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Diagonalizable Matrices & Eigenvalues!

  1. Jan 5, 2009 #1
    Hello,

    Is it sufficient to determine that a nXn matrix is not diagonalizable by showing that the number of its distinct eigenvalues is less than n?

    Thanks for your time.
     
  2. jcsd
  3. Jan 5, 2009 #2

    Hurkyl

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    You should empirically test your hypothesis; try it out on the simplest matrices you can imagine.
     
  4. Jan 5, 2009 #3

    HallsofIvy

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    For example, how many distinct eigenvalues does

    [tex]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}[/tex]
    have?
     
  5. Jan 6, 2009 #4
    Ok I got your point...I think the question should be now, how to check whether a matrix is diagonalizable or not?

    A theorem I've encountered in my textbook:
    and somewhere else I read this is sufficient to prove a matrix is diagonalizable but not the other way around.. that's why I posted the question..

    So now, all I can do is to prove that Matrix x is diagonalizable, but if it's not, I can't tell for sure....what other methods should I be using?

    Thanks for your responses.
     
  6. Jan 6, 2009 #5

    Hurkyl

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    The most direct way to tell if a matrix isn't diagonalizable is to try and diagonalize it.
     
  7. Jan 6, 2009 #6

    HallsofIvy

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    Yes- sufficient but not necessary.

    An n by n matrix is diagonalizable if and only if it has n independent eigenvectors. Since eigenvectors corresponding to distinct eigenvalues are always independent, if there are n distinct eigenvalues, then there are n independent eigenvectors and so the matrix is diagonalizable. But even if the eigenvalues are not all distinct, there may still be independent eigenvectors.

    [tex]\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}[/tex]
    as I pointed out before has the single eigenvalue 1 but is already diagonal because <1, 0, 0>, <0, 1, 0>, and <0, 0, 1> are independent eigenvectors.

    [tex]\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}[/tex]
    also has 1 as its only eigenvalue. Now, <1, 0, 0> and <0, 0, 1> are the only eigenvectors so this matrix cannot be diagonalized.

    Finally,
    [tex]\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}[/tex]
    also has 1 as its only eigenvalue but now only <1, 0, 0> and multiples of that are eigenvectors.
     
  8. Jan 6, 2009 #7
    Great! Things are pretty clear now, thanks HallsofIvy!
     
    Last edited: Jan 6, 2009
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