Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


katy
Posts:
13
Registered:
11/2/05


unitary matrix
Posted:
May 17, 2006 3:41 PM


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



