representing a vector component as column matrix. Any vector $\vec{v} \in$ a vector space V can be represented as a standard basis expansion $\vec{v}=v^{i}\vec{e_{i}}$.
The element $v^{i}$ are elements in the field $\mathbb{F}$ of reals.
Because the components $v^{i}$ of a vector wrt to basis $\beta$ is unique, once a basis $\beta$ has been chosen, any vector $\vec{v}$ can be represented by its component $v^{i}$ or equivalently as an n-tuple element $\left ( v^{1},\cdot \cdot \cdot ,v^{n} \right ) \in \mathbb{F}^{n}$.
My notes proceed to mention that the vector $\vec{v}$ can thus be represented as a column matrix made from its component denoted by $\mathbf{v} \in \mathbb{F}^{n}$.
I.e., 
v=$\begin{bmatrix}
v^{1}\\v^{2} 
\\\cdot 
\\ \cdot
\\ \cdot
v^{n}
\end{bmatrix}$
For some reason, I am unable to make the connect with representing the components of the vector as a column matrix. Perhaps, there is a lack in rigour. 
Could someone shed light on this?
 A: Maybe you will see connection if you write $\vec{v}=v^1{\vec{e_1}} + v^2{\vec{e_2}} + \dots+ v^n{\vec{e_n}} $ and you write this equation in the matrix form $\vec{v}=\begin{bmatrix}   \vec{e_1} & \vec{e_2} & \dots & \vec{e_n}\end{bmatrix}\begin{bmatrix}   v^1 \\ v^2 \\ \dots \\ v^n\end{bmatrix}$,  
but $\begin{bmatrix}   \vec{e_1} & \vec{e_2} & \dots &\vec{e_n}\end{bmatrix}=   \begin{bmatrix}   \ 1 & 0 & \dots &\ 0  \\  \ 0 & 1 & \dots &\ 0  \\ \ \dots & \dots & \dots &\ \dots  \   \\ 0 & 0 & \dots &\ 1 \end{bmatrix}= I \  \ \ \ $ - $ \ $identity matrix
  hence we have $\vec{v}=\begin{bmatrix}   v^1 \\ v^2 \\ \dots \\ v^n\end{bmatrix}$.
A: I think this can clarify a little bit:

Proposition The representation of any vector $\vec{v}$ in terms of basis vectors
  $e_1,e_2,\dots,e_n$ is unique.

Proof. Suppose that $v$ is represented as both
$$\vec{v} = \sum_{j=1}^n v^j e_j\quad\text{and}\quad\vec{v}=\sum_{j=1}^n v^{\prime j} e_j.$$
Eliminating $\vec{v}$ gives
$$0 = \sum_{j=1}^n (v^j-v^{\prime j}) e_j.$$
Since $e_1,e_2,\dots,e_n$ constitute a basis, they are linearly independent and each $(v^j-v^{\prime j}) = 0$. That is, $v^j=v^{\prime j}$, so that the representation must be unique.

Now you can use the unique components $v^1,v^2,\dots,v^n$ to represent the vector $\vec{v}$ for the given basis $e_1,e_2,\dots,e_n$ as a column vector $\begin{pmatrix} v^1 \\ v^2 \\ \vdots \\ v^n \end{pmatrix}$ if you want.
