nLab basis theorem



Higher linear algebra

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Basic facts




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 Blass (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.

See also


The original proof of the existence of Hamel bases (after which the concept was named) was for the case of the real numbers regarded as a rational vector space and used (not directly Zorn's lemma but) the well-ordering theorem:

  • Georg Hamel, pp. 460 in: Eine Basis aller Zahlen und die unstetigen Lösungen der Funktionalgleichung: f(x+y)=f(x)+f(y)f(x + y) = f(x) + f(y), Mathematische Annalen 60 (1905) 459–462 [doi:10.1007/BF01457624]

Lecture notes:

See also:

Last revised on November 11, 2023 at 19:41:19. See the history of this page for a list of all contributions to it.