*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, A

_{1}, A

_{2},…, A

_{p}be any generating set from V

_{n}. Let k be the greatest number of linearly independent vectors that can be chosen from A

_{1}, A

_{2},…,A

_{p}.

Then, we may assume that A

_{1}, A

_{2},…,A

_{k}are linearly independent but A

_{1}, A

_{2},…, A

_{k}, A

_{k+j}are linearly dependent for j = 1, 2,…, (p-k)

Thus we have-

λ

_{1}A

_{1}+ λ

_{2}+ …+ λ

_{k}A

_{k}+ λ

_{k+j}A

_{k+j}= Õ for all λ

_{k+j}≠0

Since, A

_{k+j}= -(λ

_{1}/λ

_{k+j})A

_{1}-(λ

_{2}/λ

_{k+j})A

_{2}-…-(λ

_{k}/λ

_{k+j})A

_{k}

For j = 1, 2,…, (p-k).

**Therefore**it follows that A

_{1}, A

_{2},…,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 A

_{1},A

_{2},…, 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 A

_{1}, A

_{2},…, 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, A

_{1}, A

_{2},…, A

_{n}are linearly dependent.

Hence, we have-

λA + λA

_{1}+ λA

_{2}+….+ λA

_{n}= Õ, where, λ ≠ 0

**thus we have,**

A = -(λ

_{1}/λ)A

_{1}-(λ

_{2}/λ)A

_{2}-….-(λ

_{n}/λ)A

_{n}

It follows that, every vector A ∈ V

_{n}is a linear combination of A

_{1}, A

_{2},…, 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.**

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