It is easy to diagonalize a normal matrix such that $D = P^{-1} A P$ by simply putting all the orthogonal eigenvectors as columns for $P$. But I spent hours trying a unitary diagonalization of the following Hermitian (and therefore Normal) matrix:
$$ A =
\begin{bmatrix}
0 & i & 1 \\
-i & 0 & 0 \\
1 & 0 & 0
\end{bmatrix} $$
such that $ D = U^*AU $. I know that by definition every normal matrix is unitarily diagonalizable. The eigenvalues of this matrix are $ \{ 0, -\sqrt{2}, \sqrt{2} \} $.
What did not work but was my most promising try, was to scale down the eigenvectors by their norm so the matrix $ P $ became orthonormal. The result does not give me the diagonal matrix with the desired eigenvalues though.
Also, Google search did not yield a single nicely explained way to do a unitary transform of a normal matrix. The only document that I believe to try to explain it is here, although it does not show clearly how to construct $ U $.
Best Answer
The eigenvalues of $A$ are $0, \sqrt{2}, -\sqrt{2}$. These eigenvalues correspond to the eigenvectors $$ \begin{bmatrix} 0\\ i\\ 1 \end{bmatrix},\quad \begin{bmatrix} \sqrt{2}\\ -i\\ 1 \end{bmatrix},\quad \begin{bmatrix} -\sqrt{2}\\ -i\\ 1 \end{bmatrix}, $$ respectively. You will observe that the eigenvectors are orthogonal with respect to the standard inner product on $\mathbb{C}^n$. Normalizing the eigenvectors gives the unitary matrix $$ U = \begin{bmatrix} 0 & 1/\sqrt{2} & -1/\sqrt{2}\\ i/\sqrt{2} & -i/2 & -i/2\\ 1/\sqrt{2} & 1/2 & 1/2 \end{bmatrix} $$ that diagonalizes $A$ to $D = \operatorname{diag}(0,\sqrt{2},-\sqrt{2})$.