- #1
Ad123q
- 19
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Hi,
Was wondering if anyone could give me a hand.
I need to prove that the Cayley Transform operator given by U=(A-i)(A+i)^-1 is UNITARY, ie that UU*=U*U=I where U* is the adjoint of U (I am given also that A=A* in the set of bounded operators over a Hilbert space H).
My solution so far, is this correct?
U=(A-i)(A+i)^-1 so
(U)(x) = (A-i)((A+i)^-1)x (U acting on an x)
Then (Ux,y)= {INTEGRAL}(A-i)((A+i)^-1)x y(conjugate) dx (1)
= {INTEGRAL}x(A-i)((A+i)^-1)(both conjugate)y(all three conjugate) dx (2)
=(x,U*y)
and so deduce (U*)(y) = (A+i)((A-i)^-1)y
and so the adjoint of U is U*=(A+i)(A-i)^-1
It can then be checked that UU*=U*U=I
As you can see my main query is the mechanism of finding the adjoint of U for the given U.
For clarity in step (1) it is just the y which is conjugated, and in step (2) it is (A-i)(A+i)^-1 which is conjugated and then also the whole of (A-I)((A+i)^-1)y which is also conjugated. Sorry if my notation is confusing, if unsure just ask.
Thanks for your help in advance!
Was wondering if anyone could give me a hand.
I need to prove that the Cayley Transform operator given by U=(A-i)(A+i)^-1 is UNITARY, ie that UU*=U*U=I where U* is the adjoint of U (I am given also that A=A* in the set of bounded operators over a Hilbert space H).
My solution so far, is this correct?
U=(A-i)(A+i)^-1 so
(U)(x) = (A-i)((A+i)^-1)x (U acting on an x)
Then (Ux,y)= {INTEGRAL}(A-i)((A+i)^-1)x y(conjugate) dx (1)
= {INTEGRAL}x(A-i)((A+i)^-1)(both conjugate)y(all three conjugate) dx (2)
=(x,U*y)
and so deduce (U*)(y) = (A+i)((A-i)^-1)y
and so the adjoint of U is U*=(A+i)(A-i)^-1
It can then be checked that UU*=U*U=I
As you can see my main query is the mechanism of finding the adjoint of U for the given U.
For clarity in step (1) it is just the y which is conjugated, and in step (2) it is (A-i)(A+i)^-1 which is conjugated and then also the whole of (A-I)((A+i)^-1)y which is also conjugated. Sorry if my notation is confusing, if unsure just ask.
Thanks for your help in advance!