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Gram-Shmidt getting PtAP is diagonal

  1. Nov 14, 2012 #1
    1. The problem statement, all variables and given/known data

    in light of the gram-schmidt orthogonalization process, if A: Rn -> Rn and we have a basis of Rn of eigenvectors of A, can't we just orthonormalize them and get a matrix P such that P-1=PT and thus PTAP is diagonal?

    2. Relevant equations



    3. The attempt at a solution
    I believe the answer to this is yes, but I'm not sure how to prove this. Any ideas would be appreciated, thanks!
     
  2. jcsd
  3. Nov 14, 2012 #2

    Dick

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    No, not true. If your eigenvectors have different eigenvalues and aren't already orthogonal, you can apply gram-schmidt but the results won't be eigenvectors.
     
  4. Nov 14, 2012 #3
    Why aren't those eigenvectors? Because you've changed the vectors too much?
     
  5. Nov 14, 2012 #4

    Dick

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    Sure, you changed them too much. Why would you think they would be eigenvectors? A linear combination of two eigenvectors is not necessarily an eigenvector.
     
  6. Nov 14, 2012 #5
    I thought it might be an eigenvector because i was under the impression for some reason that the orthonormalization would no necessarily mean using the Gram-Shmidt process, but rather just normalizing it (dividing by sqrt(sum of squares of eigenvector values)). Is this not the case?
     
  7. Nov 14, 2012 #6

    Dick

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    That's pretty confused. Look, take the matrix [[1,1],[0,2]]. The eigenvalues are 1 and 2. Find the eigenvectors. It's easy. Now you can normalize them, but you can't force them to be orthogonal.
     
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