Real parts of diagonal elements of $(I-U)^{-1}(I+U)$ are zero?

Question

I noticed numerically that real parts of diagonal elements of $(I-U)^{-1}(I+U)$ are zero (assuming $I-U$ is invertible), where $I$ is identity matrix and $U$ is a unitary matrix ($U^\dagger U=I$).

A good example is provided by @copper.hat: take $U={1+i\over \sqrt{2}} I$, then $(I-U)^{-1} (I+U) = (1+\sqrt{2})i I$.

I have tried to use every answer in this question (Inverse of the sum of matrices) but none of them gives me any hints how to prove this.

By the way, here is a Mathematica script I used to check this using random unitary matrices (adapted here: https://mathematica.stackexchange.com/questions/11074/creating-a-random-normal-matrix):

n = 5; (*any dimension works here*)
q = Orthogonalize[
   RandomVariate[NormalDistribution[], {n, n}] + 
    I RandomVariate[NormalDistribution[], {n, n}]];
Re[Diagonal[Inverse[IdentityMatrix[n] - q].(IdentityMatrix[n] + q)]]

I have to apologize for making too many mistakes at the beginning (even the title was wrong.). I hope some one can help me understand it.

How do you know the inverse even exists? $I$ is a unitary matrix. — copper.hat, Jan 24 '18 at 06:54
You are correct. I am referring to a situation when $I-U$ is invertible. Edited accordingly — atbug, Jan 24 '18 at 06:56
$U$ is unitary. Is $UU^* = U^*U$??? What does that say about $U$? — copper.hat, Jan 24 '18 at 07:03
By definition, $U^{\dagger}U=UU^{\dagger}=I$. See https://en.wikipedia.org/wiki/Unitary_matrix — atbug, Jan 24 '18 at 07:06

copper.hat · Accepted Answer · 2018-01-25T04:32:28.483

Since $U$ is normal it is unitarily diagonalisable. Since it is unitary all eigenvalues lie on the unit circle.

If $U$ is diagonal and $\lambda$ is an eigenvalue on the unit circle then ${1 + \lambda \over 1-\lambda}$ is an eigenvalue of $(I-U)^{-1} (I+U)$. There is no reason to presume that ${1 + \lambda \over 1-\lambda}$ is real just because $\lambda$ lies on the unit circle (in fact, it maps the unit circle onto the imaginary line!).

Take $U={1+i\over \sqrt{2}} I$, then $(I-U)^{-1} (I+U) = (1+\sqrt{2})i I$.

Addendum: The question has changed, here is an update to the answer.

Let $ U = Q\Lambda Q^* $, where $Q$ is unitary and $\Lambda$ diagonal. Then $V=(I-U)^{-1} (I+U) = Q(I-\Lambda)^{-1} (I+\Lambda)Q^*$.

Note again that the map $f(z) = {1+z \over 1-z}$ maps the unit circle into the imaginary axis, hence $(I-\Lambda)^{-1} (I+\Lambda)$ purely imaginary and so $((I-\Lambda)^{-1} (I+\Lambda))^* = - (I-\Lambda)^{-1} (I+\Lambda)$.

It follows from this that $V^* = -V$ and so $V+V^* = 0$. In particular, the real parts of the diagonal are zero.

Hi, I am really sorry for making a huge mistake in the title (and body of course). The question is actually real parts of diagonal elements of $(I-U)^{-1}(I+U)$ are zero. Sorry for the inconvenience. — atbug, Jan 25 '18 at 03:30

Real parts of diagonal elements of $(I-U)^{-1}(I+U)$ are zero?

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