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Imaginary Eigenvalues

  1. Jun 7, 2014 #1
    1. The problem statement, all variables and given/known data

    Multiply the matrices to find the resultant transformation.

    $$x\prime =2x+5y\\ y'=x+3y $$ and $$ x\prime \prime =x\prime -2y\prime \\ y\prime \prime =3x\prime -5y\prime $$

    2. Relevant equations

    $$Mr=r\prime$$

    3. The attempt at a solution
    I get imaginary eigenvalues of -i and i. I would imagine the transformation would rotate the vectors 90deg in the complex plane on either side.

    However, the answer in the text does say 90deg rotation, but it says in the real $$x\prime \prime=-y$$ and $$y\prime \prime=x $$

    Any clue what went amiss?

    Thanks,
    Chris Maness
     
  2. jcsd
  3. Jun 7, 2014 #2

    ehild

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    Apply the final matrix to the real base vectors (1,0)T and (0,1)T what do you get ?

    ehild
     
  4. Jun 8, 2014 #3
    That is pretty cool. How does this transformation work if I don't have real eigenvectors? It seems like it would rotate a vector into complex space.

    Regards,
    Chris Maness
     
  5. Jun 8, 2014 #4

    hilbert2

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    A unitary transformation generally has complex eigenvalues that lie on the unit circle of the complex plane. It should be pretty obvious that the operation of rotating a vector 90 degrees is never equal to multiplying the vector with a real number.
     
  6. Jun 8, 2014 #5

    ehild

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    x and y mean components of two-dimensional vectors. Why do you think they are real and imaginary parts of a complex number? The equations mean linear transformations of these vectors. At the end, x transforms to x'' and y transforms to y". How are x" and y" related to x,y?
    I presume you know how a linear transformation is represented by a matrix? How to multiply a matrix and a vector?
    You do not need eigenvalues to perform the transformation.
    Let it be x=1 and y =0 (unit vector along the x axis) What is x" and y"?
    What is x", y" if x=0, y=1?

    ehild
     
  7. Jun 8, 2014 #6
    No, I don't need them, but the eigenvectors should tell me the axis about which it rotates.

    Chris
     
  8. Jun 8, 2014 #7

    ehild

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    What is that "it" which rotates? And what is the rotation axis?

    ehild
     
  9. Jun 8, 2014 #8
    If you have a matrix that rotates a vector (det(M)=1) then the vector that satisfies M*(x,y,z)T=(x,y,z)T would be the axis of rotation.

    Chris
     
  10. Jun 8, 2014 #9

    ehild

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    You have two-dimensional vectors now. They are rotated in the plane. What is the vector the transformation leaves unchanged?

    ehild
     
  11. Jun 8, 2014 #10
    Good point :D

    Chris
     
  12. Jun 8, 2014 #11

    Ray Vickson

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    What is all the fuss about? Where do eigenvalues come into the issue? You have ##x', y'## expressed in terms of ##x, y## and you have ##x'', y''## expressed in terms of ##x', y'##. Therefore, it is simple algebra to express ##x'', y''## in terms of ##x, y##. You can do it with matrices, of course, but there is no need to do that. The result is that a vector ##\vec{v}## with components ##(x,y)## is transformed into a vector ##\vec{v}''## with components ##(x'',y'')##, and that transformation can be expressed through a matrix if you want that. It turns out to be a simple rotation.

    The only remaining issue is: do you rotate the vector (so that in a common coordinate system the vector ##(x,y)## is moved until it becomes ##(x'',y'')##) or do you keep the vector ##\vec{v}## fixed and rotate the coordinate system? The question does not really make that clear.
     
  13. Jun 8, 2014 #12
    Ray, the context of the question was in figuring out the rotation. I over thought it. The rotation was easy to figure out just by multiplying unit vectors, and seeing what happened to them -- which ehild pointed out early on.

    However, my trip down the rabbit hole was informative even though it was wrong headed in thinking a 2D rotation would have an axis lying in the real plane. There does appear to be a so called axis, but it is in complex space.

    Thanks,
    Chris Maness
     
  14. Jun 8, 2014 #13

    hilbert2

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    It might also be helpful to consider a 3D rotation matrix:

    ##\left[\begin{smallmatrix}0&-1&0\\1&0&0\\0&0&1\end{smallmatrix}\right]##

    That has three eigenvectors, one which corresponds to the rotation axis (the z axis) and has the real eigenvalue 1, and two of which are complex eigenvectors with complex eigenvalues.
     
  15. Jun 8, 2014 #14
    Yep, that gives that same rotation an axis to rotate around -- an axis that is in real cartesian space.

    Thanks,
    Chris Maness
     
  16. Jun 8, 2014 #15

    ehild

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    In 2D, rotation is about one point. The vector that stays unchanged during the transformation, is the nullvector.
    You do not need complex numbers to describe transformations in the real word.
    The advantage of using complex numbers for two-dimension vectors makes in-plane linear transformations represented by multiplying with a complex number.

    ehild
     
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