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Finding eigenvalues of matrix

  1. Jul 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the eigenvalues of the following matrix:

    [tex]
    \left(
    \begin{array}{ccc}
    1 & 0 & -3 \\
    1 & 2 & 1 \\
    -3 & 0 & 1
    \end{array}
    \right)
    [/tex]

    2. Relevant equations



    3. The attempt at a solution

    I think I'm forgetting a basic algebra rule or something. I know there are supposed to be 3 eigenvalues, but I am only getting 2.

    [tex]
    0=\det (\lambda I-A)=\left|
    \begin{array}{ccc}
    \lambda -1 & 0 & 3 \\
    -1 & \lambda -2 & -1 \\
    3 & 0 & \lambda -1
    \end{array}
    \right|
    [/tex]
    [tex]
    =\frac{(\lambda -1)(\lambda -2)(\lambda -1)-3(3)(\lambda -2)}{(\lambda -2)}
    [/tex]
    [tex]
    =(\lambda -1)(\lambda -1)-9
    [/tex]
    [tex]
    =(\lambda -1)^2-9
    [/tex]
    [tex]
    (\lambda -1)^2=9
    [/tex]
    [tex]
    \lambda =\pm 3+1
    [/tex]
     
  2. jcsd
  3. Jul 20, 2009 #2
    Re: Eigenvalues

    Where you cancel the (lambda - 2) term, you lose one solution: lambda = 2.

    Remember, you have to factor. By canceling the term (lambda - 2), you're basically saying "lambda cannot under any circumstances equal 2". If you want to do that, you have to consider two separate cases: lambda=2 and lambda!=2.

    So the three eigenvalues are 3+1=4, -3+1=-2, and 2.
     
  4. Jul 20, 2009 #3

    Pengwuino

    User Avatar
    Gold Member

    Re: Eigenvalues

    Why did you divide by (lambda - 2) in the first place?
     
  5. Jul 20, 2009 #4
    Re: Eigenvalues

    Thanks for your help.

    So let me get this straight. Whenever I divide by a term to cancel it, I count that term as one of my solutions, right? Does this always work? I didn't quite understand what you meant by testing two separate cases.
     
  6. Jul 20, 2009 #5
    Re: Eigenvalues

    Because it was much easier to do that than it was to multiply everything out and try to factor a third-degree polynomial.
     
  7. Jul 20, 2009 #6
    Re: Eigenvalues

    Why did you divide by lambda-2 in the second line of your equations? Do you remember the formulas for a determinant of a 3x3 matrix?
     
  8. Jul 20, 2009 #7
    Re: Eigenvalues

    Oh, duh. Since there are two terms on the RHS and they are constants then the must both be equal to zero separately. Haha.
     
  9. Jul 20, 2009 #8
    Re: Eigenvalues

    You can just factor (x-2) to get an easily solvable equation
     
  10. Jul 20, 2009 #9
    Re: Eigenvalues

    I was just doing two steps at once. Bad form on my part.
     
  11. Jul 20, 2009 #10

    Pengwuino

    User Avatar
    Gold Member

    Re: Eigenvalues

    Yes. Think of it this way, let's say you're trying to find the solutions of x for (x-1)(x-2)(x-3)=0. Obviously your solutions are x=1,2, and 3. If you divide out (x-1), you lose a solution and are left with (x-2)(x-3)=0.
     
  12. Jul 20, 2009 #11
    Re: Eigenvalues

    Very good point. Thanks!
     
  13. Jul 20, 2009 #12

    Pengwuino

    User Avatar
    Gold Member

    Re: Eigenvalues

    Your welcome. Also, to clarify AU's point, when you divide by (lambda-2), your left with that (lambda-2) in the denominator of the left hand side (the 0). Now for any number other then 2, it's going to be 0 divided by some finite number which is 0. However, with lambda = 2, you get 0/0 which is undefined, not 0.
     
  14. Jul 20, 2009 #13
    Re: Eigenvalues

    Aha. That makes sense now.
     
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