Sum of projections?! Given this basis A = $\{ (\frac{1}{2}, \frac{\sqrt{3}}{2}, 0), (-\frac{\sqrt{3}}{2}, \frac{1}{2}, 0), (0, 0, 1) \}$. How can I know the value of $[(1,2,3)]_{A}$ by finding $(1, 2, 3)$ as a sum of projections? 
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Brak}[1]{\langle #1\rangle}\newcommand{\Vec}[1]{\mathbf{#1}}\newcommand{\Basis}{\Vec{e}}$Here are some general pointers:


*

*If $S = (\Vec{v}_{1}, \dots, \Vec{v}_{n}) = (\Vec{v}_{j})_{j=1}^{n}$ is a basis of $\Reals^{n}$, the coordinate vector $[\Vec{x}]_{S} = (x_{1}, \dots, x_{n})$ of $\Vec{x}$ with respect to $S$ is the ordered $n$-tuple of numbers satisfying
$$
\Vec{x} = x_{1} \Vec{v}_{1} + \dots + x_{n} \Vec{v}_{n}
  = \sum_{j=1}^{n} x_{j} \Vec{v}_{j}.
$$

*If $A = (\Vec{u}_{j})_{j=1}^{n}$ is an orthonormal basis of $\Reals^{n}$ (each vector has length one, and any two are orthogonal), then
$$
\Vec{x} = \Brak{\Vec{x}, \Vec{u}_{1}}\, \Vec{u}_{1} + \dots + \Brak{\Vec{x}, \Vec{u}_{n}}\, \Vec{u}_{n}.
$$
That is, the components of $\Vec{x}$ with respect to $A$ are the dot products of $\Vec{x}$ with the basis elements. (The summands on the right are the orthogonal projections of $\Vec{x}$ to the lines spanned by the basis elements. It's a good exercise to see how this works for the standard basis of $\Reals^{n}$.)
A: I'll give you the general idea. Say you have a vector $x = (\alpha_1, \ldots,\alpha_n)$ represented in the usual basis of $(1,0,0,\ldots,0)$, $(0, 1, 0,\ldots, 0)$, etc. and you want to know what that vector looks like in a different basis $A$ constructed from the orthonormal vectors $v_1, v_2,\ldots, v_n$. The coefficients $\alpha_i$ should actually be thought in the following way:
$$ x = \alpha_1 (1,0,\ldots, 0) + \cdots + \alpha_n (0, \ldots, 0,1). $$
The $\alpha_i$ can be computed by taking the dot product of $x$ with the $i$th basis vector. For a general basis, the following formula holds:
$$x = \sum_{l=1}^n (x\cdot v_l)\,v_l,$$
where $x\cdot v_l$ represents the dot product of $x$ and $v_l$. The coefficients $x\cdot v_l$ represent the entries of your vector in the basis $A$. In your case $v_1 = \left(\frac{1}{2},\frac{\sqrt{3}}{2}, 0\right)$ $v_2 = \left(-\frac{\sqrt{3}}{2},\frac{1}{2}, 0\right)$ and $v_3 = (0, 0, 1)$. Computing the dot products gives you the representation of $x$ in this basis. Let me know if you have any difficulty with this.
