Find unitary matrix so that $ P^{-1}BP$ is diagonal.

Question

given is the matrix

$ B = \begin{pmatrix} 1 & i & -i \\ -i & 2 & 0 \\ i & 0 & 2 \end{pmatrix} $.

I have to find a matrix $P \in U(3)$ (in unitary group, meaning that $P^{-1}$ = transposed compex-conjugated P) so that $P^{-1}BP$ is diagonal matrix.

I'm afraid I have no idea how to do this. I thought about something with eigenvalue and eigenvectors?

Answer 1

Let $$B = \begin{pmatrix} 1 & i & -i \\ -i & 2 & 0 \\ i & 0 & 2 \end{pmatrix}$$ $$Q = \begin{pmatrix} i & -1 & 1 \\ 0 & 1 & 1 \\ -2i & -1 & 1 \end{pmatrix}$$ $$D = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

Then $$B = Q^{-1}DQ$$ and $$D=QBQ^{-1}=:P^{-1}BP$$

$$P=Q^{-1}=\frac{1}{6}\begin{pmatrix} -2i & 0 & 2i \\ -2 & 3 & -1 \\ 2 & 3 & 1 \end{pmatrix}$$