For homework, I'm given a matrix $$ A = \begin{bmatrix} 3 & 2i & -2i\\ -2i & 0 & -1\\ 2i & -1 & 0 \end{bmatrix} $$ in an hermitian space. We are trying to find an orthonormal basis from there.
So far, I've found the eigenvalues which are $-1$ and $5$ as well as the eigenspaces $$ E_{-1} = \langle \frac{1}{\sqrt{2}} \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} ; \frac{1}{\sqrt{3}} \begin{bmatrix} -i \\ 1 \\ -1 \end{bmatrix} \rangle $$ and $$ E_5 = \langle \frac{1}{\sqrt{6}} \begin{bmatrix} 2i \\ 1 \\ -1 \end{bmatrix} \rangle $$ Put together, we obtain a matrix $$ P = \begin{bmatrix} 0 & \frac{-i}{\sqrt3} & \frac{2i}{\sqrt6} \\ \frac{1}{\sqrt2} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt6} \\ \frac{1}{\sqrt2} & \frac{-1}{\sqrt{3}} & \frac{-1}{\sqrt6} \end{bmatrix} $$ where $P^T \cdot A \cdot \overline{P} = S$ should give us $S$, a diagonal matrix made of eigenvalues. However, it seems that $$ P' = \begin{bmatrix} 0 & \frac{i}{\sqrt3} & \frac{-2i}{\sqrt6} \\ \frac{1}{\sqrt2} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt6} \\ \frac{1}{\sqrt2} & \frac{-1}{\sqrt{3}} & \frac{-1}{\sqrt6} \end{bmatrix} $$ which is the matrix $P$ where the first row has been multiplied by $(-1)$ is actually the expected result. I can't figure out where this $(-1)$ comes from however.
You need $P^{-1}AP$ in order to get a diagonal matrix.
If $P$ is unitary, then $P^{-1}=P^\dagger$, the transpose conjugated.
But $P$ will only be unitary if the eigenvectors are orthonormal.
Actually I just implemented your $P$ in Maple and it really works, so you must be making some really simple mistake.