Intuition for the Dual Space in Finite Dimensional Vector space Let me give you an example of how I think of linear operators first. Consider a linear operator $A: \mathcal{V} \to \mathcal{V}$. We have $A(\mu {\bf v} + \lambda {\bf w}) = \mu A({\bf v}) + \lambda A({\bf w})$ holding identically for $\mu, \lambda \in \mathbb{K}$ and ${\bf v}, {\bf w} \in \mathcal{V}$. Therefore, the set of all linear operators on $\mathcal{V}$ forms a vector space in its own right. Suppose $\mathcal{V} = \mathbb{R}^3$. To visualize this, I typically imagine $A$ as being represented by a 9D vector. This helps me quickly see various orthogonal decompositions into its symmetric and skew pieces or its spherical and deviatoric pieces.
I have tended to do the same thing for the dual space, $\mathcal{V}^*$; that is, I have been imagining $\mathcal{V}$ and $\mathcal{V}^*$ as ${\it totally \ separate \ spaces}$. They are objects of different type after all. I have been thinking of elements of $\mathcal{V}$ as geometric vectors and elements of $\mathcal{V}^*$ as "things that operate" on those vectors. But I've just reached a hurdle. 
Say we are working in $\mathbb{R}^2$ with a set of geometric vectors. It is common to represent a vector ${\bf x} \in \mathbb{R}^2$ on either a covariant or contravariant basis: ${\bf x} = x_i {\bf a}^i = x^i {\bf a}_i$ with the property that ${\bf a}_i \cdot {\bf a}^j = \delta_i^j$. But wait a second. I've been doing some reading and is it true that ${\bf a}_i \in \mathbb{R}^2$ but ${\bf a}^j \in (\mathbb{R}^2)^*$? If so, then we are layering the two spaces on one. And the dual set is not to be thought of as a set of linear functionals but as a set of geometric vectors that are available which have a nice duality property.
So is it true that in this interpretation, if we write ${\bf x} = x_j {\bf a}^j$ it is to be thought of as a linear functional, but if we write ${\bf x} = x^j {\bf a}_j$ then it is to be thought as a geometric vector? But ${\bf x}$ is supposed to represent the ${\it same}$ object, and elements of $\mathbb{R}^2$ and $(\mathbb{R}^2)^*$ are fundamentally different! How is this possible?
 A: This is not possible in general. However, in the presence of an inner product on $V$, you can use the inner product to identify between vectors in $V$ and linear functionals on $V$ (vectors in $V^{*}$). Namely, you identify a vector $v \in V$ with the linear functional $\varphi_v$ given by
$$ \varphi_v(w) = \left< v, w \right>. $$
Without this additional structure, the equation $\mathbf{x} = x_i \mathbf{a}^i = x^i \mathbf{a}_i$ does not make sense because $x^i \mathbf{a}_i$ is an element of $V$ while $x_i \mathbf{a}^i$ is an element of $V^{*}$. When you have an inner product and $\mathbf{a}_i$ is an orthonormal basis of $V$ then if $\mathbf{x} = x^i \mathbf{a}_i$ we also have
$$ \varphi_{\mathbf{x}} = x_i \mathbf{a}^i $$
(where $x^i$ and $x_i$ are the same scalars, only labeled with upper index or a lower index). By abusing notation and identifying $\mathbf{x}$ with $\varphi_{\mathbf{x}}$ using the inner product, we can then write
$$ \mathbf{x} = x_i \mathbf{a}^i = x^i \mathbf{a}_i. $$
When $V = \mathbb{R}^n$, a common convention is to write elements of $V$ as column vectors and elements of $V^{*}$ as column vectors. When using the standard inner product on $V$, the identification between $V$ and $V^{*}$ is just the transposition operation and in this sense, it "doesn't really change" the vector. If $\mathbf{a}_i = \mathbf{e}_i$ is the standard basis of row vectors for $V$ and $\mathbf{a}^i := \mathbf{e}_i^T$ is the standard basis of column vectors for $V^{*}$ then $\mathbf{a}_i$ and $\mathbf{a}^j$ are dual ($\mathbf{e}_i^T \mathbf{e}_j = \delta_{ij}$), $\mathbf{e}_i$ is orthonormal and we have the equations
$$ \mathbf{x} = x^i \mathbf{e}_i = x^i \mathbf{a}_i, \\
\mathbf{x}^T = \left( x^i \mathbf{e}_i \right)^T = x^i \mathbf{e}_i^T = x_i \mathbf{a}^i. $$
