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Steinitz Exchange Lemma

Steinitz Exchange Lemma

Jul 12, 20251 min read

Theorem: Steinitz Exchange Lemma

Let B={v1​,⋯,vn​} be a Basis of a finitely generated.md) 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


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