Theorem: Steinitz Exchange Lemma
Let be a basis of a finitely generated vector space .
For every set of linearly independent vectors there are vectors in (without loss of generality - ) such that
is also a basis of .
PROOF
TODO
1 min read
Theorem: Steinitz Exchange Lemma
Let B={v1,⋯,vn} be a basis of a finitely generated vector space (V,K,+,⋅).
For every set {u1,⋯,um}⊂V of linearly independent vectors there are n−m vectors in B (without loss of generality - vm+1,⋯,vn) such that
{u1,⋯,um,vm+1,⋯,vn}is also a basis of (V,K,+,⋅).
PROOF
TODO