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Linear algebra question

  1. May 10, 2009 #1
    So the question is:

    For which values of a is the following matrix not diagonalizable?

    ....... 4 0 6
    A = ( 2 a 3 )
    ......-1 0 -1

    So my first question is when is a matrix diagonalizable or not diafonalizable.

    The way I approached the question was to go about finding the eigenvalues and eigenvectors.

    p(h) = (4-h)(a-h)(-1-h)-(-1)(a-h)(6)

    Distributing this out gets pretty messy. So it doesn't seem like this is the right approach.

    Can someone help me out please? How do I go about solving this?
     
  2. jcsd
  3. May 10, 2009 #2

    matt grime

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    That's a good start, though I disagree that it is messy to simplify - there is a common factor of a-h in there, and no other a's, thus making it very easy to simplify (and in particular it is clear that a is always an eigen-value, but you could see that by inspecting the original matrix).
     
  4. May 10, 2009 #3

    HallsofIvy

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    A 3 by 3 matrix is diagonalizable if and only if it has three independent eigenvectors.

    If it has 3 distinct eigenvalues, then, of course, it has three independent eigenvectors (eigenvectors corresponding to distinct eigenvalues are always independent) and so is diagonalizable. But in the case that there are not three distinct eigenvalues you need to check the eigenvectors.

    (Expanding the determinant
    [tex]\left|\begin{array}{ccc}4-\lambda & 0 & 6 \\ 2 & a-\lambda & 3 \\ -1 & 0 & -1-\lambda\end{array}\right|[/tex]
    by the middle column gives the equation a particularly simple form.)
     
  5. May 10, 2009 #4
    so I pulled out the (a-h) and ended up with:

    (a-h)(h-1)(h-2)

    so I know two eigenvalues are 1 and 2.

    so is it not diagonalizable if a = 1 or 2?
     
  6. May 10, 2009 #5

    matt grime

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    It may or may not be: the identity matrix has 1 eigenvalue but is diagonal(izable). But Halls already told you what to do. You need to find how many eigenvectors there are with eigenvalues 1 and 2 (when a=1 and 2).
     
  7. May 10, 2009 #6
    so when I plug 1 in for a I get the eigenvalues 1 and 2 with eigenvectors (0,1,0) and (3,3,-1). 1 has a multiplicity of 2 but only one eigenvector.

    when I plug 2 in for a I get the eigenvalues 1 and 2 with eigenvectors (2,-1,-1) and (0,1,0). 2 has a multiplicity of 2 but only one eigenvector.

    so there are three different eigenvectors. I'm still lost as to how this all connects though. So it's not diagonalizable for a = 1 and 2? are that the only cases?
     
  8. May 10, 2009 #7

    matt grime

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    When a=1 you say there are not 3 eigenvectors - so it's not diagonalizable.

    When a=2 you also say there are not three eigenvectors - so it's not diagonalizable.

    When a is neither 1 nor 2 then there are 3 distinct eigenvalues and hence 3 distinct eigenvectors hence it is diagonalizable.

    Then you say that there are three eigenvectors - but that is for a different a, hence a different matrix in each case.
     
  9. May 10, 2009 #8
    ok. thank you!
     
  10. May 11, 2009 #9
    my leacturer said tha I can`t multiply matrixies with different sizes,such as (2*2)(4*5) than she gave us the very complicated process!! can you make it shorter for me
     
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