Proof of Theorem on Bases of Vector Spaces 3

In a vector space V with a known dimension dim(V)=n, if {v₁,v₂,..,v_n} are a set of generators for V, then they are also a basis for V

Proof

This theorem relies on the assumption that the dimension of the vector space is already known dimension of the vector space

$$ \dim(V)= n $$

If the vector space V has dimension n, then all bases of the vector space V consist of n vectors.

Note. By definition, a basis is a set of generators of V in which all vectors are linearly independent.

In this case, the theorem states that the vectors {v₁,v₂,..,v_n} form a set of generators for V.

According to a previously proven theorem, in a set of generators {v₁,v₂,..,v_s} with a number s>n, more vectors than the base (dimension of the vector space), a basis can be formed by eliminating the s-n linearly dependent vectors.

In this instance, however, the set of generators consists of s=n vectors.

Therefore, no vector can be removed.

Note. Removing a vector would result in a set of n-1 vectors, which would no longer even be a set of generators for the vector space.

Hence, the set of generators {v₁,v₂,..,v_s} is already a basis for the vector space.

And so on.