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Diagonalization of a matrix with repeated eigenvalues

  1. Oct 1, 2006 #1
    Hey guys,
    I know its possible to diagonalize a matrix that has repeated eigenvalues, but how is it done? Do you simply just have two identical eigenvectors??
    Cheers
    Brent
     
  2. jcsd
  3. Oct 1, 2006 #2

    Hurkyl

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    Well, try an example -- the simplest you can think of. How about the identity matrix? Can you diagonalize it? What are its eigenvectors?
     
  4. Oct 1, 2006 #3
    Ok, well... i would get an eigenvalue of 1, with a multiplicity of 3. Alright, so when solving for my eigenvalues, by plugging 1 into my matrix A-(lamda)I ; i would just be left with 0's. How can i get eigenvalues from this?? I appreciate your help by the way...
     
  5. Oct 2, 2006 #4

    HallsofIvy

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    Am I misunderstanding something? If you know what eigenvalues are, then you must know how to find them in this simple case. If [itex]\lambda[/itex] is an eigenvalue of matrix A, then [itex]A-\lambda I[/itex] is NOT invertible (since the equation A- \lambda I= 0 does not have a unique solution) and so det(A- \lambda I)= 0. Since you are finding the eigenvalues, you do NOT yet know that [itex]\lambda= 1[/itex] so you are NOT "left with 0's. In this case, that equation is [itex](1- \lamba)^3= 0[/itex] which obviously has 1 as a triple root.
     
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