Let V be a vector space, {ai: i ∈ I} be a linearly independent set of elements of V, and {bj: j ∈ J} be a generating set. One has to prove that the cardinality of I is not larger than that of J.
If J is finite, this results from the Steinitz exchange lemma. (Indeed, the Steinitz exchange lemma implies every finite subset of I has cardinality not larger than that of J, hence I is finite with cardinality not larger than that of J.) If J is finite, a proof based on matrix theory is also possible.2
Assume that J is infinite. If I is finite, there is nothing to prove. Thus, we may assume that I is also infinite. Let us suppose that the cardinality of I is larger than that of J.3 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 K. This maximality implies that K spans V and is therefore a basis (the maximality implies that every element of V is linearly dependent from the elements of K, and therefore is a linear combination of elements of K). As the cardinality of K is greater than or equal to the cardinality of I, one may replace {ai: i ∈ I} with K, that is, one may suppose, without loss of generality, that {ai: i ∈ I} is a basis.
Thus, every bj can be written as a finite sum b j = ∑ i ∈ E j λ i , j a i , {\displaystyle b_{j}=\sum _{i\in E_{j}}\lambda _{i,j}a_{i},} where E j {\displaystyle E_{j}} is a finite subset of I . {\displaystyle I.} As J is infinite, ⋃ j ∈ J E j {\textstyle \bigcup _{j\in J}E_{j}} has the same cardinality as J.4 Therefore ⋃ j ∈ J E j {\textstyle \bigcup _{j\in J}E_{j}} has cardinality smaller than that of I. So there is some i 0 ∈ I {\displaystyle i_{0}\in I} which does not appear in any E j {\displaystyle E_{j}} . The corresponding a i 0 {\displaystyle a_{i_{0}}} can be expressed as a finite linear combination of b j {\displaystyle b_{j}} s, which in turn can be expressed as finite linear combination of a i {\displaystyle a_{i}} s, not involving a i 0 {\displaystyle a_{i_{0}}} . Hence a i 0 {\displaystyle a_{i_{0}}} is linearly dependent on the other a i {\displaystyle a_{i}} s, which provides the desired contradiction.
This application of the dimension theorem is sometimes itself called the dimension theorem. Let
be a linear transformation. Then
that is, the dimension of U is equal to the dimension of the transformation's range plus the dimension of the kernel. See rank–nullity theorem for a fuller discussion.
Howard, P., Rubin, J.: "Consequences of the axiom of choice" - Mathematical Surveys and Monographs, vol 59 (1998) ISSN 0076-5376. /wiki/Jean_E._Rubin ↩
Hoffman, K., Kunze, R., "Linear Algebra", 2nd ed., 1971, Prentice-Hall. (Theorem 4 of Chapter 2). ↩
This uses the axiom of choice. ↩