Editing Basis (linear algebra) (section)

== Properties ==

Many properties of finite bases result from the [[Steinitz exchange lemma]], which states that, for any vector space {{mvar|V}}, given a finite [[spanning set]] {{mvar|S}} and a [[linearly independent]] set {{mvar|L}} of {{mvar|n}} elements of {{mvar|V}}, one may replace {{mvar|n}} well-chosen elements of {{mvar|S}} by the elements of {{mvar|L}} to get a spanning set containing {{mvar|L}}, having its other elements in {{mvar|S}}, and having the same number of elements as {{mvar|S}}.

Most properties resulting from the Steinitz exchange lemma remain true when there is no finite spanning set, but their proofs in the infinite case generally require the [[axiom of choice]] or a weaker form of it, such as the [[ultrafilter lemma]].

If {{mvar|V}} is a vector space over a field {{mvar|F}}, then:
* If {{mvar|L}} is a linearly independent subset of a spanning set {{math|''S'' ⊆ ''V''}}, then there is a basis {{mvar|B}} such that <math display="block">L\subseteq B\subseteq S.</math>
* {{mvar|V}} has a basis (this is the preceding property with {{mvar|L}} being the [[empty set]], and {{math|1=''S'' = ''V''}}).
* All bases of {{mvar|V}} have the same [[cardinality]], which is called the [[Dimension (vector space)|dimension]] of {{mvar|V}}. This is the [[dimension theorem for vector spaces|dimension theorem]].
* A generating set {{mvar|S}} is a basis of {{mvar|V}} if and only if it is minimal, that is, no [[subset|proper subset]] of {{mvar|S}} is also a generating set of {{mvar|V}}.
* A linearly independent set {{mvar|L}} is a basis if and only if it is maximal, that is, it is not a proper subset of any linearly independent set.

If {{mvar|V}} is a vector space of dimension {{mvar|n}}, then:
* A subset of {{mvar|V}} with {{mvar|n}} elements is a basis if and only if it is linearly independent.
* A subset of {{mvar|V}} with {{mvar|n}} elements is a basis if and only if it is a spanning set of {{mvar|V}}.