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For a set of linearly independent vectors $\{u_1, u_2 ... u_n\}$, the Gram-Schmidt process for obtaining orthogonal vectors $\{v_1, v_2 ... v_n\}$ is:

$v_1 = u_1 \\ v_2 = u_2 - proj_{v_1} u_2 \\ ... \\ v_n = u_n - \sum_{0}^{n-1} proj_{v_n}u_n$

Is there any difference in result between this process, and:

$v_1 = u_1 \\ v_2 = u_2 - proj_{u_1} u_2 \\ ... \\ v_n = u_n - \sum_{0}^{n-1} proj_{u_n}u_n$

I view the Gram-Schmidt process as a method of removing alike components from a vector until all vectors are orthogonal, so intuitively I feel like it should result in a valid basis albeit different.

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    $\begingroup$ I know these formulas are not correctly written (indices don't make sense), but my guess is that the corrected version of the second set of formulas would not guarantee orthogonality of summands. $\endgroup$ – Tunococ Mar 13 '17 at 5:42
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Yes, this is different, and the latter process typically does not give you an orthogonal vectors. For instance, let's take the vectors $u_1=(1,0,0)$, $u_2=(1,1,0)$, and $u_3=(1,0,1)$ in $\mathbb{R}^3$. Gram-Schmidt gives $$v_1=u_1=(1,0,0),$$ $$v_2=u_2-u_1=(0,1,0),$$ and $$v_3=u_3-(u_1+0)=(0,0,1).$$

On the other hand, your proposed process gives the same values for $v_1$ and $v_2$, but instead gives $$v_3=u_3-\left(u_1+\frac{1}{\sqrt{2}}u_2\right)=\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},1\right).$$ This vector $v_3$ is not orthogonal to $v_1$ or $v_2$.

What's going on here is that to define $v_3$, you really want to take $u_3$ and subtract its projection onto the subspace spanned by $u_1$ and $u_2$ (or equivalently, the subspace spanned by $v_1$ and $v_2$). If you want to describe this as a sum of projections onto 1-dimensional subspaces, you need those subspaces to be orthogonal. That is, you need to project onto $v_1$ and $v_2$ which you have arranged to be orthogonal, rather than $u_1$ and $u_2$ which may not be orthogonal.

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  • $\begingroup$ So if I understand correctly: "What's going on here is that to define $v_3$, you really want to take $u_3$ and subtract its projection onto the subspace spanned by $u_1$ and $u_2$ (or equivalently, the subspace spanned by $v_1$ and $v_2$)." only works when $u_1$ and $u_2$ are already orthogonal? $\endgroup$ – Sentient Mar 13 '17 at 6:13
  • $\begingroup$ Gram-Schmidt is take $u_3$ and subtract its projection onto subspace spanned by $v_1$ and $v_2$. Subspace of $v_1$ and $v_2$ is not equal to $u_1$ and $u_2$ right? $\endgroup$ – Sentient Mar 13 '17 at 6:16
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    $\begingroup$ @Sentient Not quite. In order to be able to compute the projection onto the subspace as the simple sum of projections onto its basis vectors as per G-S, those basis vectors must be orthogonal. $\endgroup$ – amd Mar 13 '17 at 6:38

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