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Invertibility and Diagonalizability

  1. Nov 26, 2011 #1
    Does there exist a matrix which is both not invertible and not diagonalizable? If so, please provide an example.

    Thanks,
    David
     
  2. jcsd
  3. Nov 26, 2011 #2

    micromass

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    Is this a homework question??

    What did you try alrready??
     
  4. Nov 26, 2011 #3

    Deveno

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    hint: see if you can find a 2x2 matrix using only 0's and 1's with 0 determinant. i can think of 2 such matrices right off the bat that fulfil both your criteria.
     
  5. Nov 26, 2011 #4

    HallsofIvy

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    One that comes to mind immediately is
    [tex]\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}[/tex]

    A matrix is not invertible if and only if it has 0 as an eigenvalue. A matrix is diagonalizable if the exist a basis for the space consisting of eigenvectors. Those are not contradictory.
     
  6. Nov 26, 2011 #5

    micromass

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    Uuh, that is already a diagonal matrix...
     
  7. Nov 27, 2011 #6

    HallsofIvy

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    Yes. So? A diagonal matrix is trivially "diagonalizable".

    If you want a non-diagonal, diagonalizable, matrix that is not invertible, start with a diagonal matrix, say
    [tex]D= \begin{bmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{bmatrix}[/tex]
    and multiply by A and [itex]A^{-1}[/itex] where A is some invertible matrix.

    For example, if
    [tex]A= \begin{bmatrix}1 & 0 & 0 \\ 2 & -1 & 0 \\ 5 & -2 & 1\end{bmatrix}[/tex]
    then
    [tex]A^{-1}= \begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 0 \\ -1 & 2 & 1\end{bmatrix}[/tex]

    and then
    [tex]ADA^{-1}= \begin{bmatrix}-3 & 2 & 0 \\ -2 & -2 & 0 \\ 13 & -4 & 0\end{bmatrix}[/tex]
    which is a non-invertible matrix which can be "diagonalized" to the original matrix, D.
     
  8. Nov 27, 2011 #7

    micromass

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    It's fine. Except that the OP want a matrix that is not diagonalizable. :tongue:
     
  9. Nov 27, 2011 #8
    You can simply choose a nilpotent non-null matrix, as it's easy to see that the only diagonalizable AND nilpotent matrix is the null one.
    for example:
    01
    00
     
  10. Nov 27, 2011 #9
    How about [0 0 0]? Definitely a matrix, and definitely not invertible since it isn't even square. Also, certainly not diagonalizable.
     
  11. Nov 27, 2011 #10

    HallsofIvy

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    dang! Don't you just hate those little words like "not"?

    Okay, lets start with a Jordan Normal form, non-diagonal matrix:
    [tex]P= \begin{bmatrix}2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{bmatrix}[/tex]
    which also has 0 as an eigenvalue and so is not invertible. Using the same "A" as before,
    [tex]APA^{-1}= \begin{bmatrix}0 & 1 & 0 \\ 4 & 0 & 0 \\ 8 & 1 & 0\end{bmatrix}[/tex]

    That is neither invertible nor diagonalizable.
     
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