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Division between vectors

  1. Jun 15, 2014 #1
    I want find a value M such that given v and u, satisfies the equation v=Mu.

    Well, the vector u has a modulus u and a direction α wrt x-axis; the vector v has another modulus v and another direction β wrt x-axis. What is happened is a change of magnitude and direction of the vector u, therefore, M needs scalar the vector u and change its direction, so that:

    $$\\ M = \frac{v}{u} \begin{bmatrix}
    \;\;\;\cos(\beta-\alpha) & -\sin(\beta-\alpha)\\
    +\sin(\beta-\alpha) & \;\;\;\cos(\beta-\alpha)\\
    \end{bmatrix}$$
    So: $$\\ \frac{\vec{v}}{\vec{u}} = M$$
     
  2. jcsd
  3. Jun 15, 2014 #2
    Sorry, is this a question? Only it looks like the answer.
     
  4. Jun 15, 2014 #3
    It's a question! I just followed the tamplate of the homework section, where is necessery to post a tentative.
     
  5. Jun 15, 2014 #4

    pasmith

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    The basic linear operations available in [itex]\mathbb{R}^2[/itex] are scaling, rotation, and shears.

    There are infinitely many linear maps [itex]A \in \mathrm{M}_2(\mathbb{R})[/itex] such that [itex]\vec v = A\vec u[/itex]; to determine [itex]A[/itex] uniquely you also need to specify the image under [itex]A[/itex] of some vector which is linearly independent of [itex]\vec u[/itex].

    For example, if [itex]\vec w \cdot \vec u = 0[/itex] then you can have [itex]A\vec w = 0[/itex] or [itex]A \vec w = \vec w[/itex] or [itex]A \vec w = \vec v[/itex] or [itex]A \vec w = \vec u[/itex] or ..., and in each case it will still be the case that [itex]A\vec u = \vec v[/itex]. This is why the notation [itex]\vec v / \vec u[/itex] doesn't make sense.
     
  6. Jun 15, 2014 #5
    But my answer is valid too, correct?
     
  7. Jun 15, 2014 #6

    WWGD

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    No; the point was that if you have only a vector space structure, you cannot talk about division; you need at least a ring structure on your vectors. In a vector space alone, vectors cannot be divided.
     
  8. Jun 15, 2014 #7
    But if I have a problem like v= Mu, where u and v are known but M not, my solution isn't valid?
     
  9. Jun 15, 2014 #8

    pasmith

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    You have exhibited an [itex]M[/itex] such that [itex]\vec v = M\vec u[/itex]; the problem is that it isn't unique.
     
  10. Jun 15, 2014 #9

    WWGD

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    Jhenrique: It would be easier to answer your question if you specified what v,u,M are, and where they "live", i.e., are v,u elements in a vector space, ring, etc? Is M just any matrix, or there specific coefficients you are using, specific dimension, etc.
     
  11. Jun 15, 2014 #10
    2D, M is a tensor and u and v are vectors.

    The possible other answers will be just a change of basis. You still wants another answer better than formula in my first post?
     
  12. Jun 15, 2014 #11

    micromass

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    Division of vectors is not defined. And if it were, it sure isn't going to be a matrix.
     
  13. Jun 15, 2014 #12

    WWGD

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    The way I understand this is that you have a rotation matrix M, and you're then rotating u, to get v. So , you can rotate the plane by the necessary angle , and then rescale by the quotient of the norms of u, v ( and Dupree ). Then, yes, given vectors u,v in the plane, you can always get u from v by a combination of rotation and scaling : rotate by the difference from their angles , and then rescale by the quotient of their norms.
     
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