two possibilities i.e. k > j and k j.

(a). If k > j, then b_kj represent an entry below the main diagonal of B. Therefore, b_kj = 0 and so every term a_ik b_kj = 0.

(b). If k j, then since j < i, hence i > k. Then, a_ik represents an entry below the main diagonal of A, and a_ik = 0. Thus, a_ik b_kj = 0 . Since a_ik b_kj = 0for all k, we see that (i, j)th entry of AB = a_i1 b_1j + a_i2 b_2j + · · ·+ a_in b_nj = 0. Hence AB is upper triangular.

Now, let us assume that A and B are two lower triangular matrices of the same size. Then A^T and B^T are upper triangular, so that B^T A^T is upper triangular. Hence AB = (B^T A^T )^T is lower triangular.