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Solving for complex eigenvalues

  1. Feb 29, 2012 #1
    Quick question:

    I have a characteristic polynomial: λ2 + i = 0.........how do I solve for the eigenvalues?

    They're suppose to be + or - (√2/2)(1 - i) How'd they get those?
     
  2. jcsd
  3. Feb 29, 2012 #2

    Dick

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    λ^2=(-i)=exp(i*3*pi/2). Use deMoivre.
     
    Last edited: Feb 29, 2012
  4. Feb 29, 2012 #3
    I'm kind of shaky with deMoivre, is there another way?
     
  5. Feb 29, 2012 #4

    Dick

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    Only harder ways. deMoivre is the easy way. One square root of exp(i*t) is exp(i*t/2). What does that lead to your case? Then use exp(i*t/2)=cos(t/2)+i*sin(t/2).
     
  6. Feb 29, 2012 #5
    Only reason I ask is because we were not really shown the deMoivre way in class,...to think of it, we weren't even shown a way to solve for these complex roots. I'll give deMoivre's a try.

    Another question, just out of curiosity:

    Say I have a basis for the Ker(T). T being the transformation. Say I have an arbitrary vector v in V. Can I write v as a linear combination of the basis of the Ker(T), and if so do the co-efficients that I obtain from this linear combination make a co-ordinate vector in V, that if I apply the transformation T to this co-ordinate vector, will it map to the null-space in Im(T)?
     
  7. Feb 29, 2012 #6

    Dick

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    Ok, for the first one you need (a+bi)(a+bi)=a^2-b^2+2abi=(-i). Equating real and imaginary parts you get two equations for the real numbers a and b. Now try to solve them.

    For the second one. You can't write an arbitrary element of element of V as a combination of the basis of Ker(T) unless it's in Ker(T). And then it will map to 0 in Im(T). I'm not sure what the real question is here?
     
  8. Feb 29, 2012 #7
    Yup, that's what I was asking about.



    so I get two equations: a2-b2 = 0 and 0 + 2abi = -i Is that what you mean?

    I'm also trying the deMoivre way. So as of now, I'm converting -i into polar co-ordinates in order to use de Moivre: so currently I have 1 sin[itex]\sigma[/itex], how do you get [itex]\sigma[/itex] to be pi/2 ?
     
  9. Feb 29, 2012 #8
    well I figured out the pi/2, how it's obtained; but I got -pi/2. I got it from b = r sin[itex]\sigma[/itex] ==>

    -1 = 1 sin[itex]\sigma[/itex], and then figured out for [itex]\sigma[/itex].
     
  10. Feb 29, 2012 #9
    wanted to delete this message.......I'm actually still puzzled when it comes to the -pi/2
     
  11. Mar 1, 2012 #10

    Dick

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    exp(-i*pi/2)=cos(-pi/2)+i*sin(-pi/2)=(-i), right? A square root of that is exp(-i*pi/4)=cos(-pi/4)+i*sin(-pi/4). Which equals?
     
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