A question on dual spaces of vector spaces I've been doing a bit of self study into the formalism of dual spaces. In the book that I've been reading the author introduces the notion of a dual space, $V^{\ast}$ to a given vector space $V$ as the set of linear functionals $f:V \rightarrow \mathbb{F}$ which map $V$ to its underlying field $\mathbb{F}$ such that $$f\left(\mathbf{v} +\mathbf{w}\right) = f\left(\mathbf{v}\right) +f\left(\mathbf{w} \right)\quad\forall\;\mathbf{v},\mathbf{w}\in V$$ and $$f\left(c\mathbf{v}\right) = cf\left(\mathbf{v}\right)$$ The author then goes on to introduce an ordered basis $\mathcal{B} = \lbrace\mathbf{e}_{i}\rbrace _{i=1, \ldots , n}$ for the vector space $V$ such that for a given $\mathbf{v} \in V$ $$\mathbf{v}= \sum_{i=1}^{n}v^{i}\mathbf{e}_{i}= v^{1}\mathbf{e}_{1}+ \cdots + v^{n}\mathbf{e}_{n}$$ and then defines the $i^{th}$ dual basis vector $\mathfrak{e}^{i}$ as a linear functional that satisfies $$\mathfrak{e}^{i}\left(\mathbf{v}\right) = f\left( v^{1}\mathbf{e}_{1}+ \cdots + v^{n}\mathbf{e}_{n}\right) = v^{i}$$
such that the linear functional $\mathfrak{e}^{i}$ "picks off" the $i^{th}$ component of the vector $\mathbf{v} \in V$.
Is there any particular motivation behind this? Or do we just choose the functional this way such that the basis vectors $\mathbf{e}_{i}$ and their corresponding dual basis vectors $\mathfrak{e}^{j}$ satisfy $$\mathfrak{e}^{i}\left(\mathbf{e}_{j}\right) = \delta^{i}_{j}$$
 A: 
Do we just choose the functional this way such that the basis vectors ei and their corresponding dual basis vectors $\mathfrak e^j$ satisfy
  $$
\mathfrak e^i(e_j)= \delta _{ij}
$$

Yes, that's exactly the reason.  This, in particular, makes computation extremely easy.
If $f = \sum_{i}a_i \mathfrak e^i$ and $v = \sum_j b_j e_j$, then we have
$$
f(v) = \sum_{i} a_i b_i
$$
Which you may think of as being tantamount to a "dot product".  What we're doing here is ostensibly the same as selecting an "orthonormal basis" in a situation where we've forgone the notion of an inner product.
A: One of the reasons for doing this is to mimic the inner product
Often one writes $\langle v,f \rangle$ for $f(v)$ where $v \in V$ and $f \in V^*$. Note that for a general Hilbert space $\mathcal{H}$, all liner functionals are given by inner products, i.e. we have a correspondence of a linear functional $\phi$ with a vector $x \in \mathcal{H} $(Riesz representation theorem). So, in this case, an orthonormal basis of the Hilbert space will become an orthonormal basis of the dual space under the identification $\mathcal{H^*}$ with $\mathcal{H}$. 
In case of a general vectors space (with or without the structure of norm, inner product etc) we would still like to think of $f(v)$ as "$\langle v,f \rangle$", i.e. some sort of inner product pairing. This intuitive notion makes a lot of sense in calculation- and makes computations really easy- and one can use the intuitions from the Hilbert space setting in this case to "guess" and prove certain formula.
I can write a few more examples if you want- so let me know in comments- so that I can come back and edit the answer.
