Theorems on Rank and Basis of vector space

1. If V_n is a vector space of rank n, every generating set from V_n contains n, but not more than n linearly independent vectors.
Proof:
Let, A₁, A₂,…, A_p be any generating set from V_n. Let k be the greatest number of linearly independent vectors that can be chosen from A₁, A₂,…,A_p.
Then, we may assume that A₁, A₂,…,A_k are linearly independent but A₁, A₂,…, A_k, A _k+j are linearly dependent for j = 1, 2,…, (p-k)
Thus we have-
λ ₁A₁ + λ₂+ …+ λ_kA_k+ λ_k+jA_k+j = Õ for all λ_k+j≠0
Since, A_k+j = -(λ₁/λ_k+j)A₁-(λ₂/λ_k+j)A₂-…-(λ_k/λ_k+j)A _k
For j = 1, 2,…, (p-k).
Therefore it follows that A₁, A₂,…,A_k is generating set of V_n. Now, any set of more than k vectors is linearly dependent.
Since the rank of V_n is n, V_n contains n linearly independent vectors.
Then must have n ≤k.
Again V_n contains k linearly independent vectors A₁,A₂,…, A_k.
So, we can also have k≤n.
Hence, k = n.

2. If V_n is a vector space of rank n, every basis of V_n contains exactly n linearly independent vectors. Conversely any n linearly independent vector from V_n constitutes a basis of V_n.
Or, rank of a vector space is unique but the basis is not.
Proof :
By definition, every basis of V_n is generating set. So, it contains n but not more than n linearly independent vectors. Since, the vectors of a basis are linearly independent it contains exactly n vectors in all. Conversely, let A₁, A₂,…, A_n be any vector of V_n and let A be any other vector of V_n.
Since n is the rank of V_n, the (n+1) vectors A, A₁, A₂,…, A_n are linearly dependent.
Hence, we have-
λA + λA₁ + λA₂+….+ λA_n = Õ, where, λ ≠ 0
thus we have,
A = -(λ₁/λ)A₁-(λ₂/λ)A₂-….-(λ_n/λ)A_n
It follows that, every vector A ∈ V_n is a linear combination of A₁, A₂,…, A_n.
Hence, there n vectors from a generating system. Since they are linearly independent, they also form a basis of V_n.
Hence the proof.