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Re: unitary matrix
Posted:
May 18, 2006 4:20 PM


"katy" <mcld@mega.ist.utl.pt> wrote in message news:1147894859.922823.226600@u72g2000cwu.googlegroups.com... > Hi! > > I'm trying to prove that when A and B are positive definite (Det >0) > > U= (A^1/2 B A^1/2)^1/2 A^1/2 B^1/2 is an unitary matrix > > I simplified U: > > U= (A^1/2 B A^1/2)^1/2 A^1/2 B^1/2 > > = A^1/4 B^1/2 A^1/4 A^1/2 B^ 1/2 > = A^1/4 B^1/2 A^1/4 B^1/2 > > I calculated U^T: > > U^T = [ (A^1/2 B A^1/2)^1/2 A^1/2 B^1/2 ]^T= > > = (B^1/2)^T (A^1/2 )^T [( A^1/2 B A^1/2)^1/2]^T > > = (B^1/2)^T (A^1/2 )^T ( A^1/4 B^1/2 A^1/4)^T > > = (B^1/2)^T (A^1/2 )^T ( A^1/4) ^T (B^1/2)^T (A^1/4)^T > > = (B^1/2)^T (A^(1/21/4)) ^T (B^1/2)^T (A^1/4)^T > > = (B^1/2)^T (A^1/4) ^T (B^1/2)^T (A^1/4)^T > > > I also calculated U^1: > > U^1= [ (A^1/2 B A^1/2)^1/2 A^1/2 B^1/2 ]^1 > > = B^1/2 A^1/2 ( A^1/2 B A^1/2)^1/2 > > = B^1/2 A^1/2 A^1/4 B^1/2 A^1/4 > > = B^1/2 A^1/4 B^1/2 A^1/4 > > But i didn't achieved any conclusion :( > > > > If the space is real I've proved that U is unitary (using the fact that > A*= A^T, so A^1 = A^T) > > So i verified that > > U U^T= I > > > [A^1/4 B^1/2 A^1/4 B^1/2 ] [ (B^1/2)^T (A^1/4) ^T (B^1/2)^T > (A^1/4)^T]= I = > > <=> A^1/4 B^1/2 A^1/4 B^1/2 (B^1/2)^T (A^1/4) ^T (B^1/2)^T > (A^1/4)^T = i > > <=> A^1/4 B^1/2 A^1/4 B^1/2 (B^1/2) (A^1/4) ^T (B^1/2)^T > (A^1/4)^T = i > > <=> A^1/4 B^1/2 A^1/4 (A^1/4) ^T (B^1/2)^T (A^1/4)^T = I > > <=> A^1/4 B^1/2 A^1/4 (A^1/4) (B^1/2)^T (A^1/4)^T = I > > <=>A^1/4 B^1/2 (B^1/2)^T (A^1/4)^T = I > > <=> A^1/4 B^1/2 (B^1/2) (A^1/4)^T = I > > <=> A^1/4 (A^1/4)^T = I > > <=> A^1/4 (A^1/4) = I > > <=> I=I > > > > > Can somebody help me to prove that U is unitary in any case? > > > Thank you very much, > > Catarina Dias >
You may find this easier to do if you don't simplify U. Just compute U*U^T.
________________________________ Eric J. Wingler (wingler@math.ysu.edu) Dept. of Mathematics and Statistics Youngstown State University One University Plaza Youngstown, OH 445550001 3309411817



