Editing Dimension theorem for vector spaces

{{short description|All bases of a vector space have equally many elements}}
In [[mathematics]], the '''dimension theorem for vector spaces''' states that all [[Basis (linear algebra)|bases]] of a [[vector space]] have equally many elements. This number of elements may be finite or infinite (in the latter case, it is a [[cardinal number]]), and defines the [[Dimension (vector space)|dimension]] of the vector space.

Formally, the dimension theorem for vector spaces states that:

{{block indent | em = 1.5 | text = Given a vector space {{math|''V''}}, any two bases have the same [[cardinality]].}}

As a basis is a [[generating set]] that is [[linearly independent]], the dimension theorem is a consequence of the following [[theorem]], which is also useful:

{{block indent | em = 1.5 | text = In a vector space {{math|''V''}}, if {{mvar|G}} is a generating set, and {{mvar|I}} is a linearly independent set, then the cardinality of {{mvar|I}} is not larger than the cardinality of {{mvar|G}}.}}

In particular if {{math|''V''}} is [[finitely generated module|finitely generated]], then all its bases are finite and have the  same number of elements.

While the [[mathematical proof|proof]] of the existence of a basis for any vector space in the general case requires [[Zorn's lemma]] and is in fact equivalent to the [[axiom of choice]], the uniqueness of the cardinality of the basis requires only the [[ultrafilter lemma]],<ref>Howard, P., [[Jean E. Rubin|Rubin, J.]]: "Consequences of the axiom of choice" - Mathematical Surveys and Monographs, vol 59 (1998) {{issn|0076-5376}}.</ref> which is strictly weaker (the proof given below, however, assumes [[trichotomy (mathematics)|trichotomy]], i.e., that all [[cardinal number]]s are comparable, a statement which is also equivalent to the axiom of choice). The theorem can be generalized to arbitrary [[module (mathematics)|{{math|''R''}}-modules]] for rings {{math|''R''}} having [[invariant basis number]].

In the finitely generated case the proof uses only elementary arguments of [[algebra]], and does not require the axiom of choice nor its weaker variants.

==Proof==
Let {{mvar|V}} be a vector space, {{math|{''a''<sub>''i''</sub>: ''i'' ∈ ''I''} }} be a [[linearly independent]] set of elements of {{mvar|V}}, and {{math|{''b''<sub>''j''</sub>: ''j'' ∈ ''J''} }} be a [[generating set]]. One has to prove that the [[cardinality]] of {{mvar|I}} is not larger than that of {{mvar|J}}.

If {{mvar|J}} is finite, this results from the [[Steinitz exchange lemma]]. (Indeed, the Steinitz exchange lemma implies every finite subset of {{mvar|I}} has cardinality not larger than that of {{mvar|J}}, hence {{mvar|I}} is finite with cardinality not larger than that of {{mvar|J}}.) If {{mvar|J}} is finite, a proof based on [[matrix (mathematics)|matrix]] theory is also possible.<ref>Hoffman, K., Kunze, R., "Linear Algebra", 2nd ed., 1971, Prentice-Hall. (Theorem 4 of Chapter 2).</ref> 

Assume that {{math|''J''}} is infinite. If {{mvar|I}} is finite, there is nothing to prove. Thus, we may assume that {{mvar|I}} is also infinite. Let us suppose that the cardinality of {{mvar|I}} is larger than that of {{mvar|J}}.<ref group=note name=choice>This uses the axiom of choice.</ref> We have to prove that this leads to a contradiction. 

By [[Zorn's lemma]], every linearly independent set is contained in a maximal linearly independent set {{mvar|K}}. This maximality implies that {{mvar|K}} spans {{mvar|V}} and is therefore a basis (the maximality implies that every element of {{mvar|V}} is linearly dependent from the elements of {{mvar|K}}, and therefore is a linear combination of elements of {{mvar|K}}). As the cardinality of {{mvar|K}} is greater than or equal to the cardinality of {{mvar|I}}, one may replace {{math|{''a''<sub>''i''</sub>: ''i'' ∈ ''I''} }} with {{mvar|K}}, that is, one may suppose, without loss of generality, that {{math|{''a''<sub>''i''</sub>: ''i'' ∈ ''I''} }} is a basis. 

Thus, every {{math|''b''<sub>''j''</sub>}} can be written as a finite sum
<math display="block"> b_j = \sum_{i\in E_j} \lambda_{i,j} a_i,</math>
where <math>E_j</math> is a finite subset of <math>I.</math> As {{mvar|J}} is infinite, <math display="inline">\bigcup_{j \in J} E_j</math> has the same cardinality as {{mvar|J}}.<ref group=note name=choice /> Therefore <math display="inline">\bigcup_{j \in J} E_j</math> has cardinality smaller than that of {{mvar|I}}. So there is some <math>i_0\in I</math> which does not appear in any <math>E_j</math>.  The corresponding <math>a_{i_0}</math> can be expressed as a finite linear combination of <math>b_j</math>s, which in turn can be expressed as finite linear combination of <math>a_i</math>s, not involving <math>a_{i_0}</math>. Hence <math> a_{i_0}</math> is linearly dependent on the other <math>a_i</math>s, which provides the desired contradiction.

==Kernel extension theorem for vector spaces==
This application of the dimension theorem is sometimes itself called the ''dimension theorem''. Let

{{block indent | em = 1.5 | text = {{math|''T'': ''U'' → ''V''}}}}

be a [[linear transformation]]. Then

{{block indent | em = 1.5 | text = {{math|1=dim(range(''T'')) + dim(ker(''T'')) = dim(''U'')}},}}

that is, the dimension of ''U'' is equal to the dimension of the transformation's [[Range of a function|range]] plus the dimension of the [[Kernel (algebra)|kernel]]. See [[rank–nullity theorem]] for a fuller discussion.

==Notes==
{{reflist|group=note}}

==References==
<references />

{{DEFAULTSORT:Dimension Theorem For Vector Spaces}}
[[Category:Theorems in abstract algebra]]
[[Category:Theorems in linear algebra]]
[[Category:Articles containing proofs]]