Theorem Demonstration on Vector Space Bases 4

In a vector space V with a known dimension dim(V)=n, if {v₁,v₂,..,v_n} are a set of linearly independent vectors, they also form a basis of V

Proof

This theorem hinges on the premise that the dimension of the vector space is already known to us.

$$ \dim(V)= n $$

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

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

In this scenario, the theorem posits that the vectors {v₁,v₂,..,v_n} form a set of linearly independent vectors.

The question remains whether they also form a set of V generators.

According to a previously proven theorem, in a set of linearly independent vectors {v₁,v₂,..,v_p} with a number p

However, in this case, the set of linearly independent vectors consists of p=n vectors.

Thus, no additional vector can be added.

Note. Adding another vector to the set would yield a generator set with n+1 vectors {v₁,v₂,..,v_n+1}, including one linearly dependent on the others. Moreover, with n+1 vectors, it could not form a basis.

Therefore, the set of linearly independent vectors {v₁,v₂,..,v_n} is already a basis of the vector space.

And so on.