Sooji Shin

Theorem. Let V be a vector space and M a linearly independent subset of V. Then there exists a basis of V containing M, i.e., M can be extended to a basis.

Proof. The set [M] of all linearly independent subsets of V containing M is not empty and is ordered by set theorical indlusion. Every totally ordered subset β of [M] possesses the upper bound in [M]. Consequently there exists a maximal element B in [M], by Zorn's lemma. For ever vector x not in B, the set B∪{x} is linearly dependent as a proper superset of B. Thus there exist vectors y₁, y₂, ..., y_n in B and scalars a₀, a₁, ..., a_n which are not all zero, so that a₀x + a₁y₁ + ... + a_ny_n = 0. Because of the linear independence of B, we must have a₀ ≠ 0. Thus we obtain that x is a linear combination of the vectors y₁, y₂, ..., y_n, i.e., lies in the span(linear hull) of B. Thus B is a basis of V.
 

Corollary. If V ≠ {0}, then it contains a basis.

Proof. Let v be a nonzero vectors in V and let M = {v} then M is linearly independent. Thus there exists a basis of V that contains M.

Geschrieben von Sooji Shin