# The operator $U$ is unitary

If $$A$$ is a selft-adjoint operator, then the operator $$U$$ defined by $$\begin{eqnarray*} U=(A-iI)(A+iI)^{-1} \end{eqnarray*}$$ Is unitary.

I know if $$U$$ is unitary then $$U^{*}U=UU^{*}=I$$ But when i tried to compute: $$\begin{eqnarray*} U^{*}U&=&{((A+iI)^{-1})}^{*}(A-iI)^{*}(A-iI)(A+iI)^{-1}\\\ &=&{((A+iI)^{-1})}^{*}(A^{*}-iI^{*})(A-iI)(A+iI)^{-1} \end{eqnarray*}$$ or I tried to compute $$UU^{*}$$: $$\begin{eqnarray*} UU^{*}&=&{(A-iI)(A+iI)^{-1}((A+iI)^{-1})}^{*}(A-iI)^{*}\\\ &=&{(A-iI)(A+iI)^{-1}((A+iI)^{-1})}^{*}(A^{*}-iI^{*}) \end{eqnarray*}$$ I don't know how can I continue with the hypothesis that $$A$$ is selft-adjoin operator and the way that I have to treat the adjoint inverse. Can you give some hint to continue? Thank you

HINT: Use the fact that $$(S^*)^{-1} = (S^{-1})^*$$ and $$(iI)^* = -i I$$.
• But $(S^{*})^{-1}=(S^{-1})^{*}$ when $S$ is a selft-adjoint operator and bounded right? Feb 12, 2021 at 22:34
• Since $S S^{-1} = I$, $(S S^{-1})^* = (S^{-1})^*S^* = I^* = I$. Feb 12, 2021 at 22:44