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Homework Help: Linear Independence

  1. Aug 6, 2010 #1
    1. The problem statement, all variables and given/known data

    If a set is linear independent, proove that every of its non empty subset is linear independent.
    i'm sorry, but i'm not sure is my sentence correct or not,
    2. Relevant equations

    n/a

    3. The attempt at a solution

    let {v1,v2,v3,......,vn} be linear independent set.

    so a1v1+a2v2+a3v3+,........,+anvn=0, and ai=0 , i= 1,2,3,...,n

    let mv1+nv2=0, i can't even prove m=n=0, so help me start proving that first T_T
     
  2. jcsd
  3. Aug 6, 2010 #2

    HallsofIvy

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    Prove it by contradiction.

    Suppose there exist some subset, [itex]\{v_i| i\in A\}[/itex] for A some subset of {1, 2, ..., n} that is NOT independent. That means we can find [itex]\sum a_iv_i= 0[/itex] with every [itex]v_i[/itex] from this subset, not all [itex]a_i[/itex] equal to 0.

    Now, do you see how that immediately contradicts the fact that the entire set is independent?
     
  4. Aug 6, 2010 #3
    yeaaa i can see i clearly, but im not sure how to convert those word to mathematical form
    hmm but i remembered 1 theorem said, they are linear independent iff exist 1 vector of them can be written as linear combination of other. ngaha, my language are so terrible, nevertheless,

    Suppose there exist some subset, [itex]\{v_i| i\in A\}[/itex] for A some subset of {1, 2, ..., n} that is NOT independent that means we can find vm in the set such that

    vm=a1v1+,...,+am-1vm-1,am+1vm+1+,......+anvn
    0=a1v1,...,am-1vm-1+vm+am+1vm+1,......anvn

    contradiction

    conclusion ¬(my assumption)
    this library computer so slow, i hope you can understand it, anyway, it's correct right?
     
    Last edited: Aug 7, 2010
  5. Aug 7, 2010 #4

    HallsofIvy

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    Well, I guess my question then is "what definition of "independent" are you using?"

    You seem to be using "you cannot write one vector in terms of the other" for the subset and "if a linear combination is 0, then the coefficients must be 0" for the entire set. Use the same thing for both.

    Suppose there exist a subset, [itex]\{v_i, v_j, ..., v_k\}[/itex] of the independent set [math]\{v_1, v_2, ..., v_n\}[/math] where {i, j, ..., k} is a subset of {1, 2, ..., n} (make sure your notation does not imply that the vectors in the subset were any particular vectors in the original set, like "the first m vectors") that is not independent. Then there exist scalars [itex]a_i, a_j, ..., a_k[/itex], not all 0, such that [itex]a_iv_i+ a_jv_j+ ...+ a_kv_k= 0[/itex]. But then we have [itex]b_1v_1+ b_2v_2+ ...+ b_nv_n= 0[/itex] with [itex]b_m= 0[/itex] if m is not in {i, j, ..., k} and [itex]b_m= a_m[/itex] if it is. That is a linear combination of [itex]\{v_1, v_2, ..., v_n\}[/itex] equal to 0 with not all coefficients equal to 0.
     
  6. Aug 7, 2010 #5
    yea, sorry, i did very big mistake there,

    but thanks anyway
     
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