Suggest edit — F^n

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title: "F^n"
source: https://www.jemoka.com/posts/kbhlists_over_fields/
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\(\mathbb{F}^n\) is the set of all lists of length \(n\) with elements of \(\mathbb{F}\). These are a special case of matricies.
Formally&mdash;
\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 &amp;= (x_1,\ldots,x_n) + (y_1,\ldots,y_n)\\ &amp;= (x_1+y_1,\ldots,x_n+y_n)\\ &amp;= (y_1+x_1,\ldots,y_n+x_n)\\ &amp;= (y_1,\ldots,y_n) + (x_1,\ldots,x_n)\\ &amp;= 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.