Let ${\mathbb V}$ be a linear vector space(LVS) over a field ${\mathbb F}$ — in physical applications typically we work with ${\mathbb F}={\mathbb R}$ or ${\mathbb C}$. The dual vector space, usually denoted by ${\mathbb V}^*$, defined to be the LVS of linear maps from ${\mathbb V}$ to ${\mathbb F}$ . The dimension of the dual vector space is the same as that of the vector space, ${\mathbb V}$ — this is easily established using the definitions as shown in the lectures.
The key point is that there exists no natural map between ${\mathbb V}$ and ${\mathbb V}^*$ — this is the mathematical statement that encodes the fact that any map that can be constructed will be basis-dependent. Rather than dwell abstractly on this, we will consider an example to illustrate the issue. Choose ${\mathbb V}={\mathbb R}^2$. Consider two different bases:
Basis 1 — $(e_1,e_2)$ and Basis 2 — $(\tilde{e}_1,\tilde{e}_2)=(e_1, \sqrt3\ e_1 + e_2)$,
where we have indicated the linear transformation that relates the two bases. Recall that there is no notion of length of a vector (as yet) in the definition of a LVS. So do not picture the basis as some orthonormal basis. The basis, $(f_1,f_2)$, dual to $(e_1,e_2)$ is given by the linear maps from ${\mathbb R}^2 \rightarrow {\mathbb R}$
(1)Similarly, the dual basis, $(\tilde{f}_1,\tilde{f}_2)$, to $(\tilde{e}_1,\tilde{e}_2)$ is given by the linear maps from ${\mathbb R}^2 \rightarrow {\mathbb R}$
(2)Exercise: Since we know how the two basis for ${\mathbb V}$ are related, we can work out how the two dual bases are related. Show that
(3)Let us consider a vector $v=e_1\in {\mathbb V}$. In basis 1, it is natural to identify it with the linear map $f_1\in {\mathbb V}^*$ while in basis 2 it is natural to identify it with $\tilde{f}_1$. However, we have just seen that $\tilde{f}_1=f_1 - \sqrt3\ f_2\neq f_1$. Thus, a change in basis maps a given vector in ${\mathbb V}$ to a different vector in ${\mathbb V}^*$. Thus the answer is basis-dependent and thus is not natural (in maths terminology). We leave it as an exercise to see that this is true for an arbitrary vector $v=v^1\ e_1 + v^2\ e_2$.
Exercise: However, show that $(\mathbb{V}^*)^*=\mathbb{V}$.
However, if the LVS is endowed with an inner product, one can construct linear maps without choosing a basis. Let us denote the inner product between two vectors $v$ and $w$ by $\langle v, w\rangle$. Then, given a vector $v$, one can construct the linear map which we denote by $\langle v, \phantom{w}\rangle$ (or the bra vector $\langle v|$ in Dirac notation)
(4)It is obvious that no basis was specified in constructing the linear map.
Exercise: Verify that one does indeed get a linear map.
For ${\mathbb V}={\mathbb R}^n$ with the standard inner product, elements of ${\mathbb V}$ are called contravariant vectors and elements of the dual are called covariant vectors.
To be added Einstein summation convention, raising and lowering of indices.