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Is every matrix diagonalizable if we allow complex entries?

  1. Feb 11, 2013 #1
    I came across this statement but I am not so sure. I was stuck on this counter-example

    [tex]
    \begin{pmatrix}
    0&1\\0&0
    \end{pmatrix}
    [/tex]

    I'm not really sure what happens to this because the eigenvalues are both zero. I don't know whether this is called diagonalizable because it would just be diag(0,0) or not. And is the general statement true?

    Thank you.
     
  2. jcsd
  3. Feb 11, 2013 #2
    The answer is no, as your counterexample proves. Because of this, you generalize the diagonal decompostion to the Jordan decomposition. The matrix you have provided is called a "Jordan block", the building blocks of the Jordan decomposition.
     
  4. Feb 12, 2013 #3

    mathwonk

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    a diagonalizable matrix has its eigenvalues on the diagonal. hence a matrix whose eigenvalues are all zero, can only be diagonalizable if it is the zero matrix. thus any non matrix whose only non zero entries are above the diagonal, has all eigenvalues zero, hence if non zero, such as your example, cannot be diagonalizable.
     
  5. Feb 13, 2013 #4
    Thank you for the help!
     
  6. Feb 13, 2013 #5

    CompuChip

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    Also, referring back to the topic title, your matrix does have complrx entries. They just happen to have imaginary part 0.

    Ok that was a nitpicky remark. But goes to show that it wont 'solve' your diagonalizibility 'problem'.
     
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