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Minimal polinomial question

  1. Feb 4, 2008 #1
  2. jcsd
  3. Feb 4, 2008 #2

    Dick

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    Just writing 'there is no minimal polynomial' isn't much of a try. Form the matrix M-x*I and take it's determinant to get the characteristic polynomial and then factor to see if there are repeated factors you can drop.
     
  4. Feb 4, 2008 #3

    HallsofIvy

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    Having concluded, correctly, that MT is
    [tex]\left[\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right][/tex]
    How can you declare that MT v= 0? For all v? By the definition of M, M<1, 0, 0, 0>= <0, 0, 0, 1>. The only matrix that takes every vector to 0 is the 0 matrix! And every matrix except the 0 vector has a minimal polynomial. As Dick said, the characteristic polynomial is given by
    [tex]\left|\begin{array}{cccc} -\lambda & 0 & 0 & 1 \\ 0 & -\lambda & 1 & 0 \\ 0 & 1 & -\lambda & 0 \\ 1 & 0 & 0 & -\lambda \end{array}\right|[/tex]

    Take out any repeated factors to get the minimal polynomial.
     
  5. Feb 5, 2008 #4
  6. Feb 5, 2008 #5

    HallsofIvy

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    You have [itex]\lambda^5- \lambda^2+ \lambda- 1[/itex] and then a arrow pointing to
    [itex](\lambda^2- 1)(\lambda^3+ 1)/\lambda[/itex]. How did you go from the first to the second? Oh, and since the determinant you are evaluating only has 4 [itex]\lambda s[/itex] in it, how are you getting [itex]\lambda^5[/itex]?
     
  7. Feb 5, 2008 #6
    i was looking for numbers which make the hole thing 0

    i put 1 and -1
    and then i get
    L=lamda
    (L-1)(L+1)(ax^3+bx^2+cx+d)

    i opened the colls and equalized the coeffishents
    and fount the a b c d

    also i divided by L because i multiplied some line by L
    so the determinant must be divided by L

    what now??

    and also can you please answer on my PM about the question
    of the transformation
    i realy want to understand this thing from top to bottom
     
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