representing a vector component as column matrix.

Question

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?

Answer 1

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 $\vec{v}=\begin{bmatrix} v^1 \\ v^2 \\ \dots \\ v^n\end{bmatrix}$.

Answer 2

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.

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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.