Can Linear Independence be Proven with Given Information?

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Homework Statement



[PLAIN]http://uploadpie.com/nsXSv

Homework Equations





The Attempt at a Solution



I have no idea how to start. To be linearly independent, c1u1+c2u2+...+cnun = 0 has only trivial solution. But I don't know how can I use the given information to prove that
 
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You are given that u_i= Av_i for all i.

For any linear combination, a_1u_1+ a_2u_2+ \cdot\cdot\cdot a_nu_n= 0 we have a_1Av_1+ a_2Av_2+ \cdot\cdot\cdot a_nAv_n= A(a_1v_1+ a_2v_2+ a_nv_n)= 0.

If A is invertible, we have a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0.

On the other hand, if A is NOT invertible, there exist v_0\ne 0 such that Av_0= 0 (you should show that). Since \{v_1, v_2, \cdot\cdot\cdot, v_n\} are n independent vectors in R^n, we can write v_0= c_1v_1+ c_2v_2+ \cdot\cdot\cdot+ c_nv_n for some numbers c_n. Apply A to both sides of that equation.
 
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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