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Yet another proof

  1. Sep 23, 2004 #1
    let me know if u think i did this proof correctly...

    use the definition of linear independence to prove that if [tex]{u_1,u_2,...,u_n}[/tex] is a linearly independent set, then [tex]{u_1,u_2,...,u_k}[/tex] is also a linearly independent set for any
    [tex]k = 2,3,...,n-1[/tex]

    well heres what i did...
    if the set of vectors [tex]{u_1,u_2,...,u_n}[/tex] is linearly independent then none of the vectors in that set is linear combination of the others. therefore
    [tex]c_1 u_1 + c_2 u_2 + ... + c_n u_n = \vect{0} [/tex] is satisfied only by the set of scalars [tex] {c_1,c_2,...,c_n}[/tex] that is [tex]c_1 = 0, c_2 = 0, ..., c_n = 0[/tex] then.. [tex] c_1 u_1 + c_2 u_2 +...+c_(n-1) u_(n-1) = -c_n u_n[/tex] should only be satisfied by the same set where [tex]c_1 = 0, c_2 = 0, ..., c_n = 0[/tex] because no vector is a linear combination of the others by definition. thus this can be expanded to...[tex]c_1 u_1 + c_1 u_2 + ... + c_k u_k = -c_n u_n -c_(n-1) u_(n-1) - ... -c_(k+1) u_(k+1)[/tex] the only solution is the set where [tex]c_1 = 0, c_2 = 0, ..., c_n = 0[/tex], thus the set [tex] {u_1,u_2, ...,u_k}[/tex] is linearly independent for any [tex]k=2,3,...,n-1.[/tex]

    tell me what u think..
     
    Last edited: Sep 23, 2004
  2. jcsd
  3. Sep 23, 2004 #2

    matt grime

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    I think that this question doesn't even need proof.

    If there is a linear combination of u_1,..,u_k that is zero, then there is a linear combination of u_1,..,u_n (k<=n) that is also zero. done. You're writing far too much.
     
  4. Sep 23, 2004 #3

    HallsofIvy

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    Well, yes, it needs a proof, its just a trivial one!

    Phymath: Suppose [tex]\{u_1,u_2,...,u_{n-1}\}[/tex] were not linearly independent. That would mean that there exist constants [tex]{\kc_1,c_2,...,c_{n-1}\}[/tex], not all 0, such that [tex]c_1u_1+ c_2u_2+...+c_{n-1}u_{n-1}= 0[/tex].
    What does that tell you about the orginal set [tex]\{u_1,u_2,...,u_n\}[/tex]?
     
  5. Sep 23, 2004 #4
    well idk i guess thats why i asked the question in the first place. I suppose it could mean that u_(n-1) is a linear combination of the procedding u vectors..[tex] u_1, u_2, ...,u_{n-2}[/tex] no? but i thought by defination that can't be...i suppose that i tells excaztly what i said in my proof that there are no scalar constants to be a solution to that set other then all the constants have to be 0. because taking ur equation if they are linearly dependent (aka not linearly independent) then u can do this, which is by defintion only solved if the constants are zero, please help me out here...i don't get ur point
    [tex]c_1 u_1 +c_2 u_2 +...+c_(n-2) u_(n-2) = -c_(n-1) u_(n-1) [/tex]
     
  6. Sep 23, 2004 #5

    mathwonk

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    linear dependence means zero is a non trivial linear combination of some of the vectors.

    but if zero is a linear combination of some of the first k vectors, then zero is also (the same) linear combination of some of the full set of n vectors.

    how's this?
     
  7. Sep 23, 2004 #6

    mathwonk

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    another point of view: linear independence means the map sending the standard unit vectors to those vectors is injective. but an injective map on n space restricts to an injective map on a k dimensional subspace.

    ok ok i know that makes it worse.
     
  8. Sep 23, 2004 #7
    yes but i have to work this backwards...n then to k not k then to n
     
  9. Sep 23, 2004 #8

    mathwonk

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    you are mistaking the logic. to prove " if all n vectors are independent then so are the first k", is the same as proving: " if the first k vectors are dependent then so are all n".

    so i worked from k to n because I was proving the second statement. I.e. I was working with dependency, and proving your statement "by contradiction". proofs by contradiction work backwards, instead of assuming the hypothesis is true and then deducing the conclusion is also true, you assume instead that the conclsuion is false and then deduce that the hypothesis would also be false.

    its the same thing.
     
    Last edited: Sep 23, 2004
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