Complex projections order in inner product 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.
 A: $$ \operatorname{proj}_\vec{u} \vec{v} = \frac{\langle \vec{v},\vec{u}\rangle}{\langle\vec{u},\vec{u}\rangle} \vec{u} \quad\text{?}$$
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.
A: Projection of $\alpha\vec{u}$ on $\vec{u}$ should be the same vector. The other way makes it $\bar{\alpha}\vec{u}$
