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I'm preparing for a test and this question was a real pain to do. I get a bit confused by all the terms, so I'd appreciate if you guys peek at my attempt to solve this, and see if its correct.

What I did: First of all, $U+iI$ is self adjoint, therefore its normal and unitary diagonalizable, which means there exists a unitary matrix $P$ and a diagonal matrix $D$ so: $$U+iI = PDP^{-1}$$

Now because $U+iI$ is diagonalizable it has $n$ distinct eigenvectors, and for every such vector $v_i$ there is an eigenvalue $\lambda_k$ so: $$(U+iI)v_i = {\lambda_k}v_i$$ $$Uv_i + iIv_i = {\lambda_k}v_i$$ $$Uv_i = (\lambda_k-i)v_i$$

Therefore $(\lambda_k-i)$ is an eigenvalue of $U$, but $U$ is unitary which means that $(\lambda_k-i) (\overline{\lambda_k-i}) = 1 $ and finally $\lambda_k = 0$. This occurs for all $v_i$ eigenvectors, therefore $U+iI$ has one eigenvalue and its zero. Therefore $D$ has only zeroes on its diagonal, and $U+iI = 0$ therefore $U = -iI$.

I'm not really sure if the argument I used about the eigenvectors\values is correct, if its not, would love to hear a few tips on how to proceed, thanks!

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This solution is correct as far as I can tell. For a simpler solution:

$U + iI$ is self-adjoint, so it is equal to its complex-conjugate transpose. This means that all elements on the diagonal of $U + iI$ must be real. Thus the imaginary part of all the diagonal elements of $U$ must be $-1$. Since $U$ is unitary, each row and column must have a vector length of 1, but since $-i$ has magnitude 1, all other terms have to be zero (adding a real part to a diagonal element or any nonzero element off the diagonal will give that row and column a length greater than one). Therefore $U = -i I$.

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  • $\begingroup$ Well that makes sense as well, thank's a lot :) $\endgroup$
    – Xsy
    Jun 18, 2014 at 16:48

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