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This problem is part of exercise 2.17 in Nielsen and Chuang's textbook, and has been already answered on this site in this post.

I understand that because $A$ is normal, it can be orthogonally diagonalized

\begin{equation} A = U^*DU \end{equation}

where $U$ is orthonormal.

What I do not understand is the claim that $U^*U = UU^* = I$, where $^*$ denotes conjugate-transpose. I know that orthonormal matrices satisfy $UU^T = I$, where $^T$ is transpose. But if $U$ has complex entries, then conjugate-transpose should not in general equal transpose alone.

NNN
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1 Answers1

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You are mistaken when you write that “because $A$ is normal, it can be orthogonally diagonalized $A=U^∗DU$, where $U$ is orthonormal.” Being normal is equivalent to the assertion that $A$ is unitarily diagonalizable.

José Carlos Santos
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