Proving Cayley Transform operator is unitary

Ad123q

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!

I like Serena

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]?

Sina

[(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