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anonymous

  • 4 years ago

Linear Algebra question: let [v_1, v_2, v_3] be a set of vectors in m-space. If the transpose of v_i * v_j where i =\= j is zero, prove that the set of vectors is linearly dependent

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  1. anonymous
    • 4 years ago
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    okk .. solve this like a * v1*v2 + b*v1*v3 = 0 (given) take v1 as common then you get (a*v2+b*v3)*v1 = 0 Assuming that v1 is non zero we get a*v2 + b*v3 = 0 (linearly dependent)

  2. anonymous
    • 4 years ago
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    Yeah, forgot to mention that all vectors in the set are assumed to be non-zero

  3. anonymous
    • 4 years ago
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    I'm not following that first line of logic. It's given? We know that v1 transpose times any other vector will equal 0 but ...

  4. anonymous
    • 4 years ago
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    but...??

  5. anonymous
    • 4 years ago
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    Could you explain how you you get that first statement?

  6. anonymous
    • 4 years ago
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    its simple for any i = / j we have v_i*v_j as zero ... so I would have v1*v2 as zero. I now multilply this with a constant a and we have a*v1*v2 = 0. Similiar is the case for b*v1*v3. The concept of transpose here is that you cannot multiply two vectors directly. You have to take the transpose of one vector (think .. matrix multiplication). I hope I am clear..

  7. anonymous
    • 4 years ago
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    \[V _{i}^{T} * V _{j}\] = 0 means that \[V _{i}*V _{j}\] = 0?

  8. anonymous
    • 4 years ago
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    no dude .. its like say you have one column as (1,2,3) and the other as (2,4,6) ... and so if you have to mention v1*v2 in matrix form .. can you directly multiply? they are 3*1 matrices which cannot be multiplied .. you need to take transpose of the first to multiply them ... its more of a convention thing ..

  9. anonymous
    • 4 years ago
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    Does the fact that the transpose of one vector multiplied by another giving you zero mean anything special as far as their individual entries go? Because here's the hint they provide in the problem:

  10. anonymous
    • 4 years ago
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    "set a1v1 + a2v2 + a3v3 = zero vector and consider (zerovector)^t * zerovector

  11. anonymous
    • 4 years ago
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    consider the transpose of the zero vector multiplied by the zero vector? What is that even supposed to mean?

  12. anonymous
    • 4 years ago
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    it means that the result is 0 ... imagine a (3*1) 0 matrix and a (1*3) 0 matrix ... you can multiply both right???

  13. anonymous
    • 4 years ago
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    yup

  14. anonymous
    • 4 years ago
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    are you getting what I am trying to convey here .. ? forget about matrices ... say you have a vector ai+ bj+ck and other vector as a1i+ b1j+c1k multiplication would give you aa1 + bb1+ cc1 ... use this to prove ..

  15. anonymous
    • 4 years ago
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    Great, I just realized I have a frickin typo

  16. anonymous
    • 4 years ago
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    they want me to prove linear 'independence' not dependence @#$!

  17. anonymous
    • 4 years ago
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    this is a linearly dependent system .. if you have i not equal to j and v_i*v_j = 0

  18. anonymous
    • 4 years ago
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    Apparently not, because the question says 'show that the set is linearly independent' (and then everything else applies the same, including the hint and rest of the question)

  19. anonymous
    • 4 years ago
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    well ..then you may have another typo in the question .... is it v_i*v_j = 0 for all i not equal to j or is it zero for summation ... if it is for all i not equal to j then this CANNOT be linearly independent...

  20. anonymous
    • 4 years ago
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    v_i*v_j for all i not equal to j

  21. anonymous
    • 4 years ago
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    thanks for the help anyways man, I appreciate it

  22. anonymous
    • 4 years ago
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    Linear Algebra question: let [v_1, v_2, v_3] be a set of vectors in m-space. If the transpose of v_i * v_j where i =\= j is zero, prove that the set of vectors is linearly in dependent. All the vectors in the set are non-zero.

  23. anonymous
    • 4 years ago
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    What is the definition of linear dependence?

  24. anonymous
    • 4 years ago
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    Or independence, either one.

  25. anonymous
    • 4 years ago
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    Independence: if solution to the vector equation a1v1 + ... a_n v_n = zero vector is the trivial solution

  26. anonymous
    • 4 years ago
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    Okay, so let's assume that they are linearly dependent and then derive a contradiction. That means that \[a_1 \vec{v_1} + a_2 \vec{v_2} + a_3 \vec{v_3} = 0\] where the a's are not all zero.

  27. anonymous
    • 4 years ago
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    okay

  28. anonymous
    • 4 years ago
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    Recall that we take for granted that none of the vectors are the zero vector. What do we get when we take the transpose of a_1 times the whole equation?

  29. anonymous
    • 4 years ago
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    I'm sorry, v_1, not a_1

  30. anonymous
    • 4 years ago
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    hm...by the parameters of the problem, the only thing that should be left is a_1 v_1 * v_1 ^t which...does that just equal v_1?

  31. anonymous
    • 4 years ago
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    We should get \[a_1 (v_1^T v_1) = 0\]

  32. anonymous
    • 4 years ago
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    okay, yup

  33. anonymous
    • 4 years ago
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    furthermore, \[v_1^Tv_1 \] is just the length of the vector, is that correct? If so, what do we know this definitely CANNOT be, based on the assumptions we made?

  34. anonymous
    • 4 years ago
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    length squared.... :)

  35. anonymous
    • 4 years ago
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    Well, it can't be zero, right?

  36. anonymous
    • 4 years ago
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    Indeed. So, what must that mean?

  37. anonymous
    • 4 years ago
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    a = 0

  38. anonymous
    • 4 years ago
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    Perfect. But there's nothing special about v_1, so I could do the same thing with v_2 and v_3, and what conclusion would I be forced to draw about the a's?

  39. anonymous
    • 4 years ago
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    All the a's are zero, therefore the trivial solution is the only solution

  40. anonymous
    • 4 years ago
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    There ya go :)

  41. anonymous
    • 4 years ago
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    Dude, thanks a million

  42. anonymous
    • 4 years ago
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    No problem at all, excellent question.

  43. anonymous
    • 4 years ago
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    Well, whatever, you know, we're on a math forum, so let's just say my thanks increases without bound

  44. anonymous
    • 4 years ago
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    Hahaha good stuff. Good luck with the rest of your work, we'll help you out if you need it.

  45. anonymous
    • 4 years ago
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    Thanks again!

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