# Minimal polinomial question

1. Feb 4, 2008

2. Feb 4, 2008

### Dick

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. Feb 4, 2008

### HallsofIvy

Staff Emeritus
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.

4. Feb 5, 2008

5. Feb 5, 2008

### HallsofIvy

Staff Emeritus
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 $\lambda^5$?

6. Feb 5, 2008

### transgalactic

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??