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Linear dependence of a set

  1. May 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Let S = {[itex]v_{1}, v_{2}, \cdots , v_{n}[/itex]}
    S is linear dependent iff at least one v in S is a linear combination of the others.

    2. Relevant equations

    3. The attempt at a solution

    From here on, just take v to be a vector, and x to be some scalar please.

    I really just wanted to check my understanding of this.

    If I generalize this to the case where S contains 1 vector v, then S is linear independent iff v is not the zero vector. This is because if you write v as a linear combination xv, then xv=0 has only the trivial solution where v!=0. Likewise, if v=0, then x could be any real number in xv=0, and there are infinitely many non-trivial solutions (linearly dependent).

    This all makes sense from a geometric standpoint to me. I am more concerned about the case where S contains 1+n vectors.

    S is linearly dependent iff at least one v in S is a linear combination of the others.

    So, if S = {v,u,w}, and w is a linear combination of v and u, then w is in span{v,u} and S is linearly dependent. The same case can be made for R^3 without any issue. I am having trouble checking whether this is true for R^n.

    For instance, if S = {a,b,c,d}, and a,b,c,d each lie on a line through the different axis, then span{S} is some 4d surface thingy. a,b,c lie on a line through three different axis, and d is some linear combination of a,b,c? Clearly then, by that theorem, S is linearly dependent. So any time there is a linear dependence between any 2 vectors in a set, the set is linearly dependent? Regardless of dimension?
     
    Last edited: May 2, 2012
  2. jcsd
  3. May 2, 2012 #2

    Dick

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    A set of vectors is linearly dependent if you can write c1*v1+...cn*vn=0 with not all of the cn equal 0. Why don't you try using that?
     
  4. May 3, 2012 #3

    HallsofIvy

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    You seem to be leaving out some information here. Before, you simply noted that you had a set of n vectors. Now, making it n+ 1 doesn't change anything. In this last part, are we to assume that these vectors are in a vector space of dimension n?

    If that is the case, then the definition of "dimension" says that there exist a basis for the space containing n vectors. Writing the n+1 vectors in terms of the basis vectors gives n+1 equations in terms of n coefficients.

     
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