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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

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HINT: Use the fact that $(S^*)^{-1} = (S^{-1})^*$ and $(iI)^* = -i I$.

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  • $\begingroup$ But $(S^{*})^{-1}=(S^{-1})^{*}$ when $S$ is a selft-adjoint operator and bounded right? $\endgroup$
    – angie duque
    Feb 12, 2021 at 22:34
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    $\begingroup$ Since $S S^{-1} = I$, $(S S^{-1})^* = (S^{-1})^*S^* = I^* = I$. $\endgroup$
    – Falcon
    Feb 12, 2021 at 22:44

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