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Eigenvectors and diagnolization

  1. Oct 23, 2009 #1
    Two questions. First, I'm given a 3x3 matrix with the last row all zeroes. I'm asked to diagonalizable it, but the determinant is 0, so there are no eigenvalues. Am I reasoning correctly here? It seems an odd question to ask.

    Second, I'm asked to prove that if A n x n matrix in C space, then if A has n distinct eigenvalues, A is then diagonalizable. The way I proved it is to say that for A to be diagonalizable, it must have n distinct eigenvectors, which is only the case if it has n distinct eigenvalues. This seems far too easy, am I missing something?
     
  2. jcsd
  3. Oct 23, 2009 #2
    If the determinant is 0 then 0 is an eigenvalue for the matrix
     
  4. Oct 24, 2009 #3

    HallsofIvy

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    Pretty much nothing you have said is right. First of all, you are going the "wrong way". You were asked to prove that if A has n distinct eigenvalues, then it is diagonalizable.

    What you are trying to prove is that "if A is diagonalizable then it has n distinct eigenvalues".

    You cannot prove that because it is NOT true! Further, it is not true that a matirix having n distinct eigenvectors "is only the case if it has n distinct eigenvalues. The n by n identity matrix has n independent eigenvectors but all n of its eigenvalues is the same.


    The other way is correct. Since eigenvectors corresponding to distinct eigenvalues are independent, if an n by n matrix has n distinct eigenvalues then it must have n independent ("distinct" is not enough) eigenvectors and so is diagonalizable. (Write the linear transformation corresponding in a basis consisting of those eigenvectors.)
     
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