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Homework Help: Linear independent

  1. Aug 15, 2010 #1
    1. The problem statement, all variables and given/known data

    if v_1,v_2,...,v_k be k linear independent vector, and if

    v_1,v_2,...v_k,v be k+1 linear dependent vector, then

    v is the linear combination of v_1,v_2,...,v_k

    2. Relevant equations

    n/a

    3. The attempt at a solution

    some of my attempt,(direct proof)

    v_1,v_2,...v_k,v be k+1 linear dependent vector then when we write v_1,v_2,...v_k,v as linear combination of 0, there exist some coefficient not all of them 0. and i still no idea how to relate it to "v_1,v_2,...,v_k be k linear independent vector"
     
  2. jcsd
  3. Aug 15, 2010 #2

    hunt_mat

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    Homework Helper

    You can write:
    [tex]
    \lambda_{1}v_{1}+\cdots +\lambda_{k}v_{k}+\lambda_{k+1}v=0
    [/tex]
    As a linealy dependent vector, then write [tex]\lambda_{k+1}=-1[/tex] to find:
    [tex]
    v=\lambda_{1}v_{1}+\cdots +\lambda_{k}v_{k}
    [/tex]
     
  4. Aug 15, 2010 #3
    i don't understand, to find "[tex]

    v=\lambda_{1}v_{1}+\cdots +\lambda_{k}v_{k}

    [/tex]" ?
     
  5. Aug 15, 2010 #4

    hunt_mat

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    If the vectors are linearly dependent then the [tex]\lambda_{i}[/tex] are all nonzero.
     
  6. Aug 15, 2010 #5
    aren't some of them are non zero? not necessary all right?
     
  7. Aug 15, 2010 #6

    hunt_mat

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    Homework Helper

    There will be [tex]\lambda_{i}[/tex] which are nonzero, this will make the vector [tex]v[/tex]
     
  8. Aug 15, 2010 #7
    [tex]

    \lambda_{1}v_{1}+\cdots +\lambda_{k}v_{k}+\lambda_{k+1}v=0

    [/tex] are linearly dependent, then some of [tex]

    \lambda_{i}

    [/tex] are non zero..

    not necessary, [tex]
    \lambda_{k+1}
    [/tex] is non zero,

    T_T i'm confused
     
  9. Aug 15, 2010 #8

    hunt_mat

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    Homework Helper

    If [tex]\lambda_{k+1}\neq 0[/tex] then the vector will become linearly independent and then [tex]\lambda_{i}=0[/tex] will become zero. so [tex]\lambda_{k+1}\neq 0[/tex]
     
  10. Aug 16, 2010 #9
    you mean this right?

    If [tex]
    \lambda_{k+1}= 0
    [/tex] then the vector will become linearly independent and then [tex]
    \lambda_{i}=0
    [/tex] will become zero. so [tex]
    \lambda_{k+1}\neq 0
    [/tex]

    if yes, please check my argument here

    [tex]

    \lambda_{1}v_{1}+\cdots +\lambda_{k}v_{k}=0

    [/tex] are linear independent so, [tex]
    \lambda_{i}=0
    [/tex]

    hence

    [tex]

    0v_{1}+\cdots +0v_{k}+\lambda_{k+1}v=0

    [/tex]

    if [tex]
    \lambda_{k+1}=0
    [/tex] then the vectors are linear independent, so [tex]
    \lambda_{k+1}\neq 0
    [/tex]

    is this correct?
     
  11. Aug 16, 2010 #10

    HallsofIvy

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    Science Advisor

    You are given that [itex]\{v_1, v_2, \cdot\cdot\cdot, v_k\}[/itex] are independent so we cannot have [itex]\lambda_1 v_1+ \lambda_2 v_2+ \cdot\cdot\cdot+ \lambda_k v_k= 0[/itex] unless [itex]\lambda_1= \lambda_2= \cdot\cdot\cdot= \lambda_k= 0[/itex]

    But we are also given that [itex]\{v_1, v_2, \cdot\cdot\cdot, v_k, v\}[/itex] are dependent- there exist [itex]\lambda_1, \lambda_2, \cdot\cdot\cdot, \lambda_k, \lambda[/itex], not all 0, such that [itex]\lambda_1 v_1+ \lambda_2v_2+ \cdot\cdot\cdot+ \lambda_k v_k+ \lambda v= 0[/itex].

    Now, here is the crucial point: if [itex]\lambda= 0[/itex] [itex]v[/itex] would not be in the equation and we would have [itex]\lambda_1 v_1+ \lambda_2 v_2+ \cdot\cdot\cdot+ \lambda_k v_k= 0[/itex] with not all of [itex]\lambda_1, \lambda_2, \cdot\cdot\cdot, \lambda_k[/itex] equal to 0- which cannot happen. Thus, we must have [itex]\lambda[/itex] not 0.

    No we can rewrite [itex]\lambda_1 v_1+ \lambda_2 v_2+ \cdot\cdot\cdot+ \lambda_k v_k+ \lambda v= 0[/itex] as [itex]-\lambda v= \lambda_1v_1+ \lambda_2v_2+ \cdot\cdot\cdot+ \lambda_k vk[/itex] and because [itex]\lambda\ne 0[/itex], we can divide through by [itex]-\lambda[/itex]:
    [tex]v= -\frac{\lambda_1}{\lambda}v_1- \frac{\lambda_2}{\lambda}v_2- \cdot\cdot\cdot- \frac{\lambda_k}{\lambda}v_k[/tex]
     
    Last edited by a moderator: Aug 16, 2010
  12. Aug 16, 2010 #11
    that crucial point is realllllllyyyy helpful, thankssssssss, btw it's [itex]
    v
    [/itex]

    and thanks for all 3 quick reply, i'm still scrutinizing the other two, anyway, thanks again
     
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