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Diagonalizing a square matrix with degenerate eigenvalues

  1. Aug 2, 2008 #1

    This is more of a math question but in the context of Quantum Mechanics, hence I posted it here. Suppose I have a matrix A of order 3x3 with three eigenvalues: 0, 0, 5. I am supposed to find the diagonalizing matrix for A.

    I know that in general, if P denotes the matrix of eigenvectors of A, then [itex]PAP^{-1}[/itex] will be a diagonal matrix.

    In my particular example, for the eigenvalue 0,

    AX = 0

    gives infinitely many solutions for X, so the eigenvector with eigenvalue 0 cannot be uniquely determined.

    How do I diagonalize such a matrix?

    Thanks in advance,

  2. jcsd
  3. Aug 3, 2008 #2


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    There are infinitely many different eigenvectors for an eigenvalue, but are they all linearly independent? Remember that you only need 3 linearly independent eigenvectors for this. Just pick a simple eigenvector associated with the zero eigenvalue.
  4. Aug 3, 2008 #3
    Ok, lets say I take (1, 0, 0)' and (0, 1, 0)' as the eigenvectors for the zero eigenvalue. And then I just plug them along with the eigenvector for the eigenvalue 5...and thats it, right?
  5. Aug 3, 2008 #4


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    Well, that's provided you "plug them in" correctly, doesn't it?
  6. Aug 3, 2008 #5
    Yes, as the problem asks to find the diagonalizing matrix, I was wondering what to write as the eigenvectors are not uniquely determined here. By "plug them in", I meant that I construct a matrix P whose columns are precisely the Linearly independent eigenvectors of the matrix A...in this case I select two Linearly independent eigenvectors for the zero eigenvalue and the third column is the eigenvector for the eigenvalue 5.

    Thanks for your help Defennder :smile:
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