1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear independence again

  1. Nov 1, 2006 #1

    radou

    User Avatar
    Homework Helper

    Let V be a vector space over a field F, [tex]v_{1}, \cdots, v_{n} \in V[/tex] and [tex]\alpha_{1}, \cdots, \alpha_{n} \in F[/tex]. Further on, let the set [tex]\left\{v_{1}, \cdots, v_{n}\right\}[/tex] be linearly independent, and b be a vector defined with [tex]b=\sum_{i=1}^n \alpha_{i}v_{i}[/tex]. One has to find necessary and sufficient conditions on the scalars [tex]\alpha_{1}, \cdots, \alpha_{n}[/tex] such that the set S=[tex]\left\{b, v_{2}, \cdots, v_{n}\right\}[/tex] is linearly independent, too.

    Well, I just used a simple proposition which states that a set is dependent if there exists at least one vector from that set which can be shown as a linear combination of the rest of the vectors from the same set. So, obviously, for [tex]\alpha_{1} = 0[/tex], the set S is dependent, which makes [tex]\alpha_{1} \neq 0[/tex] a necessary condition for S to be independent. Further on, a sufficient condition would be [tex]\alpha_{1} = \cdots = \alpha_{n} = 0[/tex], which leaves us with the set S\{b}. This set must be linearly independent, since it is a subset of {v1, ..., vn}, which we know is linearly independent.

    I may be boring, but I'm just checking if my reasoning is allright.. :smile:

    Edit. I just realized, for [tex]\alpha_{1} = \cdots = \alpha_{n} = 0[/tex] we have b = 0, which makes the set dependent! So [tex]\alpha_{1} \neq 0[/tex] is a necessary condition, but what's the sufficient condition? Is it that at least one of the scalars [tex]\alpha_{2}, \cdots, \alpha_{n}[/tex]must not be equal zero?
     
    Last edited: Nov 1, 2006
  2. jcsd
  3. Nov 1, 2006 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The sufficient condition certainly includes this [tex]\alpha_{1} \neq 0[/tex]

    Basically, if a condition is necessary and sufficient, it means that the statement is true iff the condition is true. So search for a set of conditions that are true iff the set b, v 2-n is linearly independent.

    Note that if all the alphas except [tex] \alpha_1[/tex] are zero, then b is just a multiple of v1, and the new set is certainly linearly independent
     
  4. Nov 1, 2006 #3

    radou

    User Avatar
    Homework Helper

    Yes, I realized that.. So, it seems [tex]\alpha_{1} \neq 0[/tex] is both necessary and sufficient condition, since the rest of the scalars can be any elements of F, all zero, all non zero, or combined.
     
    Last edited: Nov 1, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Linear independence again
  1. Linear Independence (Replies: 8)

Loading...