Lemma: u − v = 0 iff u = v.

Proof: Assume u − v = 0

u + (−1)v = 0

u + (−1)v + 1v = 1v (add 1v to both sides)

u + ((−1) + 1)v = 1v

u + 0v = v

u + 0 = v

u = v

and conversely

This is not yet a proof!!

The set of n-tuples of the field is a vector space and any vector space with a basis with n elements is isomorphic to this space of tuples according to the obvious isomorphism:
Σα_{i}u_{i} ↔ (α_{0}, ... α_{n−1})
but to see this we must show that if
Σα_{i}u_{i} = Σβ_{i}u_{i}
then for each i α_{i} = β_{i}.

Assume Σα_{i}u_{i} = Σβ_{i}u_{i}

Σα_{i}u_{i} − Σβ_{i}u_{i} = 0

Σ(α_{i}u_{i} − β_{i}u_{i}) = 0

Σ(α_{i} − β_{i})u_{i} = 0

but by the meaning of ‘independence of the {u_{i}}’ we have:

for each i α_{i} − β_{i} = 0

for each i α_{i} = β_{i}

This establishes the one to one nature of the correspondence and thus the isomorphism.
It also shows that two vector spaces with equinumerous bases are isomorphic to each other.
A vector space with a basis of n elements is called an **n-dimensional** vector space and is unique up to isomorphism.
Indeed this produces all isomorphisms between two such spaces.