Dimension theorem for vector spaces

Template:Short description 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 infinite (in the latter case, it is a cardinal number), and defines the dimension of the vector space.

Formally, the dimension theorem for vector spaces states that:

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:

In particular if V is finitely generated, then all its bases are finite and 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, 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.

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 Template:Mvar be a vector space, {a_i : i ∈ I} be a linearly independent set of elements of Template:Mvar, and {b_j : j ∈ J} be a generating set. One has to prove that the cardinality of Template:Mvar is not larger than that of Template:Mvar.

If Template:Mvar is finite, this results from the Steinitz exchange lemma. (Indeed, the Steinitz exchange lemma implies every finite subset of Template:Mvar has cardinality not larger than that of Template:Mvar, hence Template:Mvar is finite with cardinality not larger than that of Template:Mvar.) If Template:Mvar is finite, a proof based on matrix theory is also possible.

Assume that J is infinite. If Template:Mvar is finite, there is nothing to prove. Thus, we may assume that Template:Mvar is also infinite. Let us suppose that the cardinality of Template:Mvar is larger than that of Template:Mvar. 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 Template:Mvar. This maximality implies that Template:Mvar spans Template:Mvar and is therefore a basis (the maximality implies that every element of Template:Mvar is linearly dependent from the elements of Template:Mvar, and therefore is a linear combination of elements of Template:Mvar). As the cardinality of Template:Mvar is greater than or equal to the cardinality of Template:Mvar, one may replace {a_i: i ∈ I} with Template:Mvar; that is, one may suppose, without loss of generality, that {a_i : i ∈ I} is a basis.

Thus, every b_j can be written as a finite sum $$b_j=\sum _{i\in E_j}{\lambda }_{i,j}{a}_i,$$ where $E_j$ is a finite subset of $I.$ Let Template:Tmath. Since Template:Mvar is spanned by Template:Tmath, which is itself spanned by the Template:Tmath, the latter set spans Template:Mvar. Since this set is a subset of a basis, this implies that Template:Tmath and Template:Tmath. This shows that the cardinality of Template:Tmath is at most the one of Template:Tmath, since the cardinality of an infinite indexed family of finite sets is at most the cardinality of the index set.

Kernel extension theorem for vector spaces

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.