PDE-Cayley Transform Of An Operator

1. May 5, 2012

Combinatorics

1. The problem statement, all variables and given/known data

Prove that a unitary operator $U$ acting on a Hilbert space $H$ is the Cayley transform of some self-adjoint operator if and only if $1$ is not an eigenvalue of $U$.

Hope someone will be able to help

Thanks !

2. Relevant equations

3. The attempt at a solution
At the first direction, I have tried assuming that 1 is an eigenvalue for the eigenfunction $f$. This implies $Uf=f$ ,where $U=(L-i)(L+i)^{-1}$ for some $L$.
But since $U^* U=1$ we get that also $U^* f =f$ , and thus:
$<Uf,f>=<Uf,U(U^*f)>=<f,Uf>$, which implies $U$ is symmetric, and in particular $U=U^*$ and $U^2=1$. I can't figure out how to get some contradiction out of this. Maybe we know that $U$ must be one-to-one or something?

As for the other direction, I need some detailed guidance.