Proof of Linearly independence

pyroknife
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The problem is attached. I just wanted to see if the way I proved my statement is correct.

My answer: No, because there exists more columns than rows, thus at least one free variable always exists, thus these vectors are linearly dependent.
 

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You are right but, personally, I wouldn't phrase it in terms of "columns" and "rows" since nothing is said of matrices in the problem. Rather, I would not that Rm has dimension m so there is a basis containing m vectors. Every one of the n vectors in the set can be written in terms of the the m vectors in the basis so they cannot be independent.

(A basis for an m-dimensional vector space has three properties:
1) they span the space
2) they are independent
3) there are m vectors in the set
Any set containing fewer than m vectors cannot span the space, any set with more than m vectors cannot be independent.)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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