So the complex projection is defined as $$\operatorname{proj}_\vec{u} \vec{v} = \frac{\langle \vec{v},\vec{u}\rangle}{\langle\vec{u},\vec{u}\rangle} \vec{u}$$ with complex inner product. I was wondering if there is a reason why we have to compute the inner product in the order $\langle\vec{v},\vec{u}\rangle$ instead of $\langle\vec{u},\vec{v}\rangle$. I understand that the result will change if the order is changed, however I'm trying to look for perhaps a geometric reasoning behind the order the inner product must be computed.
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I changed $< \vec{v},\vec{u}>$ to $\langle \vec{v},\vec{u}\rangle$ and did some other emendations. But I'm wondering why you use completely empty superscripts, i.e. \vec{v}^{} instead of \vec{v}? ${}\qquad{}$ – Michael Hardy Nov 28 '14 at 05:23
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@MichaelHardy Thanks! I'm not sure how to type the equations out, so I just copy and pasted it from another post :P – lllll Nov 29 '14 at 02:14
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Projection of $\alpha\vec{u}$ on $\vec{u}$ should be the same vector. The other way makes it $\bar{\alpha}\vec{u}$
Saibal
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Where you say "the same vector", you mean the same as $\alpha\vec{u}$. I'd have said so. ${}\qquad{}$ – Michael Hardy Nov 28 '14 at 05:32
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$$ \operatorname{proj}_\vec{u} \vec{v} = \frac{\langle \vec{v},\vec{u}\rangle}{\langle\vec{u},\vec{u}\rangle} \vec{u} \quad\text{?}$$
You want to write $\vec{v} = \alpha\vec{u} + (\vec{v}-\alpha\vec{u})$, and you just want to choose $\alpha$ so as to make $\vec{v}-\alpha\vec{u}$ orthogonal to $\vec{u}$. Orthogonality means $\langle \vec{v}-\alpha\vec{u},\vec{u}\rangle=0$.
$$ \langle \vec{v}-\alpha\vec{u},\vec{u}\rangle = \langle\vec{v},\vec{u}\rangle - \alpha\langle\vec{u},\vec{u}\rangle. $$ You can set that equal to $0$ and solve for $\alpha$.
Alternatively, you could say $\langle \vec{u},\vec{v}-\alpha\vec{u}\rangle=0$. Then $$ \langle \vec{u},\vec{v}-\alpha\vec{u}\rangle = \langle\vec{u},\vec{v}\rangle - \bar\alpha\langle\vec{u},\vec{u}\rangle. $$ This time solving for $\bar\alpha$ gives you a certain expression, and conjugating one side gives you $\alpha$ and conjugating the other side is done by interchanging the two vectors whose inner product is taken. You get the same result.
However, I suspect that there is still more geometric insight than what can be found either in this answer or in Saibal's. I'm not sure what it is yet.