Theorem Demonstration on the Bases of Vector Space 1
In a finitely generated vector space V of dimension n, it is always possible to derive a basis B of the vector space from any set of generators {v1,v2,..,vp} by removing some vectors from the set.
Demonstration
Consider a vector space V of dimension n
$$ \dim V = n $$
Given its dimension n, all bases of the vector space consist of n vectors.
Now, let's take a set of generators consisting of p vectors
$$ \{ \vec{v}_1 , \vec{v}_2 , ... , \vec{v}_p \} $$
As a set of generators, it comprises a number of vectors (p) that is greater than or equal to the number of vectors (n) in the basis
$$ p \ge n $$
Next, I'll determine if the set of generators {v1,v2,..,vp} forms a basis for the vector space V.
To do this, I'll check if the vectors {v1,v2,..,vp} are linearly independent.
- If the vectors {v1,v2,..,vp} are linearly independent, then the set of generators also forms a basis for the vector space V. Knowing that all bases have the same number of vectors, it follows that p=n. This concludes the demonstration.
- If the vectors {v1,v2,..,vp} are not linearly independent, then at least one vector in the set is linearly dependent on the others.
I'll reposition the linearly dependent vector to the last place in the linear combination (p).
Then, I'll express the linearly dependent vector vp as a linear combination of the other vectors {v1,v2,..,vp-1}
$$ v_p = \alpha_1 \vec{v}_1 + \alpha_2 \vec{v}_2 + ... + \alpha_1 \vec{v}_{p-1} $$
Since {v1,v2,..,vp} is a set of generators of V, every vector in the vector space is generated by a linear combination of the vectors {v1,v2,..,vp}
$$ \forall \vec{v} \in V \ \ \ \vec{v} = \lambda_1 \vec{v}_1 + \lambda_2 \vec{v}_2 + ... + \lambda_n \vec{v}_{p-1} + \lambda_n \vec{v}_p $$
I'll now substitute vp in the initial linear combination that generates every vector v in the vector space V.
$$ \forall \vec{v} \in V \ \ \ \vec{v} = \lambda_1 \vec{v}_1 + \lambda_2 \vec{v}_2 + ... + \lambda_n \vec{v}_{p-1} + \lambda_n ( \alpha_1 \vec{v}_1 + \alpha_2 \vec{v}_2 + ... + \alpha_1 \vec{v}_{p-1} ) $$
Now, the set of generators consists of p-1 vectors
$$ \{ \vec{v}_1 , \vec{v}_2 , ... , \vec{v }_{p-1} \} $$
The vectors {v1,v2,..,vp-1} still form a set of generators because I've replaced one of the vectors (vp) with its linear combination.
Next, I'll check if the p-1 vectors {v1,v2,..,vp-1} are also linearly independent.
- If the vectors {v1,v2,..,vp-1} are linearly independent, then the set of generators also forms a basis. Knowing that all bases are of the same dimension, it follows that p-1=n.
- If the vectors {v1,v2,..,vp-1} are not linearly independent, then there's at least one vector linearly dependent on the others. I select it and repeat the same process from the start.
This process continues until the set of generators consists of n linearly independent vectors {v1,v2,..,vn}, because all bases of the vector space V have the same number n=dim(V) of vectors.
And so on.
