I've to find a matrix unitary matrix U, so that $U^* A U= diag(A)$. Or in other words: I've to find a unitary Matrix U that diagonalizes A.
$A = \begin{bmatrix}i&1&i&-1\\-1&i&1&i\\i&-1&i&1\\1&i&-1&i\end{bmatrix} $
I've calculated the Eigenvalues ($\lambda_1 = -2i \wedge \lambda_{2,3,4} =2i$) and the Eigenvectors...
$V_{\lambda_1} = \begin{bmatrix}-i\\-1\\i\\1\end{bmatrix}$, $V_{\lambda_2} = \begin{bmatrix}-i\\1\\0\\0\end{bmatrix}$, $V_{\lambda_3} = \begin{bmatrix}1\\0\\1\\0\end{bmatrix}$, $V_{\lambda_4} = \begin{bmatrix}i\\0\\0\\1\end{bmatrix}$
After Normalization the Matrix U should be...
$U=\begin{bmatrix}\frac{-i}{2}&\frac{-i}{\sqrt2}&\frac{1}{\sqrt2}&\frac{i}{\sqrt2}\\\frac{-1}{2}& \frac{1}{\sqrt2}&0&0\\\frac{i}{2}&0&\frac{1}{\sqrt2}&0\\\frac{1}{2}&0&0&\frac{1}{\sqrt2}\end{bmatrix}$
But that's not true, because $U^*U \neq I$
Any suggestions what went wrong?