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Let $S = \{(1, −1, 0, 2),(2, 1, −3, 1),(3, 0, −3, 3),(0, 1, 1, 1)\} ⊂ \mathbb R^4 .$ Use the Gram-Schmidt procedure to find an orthonormal basis for $\operatorname{span}(S)$ (using the dot product).

This is what I have tried:

Let:

$V_1 = (1,-1,0,2 ), \ V_2 = (2,1,-3,1 ),\ V_3 = (3,0,-3,3 ),\ V_4 = (0,1,1,1 )$

$$U_1 ={V_1 \over \|V_1\|}= {(1,-1,0,2)\over \sqrt 6}= \left({1 \over \sqrt6},-{1 \over \sqrt 6},0,{2 \over \sqrt6}\right)$$

I trying to get $U_2$ but I'm getting a really weird answer. Could anybody show me how to do it?

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To make the derivation more effective note that $V_3$ and $V_4$ are orthogonal then you can use these two as starting vectors, moreover I suggest to normalize only at the end of the process in order to avoid square roots and make calculations easier.

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  • $\begingroup$ +1 for normalizing at the end. But:: A basis is considered as an ordered basis so you must not change the order of its element. The key point of Gram-Schmidt is that if $\{v’_k\}_{k=1,\dots, n}$ is the orthonormal basis of $\{v_k\}_{k=1,\dots, n}$ we have $\mathrm{span}\{v_1,\dots v_j\}=\mathrm{span}\{v’_1,\dots v’_j\}$ for all $j\in\{1\,\dots,n\}$. $\endgroup$ – Michael Hoppe May 5 '18 at 11:55
  • $\begingroup$ @MichaelHoppe Thanks, I was thinking to the possibility to rerder the basis vectors, but I agree with your interpretation. $\endgroup$ – user May 5 '18 at 12:04
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You started well. Now, let$$a_2=v_2-\langle v_2,u_1\rangle u_1=\left(\frac32,\frac32,-3,0\right)$$and$$u_2=\frac{a_2}{\|a_2\|}=\frac1{\sqrt6}\left(1,1,-2,0\right).$$And so on…

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  • $\begingroup$ Hi José, how did you derive to 1/sqrt 6 as the factor? I keep getting Sqrt (2/27) as my factor out.. $\endgroup$ – Jenny May 5 '18 at 10:49
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    $\begingroup$ @Jenny $\left(\frac32\right)^2+\left(\frac32\right)^2+(-3)^2=\frac{27}2=\left(3\frac{\sqrt3}{\sqrt2}\right)^2$. $\endgroup$ – José Carlos Santos May 5 '18 at 11:04

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