Vector Space Basis Theorem Demonstration 2
In a finitely generated vector space V of dimension n, from any set of linearly independent vectors {v1,v2,..,vp} with p≤n, it's always possible to form a basis B of the vector space by adding linearly independent vectors from the set of vectors (proof).
Proof
Consider a finite dimensional vector space V of dimension n
$$ \dim V = n$$
Being of dimension n, the vector space has bases composed of n vectors
$$ B = \{ \vec{v}_1. \vec{v}_2, ... , \vec{v}_n \} $$
Let's consider a set composed of p linearly independent vectors of V
$$ \{ \vec{v}_1. \vec{v}_2, ... , \vec{v}_p \} $$
Where p is an integer less than or equal to n
$$ p \le n $$
Now, check if {v1,v2,...,vp} is a generating set of V.
- If the set of vectors {v1,v2,...,vp} is a generating set of the vector space V, then it's also a basis. The proof ends here.
- If the set of vectors {v1,v2,...,vp} is not a generating set of the vector space V, then there exists some vector vp+1 that is not a linear combination of {v1,v2,...,vp}. Once found, add it to the set of vectors $$ \{ \vec{v}_1. \vec{v}_2, ... , \vec{v}_p , \vec{v}_{p+1} \} $$ Given that the vectors {v1,v2,...,vp} are linearly independent by initial hypothesis and vp+1 is also linearly independent from {v1,v2,...,vp}, the set of vectors {v1,v2,...,vp,vp+1} is also a set of linearly independent vectors. Thus, repeat the process and check if the set {v1,v2,...,vp,vp+1} is a generating set.
The process is repeated, adding more linearly independent vectors until the number p+k of vectors {v1,v2,...,vp,vp+1, ..., vp+k} equals the dimension n of the vector space
$$ p+k = n $$
When p+k equals n, the set of linearly independent vectors is a basis of V because all bases of a vector space have the same dimension (cardinality), that is, an equal number of vectors.
And so on.
