|
In mathematics, the dimension theorem for vector spaces states that all bases of a vector space have equally many elements. This number of elements may be finite, or given by an infinite cardinal number, and defines the dimension of the space. Formally, the dimension theorem for vector spaces states that :Given a vector space ''V'', any two linearly independent generating sets (in other words, any two bases) have the same cardinality. If ''V'' is finitely generated, then it has a finite basis, and the result says that any two bases have the same number of elements. While the 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,〔Howard, P., Rubin, J.: "Consequences of the axiom of choice" - Mathematical Surveys and Monographs, vol 59 (1998) ISSN 0076-5376.〕 which is strictly weaker (the proof given below, however, assumes trichotomy, i.e., that all cardinal numbers are comparable, a statement which is also equivalent to the axiom of choice). The theorem can be generalized to arbitrary ''R''-modules for rings ''R'' having invariant basis number. The theorem for finitely generated case can be proved with elementary arguments of linear algebra, and requires no forms of the axiom of choice. ==Proof== Assume that and are both bases, with the cardinality of ''I'' bigger than the cardinality of ''J''. From this assumption we will derive a contradiction. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「dimension theorem for vector spaces」の詳細全文を読む スポンサード リンク
|