TITLE: why scalar projection does not yield coordinates?
QUESTION [11 upvotes]: Suppose we have an ordered basis $\{v_1,\dots,v_n\}$ in some inner product space. Let us project a vector $v$ on each $v_i$ by multiplying $v_i$ by the "scalar projection" $(v,v_i)/\|v_i\|$. Intuitively, it seems that each scalar projection $(v,v_i)/\|v_i\|$ indicates the amount of $v$ that goes in $v_i$ and therefore the $i^{th}$ coordinate of $v$ should be $(v,v_i)/\|v_i\|$. But that does not happen unless the basis is orthogonal. 
Mathematically I can justify this but can someone give an intuitive reason as to what goes wrong. For example with $B=\{(1,0),(1,1)\}$ in the Euclidean space $\mathbb R^2$ where $v=(0,1)$?

REPLY [4 votes]: I think your intuition should be that there is a huge difference between "the portion of $v$ that points in the direction of $v_1$" and "the portion of $v_1$ that is necessary to compose $v$".  You may be carrying the intuition that they are the same from the orthogonal case, but you really shouldn't be.  Here are a few reasons why:

$v_2$ may also point significantly in the direction of $v_1$, but it makes no sense to include this in the coefficient of $v_1$.  E.g. consider $v = 1v_1 + 1v_2$ and dot both sides with $v_1$.
The portion of $v$ pointing towards $v_1$ is a quantity that depends purely on $v$ and $v_1$, but the basis representation of $v$ depends critically on all the basis vectors.  It's unreasonable to expect the former to be able to tell you the latter, even approximately.
Fundamentally, representing a vector in terms of a basis is an inversion problem (it involves solving a matrix equation).  The process you are proposing is fundamentally a multiplication process (let's simplify to the case of a basis of unit vectors, and it literally is multiplication).  So in a very general, abstract sense you are doing it backwards.  But in the special case of orthogonal basis $B$ we have $B^{-1} = B^T$ so we can indeed take inverses by multiplying.  The fallacy here is believing that one can, in the generic case compute $B^{-1}$ just by rescaling $B$ by some $\|B\|^2$: that's not how matrix inverses work.