1st December 2004

Munkres §11

Ex. 11.8 (Morten Poulsen). First recall some definitions: Let V be a vector space over a field K. Let A be a (possibly empty) subset of V. The subspace spanned by A is denoted span_{K}A and is defined by

span_{K}A = { k_{1}a_{1} + ⋯ + k_{n}a_{n} | n ∈ Z_{+} ∪ {0}, k_{1}, …, k_{n} ∈ K, a_{1}, …, a_{n} ∈ A }

i.e. the set of all finite linear combinations of elements from W. If n = 0 then the linear combination is defined to be the zero element in V. If A is the empty set then span_{K}A is the subspace {0}.

The subset A of V is said to be (linearly) independent if n ∈ Z_{+} ∪ {0}, k_{1}, …, k_{n} ∈ K and a_{1}, …, a_{n} ∈ A satisfy

k_{1}a_{1} + ⋯ + k_{n}a_{n} = 0

then

k_{1} = 0, …, k_{n} = 0.

The empty set is (by definition) independent.

The subset A is said to be a basis if A is independent and span_{K}A = V. Note that if A is a basis for V then every element has an unique representation as a finite linear combination of elements of A.

If V = {0} then the empty set is a basis, thus assume in the following that V ≠ {0}.

(a). Assume v ∉ span_{K}A. Suppose

k_{1}a_{1} + ⋯ + k_{n}a_{n} + k_{n+1}v = 0,

where n ∈ Z_{+} ∪ {0}, k_{1}, …, k_{n+1} ∈ K and a_{1}, …, a_{n} ∈ A.

If k_{n+1} = 0 then k_{1} = 0, …, k_{n} = 0, since A is independent. If k_{n+1} ≠ 0 then v = \frac{1}{k_{n+1}} (k_{1}a_{1} + ⋯ + k_{n}a_{n}), contradicting v ∉ span_{K}A. It follows that A ∪ \{v\} is independent.

(b). Let W be the set of independent subsets of V. Define a relation ≺ on W by

∀ w_{1}, w_{2} ∈ W : w_{1} ≺ w_{2} ⇔ w_{1} ⊊ w_{2}.

The relation ≺ is clearly a strict partial order on W.

Let W_{0} be a simply ordered subset of W then

U = ⋃_{w_{0}∈W_{0}} w_{0}

is an independent subset of V: Suppose

k_{1}u_{1} + ⋯ + k_{n}u_{n} = 0,

where n ∈ Z_{+} ∪ {0}, k_{1}, …, k_{n} ∈ K and u_{1}, …, u_{n} ∈ U. Then there are w_{1}, …, w_{n} ∈ W_{0}, such that u_{i} ∈ w_{i} for 1 ≤ i ≤ n. Since W_{0} is simply ordered the subset {w_{1}, …, w_{n}} ⊂ W_{0} has a largest element w, hence u_{1}, …, u_{n} ∈ w. Since w is independent it follows that k_{1} = 0, …, k_{n} = 0, hence U is independent, i.e. U ∈ W.

The set U is clearly an upper bound for W_{0}, hence every simply ordered subset of W has an upper bound in W. Zorn’s Lemma gives that W has a maximal element B.

(c). The maximal element B in W is basis for V: Suppose span_{K}B is a proper subspace of V. Thus there is an element b ∈ V − span_{K}B, hence, by (a), B ∪ {b} ∈ W, contradicting the maximality of B. It follows that B is a basis for V.

Thus we have proved:

Theorem 1. Every vector space has a basis.

A few remarks: We know that if a vector space V over the field K has a basis with n ∈ Z_{+} elements then every basis for V has n elements and n is called the dimension of V over K, and is denoted dim_{K} V = n. The notion of dimension extends to vector spaces with infinite bases. First one proves the following theorem.

Theorem 2. If A and B are bases for V then A and B have the same cardinality.

In view of the previous theorem we define the dimension of a vector space V to be the cardinality of some basis A for V, i.e. dim_{K} V = card A.

Hamel bases. Regard the real numbers as a vector space over the rationals. Then this vector space has a basis, any basis for this vector space is called a Hamel basis. Furthermore one shows that dim_{Q} R = card R.

References