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!
Hi Ad123q! How do you conclude this from your expression for U*? Btw, instead of using the integral, can't you simply use the properties of the adjoint operator? That is, [itex](AB)^*=B^*A^*[/itex] and [itex](A^{-1})^*=(A^*)^{-1}[/itex]?
[(A-i)(A+i)^{-1}]* = [(A+i)^{-1}]*(A-i)* = [(A+i)*]^{-1}(A-i)* =(A* - i)^{-1}(A*+i) = (A - i)^{-1}(A+i) Now for instance multiply this with the original operator (A - i)^{-1}(A+i)(A-i)(A+i)^{-1} Note that A+i and A-i commute hence you get the result. Similiarly for the other way around