given are
$$v_1 = (i, \sqrt{2}i, -i) ,v_2 = (i, 0, -1) ,v_3 = (0, -i, \sqrt{2}i)$$
I need to calculate a orthogonalized basis and then norm then. I am not allowed to immediately use the orthonormalization of Gram Schmidt, I first have to to the orthogonalization of Gram Schmidt.
Here's what I did:
1.) I need Basis $B = (w_1, w_2, w_3)$
2.) $w_1 = v_1$
3.) $w_2 = v_2 - \frac{\langle w_1,v_2\rangle}{\langle w_1,w_1\rangle} \cdot w_1$
4.) $w_3 = v_3 - \frac{\langle w_1,v_3\rangle}{\langle w_1,w_1\rangle} \cdot w_1 - \frac{\langle w_2, v_3\rangle}{\langle w_2,w_2\rangle} \cdot w_2 $
I got the following results and somewhere must be a big mistake:
a) $ w_1 = v_1 $ is clear.
b) $\langle w_1, v_2\rangle = 2$ and $\langle w_1, w_1\rangle = 4$ so I have $w_2 = \frac{1}{2} \cdot \begin{pmatrix} i \\ -i\sqrt{2} \\ -2+i \end{pmatrix}$
c) $\langle w_2, v_3\rangle = \sqrt{2} (1+i)$ and $\langle w_2, w_2\rangle = 2$ Together I have $$w_3 = \frac{1}{4} \left[ \begin{pmatrix} 2i\sqrt{2} \\ 0 \\ 2i\sqrt{2} \end{pmatrix} - \begin{pmatrix} i\sqrt{2}-\sqrt{2} \\ 2 -2i \\ -3\sqrt{2} - \sqrt{2}i\end{pmatrix}\right]$$
I mentioned the complex-conjugation but I simply cannot find my mistake. There must be something wrong since they're not orthogonal =/
You seem to be using the usual, Euclidean inner product, and then you have some mistakes:
$$\langle w_1,v_2\rangle=\left\langle(i,\sqrt2\,i,-i)\,,\,(i,0,-1)\right\rangle=-i^2+i=1+i\neq 2$$
so in fact
$$w_2=(i,0,-1)-\frac12(1+i)(i\,,\,\sqrt2\,i\,,\,-i)=\ldots$$