Finding a dual basis This is one of my homework questions - I'm pretty sure I understand part of it.
Let $V=\Bbb R^3$, and define $f_1, f_2, f_3 \in V^*$ as follows:
$$f_1(x,y,z) = x - 2y;\quad f_2(x,y,z) = x + y + z; \quad f_3(x,y,z) = y - 3z.$$
Prove that $\{f_1, f_2, f_3\}$ is a basis for $V^*$ (they are linearly independent, so this part is true), and then find a basis for $V$ for which it is the dual basis.  The textbook does a horrible job as explaining dual bases in general.  Can someone explain me the methods behind formulating the dual basis here?
Thanks.
 A: You need to find vectors
$$
e_1 = (x_1,y_1,z_1), e_2 = (x_2,y_2,z_2), e_3 = (x_3,y_3,z_3),
$$
so that 
$$
f_i(e_j) = \begin{cases}
1 & i = j \\
0 & i \neq j
\end{cases}
$$
Write down what this means for $e_1$:
$$
x_1 - 2y_1 = 1 \\
x_1 + y_1 + z_1 = 0 \\
y_1 - 3z_1 = 0
$$
and solve for $x_1,y_1,z_1$. Then do the same for $e_2$ and $e_3$.
A: First, we must show that $\beta^{*} = \{ f_{1},f_{2},f_{3}\}$ is a basis for the dual space $V^{*} = \mathcal{L}(V,F) = \mathcal{L}(\mathbb{R}^{3},\mathbb{R}):$
In the finite-dimensional setting, there is a one-to-one correspondence between linear transformations and matrix representations. Hence, we can compute the dimension of the dual space, as follows:
$$dim(V^{*}) = dim(\mathcal{L}(V,F)) = dim(V) \cdot dim(F) = dim(V) = dim(\mathbb{R}^{3}).$$
Thus, $dim(V^{*}) = 3.$
Now, if we find a linearly independent subset of $V^{*}$ that contains exactly $3$ vectors, then it is a basis for $V^{*}.$ In particular, consider $\beta^{*} = \{ f_{1},f_{2},f_{3}\},$ and suppose that $\sum_{i=1}^{3} a_{i} f_{i} = 0,$ where $a_{i} \in \mathbb{R}$ for $i=1,2,3.$ Then, pick an arbitrary vector $(x,y,z) \in \mathbb{R}^{3},$ and show that
$$0 = 0(x,y,z) = \sum_{i=1}^{3} a_{i} f_{i}(x,y,z)$$ implies $a_{i}=0$ for $i=1,2,3$ (note the multiple uses of the symbol $0$). From this, it follows that $\beta^{*}$ is a linearly independent subset of $V^{*}$ with exactly $3$ vectors.
Second, we follow the solution of BaronVT to find a basis $\beta$ for $\mathbb{R}^{3}$ such that $\beta^{*}$ is the dual basis of $\beta:$
Namely, we encode the three separate systems of equations in the following augmented matrix
$$\left(\begin{array}{ccc|ccc} 1 & -2 & \phantom{-} 0 & 1 & 0 & 0 \\ 1 & \phantom{-} 1 & \phantom{-} 1 & 0 & 1 & 0 \\ 0 & \phantom{-} 1 & -3 & 0 & 0 & 1 \end{array}\right),$$
then we perform Gauss-Jordan elimination to find $\beta.$
