Unitary + Upper Triangular => Diagonal
That is, if 
and 
is upper triangular, then is diagonal. Can you prove it?
You can do it by induction easily– the idea is the same as that used in proving that every normal matrix is diagonalizable (in complex space), using Schur’s theorem– but try for a way that uses less calculation. After all, one of the nice aspects of positivity, normality, self-adjointness, etc. is that often times arguments about matrices don’t need to appeal directly to their elements.
Possibly relevant posts:
- Cholesky decomposition (7/11/2006)
- Positive definiteness of a certain matrix (6/15/2006)
- El Fin (9/28/2007)
(a) U unitary U* = inv(U)
(b) U unitary sum_j |U_ij|^2 = 1
(c) Inverse of an upper-triangular matrix U with diagonal elements U_ii is also upper-triangular with diagonal elements 1/U_ii
(a),(c) => 1/U_ii = U*_ii => |U_ii|=1
with (b) => U diagonal
Comment by Fanfan — 2/23/2007 @ 2:39 pm
Damn, I didn’t realize it was so simple, mainly because I didn’t exploit (b)– I usually think about the orthonormality condition on the columns in terms of an ‘inner product’, not the most concrete euclidean one. That makes this problem so trivial!
An alternate way to get
is 
, so already you know the eigenvalues are all modulus one. Easier for me than arguing that the inverse of an upper-triangular matrix is upper-triangular (sounds sensible, but off the top of my head, I can’t agree).
My way of doing it, which I’m now slightly less proud of, is to note that
for some poly 
, so 
is also upper-triangular. But 
is upper-triangular, so 
is lower triangular, so 
are diagonal.
Comment by Alex — 2/25/2007 @ 12:24 am
Indeed, combining your remark that the eigenvalues have modulus one, the fact that they are equal to the diagonal entries on a triangular matrix and the norm of any row equal to one gives a not very difficult solution, but yours is even simpler: knowing that the inverse of upper-triangular is upper-triangular, the answer is immediate.
Comment by Fanfan — 2/25/2007 @ 1:48 pm