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I have three vectors $$v_1=(1,1,1)^T$$ $$v_2=(1,1,0)^T$$ $$v_3=(1,0,0)^T$$ and special dot product definition $$(\overline{(x_1,x_2,x_3)},\overline{(y_1,y_2,y_3)})=2x_1y_2+x_1y_1+2x_2y_1+x_3y_3$$ I use standard Gram-Schmidt orthogonalization process. So $$e_1=v_1=(1,1,1)^T$$ $$e_2 = v_2 - \frac{(v_2, e_1)}{(e_1, e_1)} \cdot e_1 = (1,1,-5)^T$$ and $$e_3 = v_3 - \frac{(v_3,e_1)}{(e_1,e_1)}\cdot e_1 - \frac{(v_3,e_2)}{(e_2,e_2)} \cdot e_2 = (2,-3,0)^T$$ But $(e_2, e_3) = -9$ instead of $0$. I can't recognize my mistake. Thank you in advance!

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    $\begingroup$ The "dot product" isn't symmmetric $\endgroup$ – user251257 Mar 19 at 19:40
  • $\begingroup$ @user251257 Sorry, my mistake, I fixed it. But it changes nothing. $\endgroup$ – Егор Пономарёв Mar 19 at 19:50
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    $\begingroup$ Your dot product is not positive definite: $\langle (0,1,0),(0,1,0)\rangle = 2(0)(1) + (0)(0) + 2(1)(0) + (0)(0) = 0$, but $(0,1,0)\neq (0,1,0)$. Also, it's not positive semidefinite: $\langle (1,-1,0), (1,-1,0)\rangle = 2(1)(-1) + (1)(1) + 2(-1)(1) = -3\lt 0$. $\endgroup$ – Arturo Magidin Mar 19 at 19:52
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    $\begingroup$ It isn't positive definite. So the gs-process may fail. $\endgroup$ – user251257 Mar 19 at 19:52
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    $\begingroup$ Note 1: Don't use $=$ when you mean "and after I multiply by a constant, then I get"; the symbol = means equal, not "and then you do some other stuff and you get this". Note 2: Gram-Schmidt with a product that is not definite can fail, because your denominators can be zero. In this case that doesn't happen, but you seem to have miscalculated the last inner product: $\langle (2,-3,0),(1,1,-5)\rangle = 2(2)(1) + (2)(1) + 2(-3)(1) + (0)(-5) = 4+2-6 = 0$. Why do you say it is $-9$? $\endgroup$ – Arturo Magidin Mar 19 at 20:29

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