Let $\mathcal{B} = \{\alpha_{1},\alpha_{2},\alpha_{3}\}$ be the basis of $\textbf{R}^{3}$ defined by \begin{align*} \alpha_{1} = (1,0,-1),\quad\alpha_{2} = (1,1,1),\quad \alpha_{3} = (2,2,0). \end{align*}
Find the dual basis of $\mathcal{B}$.
MY ATTEMPT
The dual basis $\mathcal{B}^{*}$ of $\mathcal{B}$ is precisely the set of linear functionals $\{f_{1},f_{2},f_{3}\}$ such that $f_{i}(\alpha_{j}) = \delta_{ij}$, since $\dim\textbf{R}^{3} = 3$ over the field of real numbers. To start with, let us consider the equations given by $f_{1}(\alpha_{1}) = 1$, $f_{1}(\alpha_{2}) = 0$ and $f_{1}(\alpha_{3}) = 0$: \begin{align*} & f_{1}(\alpha_{1}) = a_{1}\times 1 + b_{1}\times 0 - c_{1}\times 1 = 1 \Longleftrightarrow a_{1} = 1 + c_{1}\\\\ & f_{1}(\alpha_{2}) = a_{1}\times 1 + b_{1}\times 1 + c_{1}\times 1 = 0 \Longleftrightarrow a_{1} + b_{1} + c_{1} = 0\\\\ & f_{1}(\alpha_{3}) = a_{1}\times 2 + b_{1}\times 2 + c_{1}\times 0 = 0 \Longleftrightarrow b_{1} = -a_{1} \end{align*} From this systems of equations, one obtains that $a_{1} = 1$, $b_{1} = -1$ and $c_{1} = 0$. Consequently, the matrix representation of the linear functional $f_{1}$ can be expressed as $[1,-1,0]$. Similarly, one gets the matrix representation $[1,-1,1]$ for $f_{2}$ and $[-1/2,1,-1/2]$ for $f_{3}$.
What concerns me is the basis in which such functionals are represented. As far as I have understood, they are represented in the canonical basis of $\textbf{R}^{3}$ and the basis $\{1\}$ of $\textbf{R}$. Thus, if we want to obtain the coordinates of $\alpha$ in the given basis $\mathcal{B}$, we must calculate $f_{i}(\alpha)$. Hence, if $\alpha = (x,y,z)$ is written in the canonical basis, then its first coordinate in the basis $\mathcal{B}$ is given by $f_{1}(\alpha) = x - y$, its second coordinate is given by $f_{2}(\alpha) = x - y + z$ and its third coordinate is $f_{3}(\alpha) = -x/2 + y - z/2$. When we do the corresponding summation, we get \begin{align*} \sum_{i=1}^{3}f_{i}(\alpha)\alpha_{i} & = (x-y,0,y-x) + (x-y+z,x-y+z,x-y+z)\\ & + (-x + 2y - z,-x + 2y - z,0) = (x,y,z) \end{align*}
which are exactly the coordinates of $\alpha$ in the canonical basis.
I know this question has already been answered here, but since I have additioned some queries, I insist in posting it. Any theoretical clarification would be appreciated.