basis theorem



The basis theorem for vector spaces states that every vector space VV admits a basis, or in other words is a free module over its ground field of scalars. It is a famous classical consequence of the axiom of choice (and is equivalent to it by a result of Andreas Blass, proved in 1984).

Statement and proofs

Basis theorem

If VV is a vector space over any field (or, say, a left vector space over a skewfield) KK, then VV has a basis.


We apply Zorn's lemma as follows: consider the poset consisting of the linearly independent subsets of VV, ordered by inclusion (so SSS \leq S' if and only if SSS \subseteq S'). If (S α)(S_\alpha) is a chain in the poset, then

S= αS αS = \bigcup_\alpha S_\alpha

is an upper bound, for each vv in the span of SS belongs to some S αS_\alpha and is uniquely expressible as a finite linear combination of elements in S αS_\alpha and in any S βS_\beta containing S αS_\alpha, hence uniquely expressible as a finite linear combination of elements in SS. Thus the hypothesis of Zorn’s lemma obtains for this poset; therefore this poset has a maximal element, say BB.

Let WW be the span of BB. If WW were a proper subspace of VV, then for any vv in the set-theoretic complement of WW, B{v}B \cup \{v\} is a linearly independent set (for if

a 0v+a 1w 1++a nw n=0(w 1,,w nB)a_0 v + a_1 w_1 + \ldots + a_n w_n = 0 \qquad (w_1, \ldots, w_n \in B)

we must have a 0=0a_0 = 0 – else we can multiply by 1/a 01/a_0 and express vv as a linear combination of the w iw_i, contradicting ¬(vW)\neg(v \in W) – and then the remaining a ia_i are 0 since the w iw_i are linearly independent). This contradicts the maximality of BB. We therefore conclude that W=VW = V, and BB is a basis for VV.

I’ll write out a proof of the converse, that the axiom of choice follows from the basis theorem, as soon as I’ve digested it – Todd.

Equipotence of vector space bases (Steinitz)

Any two bases of a vector space are of the same cardinality.


Given any linearly independent set AA and spanning set CC, if ACA \subseteq C, then there is a basis BB with ABCA \subseteq B \subseteq C; the theorem above is the special case where A=A = \empty and C=VC = V. The proof of this more general theorem is a straightforward generalisation of the proof above.

Last revised on March 30, 2016 at 05:35:35. See the history of this page for a list of all contributions to it.