\(\mathbb{F}^n\) is the set of all lists of length \(n\) with elements of \(\mathbb{F}\). These are a special case of matricies.

Formally—

\begin{equation} \mathbb{F}^n = \{(x1,\ldots,x_n):x_j\in\mathbb{F}, \forall j =1,\ldots,n\} \end{equation}

For some \((x_1,\ldots,x_n) \in \mathbb{F}^n\) and \(j \in \{1,\ldots,n\}\), we say \(x_j\) is the \(j^{th}\) **coordinate** in \((x_1,\ldots,x_n)\).

## additional information

### addition in \(\mathbb{F}^n\)

*Addition* is defined by adding corresponding coordinates:

\begin{equation} (x1,\ldots,x_n) + (y_1,\ldots,y_n) = (x_1+y_1, \ldots,x_n+y_n) \end{equation}

#### addition in \(\mathbb{F}^n\) is commutative

If we have \(x,y\in \mathbb{F}^n\), then \(x+y = y+x\).

The proof of this holds because of how addition works and the fact that you can pairwise commute addition in \(\mathbb{F}\).

\begin{align} x+y &= (x_1,\ldots,x_n) + (y_1,\ldots,y_n)\\ &= (x_1+y_1,\ldots,x_n+y_n)\\ &= (y_1+x_1,\ldots,y_n+x_n)\\ &= (y_1,\ldots,y_n) + (x_1,\ldots,x_n)\\ &= y+x \end{align}

This is a lesson is why avoiding explicit coordinates is good.

### additive inverse of \(\mathbb{F}^n\)

For \(x \in \mathbb{F}^n\), the **additive inverse** of \(x\), written as \(-x\) is the vector \(-x\in \mathbb{F}^n\) such that:

\begin{equation} x+(-x) = 0 \end{equation}

Which really means that its the additive inverse of each of the coordinates.

### scalar multiplication in \(\mathbb{F}^n\)

At present, we are only going to concern ourselves with the product of a number \(\lambda\) and a vector \(\mathbb{F}^n\). This is done by multiplying each coordinate of the vector by \(\lambda\).

\begin{equation} \lambda (x_1,\ldots,x_n) = (\lambda x_1, \lambda, \lambda x_n) \end{equation}

where, \(\lambda \in \mathbb{F}\), and \((x_1,\ldots,x_n) \in \mathbb{F}^n\).

The geometric interpretation of this is a scaling operation of vectors.