EDB — 02D

[02D] Let \(V\) be a real vector space. Let \(B⊆ V\) be a subset. A finite linear combination \(v\) of elements of \(B\) is equivalently defined as

• \(v=∑_{i=1}^ nℓ_ i b_ i\) where \(n=n(v)∈ℕ\), \(ℓ_ 1,\ldots , ℓ_ n∈ℝ\) and \(b_ 1,\ldots ,b_ n\) are elements of \(B\);

• \(v=∑_{b∈ B} 𝜆(b) b\) where \(𝜆:B→ℝ\) but also \(𝜆(b)≠ 0\) only for a finite number of \(b∈ B\).

We call \(Λ⊆ ℝ^ B\) the set of functions \(𝜆\) as above, which are non-null only for a finite number of arguments; \(Λ\) is a vector space: so the second definition is less intuitive but is easier to handle.

We will say that \(B\) generates (or, spans) \(V\) if every \(v∈ V\) is written as finite linear combination of elements of \(B\).

We will say that the vectors of \(B\) are linearly independent if \(0=∑_{b∈ B} 𝜆(b) b\) implies \(𝜆≡ 0\); or equivalently that, given \(n≥ 1\), \(ℓ_ 1,\ldots , ℓ_ n∈ℝ\) and \(b_ 1,\ldots ,b_ n∈ B\) all different, the relation \(∑_{i=1}^ nℓ_ i b_ i=0\) implies \(∀ i≤ n,ℓ_ i=0\).

We will say that \(B\) is an algebraic basis (also known as Hamel basis) if both properties apply.

If \(B\) is a basis then the linear combination that generates \(v\) is unique (i.e. there is only one function \(𝜆∈Λ\) such that \(v=∑_{b∈ B} 𝜆(b) b\)).

Show that any vector space has an algebraic basis. Show more in general that for each \(A,G⊆ V\), with \(A\) family of linearly independent vectors and \(G\) generators, there is an algebraic basis \(B\) with \(A⊆ B⊆ G\).

The proof in general requires Zorn’s Lemma; indeed this statement is equivalent to the Axiom of Choice; this was proved by A. Blass in [ 8 ] ; see also Part 1 §6 [ 24 ] .