Comments on: Unitary + Upper Triangular => Diagonal http://www.tangentspace.net/cz/archives/2007/02/unitary-upper-triangular-diagonal/ somewhere near the beginning. Mon, 01 Dec 2008 21:29:52 +0000 http://wordpress.org/?v=2.5 By: Fanfan http://www.tangentspace.net/cz/archives/2007/02/unitary-upper-triangular-diagonal/#comment-97072 Fanfan Sun, 25 Feb 2007 20:48:31 +0000 http://www.tangentspace.net/cz/archives/2007/02/unitary-upper-triangular-diagonal#comment-97072 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. 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.

]]> By: Alex http://www.tangentspace.net/cz/archives/2007/02/unitary-upper-triangular-diagonal/#comment-96880 Alex Sun, 25 Feb 2007 07:24:33 +0000 http://www.tangentspace.net/cz/archives/2007/02/unitary-upper-triangular-diagonal#comment-96880 <p>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!</p> <p>An alternate way to get [tex]|U_{ii}|=1 [/tex] is [tex]\|Ux\|=\|x\|[/tex], 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).</p> <p>My way of doing it, which I'm now slightly less proud of, is to note that [tex]U^* = U^{-1} = p(U)[/tex] for some poly [tex]p[/tex], so [tex]U^*[/tex] is also upper-triangular. But [tex]U[/tex] is upper-triangular, so [tex]U*[/tex] is lower triangular, so [tex]U^*, U[/tex] are diagonal.</p> 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 |U_{ii}|=1 is \|Ux\|=\|x\|, 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 U^* = U^{-1} = p(U) for some poly p, so U^* is also upper-triangular. But U is upper-triangular, so U* is lower triangular, so U^*, U are diagonal.

]]> By: Fanfan http://www.tangentspace.net/cz/archives/2007/02/unitary-upper-triangular-diagonal/#comment-96551 Fanfan Fri, 23 Feb 2007 21:39:59 +0000 http://www.tangentspace.net/cz/archives/2007/02/unitary-upper-triangular-diagonal#comment-96551 (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 (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

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