Examples:: [[basis#^921ff1|Basic Examples]], Nonexamples:: [[basis#^6f75a8|Basic Nonexamples]] Constructions:: [[change of basis formula]], [[orthonormal basis]], [[eigenbasis]], [[dual basis]] Generalizations:: *[[Generalizations]]* Justifications and Intuition:: *[[Justifications and Intuition]]* Properties:: [[finite-dimensional vector space basis length does not depend on basis]], [[every vector space has a basis]], [[every linearly independent list extends to a basis]], [[every generating set contains a basis]] Sufficiencies:: [[spanning list of the right length is a basis]], [[orthonormal list of the right length is a basis]], [[linearly independent list of the right length is a basis]] Equivalences:: [[vector space basis iff minimal spanning set]], [[vector space basis iff maximal linearly independent]] ---- > [!definition] Definition. ([[basis]]) > Let $R$ be a [[commutative ring]] and $M$ an $R$-[[module]]. An indexed set $B \to M$ is a **(Hamel) basis** if it [[submodule generated by a subset|generates]] $M$ and is [[linearly independent]]. > One can harmlessly refer to bases as 'subsets' of $M$, since the images of all $b \in B$ are distinct (else [[linearly independent|linear independence]] fails). > [!note] Remark. > - Bases are necessarily [[maximal]] [[linearly independent]] subsets and [[maximal|minimal]] [[submodule generated by a subset|generating subsets]]. What makes [[module|modules]] over [[field|fields]] — i.e., [[vector space|vector spaces]] — so special is that the converse holds. > - In functional analysis, the main use for Hamel bases is for producing [[operator norm|unbounded]] [[linear map|linear maps]]; one of the reasons they're not so useful elsewhere in functional analysis is because they're always too large: as a consequence of [[Baire category theorem|Baire's Theorem]], any Hamel basis for an [[dimension|infinite-dimensional]] [[Banach space]] is necessarily [[uncountably infinite|uncountable]].[^1] ^note [^1]: See [[Hamel basis of infinite-dimensional Banach space is uncountable]] > [!specialization] Definition for finite-dimensional vector spaces. > Let $V$ be a [[vector space]]. If the list of [[vector]]s $a_{1},\dots, a_{m}$ [[spans]] $V$ and is [[linearly independent]], then we say they form a **basis** of $V$. \ **Equivalently**, A list of [[vector]]s $v_1, \dots, v_n$ in $V$ forms a **basis** of $V$ **if and only if** every $v \in V$ can be represented as a *unique* [[linear combination]] of $v_1, \dots, v_n$. > ^specialization ---- > [!justification]- Justification. (Proof of Equivalence) > First suppose that $v_1, \dots, v_n$ is a **basis** for $V$. Let $v \in V$. Because $v_1, \dots, v_n$ [[spans]] $V$, there exist $a_1, \dots, a_n \in \ff$ such that $v = a_1v_1 + \dots + a_nv_n$. To show uniqueness, suppose that $c_1, \dots, c_n$ are scalars in $\ff$ for which $v = c_1v_1 + \dots + c_nv_n$. Then we have $0=(a_1-c_1)v_1 + \dots + (a_n-c_n)v_n$ and by [[linearly independent|linear independence]] of $v_1, \dots, v_n$ this implies $a_i-c_i=0$ for $1 \leq i \leq n$. > \ > Conversely suppose that every $v \in \ff$ can be written uniquely in the form $v=a_1v_1 + \dots + a_nv_n, \ \ a_1, \dots,a_n \in \ff.$ Then $v_1, \dots, v_n$ [[spans]] $V$ by definition. For the case of $v=0$ we see the uniqueness of the representation imply [[linearly independent|linear independence]]. > [!basicexample]- > - The list $(1,1,0), (0,0,1)$ is a **basis** for the [[linear subspace]] of $\ff^3$ given by $\lb (x,x,y) \in \ff^3 :x,y \in \ff \rb.$ > - The list $1,z,\dots,z^m$ is a **basis** of [[set of all polynomials with coefficients in F and degree at most m]] ^921ff1 ^cd707a > [!basicnonexample]- > - The list $(1,2,-4), (7,-5,6)$ is [[linearly independent]] in $\ff^3$ but is not a **basis** of $\ff^3$. This is due to [[length of linearly independent list is at most length of spanning list]]: since $e_1,e_2,e_3$ in $\ff^3$ are [[linearly independent]], we know that no list of length less than $3$ may [[submodule generated by a subset]] $\ff^3$. > - The list $(1,2),(3,5),(4,13)$ [[spans]] $\ff^2$ but is not a [[basis]] of $\ff^2$ because it is not [[linearly independent]]: by [[length of linearly independent list is at most length of spanning list]] we have that no list of length larger than $2$ [[spans]] $\ff^2$ since $e_1,e_2$ [[submodule generated by a subset]] $\ff^2$. ^6f75a8 ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```