# Minimal polinomial question

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Dick
Homework Helper
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.

HallsofIvy
Homework Helper
Having concluded, correctly, that MT is
$$\left[\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right]$$
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
$$\left|\begin{array}{cccc} -\lambda & 0 & 0 & 1 \\ 0 & -\lambda & 1 & 0 \\ 0 & 1 & -\lambda & 0 \\ 1 & 0 & 0 & -\lambda \end{array}\right|$$

Take out any repeated factors to get the minimal polynomial.

HallsofIvy
Homework Helper
You have $\lambda^5- \lambda^2+ \lambda- 1$ and then a arrow pointing to
$(\lambda^2- 1)(\lambda^3+ 1)/\lambda$. How did you go from the first to the second? Oh, and since the determinant you are evaluating only has 4 $\lambda s$ in it, how are you getting $\lambda^5$?

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??