Justin Tatch Moore

This note gives a “Zorn’s Lemma” style proof that any two bases in a vector space have the same cardinality. One of the most fundamental notions in linear algebra is that of a basis: A subset B of a vector space V is a basis if every element of V is a unique linear combination of elements of B. The following theorem of Georg Hamel expresses one of the most important aspects of this definition. Dimension Theorem. If V is a vector space and A and B each form a basis for V , then |A| = |B|. This common cardinality of a basis for V is of course known as the dimension of V . Typically, this result is proved in an undergraduate course, but only under the assumption that A is finite. The proof of the general result is usually considered “beyond the scope of the text.” One reason for this is presumably that the best-known proofs use some nontrivial facts about cardinal arithmetic for infinite sets (e.g., any infinite set is equinumerous with the set of all of its finite subsets). While important in their own right, these facts are nevertheless tangential from the point of view of linear algebra. The purpose of this note is to give a self-contained Zorn’s Lemma-style proof of the Dimension Theorem, in the same spirit that it is so often used in more advanced undergraduate courses. We will take the finite instance of the Dimension Theorem for granted and treat it as a black box. Let’s begin with the definition of the partial ordering to which we will apply Zorn’s lemma; once this simple definition is in place, the rest of the proof follows naturally. Define P to be the set consisting of all linear transformations T from a subspace W of V onto W such that: 1. A ∩W and B ∩W are each a basis for W , and 2. the restriction of T to A ∩W is a bijection between A ∩W and B ∩W . This set is equipped with a natural partial ordering: Define S ≤ T to mean that the domain of S is included in the domain of T and the restriction of T to the domain of S equals S. Notice that P is nonempty, since the transformation 0 7→ 0 defined on the trivial subspace {0} of V is in P. It is not immediately obvious, however, that P contains any other elements even if V is nontrivial; we will remark more on this at the end of the proof. Recall that Zorn’s Lemma is the following assertion. Zorn’s Lemma. If P is a partially-ordered set in which every totally ordered subset of P has an upper bound in P, then P has a maximal element. http://dx.doi.org/10.4169/amer.math.monthly.121.03.260 MSC: Primary 15A03, Secondary 03E25 260 c © THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121 This content downloaded from 80.62.116.155 on Sun, 18 May 2014 09:21:47 AM All use subject to JSTOR Terms and Conditions The proof that the partial order P we defined above isn’t trivial is closely related to our main task: demonstrating that if T : W → W is a maximal element of P, then W must equal V . First we will prove, however, that P has a maximal element. This is easy and completely standard. If C ⊆ P is totally ordered, let T be the function whose domain W is the union of the domains of elements of C and that satisfies T u = v if there is an element S of C such that Su = v. It is easy to verify that, since C is totally ordered, W is a subspace of V spanned by both A ∩W and B ∩W , T is well defined (i.e., the definition of T u does not depend on which element S of C is used in its definition), and that T defines a linear transformation from W to W , which maps A ∩W bijectively to B ∩W . Thus, we may apply Zorn’s Lemma and find a T : W → W , which is a maximal element of P. Our task now will be to prove that W = V . Suppose that this is not the case and construct increasing sequences A1, A2, . . . and B1, B2, . . . of finite subsets of A and B respectively, such that for all natural numbers n: • A1 is not contained in W , • either An ∪W spans V or else An is properly contained in An+1, • An is contained in the span of Bn , and • Bn is contained in the span of An+1. Define A∞ = (A1 ∪ A2 ∪ . . .) \W