# Unitary matrix U to diagonalize matrix A

I'm working on this exercise and I got stuck.

Find a unitary matrix $$U$$ and a diagonal matrix $$D$$ such that $$A=U^{*}DU$$ for

$$A=\begin{pmatrix}1&i\\-i&1\end{pmatrix}$$

So what I decided to do was:

1) Find the characteristic polynomial and the eigenvalues:

$$p(\lambda)= det(A-\lambda I) = (1-\lambda^2)-1 = \lambda^2-2\lambda = \lambda(\lambda-2)$$ $$\lambda_1=0; \lambda_2=2$$

2) Find the eigenvectors:

$$E_1$$ for $$\lambda_1=0$$:

$$\begin{pmatrix}1&i\\-i&1\end{pmatrix}$$ $$\begin{pmatrix}x\\y\end{pmatrix}$$ $$= \begin{pmatrix}0\\0\end{pmatrix}$$

Endind up with $$E_1=\begin{pmatrix}1\\-i\end{pmatrix}$$

Similarly for $$\lambda_2=2$$, $$E_2=\begin{pmatrix}1\\i\end{pmatrix}$$

Finally getting $$P=\begin{pmatrix}1&1\\-i&i\end{pmatrix}$$

Then I decided to check if $$P^{-1}AP=D$$ holds true:

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

So with this computation, my next step would be apply the Gram-Schmidt process to the columns of $$P$$ to make it unitary, which is where I got stuck at the following:

Taking $$v_1=\begin{pmatrix}1\\-i\end{pmatrix}$$ I thought I had to normalize it, but then as I tried: $$\parallel v_1\parallel = \sqrt{1+(i)^2} = \sqrt{0}$$.

Did I do a wrong computation? Is my process to find $$U$$ incorrect? Are there easier ways to do it?

Any help would be appreciated.

I think everything is fine, except that for a vector $$v \in \mathbb{C}^n$$, the definition of the norm is $$||v|| = \sqrt{v^* v}$$. So $$||v_1|| = \sqrt{\bar 1 \cdot 1 + \bar i \cdot i} = \sqrt{2}$$