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Suppose I'm given $2$ random vectors $(v_1,v_2)$

I want to find orthonormal basis $(w_1,w_2)$

Are the following equivalent? for the $w_2$ case

$$w_1=\frac{v_1}{\|v_1\|}$$ $$x_2=v_2-\frac{(v_1,v_2)}{(v_1,v_1)}v_1$$

or $$x_2=v_2-(v_1,v_2)v_1 $$ or $$x_2=v_2-\frac{(w_1,v_2)}{(w_1,w_1)}w_1$$

or $$x_2=v_2-(w_1,v_2)w_1 $$

and then $$w_2=\frac{x_2}{\|x_2\|}$$

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  • $\begingroup$ What are your thoughts? Do you know that $\|v\|^2 = (v,v)$? No, not all of them are equivalent. $\endgroup$ – Roland Apr 30 '16 at 18:08
  • $\begingroup$ yes, well I usuall use the first case (for $x_2$), but they all seem to be projections $\endgroup$ – GRS Apr 30 '16 at 18:09
  • $\begingroup$ Is $x \mapsto v - (v,x)v$ indeed a projection if $\|v\|\neq 1$? What's $\|v -(v,v)v\|$ in this case? $\endgroup$ – Roland Apr 30 '16 at 18:10
  • $\begingroup$ It's a projection of $x$ along $v$, but since $v$ and $w$ are the same vectors, I thought it would be the same. $\| v(v,v)\|=(v,v)^{3/2}$, but $\|v-v(v,v)\|=0$ $\endgroup$ – GRS Apr 30 '16 at 18:15
  • $\begingroup$ I don't see how this implies it's not a projection. We simply projected $v$ along itself. Could you tell me the correct ones I could use for Gram Schmidt $\endgroup$ – GRS Apr 30 '16 at 18:20

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