Let S be a Vector Space which consists of V (set of vectors) and F (field). Fifth axiom of vector spaces states that
1∗α=α
where 1 ∈ F and α ∈ V.
We can change this axiom as follows; there is an element, b, in F such that
b*α=α
Now, we can prove that for any vector space (with slightly different axiom), b can be replaced by 1.
Proposition: b can be replaced by 1. In fact, if V has some non-zero vectors and S is a vector space, then b must be 1.
Proof: If V has only zero vector, then
b*0=0
for every b ∈ F, particularly for b=1
If V has some non-zero vectors and for every α ∈ V
b*α=α
then b must be equal to 1. Since there is at least one non-zero vector,
b≠0
From another axiom of vector spaces, we know that
c1∗c2(α)=c1(c2α)
Since b≠0, below equalities are valid
(b-1∗b)α=1α
b-1(bα)=b-1α
then,
1α=b-1α
(1-b-1)α=0
since this is true for even non-zero vectors in S,
1-b-1=0
1=b-1
b=1
What does this proposition mean? It means that every vector space in which fifth axiom is slightly different is “normal” vector space.