Saturday, June 27, 2015

Theorems on Rank and Basis of vector space

1. If Vn is a vector space of rank n, every generating set from Vn contains n, but not more than n linearly independent vectors.
Let, A1, A2,…, Ap be any generating set from Vn. Let k be the greatest number of linearly independent vectors that can be chosen from A1, A2,…,Ap.
Then, we may assume that A1, A2,…,Ak are linearly independent but A1, A2,…, Ak, A k+j are linearly dependent for j = 1, 2,…, (p-k)
Thus we have-
λ 1A1 + λ2+ …+ λkAk+ λk+jAk+j = Õ for all λk+j≠0
Since, Ak+j = -(λ1k+j)A1-(λ2k+j)A2-…-(λkk+j)A k
For j = 1, 2,…, (p-k).
Therefore it follows that A1, A2,…,Ak is generating set of Vn. Now, any set of more than k vectors is linearly dependent.
Since the rank of Vn is n, Vn contains n linearly independent vectors.
Then must have n ≤k.
Again Vn contains k linearly independent vectors A1,A2,…, Ak.
So, we can also have k≤n.
Hence, k = n.

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

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