How do I show that the inverse of an upper triangular matrix is also upper triangular? Thanks
Another way: write , where D is the diagonal part of A and N is the strictly upper-diagonal part of A. If A is invertible then so is D (because A and D have the same—nonzero—determinant); and is diagonal and hence upper-triangular. The strictly upper-diagonal matrix N is nilpotent, with if A is an matrix. You can check that
(just multiply that by and check that you get the identity).
The result then follows from the fact that sums and products of upper-diagonal matrices are upper-diagonal.