Standard Basis

Last updated Nov 1, 2022

Let $F$ be a Field and let $n \in \mathbb{N}$. Denote $\epsilon_{i} = (0, \dots, 0, 1, 0, \dots, 0) \in F^{n}$ to be the element of all $0$’s except for a $1$ in the $i$th position for some $i \in [n]$. The Standard Basis is the set: $$S = {\epsilon_{i}}_{i=1}^{n}$$

# The Standard Basis is a Vector Space Basis for $F^{n}$

$S$ spans $F^{n}$: for any $\mathbf{x} = (x_{1}, \dots, x_{n}) \in F^{n}$, we have that $$x_{1} \epsilon_{1} + \cdots + x_{n} \epsilon_{n} = \mathbf{x}.$$ The Subspace Span is the Set of all Linear Combinations, so $\text{span} S \supset F^{n}$. But, $\text{span} S \subset F^{n}$ by definition, so we must have that $\text{span} S = F^{n}$. $\checkmark$

$S$ is Linearly Independent: Suppose $c_{1}, \dots, c_{n} \in F$ and that $$c_{1} \epsilon_{1} + \cdots + c_{n} \epsilon_{n} = \mathbf{0}.$$ But $$c_{1} \epsilon_{1} + \cdots + c_{n} \epsilon_{n} = (c_{1}, \dots, c_{n}) = \mathbf{0} = (0, \dots, 0),$$ so $c_{1} = \cdots = c_{n} = 0$, and we have that $S$ is Linearly Independent. $\checkmark$ $\blacksquare$