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Minimal polinomial question

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  • #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.
 
  • #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
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]?
 
  • #6
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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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