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Diagonalisable matrix

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Is the following matrix diagonalisable?

    [0 0 0 0 0 ....0
    0 1 0 0 0 ....0
    0 0 2 0 0 .... 0
    0 0 0 3 0 ....0
    . . .
    . . .
    . . .
    . .0
    0 0 0 0 0 0....n]

    (having trouble showing the mtarix, basically 0,1,2,3...n down the diagonal and zeroes off-diagonal)

    2. Relevant equations

    Given.

    3. The attempt at a solution

    I thought that because it contains the zero vector, the matrix doesn't have n linearly independent columns, thus not diagonalisable? I'm not entirely sure. Plus, is it possible to find the eigenvalues and eigenvectors for the matrix?
     
  2. jcsd
  3. Sep 19, 2009 #2

    gabbagabbahey

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    If you mean

    [tex]\begin{pmatrix}0&0&0&0&\ldots&0\\0&1&0&0&\ldots&0\\0&0&2&0&\ldots&0\\0&0&0&3&\ldots&0\\\vdots&\vdots&\vdots&\vdots&\ldots&\vdots\\0&0&0&0&\ldots&n\end{pmatrix}[/tex]

    then isn't the matrix already diagonalized?
     
  4. Sep 19, 2009 #3
    Yes, that's the matrix that I mean. :)

    So, if the matrix has a zero column, it can still be diagonalisable?
    I thought that if it didn't have n linearly independent vectors, it cannot diagonalisable...
     
  5. Sep 19, 2009 #4

    gabbagabbahey

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    Obviously, since your matrix is diagonal, choosing [itex]P=I_n[/itex] (the [itex]n\times n[/itex] identity matrix) will show that it is diagonalizable.

    I think you are mistaking a matrix's columns for its eigenvectors; if an [itex]n\times n[/itex] matrix doesn't have [itex]n[/itex] linearly independent eigenvectors it isn't diagonalizable.
     
    Last edited by a moderator: May 4, 2017
  6. Sep 19, 2009 #5
    Okay, thank you! I should obviously read my notes better next time! :)

    You've been a great help!
     
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