dimension theorem for vector spaces

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 theorem is a consequence of the following theorem, which is also useful:

In particular if 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 -modules for rings 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 be a vector space, be a linearly independent set of elements of , and be a generating set. One has to prove that the cardinality of is not larger than that of .

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

Assume that is infinite. If is finite, there is nothing to prove. Thus, we may assume that is also infinite. Let us suppose that the cardinality of is larger than that of .This uses the axiom of choice. 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 . This maximality implies that spans and is therefore a basis (the maximality implies that every element of is linearly dependent from the elements of , and therefore is a linear combination of elements of ). As the cardinality of is greater than or equal to the cardinality of , one may replace with , that is, one may suppose, without loss of generality, that is a basis.

Thus, every can be written as a finite sum
$b_j = \sum_ \lambda_ a_i,$
where $E_j$ is a finite subset of $I.$ As is infinite, $\bigcup_ E_j$ has the same cardinality as . Therefore $\bigcup_ E_j$ has cardinality smaller than that of . So there is some $i_0\in I$ which does not appear in any $E_j$. The corresponding $a_$ can be expressed as a finite linear combination of $b_j$s, which in turn can be expressed as finite linear combination of $a_i$s, not involving $a_$. Hence $a_$ is linearly dependent on the other $a_i$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

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.

Notes

References

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