Standard Basis

Last updated Nov 1, 2022

# Definition

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}

# Remarks

# The Standard Basis

Standard Basis

Definition Let $F$ be a and let $n \in \mathbb{N}$ . Denote $\epsilon_{i} = (0, \dots, 0, 1, 0, \dots, 0)...

11/7/2022

is a Vector Space Basis

Vector Space Basis

Definition Let $V$ be a . Then $S \subset V$ is a basi for $V$ if $\text{span} S = V$ $S$ is...

11/7/2022

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),

c_{1} = \cdots = c_{n} = 0

, and we have that

S

is Linearly Independent.

\checkmark

\blacksquare

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Standard Basis

# Definition

Field

Standard Basis

# Remarks

Standard Basis

Vector Space Basis

Subspace Span

The Subspace Span is the Set of all Linear Combinations

Linearly Independent

Linearly Independent

Backlinks

Standard Basis

Interactive Graph

Standard Basis

# Definition

Field

Standard Basis

# Remarks

# The Standard Basis Standard Basis is a Vector Space Basis Vector Space Basis for FnF^{n}Fn

Standard Basis

Vector Space Basis

Subspace Span

The Subspace Span is the Set of all Linear Combinations

Linearly Independent

Linearly Independent

Backlinks

Standard Basis

Interactive Graph