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Eigenvalue question

  • 10-04-2008 10:09pm
    #1
    JavaBear
    Closed Accounts Posts: 209 ✭✭


    Problem

    eigenvalueshy2.jpg

    Having problems with this question. I know how it should be done, but can't get it out :confused:


Comments

  • #2
    rejkin
    Registered Users, Registered Users 2 Posts: 891 ✭✭✭rejkin


    I got 1,1,3 when i worked it out


  • #3
    JavaBear
    Closed Accounts Posts: 209 ✭✭JavaBear


    I have the answers (it's a q from a book) and those aren't them. If possible, could others post their answers?


  • #4
    Pherekydes
    Registered Users, Registered Users 2 Posts: 16,201 ✭✭✭✭Pherekydes


    -1,1,2


  • #5
    JavaBear
    Closed Accounts Posts: 209 ✭✭JavaBear


    Yep that's it.

    Not sure if it has a name, but the way I tried it involved lambda. The expanding it the same as you would to get the determinant. Got to: (P = lambda)
    (1 - P)(P^2 - P - 3) + 1 - P = 0


  • #6
    Michael Collins
    Moderators, Science, Health & Environment Moderators Posts: 1,852 Mod ✭✭✭✭Michael Collins


    Yeh you have it now. If you expand the determinant along the top row (the zero in the top right makes it easier), then you should get a cubic in the form of a quadratic times a linear factor in lamda - which is easier to solve than an unfactored cubic.


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