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Is the Set Linearly Independant?

  1. Oct 26, 2011 #1
    1. The problem statement, all variables and given/known data
    Let V be a real vector space and [tex]{b_1,b_2,b_3,b_4}[/tex] a linearly independent set of vectors in V
    Is the set [tex]\left \{ b_1,b_2,b_3,b_1+b_4,b_2+b_4 \right \}[/tex]

    3. The attempt at a solution

    [tex]\alpha_1b_1+\alpha_2b_2+\alpha_3b_3+\alpha_4\left \{ b_1+b_4 \right \}+\alpha_5\left \{ b_2+b_4 \right \} =0[/tex]

    Rearranging

    [tex]b_1(\alpha_1+\alpha_4)+b_2(\alpha_2+\alpha_5)+b_3(\alpha_3)+b_4(\alpha_4+\alpha_5)=0[/tex]
    Hence it is NOT linearly independant since the scalars are all not 0.....?
     
  2. jcsd
  3. Oct 26, 2011 #2
    Hi there,

    It's been a while since I've done this, and I'm sure that my methods are not sound but I think I have a pretty good idea of the answer here.

    A set of vectors are linearly independent if you cannot write any on of the vectors in terms of any finite number of other vectors in the set. I can see right now that b2 + b4 = b2 + (b1 + b4) - b1

    in other words if b1 = V1, b2 = V2....b2 +b4 = V5 then V5 = V2 + v4 - V1

    Does that make sense to you?
     
  4. Oct 26, 2011 #3
    And I think that if a set of vectors are linearly independent they span the space correct? And it seems illogical that a set of 4 vectors can span 5 space :)
     
  5. Oct 26, 2011 #4
    ok, that makes sense. Thanks :-)
     
  6. Oct 26, 2011 #5

    vela

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    You're not quite finished. You need to show that all of the alpha's are necessarily equal to 0 if you want to show those vectors are linearly independent. Since the b's are independent, you know that equation can only be satisfied if
    \begin{align*}
    \alpha_1+\alpha_4 &= 0 \\
    \alpha_2+\alpha_5&= 0 \\
    \alpha_3&= 0 \\
    \alpha_4+\alpha_5&= 0
    \end{align*}If the only solution to this system is [itex]\alpha_i = 0[/itex] for i=1, 2, 3, 4, 5, the vectors are linearly independent. In this case, you'll find there are non-zero solutions to the system, so they are dependent.
    You're thinking of a basis, which consists of linearly independent vectors that span the space. Any set of vectors, independent or not, will span some space or subspace.

    You're close though. There's a theorem that says if you have more vectors than the dimension of the space they're in, they have to be dependent. If you've already covered this theorem, you can use it as well to solve this problem.
     
  7. Oct 26, 2011 #6
    Ok, do I use the Wronskian to find if the determinant is 0. if it is 0 then the vectors are linearly dependant? Thanks
     
  8. Oct 26, 2011 #7

    vela

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    No, the Wronskian has nothing to do with this problem. I think you simply mean you want to find the determinant of the coefficient matrix. If it's 0, the system is dependent. It's probably less work just to go ahead and solve the system of equations though.
     
  9. Oct 26, 2011 #8
    but can one not use the Wonrskian in this problem? Anyhow, will I solve this using gaussian elimination?
     
  10. Oct 26, 2011 #9

    vela

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