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What is wrong with this diagonalization problem.

  1. Nov 26, 2008 #1
    1. The problem statement, all variables and given/known data
    A is

    [4 0 1
    2 3 2
    1 0 4]

    Find an invertible P and a diagonal D so that D=P-1AP.

    I keep getting two linearly dependent eigenvalues which means it's not diagonal but this problem doesn't state "If it can't be done explain why" or anything like that. So I just want to verify with some of you.

    3. The attempt at a solution
    I subtract with LI and take the determinant and get:

    (L-3)((L-4)^2 - 1)
    (L-3)(L^2-8L+16)-(L-3)
    L^3-8L^2+16L-3L^2+24L-48-L+3
    L^3-11L^2+39L-45

    Which I then factor out to be 5, 3, 3.

    Am I doing something wrong/missing something?
     
  2. jcsd
  3. Nov 26, 2008 #2
    The eigenvalues are correct. Now, did you plug each eigenvalue, one at a time, into your matrix to find the corresponding eigenvectors? There are two eigenvectors for eigenvalue 3, and one for eigenvalue 5.
     
  4. Nov 26, 2008 #3
    Right but you can't find the diagonalization when the two eigenvalues are the same. They are linearly dependent. Right? I just wanted to verify that I wasn't going crazy.

    Once you have two identical eigenvectors for P you can't invert it to eventually find D. Or am I supposed to just assume that D is the diagonal matrix of 5,3,3 even though you can't actually get to that point.
     
  5. Nov 26, 2008 #4
    The minimal polynomial of your matrix (x - 5)(x - 3), so it is diagonalizable.
     
  6. Nov 26, 2008 #5
    It is diagonalizable if and only if the characteristic polynomial of your matrix, A, splits and for each eigenvalue X of A, the multiplicity of X equals n - rank (A-XI). So check to see if these criteria apply here (you have almost done the first step)
     
  7. Nov 26, 2008 #6

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    When two or more of your eigenvalues are the same, you need to search for generalized eigenvectors. These generalized eigenvectors will be linearly independent, and you can construct your P-matrix from them.
     
  8. Nov 26, 2008 #7
    Or more simply, if and only if the minimal polynomial has the form (x - a1)(x - a2)...(x - an) where ai are the distinct eigenvalues of A. D will be the diagonal matrix having a1, a2, etc. down the diagonal and P will be the matrix whose columns are a linearly independent set of eigenvectors corresponding to the eigenvalues of A.
     
  9. Nov 26, 2008 #8
    Can one of you help me find the generalized vectors I have no idea what to do.
     
  10. Nov 26, 2008 #9
    Nm, I figured it out. when looking for the eigenvector for 3 it actually splits into 2. I think I did the math wrong the first time. Thanks you guys.
     
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