So I've been trying to define the orthogonal projection as the linear approximation of a vector by another, such that the error is minimised, but for an arbitrary field...
Formally, in an inner product space over a field F $(V, +, . , F, <.,.>)$, the orthogonal projection of a vector f on a vector x is
$proj_x(f) = \frac{<f,x>}{<x,x>}x$
now whenever the field is $\mathbb{R}^n$, you can argue that the formula stems from the best linear approximation of f by x. Indeed we can write
$f = \alpha x + e = proj_x(f) + e$ (the linear approximation and it's error)
Then the best linear approximation is the one with the smallest error, i.e., with $<e,e> = |e|^2$ minimised. And since:
$|e|^2 = <f-\alpha x, f-\alpha x> = |f|^2+\alpha\alpha^*|x|^2-\alpha^*<f,x>^* -\alpha<f,x>\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$
it is then easy to see that if the field is Real, we get
$|e|^2 = |f|^2+\alpha^2|x|^2-2\alpha<f,x>$
and that minimizing $e$ with $\alpha$ yields
$\frac{\partial |e|^2}{\partial\alpha} = 2\alpha|x|^2 -2<f,x> = 0$
effectively retrieveing the projection coefficient.
I however run into trouble when trying to generalize this notion to an arbitrary field such as the Complex numbers with the classic complex conjugate, but also to other arbitrary field and their own conjugation rule. I haven't taken any particularly advanced abstract algebra classes and don't know how to formally take the derivative of equation (*) w.r.t. alpha, in order to make the same argument.
Thanks in advance for any help