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I'm given the inner product:

$\bigl\langle(x_1,x_2,x_3),(y_1,y_2,y_3)\bigr\rangle:=3x_1y_1+x_1y_3+y_1x_3+x_2y_2+2x_3y_3$

And I'm asked to find the orthonormal basis in respect of the above inner product that results from the normal basis $B=(e_1,e_2,e_3),$ $e_1=(1,0,0),e_2=(0,1,0), e_3=(0,0,1)$

My thought was to use the Gram-Schmidt process but use the given inner product to calculate the projections. However when I did so I ended up with the same vectors $e_1,e_2,e_3$ so the exact same basis...

I'm I doing somthing wrong or does the basis just stay the same?

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    $\begingroup$ No, e.g. $\langle e_1,e_1\rangle=3$ so $e_1$ will have to induced norm $\sqrt3$ instead of $1$. Also, we won't have $\langle e_1,e_3\rangle=0$, so they are not $\langle, \rangle$-orthogonal. $\endgroup$ – Berci May 6 '18 at 10:34
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You are doing something wrong, and you are doing it from the start. Note that $\bigl\langle(1,0,0),(1,0,0)\bigr\rangle=3$ and that therefore the first vector that you should have got was $\frac1{\sqrt3}(1,0,0)$.

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  • $\begingroup$ Ooooh yes! You are absolutely right I totally forgot to apply it to the first vector... Thank you $\endgroup$ – VakiPitsi May 6 '18 at 10:35

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