{u1, ,uk} is linearly dependent iff {[u1]_B, [uk]B} is.

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Hi,

Homework Statement



Given a vector space V and its basis B = {v1, v2, ..., vn}, I was asked to prove that:
a group of vectors {u1,...,uk} in V is linearly dependent if and only if {[u1]B,...[uk]B} is linearly dependent.

Homework Equations





The Attempt at a Solution



I proved that a group of vectors {u1,...,uk} in V is linearly independent if and only if {[u1]B,...[uk]B} is linearly independent. Following the principles of logic, that would be the same as proving dependence, right?
 
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Yes, and it is an easy one-liner to prove it instead of just declare it.
 
So, in general, proving (A if and only if B) is equivalent to, and hence may be proven by, proving (-A if and only if -B), correct?
 
peripatein said:
So, in general, proving (A if and only if B) is equivalent to, and hence may be proven by, proving (-A if and only if -B), correct?

Yes. But declaring it doesn't make it so. If I were handing in a homework proof of A iff B and I instead handed in a proof that -A iff -B, I wouldn't just stop there even though you might consider the conclusion to be trivial from there.
 
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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