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Is there way to prove that every unitary matrix $A$ can be expressed in form of a Hermitian matrix $H$ such that eigenvalue of A is not 1?

I know that the form is $A=(I-iH)(I+iH)^{-1}$ And $A$ can be proved to be unitary by taking transjugate of $A$. But can this form be arrived starting from $A$?

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    Your identity does not hold for $A= I$ – 2017-02-10
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    I think you mean $(I-i H)(I+i H)^{-1}$ – 2017-02-10
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    @Omnomnomnom It came as is in one of the previous year papers of an exam. But clearly there seems to be an error. I will edit it. Thank you. – 2017-02-13

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To show that every unitary matrix can be expressed in this form, it suffices to make the following observations:

  • A unitary matrix is determined completely by its eigenvalues
  • $H$ may take any set of real numbers as its eigenvalues
  • For every eigenvalue $\lambda$ of $H$, $\frac{1 - i\lambda}{1 + i \lambda}$ is an eigenvalue of $A$
  • The map $\lambda \in \Bbb R \mapsto \frac{1 - i\lambda}{1 + i \lambda}$ has, as its image, the entirety of the unit circle in the complex plane.
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    I know how to prove the identity. I need help with proving every unitary matrix can be expressed in this form – 2017-02-13
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    Excuse me, I'll edit accordingly – 2017-02-13
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    See my latest edit – 2017-02-13
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    yeah, thanks a lot! – 2017-02-13