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Homework Help: Matrices and determinant

  1. Mar 25, 2012 #1
    please look at the attachement and my attempt at the solution - hope you can help.

    Last edited: Mar 25, 2012
  2. jcsd
  3. Mar 25, 2012 #2
    I see no attachment.
  4. Mar 26, 2012 #3
    sorry I amended the post and forgot to reattached my document - here it is please look at it.
    Last edited: Mar 26, 2012
  5. Mar 26, 2012 #4
    After "This gives:", you need to check your math. I see [tex](1-\lambda)^2(-1-\lambda)-4(1-\lambda)-4(1-\lambda)[/tex]

    To find eigenvectors, see if this resource helps: http://www.sosmath.com/matrix/eigen2/eigen2.html

    Essentially, you will set up [tex]A \left(\begin{matrix} x \\ y \\ z \end{matrix}\right) = \lambda \left(\begin{matrix} x \\ y \\ z \end{matrix}\right)[/tex] where you substitute in each [itex]\lambda[/itex] found earlier. Then you'll have 3 equations and 3 unknowns for each [itex]\lambda[/itex].
  6. Mar 26, 2012 #5


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    What does it mean for there to be an eigenvalue of -1?
    If you write down the equation that the eigenvector satisfies, you should get three equations and three unknowns - note in general the solution for the eigenvector will be non-unique since you can scale it to get a new eigenvector, but you should be able to get a one-dimensional set of vectors that works
  7. Mar 26, 2012 #6
    ok I had anotherlook at my simplification and I have spotted some error while finding eigen values using 3x3 matrix

    what I get is :
    ( i am using y instaed of lambda as I am typing here so apologies)

    (1-y){(1-y)(-1-y)-(-2x-2)} -2{(0x-2) -(-2)(1-y)}
    =(1-y){-1-y+y+y^2-4} -2 { 0-(-2)(1-y)}
    =(1-y) (y-3) (y+3)
    hence giving eigen values of y=1, y=-3 and y=3

    is this correct?
  8. Mar 26, 2012 #7
    That looks correct.
  9. Mar 26, 2012 #8
    thank you so much for all your help.
  10. Mar 26, 2012 #9

    Ray Vickson

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    I still do not see an attachment.

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