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Fiding Eigenvalues

  1. Oct 19, 2011 #1

    Ush

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    attachment.php?attachmentid=40167&stc=1&d=1319056232.png
    helppp please
     

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  3. Oct 19, 2011 #2

    phyzguy

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    I think what you've done is correct so far. That polynomial can be factored. Give it a try.
     
  4. Oct 19, 2011 #3

    Ush

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    i have no idea how to do that..
    ex
    λ (λ^2 – 21λ + 147k^2) – 343k^3
    leaves a variable out
    =(
     
  5. Oct 19, 2011 #4

    Mark44

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    [itex]\lambda[/itex] - 7k is a factor.
     
  6. Oct 19, 2011 #5

    vela

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    There's really no cookbook way of factoring. You just have to futz around with it, look for patterns, and make educated guesses.

    If you're going to have any hope of factoring by hand, the root has to divide into 343k3, so the first thing you need to do is find the factors of 343, which turns out to be 73. So now you can write
    [tex]\lambda^3-21k\lambda^2+147k^2\lambda-(7k)^3=0[/tex]Can you see or discover anything else?
     
  7. Oct 19, 2011 #6

    Ush

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    how did you get that as a factor? and how would i use that to find the other factors?
    i tried dividing the equation with λ-7k and i got

    3-21λ2k+147λk2-343k3) / λ-7k = λ2 + (147λk2-343k3 / λ-7k)
    (using polynomial division)

    ...and that looks more complicated then the original =[
    I'm totally off track =/
     
  8. Oct 19, 2011 #7

    vela

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    I would suggest you factor all of the numerical coefficients and look for patterns.
     
  9. Oct 19, 2011 #8

    Ush

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    i know they are all divisible by 7 =/
     
  10. Oct 19, 2011 #9

    Mark44

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    Your division is incorrect. λ - 7k divides the cubic with no remainder, and the resulting polynomial can be easily factored.
     
  11. Oct 19, 2011 #10

    Ush

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    Mark44
    I re-did the division, and I got
    (λ-7k)(λ2 - 14λk + 49k2)
    =(λ-7k)(λ-7k)(λ-7k)

    the three eigenvalues are 7k?
     
  12. Oct 19, 2011 #11

    phyzguy

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    You got it!
     
  13. Oct 19, 2011 #12

    Ush

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    thank you everyone! =)
     
  14. Oct 19, 2011 #13

    Mark44

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    Bingo!
     
  15. Oct 19, 2011 #14

    vela

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    Well, you can say a bit more than that. If you fully factor the coefficients, you can see that you can write the polynomial as
    [tex]\lambda^3 - 3\lambda^2(7k)^1+3\lambda^1(7k)^2-(7k)^3[/tex]which you hopefully should recognize as the binomial expansion of [itex](\lambda-7k)^3[/itex].
     
  16. Oct 20, 2011 #15

    Ush

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    vela, you're a genius =) thank you for showing me that; that's much much much quicker then guessing a factor, and then doing long division. Useful to know for the next test!
     
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