ON THE FUNDAMENTAL THEOREM OF LINEAR
ALGEBRA
MARTIN FLUCH
Abstract. The Fundamental Theorem of Linear Algebra states that
every vector space has a basis. In its full generality the proof of this
theorem requires the use of Zorn’s lemma which is known to be equivalent to the Axiom of Choice in Set Theory. In 1984 it has been proven
by Andreas Blass that the statement of the fundamental theorem is so
strong that one can deduce from it the Axiom of Choice. The aim of this
text is to present this relation ship between the Fundamental Theorem
of Linear Algebra and the Axiom of Choice.
1. The Fundamental Theorem of Linear Algebra
The Fundamental Theorem of Linear Algebra is often introduced in a
slightly weaker form for finitely generated vector spaces in the first course of
Linear Algebra. In its most general form it just takes six words to write the
theorem down:
Fundamental Theorem of Linear Algebra. Every vector space has a
basis.
Proof. Let V be a vector space. Then the set L of all linear independent
subsets of V is partially ordered1 by inclusion.2 Let
X0 ⊂ X1 ⊂ X2 ⊂ X3 ⊂ . . .
be a ascending chain in L, that is a ascending chain of linear independent
subsets of V . Then the union
X :=
[
Xi
2008-11
i∈N
is a linear independent subset of V , hence an element of L. By construction
X is an upper bound for the Xi , that is Xi ⊂ X for all i ∈ N. Therefore
every totaly ordered chain of elements in L has an upper bound in L and
Date: January 20, 2011.
1A partial order is a relation which is reflexiv, anti-symmetric and transitive.
2Any familiy of sets is always partially ordered by inclusion.
2
MARTIN FLUCH
Zorn’s Lemma (see next section) implies that there exists a maximal element
M ∈ L. A maximal element in L is a maximal linear independent subset of
V and this characterices a basis of a vector space. Thus we have found a
basis for V , namely M .
Note the use of Zorn’s Lemma in the proof as it is the essential step of
the proof. For finitely generated vector spaces we do not need to use this
heavy machinery, here the existence of a basis follows from the observation
that every finite generating system contains a minimal generating system,
hence a basis.
Example. As a consequence of the Fundamental Theorem of Linear Algebra, the set of real numbers R has a basis B as a vector space over the
rational numbers Q. This means that for any real number x ∈ R there exists
a unique and finite subset B(x) ⊂ B such that
x=
X
ab (x) · b
b∈B(x)
with ab (x) being unique non-zero rational numbers for every b ∈ B(x)! Since
Q is countable and R is not the set B must be uncountable. But a concrete
algebraic description of such a set is unknown. The Fundamental Theorem
of Linear Algebra just states the existence of the set B with this properties.
2. The Axiom of Choice
The development of Axiomatic Set Theory was a necessity to create a
proper foundation for mathematics which did not include paradoxes like
for example the famous Russell’s Paradox [Sup72].3 There exists differnet
variants of theories for sets. In the following we will us the theory of sets
developed by Zermelo–Fraenkel (which we shall abreviate in what follows by
ZF) in the first half of the 20th century. One of the nine axioms of ZF is the
Axmiom of Choice.
3If R := {x : x ∈
/ x}, then R ∈ R ⇐⇒ R ∈
/ R.
ON THE FUNDAMENTAL THEOREM OF LINEAR ALGEBRA
3
Axiom of Choice. For every set A of non-empty sets there exists a function
f : A → ∪A
from A to the union ∪A of the elements of A such that f (B) ∈ B for every
B ∈ A.
In short short it says that for any set A of non-empty sets we can choose
one element f (B) ∈ B for each B ∈ A and this simultaneously.
There exist many statements which are equivalent to the Axiom of Choice.
One of them is Zorn’s Lemma, the other is the Axiom of Multiple Choice.
Zorn’s Lemma. Assume that X is a non-empty set which is partially order
by a relation “≤”. Assume that for every linearly ordered chain
a0 ≤ a1 ≤ a2 ≤ a3 ≤ . . .
in X there exists a upper bound a ∈ X for the elements ai , i ∈ N. Then there
exists a maximal element m ∈ X, that is for every x ∈ X, m ≤ x implies
x = m.
Note that the maximal element m ∈ X is not necessarily unique! The
proof that the Axiom of Choice and Zorn’s Lemma are equivalent statements
can be found in any book on Axiomatic Set Theory.
Axiom of Multiple Choice. For every set A of non-empty sets there exists
a function
f : A → ∪A
from A to the union ∪A of the elements of A such that f (B) is a finite and
non-empty subset of B for every B ∈ A.
Note here, that the emphasisement lies on the finiteness of the subsets.
The fact that the Axiom of Choice implies the Axiom of Multiple Choice is
evident. For the converse implication see for example [Jec73].
Compare the following with Figure 1: In the first part of this text we
have shown that Zorn’s Lemma implies the existence of a basis for arbitrary
vector spaces (implication 1). The equivalences 2 and 3 have been stated
4
MARTIN FLUCH
Zorn’s
Lemma
1
Fundametal
Theorem of
Linear Algebra
2
4
Axiom of
Choice
3
Axiom of
Multiple Choice
Figure 1. The relation ship between the different statements in this text.
just above in this section without proof. And the remaining part of this text
aims to show how the the existence of a basis for arbitrary vector spaces
implies the Axiom of Multiple Choice (implication 4).
