Let \(V\) be a finite-dimensional complex inner product space and \(U\) a unitary operator on \(V\) such that \(U\alpha = \alpha\) implies \(\alpha = 0\). Let \[ f(z) = i\frac{(1 + z)}{(1 - z)}, \quad z \neq 1 \] and show that
\(f(U) = i(I + U)(I - U)^{-1}\);
\(f(U)\) is self-adjoint;
for every self-adjoint operator \(T\) on \(V\), the operator \[ U = (T - iI)(T + iI)^{-1} \] is unitary and such that \(T = f(U)\).
Solution
Theorem.
For the map \(f(z)=i(1+z)(1-z)^{-1}\),
the identities in (a),(b),(c) hold.
For unitary \(U\) with no fixed vector at eigenvalue \(1\), \(I-U\) is invertible, so \(f(U)=i(I+U)(I-U)^{-1}\) is well-defined. This is exactly part (a).
Since \(U^*=U^{-1}\) and \(I\) commutes with \(U\), \[ f(U)^*=-i(I-U)^{-1}(I+U)=i(I+U)(I-U)^{-1}=f(U), \] so \(f(U)\) is self-adjoint.
Let \(T\) be self-adjoint and define \(U=(T-iI)(T+iI)^{-1}\). Because \(T+iI\) is invertible, this is a valid operator.
We compute \[ U^*=(T+iI)^{-1}(T-iI)=U^{-1}, \] hence \(U\) is unitary.
Also \[ I+U=\frac{2T}{T+iI},\qquad I-U=\frac{2iI}{T+iI}. \] Then \[ i(I+U)(I-U)^{-1}=i\left(\frac{2T}{T+iI}\right)\left(\frac{2iI}{T+iI}\right)^{-1}=T. \]
Hence \(T=f(U)\), proving part (c).