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Proving Cayley Transform operator is unitary

  1. Dec 7, 2011 #1

    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)


    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!
  2. jcsd
  3. Dec 7, 2011 #2

    I like Serena

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    Homework Helper

    Hi Ad123q! :smile:

    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]?

  4. Dec 7, 2011 #3
    [(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
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