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I was looking for a geometrical explanation for the Gram-Schmidt process and came across this animation on Wikipedia:

The animation

In the description, the owner wrote that he found an intermediate vector $v'_3$ by subtracting from $v_3$ its projection onto $u_1$. Then, $u_3$ is obtained by subtracting from $v'_3$ its projection onto $u_2$.

But given the formula $u_3=v_3-\operatorname{proj}_{u_1}(v_3)-\operatorname{proj}_{u_2}(v_3)$, does the description imply that $\operatorname{proj}_{u_2}(v_3)$ is the same as $proj_{u_2}(v'_3)$?

Edit: Also, why is vector $u_3$ ,which is a combination of vector $v'_3$ and a vector on $u_2$, still orthogonal to $u_1$? thank!

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Yes, that's right. Note in particular that $$ \DeclareMathOperator{\proj}{proj} \proj_{u_2}(v_3')=\proj_{u_2}(v_3-\proj_{u_1}(v_3)) = \\ \proj_{u_2}(v_3)- \proj_{u_2}(\proj_{u_1}(v_3)) $$ Now, can you see why the second term above should be zero?

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Yes, they are the same because $u_1$ and $u_2$ are orthogonal by construction.

Recall that projection maps are linear. Hence $$\text{proj}_{u_2}(v_3')=\text{proj}_{u_2}(v_3-\text{proj}_{u_1}(v_3)) = \text{proj}_{u_2}(v_3) - \text{proj}_{u_2}[\text{proj}_{u_1}(v_3)].$$

Orthogonality of $u_1$ and $u_2$ ensures that $\text{proj}_{u_2}[\text{proj}_{u_1}(v)] = 0 $ for all $v$, leaving us with $$\text{proj}_{u_2}(v_3')=\text{proj}_{u_2}(v_3).$$

Note that this same argument works when performing the Gram-Schmidt process in 4 or more dimensions.

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