Assuming the basis of $\mathbb{R}^3$ is $e_1, e_2, e_3$ and the corresponding dual basis vectors are $\theta^1, \theta^2, \theta^3$ (by the standard identification $\theta^i(e_j)=\delta^i_j$).
What is the dual of $2e_1$? My guess would be $2\theta^1$, however, I'm unsure of whether the coefficient $2$ would become $\frac{1}{2}$ under the isomorphism from $\mathbb{R}^3$ to its dual space.
Any clarification would be appreciated, thanks!
Edit: I've realised I am referring to the outcome of applying a musical isomorphism (given all the necessary structure of a pseudo-Riemannian manifold (i.e. metric) required). In this case, the isomorphism is conventionally known as its flat, $$\flat:TM\to T^*M$$ where $M=\mathbb{R}$.
There is not a dual linear form of a vector.
You can complement $2 e_1$ into a basis $\{2 e_1, f_2, f_3\}$. Providing that $f_2,f_3$ belong to the subspace spanned by $\{e_2,e_3\}$, the dual basis will then be $\{1/2 \theta_1, f_2^*, f_3^*\}$.
However again, there is not a notion of the dual of a single vector.