Linear algebra proof on linearly independence

gavin1989
Messages
8
Reaction score
0

Homework Statement


Show that if vectors v1 , . . . , vk in a vector space V have the properties that v1
does not = 0, and each vi is not in the span of the preceding ones, then the vectors are linearly independent.
Conversely, show that if v1 , . . . , vk is an ordered list of linearly independent vectors, then it has the above properties.

Homework Equations


The Attempt at a Solution



I know it is kinda easy to prove the set is linearly independent, but with the property there, how would I start the proof? I think since v2 is not a multiple of v1, and v3 is not a multiple of vi and v2 ... as well as v1 does not = 0. so i need to show: av1+bv2+cv3...nvn=0 becuz v1 does not = 0, and vi is not the span of the preceding ones(so they are all not =0), does it mean a=b=c...=0?

thanks
 
Last edited:
Physics news on Phys.org
You would write a formal proof using induction. There's not much to prove for the k=1 case. Now assume it's true for k and show it's true for k+1. How would that proof look?
 
is proof by induction the only way to do it? my prof has not really taught how to prove by induction
 
Well, you could also do proof by contradiction. For the first case, assume the vectors are NOT linearly independent. Then prove there IS some vi that is in the span of the preceding ones.
 
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...
Back
Top