3. Existence of Bases Implies Axiom of Multiple Choice
Theorem (Blass, 1984). The Fundamental Theorem of Linear Algebra implies the Axiom of Choice in ZF.
Before we begin with the proof of the theorem we introduce some notation.
Let k be an arbitrary field and X a set. We can form the a new field k(X) by
adjoining as all the elements of X as indeterminants. The field k(X) is then
the field of all rational functions in the indeterminats X. We can identify k
as a subfield of k(X) and X as a subset of k(X) in the obvious way.
Consider a monomial
y = axr11 · · · xrnn ,
a ∈ k, x1 , . . . , xn ∈ X, r1 , . . . , rn ∈ N.
(1)
If A is a subset of X then the A-degree of y is the sum of all the exponents
ri for which xi ∈ A. If y is an arbitrary element of k(X), then we say that
y is of A-homogeneous of degree d if y is the quotient
y=
f (x1 , . . . , xn )
g(x1 , . . . , xn )
ON THE FUNDAMENTAL THEOREM OF LINEAR ALGEBRA
5
of two polynomials such that every monomial of f has the same A-degree d1
and such that every monomial of g has the same A-degree d2 with
d = d1 − d2 .
If Xi , i ∈ I, is a family of subsets of X indexed by some index set I, then
for i ∈ I we understand by the i-degree of a monomial as in (1) its Xi -degree.
Similarly we say that an element y ∈ k(X) is of i-homogeneous of degree d
if it is Xi -homogeneous of degree d.
Note that the elements of i-homogeneous degree 0 for all i ∈ I form a
subfield of k(X) which we denote by kI (X).
Proof of the Theorem [Bla84]. Let Xi , i ∈ I, be a family of non-empty sets
indexed by an abstract index set I. We have to show in ZF without the use
of the Axiom of Choice that we can construct a family Fi , i ∈ I, of finite
and non-empty sets such that Fi ⊂ Xi for every i ∈ I.
Without any loss of generality we may assume that the sets Xi are pairwise
disjoint. We set
[
X :=
Xi .
i∈I
Let k be an arbitrary field. Consider the field k(X). It is a vector space
over the subfield K := kI (X) and contains X as a subset. Let V be the
K-vector space generated by the set X. By assumption V has a basis B.
For every i ∈ I and x ∈ Xi we have that x 6= 0 and thus by the property
of a basis there exists a unique non-empty finite subset B(x) ⊂ B such that
x=
X
ab (x) · b
(2)
b∈B(x)
with ab (x) being unique non-zero elements of K. If x, y ∈ Xi , then y/x is
j-homogeneous of degree 0 for every j ∈ I and thus y/x ∈ K. Thus we can
multiply (2) by y/x and obtain
y=
X
(y/x)ab (x) · b
b∈B(x)
6
MARTIN FLUCH
which is on the other hand also equal to
X
=
ab (y) · b
b∈B(y)
Comparing the coefficients we obtain that B(x) = B(y) and (y/x)ab (x) =
ab (y) for all b ∈ B(x) = B(y). Thus B(x) is independent of the choice of
x ∈ Xi and the numbers ab (x)/x depend only on b ∈ B(x). For every i ∈ I
we set
Bi := B(x)
for some element x ∈ Xi and for every b ∈ Bi we set
βbi := ab (x)/x
which are then elements in the field k(X).
Since the numbers ab (x) have i-homogeneous degree 0 (they are elements
of the field K) we have that βbi has i-homogeneous degree −1. Thus, whenever we write βbi as a quotient
βbi =
fbi (x1 , . . . , xn )
gbi (x1 , . . . , xn )
of two polynomials, then some of the (finitely many) variables x1 , . . . , xn ∈ X
are elements in Xi and therefore these must appear in the polynomial gbi .
This is in particular true if we write βbi in reduced form. Since the reduced
form of an element in k(X) is unique we obtain well defined subsets Fi ⊂ Xi
when we set
Fi := {x ∈ Xi : x appears in gbi in reduced form for some b ∈ Bi }.
These sets are clearly non-empty and finite.
Thus we have constructed, without the use of the Axiom of Choice, for
every i ∈ I a finite, non-empty subset Fi ⊂ Xi and this proves the the
theorem.
References
[Bla84] Andreas Blass. Existence of bases implies the axiom of choice. In Axiomatic set
theory (Boulder, Colo., 1983), volume 31 of Contemp. Math., pages 31–33. Amer.
Math. Soc., Providence, RI, 1984.
[End77] Herbert B. Enderton. Elements of set theory. Academic Press [Harcourt Brace
Jovanovich Publishers], New York, 1977.
ON THE FUNDAMENTAL THEOREM OF LINEAR ALGEBRA
7
[Fra53] Abraham A. Fraenkel. Abstract set theory. Studies in logic and the foundations
of mathematics. North-Holland Publishing Co., Amsterdam, 1953.
[Jec73] Thomas J. Jech. The axiom of choice. North-Holland Publishing Co., Amsterdam,
1973. Studies in Logic and the Foundations of Mathematics, Vol. 75.
[Sup72] Patrick Suppes. Axiomatic set theory. Dover Publications Inc., New York, 1972.