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1.4.6. Proposition. Let ℬ = {v1, v2, ..., vn} a basis of a vector space V, then ℬ is a maximal set of linear independent vectors of V.

Proof. By definition the vectors {v1, v2, ..., vn} are linear independent; we must show that for any other vector v ∈ V, v ∉ ℬ the vectors v1, v2, ..., vn, v are linear dependent. Since ℬ spans V, there exist λ1, ..., λn ∈ 𝕂, such that

v = λ1v1 + ... + λnvn

thus we have

v − λ1v1 − ... − λnvn = O

which is a linear dependence relation between the vectors v1, v2, ...,vn (the coefficient of v is 1, which is not null).   □

The converse is also true as we shall prove after introducing the following lemma.

1.4.7. Lemma. Let ℬ = {v1, v2, ..., vr} a subset of a vector space V. Suppose that Span(ℬ) contains a set of generators of V. Then V = Span(ℬ), i.e. ℬ is also a set of generators.

Proof. Let 𝒜 ⊂ Span(ℬ) a set of generators of V. We can write a generic vector v ∈ V as linear combination of the vectors of 𝒜. Since 𝒜 ⊂ Span(ℬ), we can also write an element of 𝒜 as linear combination of elements of ℬ, that is v ∈ Span(ℬ). Being v arbitrary, ℬ must be a basis of V.  □

1.4.8. Proposition. Let 𝒜 = {v1, v2, ..., vk} a set of generators of V. Let ℬ ⊆ 𝒜 a maximal subset of A of linearly independent vectors. Then ℬ is basis of V.

Proof. Let ℬ = {v1, v2, ..., vr} with r ≤ k. Since the elements of ℬ are linearly independent we are left to prove that V = Span (ℬ). By Lemma 1.4.7, it suffices to show that 𝒜 ⊂ span(ℬ). Let w ∈ 𝒜, which is not in ℬ. Then v1, v2, ..., vr, w are linearly dependent, thus there exist α1, αr ∈ ℝ such that

α1v1 + ... + λrvr + βw = O

β ≠ 0, otherwise, we would have written a relation of linear dependence among the vectors of ℬ. Then

$$ \mathbf{w} = -\frac{\alpha_1}{\beta}\mathbf{v}_1 - \ldots - \frac{\alpha_r}{\beta} \mathbf{v_r} \in \text{Span{ℬ}} $$

Since w is an arbitrary element of 𝒜, then 𝒜 ⊂ Span(ℬ).  □

Sat, 15 Apr 2023 09:00 GMT