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Geometric Properties from Eigenvectors

  1. Oct 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi! I just used MATLAB to find the eigenvalues and eigenvectors of A=[0 -1; 1 0]
    I obtained the eigenvalues of 0 +/- i
    and eigenvectors of v(1) = [ 0.7071; 0 - 0.7071i] and v(2) = [ 0.7071; 0 + 0.7071i]

    2. Relevant equations
    I'm having trouble interpreting these results in relation to the geometric properties of the linear transformation T(x; y) = [0 -1; 1 0] [x; y]


    3. The attempt at a solution
    As far as what the transformation does this is my explanation: It changes the sign of the y values i.e. positive to negative or negative to positive.
    As far as interpreting the MATLAB results in relation to the geometric properties of the linear transformation T, I can only explain it as: the eigenvalues illustrate that the x values remain the same but the y values change in sign.

    Is such an explanation correct. Is this what the transformation does.
    Sorry I haven't done much mathematical explainations and not quite sure if this is how to go about it.

    Thank you in advance.
     
  2. jcsd
  3. Oct 20, 2008 #2

    tiny-tim

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    Hi Daniel323! :smile:

    (have plus-or-minus: ± and a square-root: √ :smile:)

    i] You can multiply eiegenvectors (but not eigenvalues, of course) by any factor … in this case, it's much neater if you divide by .7071 (= 1/√2 :wink:), to give eigenvectors of v(1) = [ 1; -i] and v(2) = [1; i]. :smile:

    ii] If a complex number is purely imaginary, you don't have to write it as "0 + …": in this case, you can write that the eigenvalues are ±i.

    iii] Hint: draw the lines through the origin along [ 1; -i] and [1; i]. What does T do to these lines? Is there a rotation? Is there a reflection? :wink:
     
  4. Oct 20, 2008 #3
    Hi tiny-tim, thanks for that.

    I drew the lines on a graph and I think this is the effect (and this would be a much better explanation then my previous one);
    The linear transformation T reflects the line or each point through the x axis.

    Hopefully I'm right, thanks again! :)
     
  5. Oct 20, 2008 #4

    tiny-tim

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    Hi Daniel323! :smile:
    Nooo … y = x, x = -y is a … ? :smile:

    Hint: a reflection will have a negative determinant.

    A rotation (in 2D) has no fixed line, and therefore no real eigenvalues. :wink:
     
  6. Oct 20, 2008 #5
    Ahhhh.... here's goes nothing:
    The linear transformation T rotates the line or each point through the x axis?

    Thanks again.
     
  7. Oct 20, 2008 #6

    tiny-tim

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    "nothing shall come of nothing …"

    I'm not following you … what's a rotation through the x axis? :confused:

    (this is only 2D …)
     
  8. Oct 20, 2008 #7
    Sorry about that, my mistake. I'm just horrible at mathematical explanations.

    The linear transformation T causes a rotation for each vector. Is this generally a good sort of explanation for such findings?
     
  9. Oct 20, 2008 #8

    tiny-tim

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    Yes, that's fine! :smile:

    Rotation reflection and translation are the correct mathematical terms for what all 2D or 3D transformations are made of.

    But can I just check … a rotation of how much, and about what axis?
     
  10. Oct 20, 2008 #9
    Of course you can! I believe it is a rotation about the x-axis, but as far as how much I'm not completely sure how to say. If I was to guess I'd say a rotation of 180 degrees about the x-axis.
     
  11. Oct 20, 2008 #10

    tiny-tim

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    hmm … this is 2D, so any rotation must be about the z-axis (or a line parallel to the z-axis).

    Just try a couple of test points, to see where they go. :smile:
    (Actually, I should have added expansion, either generally or in one direction.)
     
  12. Oct 20, 2008 #11
    Hmm I'm kind of confused now, because it is 2D I thought we were only dealing with a x and y axis.

    So now the rotation is about the z-axis, but again not sure how to describe how much of a rotation.

    I tried out some points and generally understand what the transformation is doing, just can't explain it mathematically, so to speak.
     
  13. Oct 20, 2008 #12

    tiny-tim

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    ok, I agree, technically there's no z-axis, but you can rotate about the origin (or any other point), which is the same as a rotation about the z-axis.

    In other words, stick a pin in the paper, to fix it to the desk, and rotate!

    As for a proper mathematical explanation, any rotation about a particular axis (or point, in 2D) is defined by its angle … so how through many degrees is everything rotated? :smile:
     
  14. Oct 20, 2008 #13
    Ok so I will say it is a rotation about the origin or point by 180 degrees.
    I suspect it is 180 degrees, but maybe I'm looking at it wrong and it is only 90 degrees.
     
  15. Oct 20, 2008 #14

    tiny-tim

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    Yes, it's one or the other! :biggrin:

    Hint: stop trying to be abstract …

    do a concrete example …

    what happens to (1,0)? :wink:
     
  16. Oct 21, 2008 #15
    Umm I'm gonna go with a rotation about the origin by 180 degrees. Please tell me I'm finally right!
     
  17. Oct 21, 2008 #16

    tiny-tim

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    Well, 180º would send (1,0) to (-1,0), but A(1,0) = (0,1), so … ? :smile:
     
  18. Oct 21, 2008 #17

    HallsofIvy

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    No need to guess. What does that transformation change <1, 0> to?
     
  19. Oct 26, 2008 #18
    Ahh the transformation changes (1,0) to (0,1) hence the x and y value swaps at this point and therefore a rotation about the origin by 90 degrees.

    I believe I am right here, thanks very much!
     
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