How do I find a dual basis given the following basis?

Solution 1:

Notice that the definition of a dual basis is that, given $\beta = \{v_1, ..., v_n\}$, its dual is $\beta^* = \{f_1, ..., f_n\}$ such that $f_i(v_j) = \delta_{ij}$.

Given $\beta = \{\vec{e_1}-\vec{e_2},\vec{e_1}+\vec{e_2},\vec{e_3}\}$, we want such a basis. Also, since we know that the $f_i$ are linear functionals, we have that $f_i(x_1, x_2, x_3) = ax_1 + bx_2 + cx_3$. As you probably know, if we define the behaviour of $f_i$ in terms of our basis, we completely determine the function. I'll do the first example, the requirements are:

$$f_1(e_1 - e_2) = a - b = 1$$ $$f_1(e_1 + e_2) = a + b = 0$$ $$f_1(e_3) = c = 0$$

Hence, $a = -b$, which implies that $-2b = 1$, hence $b = \frac{-1}{2}$, and $a = \frac{1}{2}$, while $c = 0$. Thus:

$$f_1(x_1, x_2, x_3) = \frac{1}{2} x_1 - \frac{1}{2} x_2$$

Which, as desired, satisfies all the constraints. Just repeat this process for the other $f_i$s and that will give you the dual basis!

Solution 2:

Let $P$ be the change of basis matrix from the canonical basis $\mathcal C$ to basis $\mathcal B$. It is the matrix of the identity map from $(V,\mathcal B)$ to $(V,\mathcal C)$, and $P^{-1}$ is the matrix of the identity map from $(V,\mathcal C)$ to $(V,\mathcal B)$.

By duality, $\color{red}{{}^\mathrm t\mkern-1.5muP^{-1}}$ is the matrix of the identity map from $(V^*,\mathcal C^*)$ to $(V^*,\mathcal B^*)$. The column vectors of this matrix are the coordinates of vectors of $\mathcal B^*$ in the canonical basis of the dual space $(e_1^*,e_2^*,e_3^*)$.