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.
Having concluded, correctly, that M_{T} 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 M_{T} 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.
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]?