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A vector space and linear independent set.

  1. Feb 2, 2007 #1
    let V be a vector space and K a nonempty subset of V prove/disprove :
    K is linear independent set iff for every T such that T is a proper subset of K, span(T) is a proper subset of spanK.

    im having difficulty finding a counter example, so i think this statement is correct, but how to prove it?

    if K is an independent set, then i can show the for every susbset of it its span includes the span of the subset, but if im not mistaken this is correct also when K isnt an independent set.
    so my question is how to show that if for every T a proper subset of K, and span{T} a proper subset of span{K}?
     
  2. jcsd
  3. Feb 2, 2007 #2

    StatusX

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    If K is linearly independent, then the span of any proper subset doesn't even include all of K. Conversely, assume K is not linearly independent, and see if you can find a proper subset whose span is all of span(K).
     
  4. Feb 3, 2007 #3
    i dont think you understand, K is linear independent set, i need to prove/disprove that K is linear independent iff for every proper subset of K,T SpT is a proper subset of SpK.

    perhaps my first post was confusing, sorry for that.
     
  5. Feb 3, 2007 #4

    HallsofIvy

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    Yes, StatusX did understand! He said "the span of any proper subset doesn't even include all of K." What is the relationship between K and SpK?
     
  6. Feb 3, 2007 #5

    mathwonk

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    this is a tautology from the definitions, (plus subtraction and scalar division), so be sure you grok the definitions.
     
  7. Feb 3, 2007 #6
    K is obviously included in spanK, but i dont quite follow, does this mean the statement is correct, and if so then how to prove it, i mean,
    even if K is included in spanK.
    if K is independent and T is a proper subset of K, then T is also independent, and every combination of the vector of T are included in spanT.
    now how do i prove the other side of the statement, i.e when im given a proper subset, T of K such that spanT is a subset of spanK.
    i.e if T={v1,...,vk} K={v1,...,vn} n>k then
    a1v1+...anvn=0
    a1v1+...akvk=-(ak+1vk+1+...+anvn)
    so this is in spanT and thus also in spanK, now im kind of stuck.
    any help would be appreciated.
     
  8. Feb 3, 2007 #7

    StatusX

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    I meant, if you take some element v out of K, then, since K is linearly independent, there is no way to write v as a linear combination of the remaining elements. Thus the span of the remaining elements doesn't include all of K, and so can't possibly include all of span(K) (which contains K).
     
  9. Feb 4, 2007 #8

    HallsofIvy

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    Now we're talking about spanKing? This is getting too kinky for me!:rofl:

    Yes, K is included in span(k)- that's the whole point. If T is a proper subset of K, since K is independent, K is NOT in span(T) and therefore neither is span(K)!
     
  10. Feb 4, 2007 #9
    I'll give it a try. Consider vector space V.

    Given any nonempty subset K of V, given any proper subset T of K
    [K linearly independent -> Span(T) is a proper subset of Span(K)].

    I'd say this assertion has already pretty well been established.


    Claim: The converse is false. So we should be able to prove its negation.


    It is false that, given any nonempty subset K of V, given any proper subset T of K [Span(T) is a proper subset of Span(K) -> K is linearly independent].

    This is equivalent to:

    There exits a nonempty subset K of V, there exists a proper subset T of K such that [Span(T) is a proper subset of Span(K) and K is not a linearly independent].

    This is now an existence assertion. We need only find an example.

    Let T = {(1,0)}, K = {(1,0),(1,1),(0,1)}. Clearly Span(T) is a proper subset of Span(K), and K is not linearly independent.


    Maybe I misread the OP. Anyway, just my (perhaps a bit pedantic) two cents.
     
  11. Feb 4, 2007 #10
    ok, thanks fopc.
    those little simple counter examples. (-:
    and halls, you kinky geezer!
    (-:
     
  12. Feb 4, 2007 #11

    mathwonk

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    well paddles were always a standard tool in the old school house as i understand. maybe there is a link from the lower chakras to the brain.
     
  13. Feb 5, 2007 #12

    HallsofIvy

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    Personally, I prefer the 2 by 4.
     
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