- 1.6
- 2.1

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*= {*v*_{1}, …,*v*_{m + 1}} be a L.I. set.The set {

*v*_{1}, …,*v*_{m}} is also linearly independent. So we can apply the induction hypothesis and deduce that there is a subset {*u*_{1}, …,*u*_{n − m}} of*G*such that together they span*V*This means

*v*_{m + 1}is the span of the 2 sets. So we can deduced*n*−*m*> 0Also, we can deduce that some

*u*_{i}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.

- Define linear maps and their associated subspaces.