----- > [!proposition] Proposition. ([[vector space basis iff maximal linearly independent]]) > A [[basis]] of a [[free module]] $M$ over *any* [[ring]] $R$ is necessarily [[maximal|maximally]] [[linearly independent]] and [[maximal|minimally]] [[submodule generated by a subset|generating]]. If $R=k$ is a [[field]] (so that $M=V$ is a [[vector space]]), then the converse holds: $B$ is a [[basis]] of $V$ if and only if $B$ is a [[maximal|maximally]] [[linearly independent]] subset of $V$. ^proposition > [!basicnonexample] Warning. > This lemma fails over general [[ring|rings]]. Viewing $\mathbb{Z}$ as a $\mathbb{Z}$-[[module]], $\{ 2 \} \subset \mathbb{Z}$ does not [[submodule generated by a subset|generate]] $\mathbb{Z}$ but is [[maximal]] [[linearly independent]] [^1]. ^nonexample [^1]: Indeed, if we were to append $n \in \mathbb{Z}$ to $\{ 2 \}$, the equation $2r+ns=0$would be nontrivially solved by taking $s=-2$ and $r=n$... > [!specialization] Statement for finite-dimensional vector spaces. > Suppose $V$ is [[vector space#Finite-Dimensional Vector Space|a finite-dimensional vector space]]. A list $B \subset V$ is a [[basis]] of $V$ if and only if it is a maximal [[linearly independent]] set. > > (Here maximal [[linearly independent]] means that appending any $v \in V$ to $B$ will cause $B$ to fail to be [[linearly independent]].) > [!proof]- Proof. ([[vector space basis iff maximal linearly independent]]) > > Let $v \in V$, $v \notin B$. Then $B \cup \{ v \}$ is not [[linearly independent]], by the [[maximal|maximality]] of $B$; thus there exist $c_{0},\dots,c_{t} \in k$ and (distinct) $b_{1},\dots,b_{t}$ such that $c_{0}v+c_{1}b_{1}+ \dots + c_{t}b_{t}=0,$ > with not all $c_{0},\dots,c_{t}$ zero. Rearranging: $c_{0}v=-(c_{1}b_{1} + \dots + c_{t}b_{t}).$ > If $v=0$ there is nothing to show. So assume $v \neq 0$. Then $c_{0} \neq 0$, for if $c_{0}=0$ then $0=-(c_{1}b_{1} + \dots +c_{t}b_{t})=c_{1}b_{1}+\dots+c_{t}b_{t}$ and [[linearly independent|linear independence]] implies $c_{1}=\dots=c_{t}=0$. Thus using that $c_{0}$ is a [[unit]] we can represent $v=-c_{0}^{-1}(c_{1}b_{1}+\dots+b_{t})$ > and since $v$ is arbitrary this shows that $B$ [[submodule generated by a subset|generates]] $V$. > [!proof] Proof for finite-dimensional vector spaces. > Suppose $\dim V=n$. Let $B \subset V$. > > Suppose $B$ be is **maximal [[linearly independent]] subset** of $V$. By [[linearly independent list of the right length is a basis]] we're done with one direction if we can show $\len \ B = n$. $B$ [[every linearly independent list extends to a basis|can be extended to a]] [[basis]] of $V$ by adjoining to $B$ some number $k \in \nn$ of [[vector]]s. Since $B$ is maximally [[linearly independent]], $k=0$ (otherwise the resultant list would not be a [[basis]]). So $\len \ B=n$ as required. > > Conversely suppose $B$ is a [[basis]] for $V$. Adjoin to $B$ a [[vector]] $v \in V$, note that $B \cup \{ v \}$ [[spans]] $V$ because $B$ does. Then by [[length of linearly independent list is at most length of spanning list]], we have that because the original $B$ with length $n$ [[spans]] $V$, then the new $B \cup \{ v \}$ with length $n+1$ cannot be [[linearly independent]]. Hence $B$ is maximal [[linearly independent]] as required. > ----- #### ---- #### 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 > ```