Let's say I have the characteristic polynomial of an operator:
Wouldn't then the minimal polynomial be exactly:
It seems to me like it should be, since it is the polynomial of least power that has all the eigenvalues as roots. Still, people seem to be using more complicated processes to calculate the minimal polynomial. If this holds, won't it be trivial given a characteristic polynomial? The characteristic polynomial can itself be obtained from an upper triangular matrix of the operator.
Or is it simply the case that there exists a simpler polynomial than the one having the eigenvalues as roots, which still gives $q(T)=0$?