Infinite dimensional vector spaces

This is a quick note to prove that two bases of an infinite dimensional vector space have the same cardinality. We freely use the axiom of choice and other standard facts about cardinalities of infinite sets. We will in fact prove the following:

Theorem: Let V be a vector space with basis \{v_i\}_{i\in I} with I an infinite set. Let \{w_j\}_{j\in J} be a linearly independent subset of V. (e.g. a basis of a subspace). Then |J|\leq |I|.

To prove this, WLOG J is a basis of V (by extending \{w_j\}_{j\in J} to a basis of V if necessary). For all i\in I, write

    \[v_i=\sum_j c_{ij}w_j.\]

Let E\subset I\times J be the set of pairs (i,j) with c_{ij}\neq 0. Then E\to I has finite fibres, since the sum above is finite, and E\to J is surjective, since the v_i lie in the span of the w_j with j in the image of V, but also the v_i generate V. Since I is assumed infinite, this is enough to prove that |I|\geq |J|, as required.

Leave a Comment

Filed under maths

Leave a Reply

Your email address will not be published. Required fields are marked *