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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 =/

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  • $\begingroup$ $\langle w_1,v_2\rangle=i\cdot(-i)+i\cdot 1 \ne 2$, not sure if there are other errors $\endgroup$ – angryavian Jun 13 '14 at 18:14
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    $\begingroup$ You cannot use immediately G-S, you first have to... what? $\endgroup$ – DonAntonio Jun 13 '14 at 18:14
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    $\begingroup$ Don't stress yourself with these names, "orthonormalization" consists of only one more step after "orthogonalization". I honestly don't know what happens in the head of some teachers. For Gauss's sake, they should focus first on the ideas, then on names. $\endgroup$ – Ivo Terek Jun 13 '14 at 18:34
  • $\begingroup$ Nice, a mathematical expletive. Thousand physicists and engineers! $\endgroup$ – mvw Jun 13 '14 at 18:35
  • $\begingroup$ Ok, I get it...but it's almost absurd: once one have an orthoghonal set (without zero) one only has to normalize each vector to get an orthonormal set... $\endgroup$ – DonAntonio Jun 13 '14 at 18:37
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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

$$w_2=(i,0,-1)-\frac12(1+i)(i\,,\,\sqrt2\,i\,,\,-i)=\ldots$$

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