Theorem (Replacement Theorem): Let V be a vectir spaces spanned by a set G of cardinality n, and let L be a linearly independent subset of V containing exactly m vectors. Then m ≤ n and there exists a subset H of G containing exactly n − m vectors such that L ∪ H generates V.
Proof by induction on m. Base case: m = 0….
Suppose the replacement theorem is true for some m ≥ 0. We prove the theorem is true for m + 1. Let L = {v1, …, vm + 1} be a L.I. set.
The set {v1, …, vm} is also linearly independent. So we can apply the induction hypothesis and deduce that there is a subset {u1, …, un − m} of G such that together they span V
This means vm + 1 is the span of the 2 sets. So we can deduced n − m > 0
Also, we can deduce that some ui is not needed.
Corollary: Suppose V is a vector space having a finite basis. Then every basis for V has the same number of vectors.
Corollary: A L.I. (or spanning) set of top cardinality is a basis. L.I. sets can be extended.
Subspaces have a lower dimension.