> [!summary]
Span describes the "reach" that a set of linear combinations of vectors can reach
>
>
**Definition of Span (Euclidean):**
> $\text{Span} \space \left\{ \vec{v_{1}},\vec{v_{2}},\dots,\vec{v_{k}} \right\} = \{ c_1 \vec {{v_{1}} }+ \dots + c_k \vec{v_{k}} \space | c_1 \dots c_k \in \mathbb{R} \}$
>
**Definition of Span (Vector Space):**
> $\text{Span} \space \left\{ \vec{v_{1}},\vec{v_{2}},\dots,\vec{v_{k}} \right\} = \{ c_1 \vec {{v_{1}} }+ \dots + c_k \vec{v_{k}} \space | c_1 \dots c_k \in \mathbb{F}, \vec{v}\in S\}$
>
**Requirements in General:**
> 1. There are at least ${n}$ vectors
> 2. At least ${n}$ amount of vectors in the set are linearly independent
>[!info]+ Read Time
**⏱ 3 mins**
# Euclidean Definition
Span is the [[Sets|set]] of all [[Linear Combinations|linear combinations]] of vectors in $\mathbb{R}^n$. It states what vectors can be reached by combining given vectors in a set. As well as what kind of geometric object (line, plane, or higher-dimensional space) they form. The formal definitions are the following
$
\text{Span} \space \left\{ \vec{v_{1}},\vec{v_{2}},\dots,\vec{v_{k}} \right\} = \{ c_1 \vec{{v_{1}}}+ \dots + c_k \vec{v_{k}} \space | c_1 \dots c_k \in \mathbb{R} \}
$
> [!note]
If a set of vectors $\left\{ \vec{v_{1}}, \vec{v_{2}},\dots,\vec{v_{k}} \right\}$ spans a space, then the spanning set spans the space. Meaning the set of vectors span the space.
## Requirements
If $\left\{ \vec{v_{1}}, \vec{v_{2}}, \dots, \vec{v_{k}} \right\}$ are in $\mathbb{R}^{\textcolor{orange}{n}}$, they only span $\mathbb{R}^{\textcolor{orange}{n}}$ if and only if the following is satisfied:
1. There are at least $\textcolor{orange}{n}$ vectors
2. At least $\textcolor{orange}{n}$ amount of vector in the set are [[Linear Independence & Dependence|linearly independent]]
## Matrix Requirements
If a matrix $A = \left[ \vec{v_{1}},\dots, \vec{v_{k}} \right]$ whose columns are the vectors $\vec{v_{1}},\dots,\vec{v_{k}}$
Then set of $k$ vectors $\left\{ v_{1},\dots,v_{k} \right\}$ span $\mathbb{R}^n$ if the [[Rank|rank]] of the [[Matrix Notation|coefficient matrix]] of the system ($t_{1}v_{1}+\dots+ t_{k}v_{k}=\vec{b}$) is $n$, so that $k \geq n$
# Vector Space Definition
If the set of vectors $\left\{ \vec{v_{1}}, \vec{v_{2}},\dots,\vec{v_{k}} \right\}$ are in a [[Vector Spaces|vector space]] $V$. Then the [[Sets|set]] of all [[Linear Combinations|linear combinations]] of those [[Scalar & Vectors|vectors]] form a [[Subspace|subspace]] of $V$, defined as $S$. So the span would a [[Subspace|subspace]] of $V$, formally defined below
$
\text{Span} \space \left\{ \vec{v_{1}},\vec{v_{2}},\dots,\vec{v_{k}} \right\} = \{ c_1 \vec{{v_{1}}}+ \dots + c_k \vec{v_{k}} \space | c_1 \dots c_k \in \mathbb{F}, \vec{v}\in S\}
$
## Proof of Subspace
Show $S$ is a subspace of $V$
$
\begin{array}{c}
\text{1. Zero Vector is in } S \\
0 \cdot \vec{v}_1 + 0 \cdot \vec{v}_2 + \dots + 0 \cdot \vec{v}_k = \vec{0} \in S \\ \\
\text{2. Closed under Addition} \\
\text{Let } \vec{u} = c_1 \vec{v}_1 + c_2 \vec{v}_2 + \dots + c_k \vec{v}_k \in S \\
\text{and } \vec{w} = d_1 \vec{v}_1 + d_2 \vec{v}_2 + \dots + d_k \vec{v}_k \in S \\
\text{Then: } \vec{u} + \vec{w} = (c_1 + d_1)\vec{v}_1 + (c_2 + d_2)\vec{v}_2 + \dots + (c_k + d_k)\vec{v}_k \in S \\
\\
\text{3. Closed under Scalar Multiplication} \\
\text{Let } \vec{u} = c_1 \vec{v}_1 + c_2 \vec{v}_2 + \dots + c_k \vec{v}_k \in S \\
\text{Let } a \in \mathbb{F} \text{ be any scalar} \\
\text{Then: } a\vec{u} = a(c_1 \vec{v}_1 + c_2 \vec{v}_2 + \dots + c_k \vec{v}_k) \\
= (ac_1)\vec{v}_1 + (ac_2)\vec{v}_2 + \dots + (ac_k)\vec{v}_k \in S
\end{array}
$
# Resources
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